THE Seaman's Companion, BEING A Plain Guide to the Understanding OF 

ARITHMETIC,  
GEOMETRY,  
TRIGONOMETRY,  
NAVIGATION, and  
ASTRONOMY.    
Applied Chief to NAVIGATION: AND Furnished with a Table of Meridional Parts, to every third Minute: With excellent and easy ways of keeping a Reckoning at Sea, never in Print before.  
ALSO, A Catalogue of the Longitude and Latitude of the principal Places in the World: With other useful things.  
The Third Edition corrected and amended.  
By MATTHEW NORWOOD, Mariner.  
LONDON, ●●●●●ed by Anne Godbid, and John Playford, for William Fisher, at the Postern-Gate near Tower-Hill; Robert Boulter, at the Turks-Head, and Ralph Smith, at the Bible in Cornhill; Thomas Passenger, at the Three Bibles on London-Bridge; and Richard Northcot, next St. Peter's- Alley in Cornhill, and at the Anchor and Mariner, on Fishstreet-Hill.   

TO THE READER.  
THE famous and ever to be admired Art of Navigation, having been so learnedly handled and written of, not only in all other Languages, but also in our Mother-Tongue, by so many learned and able men, both of former and of our present Age, that it may seem impossible to write any thing more thereof that hath not already been done by others; yet in my experience which I have seen, being at Sea in several Vessels, where divers young Mariners have been: I have heard this general Complaint among them, that though the things that are chief useful for them in their Art may be found in several Books here and there dispersed; yet they could wish that there were such a Book contrived, that might be soley useful for them entire by itself, which then would be more convenient for them, and might be purchased at a more reasonable rate than otherwise they could be, by buying of so many sorts of Books, which they must be constrained to do, if ever they intent to be able Proficients in that most noble Profession of Navigation, at which they chief aimed.  
This Complaint of theirs, was one chief Motive which induced me to collect and compose the subsequent Treatise, which I have endeavoured to handle in such a methodical manner, as it ought to be read and practised by the young Seaman. For  
First, There is a Treatise of ARITHMETIC, containing all the Rules thereof, which are necessary for the Seaman to know and practise, all or most of the Questions thereof being made applicable in one kind or other to Nautical Affairs.  
Secondly, There is a Treatise of GEOMETRY, containing the first Grounds and Principles thereof, with the making and dividing of the Mariner's Scale and Compass, with the projecting of the Sphere in Plano, and the resolving of many Questions in Astronomy, which are useful in Navigation thereby.  
Thirdly, you have a Treatise of the practic part of NAVIGATION, wherein is showed, after a new experienced way used by the Author, how to keep a Reckoning at Sea: the making and use of the Plain Sea-Chart: the Doctrine of plain Triangles made applicable to such Questions in Navigation as concern Course, Distance, Rumb, Difference of Longitude, and Departure: likewise, a Table of Meridional parts to every third Minute, and the application and use thereof exemplified in Questions of Sailing by Mercator's Chart: with Tables of Longitude and Latitude of Places, of right Ascension, and Seasons of certain Fixed Stars, with Rules to keep a Reckoning, and to find the Latitude by the Meridian Altitude of the Sun or Stars.  
Fourthly, There is a short Treatise of ASTRONOMY, wherein you have the Doctrine of Spherical Triangles applied to Questions in Astronomy and Navigation.  
This is the brief Sum and Substance of the following Treatise, which I commend to the Practice of all young Seamen, desiring their kind acceptance of these my first Labours, which if I shall find to be kindly entertained by them, it will encourage me to launch farther into the more nice and critical part of this most noble Science. In the mean time I commend this to them, wishing good success in all their honest and laudable undertake, and in the interim bid them Farewell.  
MATTHEW NORWOOD.   

THE CONTENTS.  

Of ARITHMETIC.  
NUmeration 1  
Addition 2  
Subtraction 5  
Multiplication 7  
Division 10  
Reduction 16  
The Golden Rule or Rule of Three 19  
The Rule of Interest and Interest upon Interest 24  
The Rule of Fellowship 25   

Of GEOMETRY.  
Geometrical Definitions 28  
How to raise a Perpendicular 30  
How to divide a line into two equal parts 31  
How to raise a Perpendicular on the end of a line ibid.  
From a Point given to let fall a Perpendicular 32  
How to draw Parallel lines 33  
How to make a Geometrical Square 34  
How to make an Oblong or Square, whose length and breadth is given 34  
How to make a Diamond-figure, whose Side and Angles shall be limited 35  
To make a Rhomboiades, whose Sides and Angles are given 36  
To find the Centre of a Circle 37  
To find the Centre of that Circle which shall pass through any three given points, which are not situate in a straight line 37  
How to divide a Circle into 2, 4, 8, 16, 32 equal parts 40  
The Projection of the Mariner's Compass 41  
The Projection of the Plain Scale 42  
The Projection of the Sphere in Plano 45  
The Names and Characters of the Signs with the Months they belong to 48  
How to project the Sphere 49  
The meaning of certain Terms of Art 51  
How to find the Sun's Meridian Altitude 55  
How to find the Amplitude of rising or setting ibid.  
How to find the Azimuth at six of the Clock 57  
To find the Altitude at six 58  
To find the Altitude the Sun being due East or West 59  
How to find the Ascensional Difference ibid.  
To find the time of the Suns rising or setting 60  
To find what hour it is, the Sun being due East or West 61  
To find the time of Day breaking and Twilight ending 62  
To find the Continuance of Twilight ibid.  
To find the length of the longest day in any Latitude 63  
To find the Sun's Place and right Ascension 68  
To find the Sun's declination, etc. 69  
Terms of Art used in Navigation explained 70  
Propositions of Sailing by the Plain Scale 72  
Questions in Navigation resolved, from page 73 to page 92  
Of a Travis 92  
The manner of keeping a Reckoning at Sea 97  
Concerning the Variation of the Compass 102  
The Use of a Plain Sea-Chart 108   

Of Obliqne TRIANGLES.  
THe application whereof are in ten Questions perfectly explained, all which Questions are applied to Questions in Sailing, and are wrought both Geometrically by the Plain Scale, and also by the Tables of Sines, Tangents, and Logarithms from page 113 to page 146  
A Table of Meridional parts to every third minute 146  
A Declaration of the Table 158  
The Use of the Table of Meridional parts exemplified in five Questions appertaining to Navigation, from page 160 to page 166  
Of the Longitude and Latitude of Places. A Table of them 167  
The Use of them 171  
How to keep a Reckoning of the Longitude and Latitude a Ship makes at Sea 172  
The Names, Declinations, and Seasons of certain Fixed Stars near the Equinoctial 179  
The like for Stars near the North Pole 180  
A Table showing how much the North Star is above or below the Pole for every several Position of the former or greater Guard 181  
Rules to find the Latitude by the Meridian Altitude of the Sun or Stars 182   

Of ASTRONOMY.  
Wherein the Application and Use of the Doctrine of Spherical Triangles is exemplified in the Resolution of several Propositions of the Sphere, which appertain to Astronomy and Navigation, from page 185 to page 203    

OF ARITHMETIC.  


THE TABLE.  7000000000000 Millions of millions. 800000000000 Hundred thousand millions. 90000000000 Ten thousand millions. 6000000000 Thousand millions. 500000000 Hundred millions. 30000000 Ten millions. 2000000 Millions. 800000 Hundred thousands. 40000 Ten thousands. 5000 Thousands. 700 Hundreds. 10 Ten. 1 Vnits.   

The Use of the Table to number.  
THis Table signifies thus much, That if there be one Figure alone, it is but so many units as there is in its name, as 7 is seven ones or units. If there be two Figures, the first Figure towards the left hand is as many ten as there is units in its name, and the other Figure is units: As if it were 10, that is ten units: if it were 23, the first Figure is two ten, that is twenty units: and the next Figure is 3, that is three units. The next place is hundreds, and consists of three Figures, as 700 is seven hundred: 723 is seven hundred twenty three. And thus you may count any Number by observing the place of any Figure, and in their places giving them their names as you count: As suppose I have a Number, namely 7864319, if you tell from 9 backwards to 7, you will find 7 in the place of millions, then begin and say seven millions: then if you count again, you will find 8 in the place of hundred thousands, 6 in the place of ten thousands, and 4 in the place of thousands, then to give them their proper Number, you will read thus: Seven millions for 7, eight hundred thousand for 8, sixty thousand for 6, four thousand for 4, that is, Seven millions eight hundred sixty four thousand; then for 3 three hundred, for 1 ten, and for 9 nine units, that is then altogether 7 millions 8 hundred 64 thousand 3 hundred and ninteen, which is the quantity of units that is in the given Number 7864319. And thus much for Numeration.   

Addition.  
ADdition teacheth how to bring several Sums into one total, and is done as in the Example, thus: Suppose there be several Persons indebted to me, and they own as followeth: 
Example.  A oweth 7272 B oweth 2732 C oweth 3399 D oweth 3999 Total is, or debt 17402  
I demand what the debt is? Add the numbers together thus, begin at the right hand, and for every ten carry one to the next row, adding them up in their several rows, till you come to the last row on the left hand, and as many ten as is in that after it is added up, so many set down by it in the next place on the left hand.  
Example.  
In the first row towards the right hand, say 9 and 9 is 18, and 2 is twenty, and 2 is twenty two, which should be expressed thus 22: but because the next place is the place of ten, you set down 2 units, and carry the other two (which is two ten) and add them to the next row of ten, by saying 2 that I carried and 9 is eleven, and 9 is twenty, and so forwards, still doing so till you come to the last row, and as many ten as there is in that, so many set down to the left hand: as here it came but to 17, that is but one ten besides the 7: if they come to even ten (as here one row came to thirty) set down 0, and carry the ten to the next place, except it be in the last ro● to the left hand, and then set down 0, and your number of ten to the left hand of it.   

The Characters used in Arithmetic.  

For pounds l.  
For shillings s.  
For pence d.  
For farthings q.  
For degrees deg.  
For minute's ′  
For seconds ″  
For hours ho.  
Minutes of time ′  
For hundreds C.  
For quarter's qu.  
For ounces oun.    

Addition of Degrees and Minutes.  
NOte that sixty Minutes is one Degree, three Miles is one League, sixty Seconds is one Mile or Minute, sixty Minutes of time is one Hour in time: fifteen Minutes of a Degree is one Minute in time, twelve Hours is an artificial Day, twenty four Hours is a natural Day, fifteen Degrees is an Hour in time, four Minutes of time is one Degree, three hundred and sixty Degrees is the Circumference of a Circle, one hundred and eighty Degrees is half a Circle, or a Semicircle; ninety Degrees is a Quadrant, or a quarter of a Circle, and eleven Degrees fifteen Minutes is one point of the Compass, two and thirty times eleven Degrees fifteen Minutes is the Circumference of a Circle; so that there is two and thirty points of the Compass in every Circle: one point of the Compass is three quarters of an Hour in time, or 45 Minutes.  
Admit I had kept an account of a Ships Difference of Longitude in Degrees and Minutes, that had been out eight days, and had made Difference of Longitude each day, as followeth: 

The first day 2 deg. 20′  
The second day 2 deg. 45  
The third day 1 deg. 30  
The fourth day 1 deg. 11  
The fifth day 2 deg. 39′  
The sixth day 1 deg. 29  
The seventh day 2 deg. 19  
The eighth day 2 deg. 10    
I desire to know (if it be all one way) what number of degrees and minutes it is in one sum.  
Begin as you did in Addition, and say 9 and 9 is 18, and 9 is 27, and 1 is 28, and 5 is 33, set down 3 and carry 3: then say 3 that I carried and 1 is 4, and 1 is 5, and 2 is 7, and 3 is ●0, and 1 is a 11, and 3 is 14, and 4 is 18, and 2 is 20: cast out all the six from 20, and there remains 2, set it down, and as many six as you cast away, so many units carry to the place of Degrees (which was three six) and say 3 that I carried and 2 is 5 and so forwards in that as in any other Sum in Addition: it being thus added, I find that it comes to 16 d. 23 m. the whole. 2 deg. 20′ 2 deg. 45 1 deg. 30 1 deg. 11 2 deg. 39 1 deg. 29 2 deg. 19 2 deg. 10 16 deg. 23  
Now you may ask a reason why I cast away the six, the reason is because 6 ten make a degree: now with the three ten that I carried, and the ten that where in the row added up last, there were 20 ten, that must be 3 degrees and 2 ten over: the 3 degrees I carry to the place of degrees, and the two ten I set down in the place of ten: and thus in any other Sum.   

Addition of Hours and Minutes.  
ADmit I have these several numbers of hours and minutes to cast up in one sum, namely, then because 60 minutes is one hour, I do just as I did before: be sure in the place of minutes set never above 60, for if you do, it is more than a degree. ho. min. 41 07 22 11 12 29 18 32 15 15 109 34  
The same is to be done in Addition of Money, only remembering in the place of pence to go no farther than 12, because twelve pence make a shilling, as you do here to 60, because 60′ makes a degree, and in shillings no farther than twenty, because twenty shillings make a pound. And thus in any thing else (provided that you know how many of the one makes one of the other) but for such things they are not so necessary to the thing we purpose to handle, and therefore I refer you to Mr. Hodder's Book, it being of small price, and much matter, and likewise very plain.   

Subtraction.  
SUbtraction is the taking a lesser number from a greater, and to see what remains: As suppose a man was bound from Cales to Dover, which is 21 miles: now if he have sailed 11 miles of the way, I demand what he hath yet to sail? Take 11 from 21, and the remainder is 10, so I conclude he hath 10 miles still to sail before he come to Dover: but in greater numbers they are ordinarily set so that you are forced to borrow, so that they are something harder.  
Suppose the distance between two places be 1799 miles. Now I have sailed of these miles 1227 I demand what number of miles I have still to sail. Still to sail 0572  
Set the greatest number uppermost, and the least under it, and draw a line between them, and a line under them both, and say, 7 from 9 and the remainder is 2, set down 2: then 2 from 9, there remains 7 set it down: then 2 from 7, and the remainder is 5: whatsoever number is underneath so much is still to sail, as here 572 miles.  
But now I'll give you an Example wherein the figures in the lesser Sum, are some of them bigger than in the greater, from which they must be taken: now in such a case borrow ten, and add to the upper figure.  
Example.  
A oweth to B 7822 pound of Tobacco, now he hath paid 4998 l. of it: I demand what is still due. Whole Debt 7822 l Received 4998 Still due 2824 Set down your sums as in the former directions, and say 8 from 2 I cannot, but 8 from 12, and the remainder is 4: set it down, 1 that I borrowed and 9 is 10, from 2 I cannot, but 10, from 12, and the remainder is 2, set down 2 and carry 1, then 9 and 1 that I carried is 10, from 8 I cannot, but 10 from 18, and the remainder is 8 set down 8 and carry 1: then 1 that I carried and 4 is 5: 5 from 7 and the remainder is 2, set it down, and thus you may do any sum of one denomination.   

Subtraction of Degrees and Minutes.  
THis being of several Denominations, observe as before in Addition (of degrees and minutes) how many of the one makes one of the other: and if the under figures be bigger than the uppermost (in the place of minutes) borrow from 60 the denominator of a degree, and subtract from it, and that remainder add to the upper figure, and set it down: but in the place of degrees, if the under number be bigger, do as you did before.  
I have made in my Reckoning— Difference of Latitude North 32 deg. 48′ Difference of Latitude South 29 deg. 59  02 deg. 49 I demand what way I have differed my Lat. most, and how much? I have differed my Latitude and it is Northerly, because my Northing is more than my Southing.  
Say 9 from 8 I cannot, but 9 from 18 there remains 9, set it down; 5 and 1 that I borrowed is 6, 6 from 4 I cannot, but 6 from 6, and the remainder is 0, that 0 and 4 is 4: set it down, then 1 that I borrowed and 9 is 10, from 2 I cannot, but 10 from 12, and there remains 2, set it down: 1 that I carried and 2 is 3, 3 from 3 and there remains 0, so that 32 deg. 48 min. is 02 deg. 49 more than 29 deg. 59  
The same may be done in Subtraction of pounds, shillings, and pence, observing but how many of the one goes to make one of the other: be sure in setting down a sum, you do not set more minutes in the place of minutes than 60, or pence in the place of pence than 12, or so for any thing else set no more than goeth to make one in the next denomination.   

Subtraction of Time.  
LEt the greatest number of time be 272 hor. 22 min. Let the lesser number of time be 199 hor. 49 min. The Remainder of time is 072 hor. 33 min.  
You may try this over: it is done as the former, and needs no demonstration (as I said before) if you observe in any case how many of one makes one of the other, it is but borrowing from that number, as we have done here.   

The TABLE.  
2 times 2 is 4 2 times 3 is 6 2 times 4 is 8 2 times 5 is 10 2 times 6 is 12 2 times 7 is 14 2 times 8 is 16 2 times 9 is 18 3 times 3 is 9 3 times 4 is 12 3 times 5 is 15 3 times 6 is 18 3 times 7 is 21 3 times 8 is 24 3 times 9 is 27 4 times 4 is 16 4 times 5 is 20 4 times 6 is 24 4 times 7 is 28 4 times 8 is 32 4 times 9 is 36 5 times 5 is 25 5 times 6 is 30 5 times 7 is 35 5 times 8 is 40 5 times 9 is 45 6 times 6 is 36 6 times 7 is 42 6 times 8 is 48 6 times 9 is 54 7 times 7 is 49 7 times 8 is 56 7 times 9 is 63 8 times 8 is 64 8 times 9 is 72 9 times 9 is 81  
This Table is so familiar to any one, that I need to say but little concerning it; when you read it, read thus; 2 times 2 is 4, 2 times 3 is 6, and so along to 2 times 9 which is 18, then go to the next, and say 3 times 3 is 9, than 3 times 4 is 12, and so forwards. Now I wish the Learner to have these Tables perfect by heart, before he proceed further.   

Multiplication.  
IT doth Addition many times, for when you have many Additions it makes one work of it, and also of two given Numbers it increaseth the greater as often as there is units in the lesser, or the lesser as often as there is units in the greater. Now in multiplying there are three things considerable.  
First, The Multiplicand, or Number that you would multiply.  
Secondly, The Multiplier, or Number you multiply by.  
Thirdly, The Product or Sum produced.  
Quest. 1. I demand how many Feet there are in 320 Yards, every Yard being three Feet?  
If you set down 320 yards one under the other three times, and add them, you will have the sum of 320 yards in feet: but that is a tedious way, for than you must set down your 320 yards an hundred times one under the other, if there were one hundred times so much (of any thing) in it: but now Multiplication doth it at once; because there is 3 foot in a yard, therefore there must be three times 320 foot in 320 yards.  
Multiplicand 320 Multiplier 3 Product 960 Example. I set down 320 and 3 under the first figure on the right hand, and say 3 times 0 is 0, set down 0, than 3 times 2 is 6, set down 6: then 3 times 3 is 9, set down 9, so the Product is 960 feet, which answers your demand. Note, that when your Multiplication cometh to more than 9, as many ten as are in it, so many you carry to the next place, (as you did in Addition) and set down the rest, as you shall in this Example following.  
Quest. 2 I demand how many Minutes is in 360 Degrees?  
Note that 60 minutes is one degree: it is less work always to make your greatest Number your Multiplicand, and then it stands as in the Margin: then say 0 nothing is 0, set down 0; then 0 six is 0, set it down; then 0 three is 0, set it down: then go to 6, and say 6 times 0 is 0, set it down, one figure to the left hand under the former; then say 6 times 6 is 36, set down 6 and carry 3; then say 6 times 3 is 18, and 3 that I carried is 21, set down 1 and carry 2; now because there is no more figures, set down the two (as I have showed in Addition) and thus 360 degrees is 21600 minutes, if both the Products be added together. Then to prove your work, add up your Multiplicand, and cast away the nine from it, and set the remainder down over the Cross; then cast away the nine of the Multiplier if it comes to 9, but if not, set it down under the Cross; then multiply 6 by 0, and it makes 0, set down that on the right side of the Cross; then cast up your Product, and cast away the nine from it, and if it comes to the same that the Multiplicand and Multiplier did, (the nine being cast away) then the sum is done right, and stands as you see in the Margin. I shall here following show two Examples, and let down two more for your practice: if you do not understand what is said, examine these Examples which are wrought, and they will satisfy you.  
Quest. 3. How many units is 612345 times 62453125 units?  
Quest. 4. How much is 123456 times 45321672?  
Let these be the two Questions which here following are wrought one after the other.  
 
The Product of the first Question 38242858828125  
 
The Product of the second Question 55952640636024  
Quest. 5. How many units is in 3438239 times 5289231423?  
Answ. 18185641758584097.  
Quest. 6. How many units is in 289764532172 times 343217812?  
Answ. 199452348727277447664.   

Division.  
DIvision is that by which one may know (at one operation) any part of a whole Number, provided that the part be not under an unit: by part I mean thus, the third of a number, or the fourth of a number, or the twentieth part of a number, or the like.  
In Division there are four things considerable: First, the Dividend; Secondly, the Divisor; Thirdly, the Quotient; And fourthly, the Remainder after the Division.  
The Dividend is the number which you desire to have divided, or to have a part known.  
The Divisor is that which you divide it by, or the name of the part which you would have known.  
The Quotient is the number of units that is contained in the part required.  
And the Remainder of the Division is that part under an unit, which is wanting to the true portion which is left in the Quotient.  
Your Sum when you set it, stands thus: 3478 (224  
Quest. 1. A man promised to do a Piece of work for me in a Quarter of a Year: Now if there be 365 Days in a Year, I desire to know how many Days is in this Quarter?  
The fourth part of 365 days must needs be the number of days which this man will be doing the Piece of Work, therefore divide 365 days by 4, and the Quotient shall be the whole days, and the remainder of the Division shall be the parts of a day above the whole days. 91 days, and one fourth of a day is a quarter of a year, or the quantity of time he will be doing this Work. We express a Fraction thus: one fourth ¼, two fifths ⅖, or so for any Fraction.  
But you will ask how this is done? 365 (4 Which is thus: First, set down your Dividend 365; then to the left hand set down the Divisor under your Dividend. but you will ask why I did not set it under the 3? And the reason is, because 4  more than 3, so that I could have had nothing from 3 till I had removed it where it is: if 3 had been a number that I could have took my Divisor from, I would have set the 4 under the 3.  
Your Sum being set, say how many times 4 can you have in 36? Nine times. Set down 9 in the Quotient, and multiply 4 by 9, saying 9 times 4 is 36, from 36 and the remainder is 0: cancel 4 and 36 as you go (by a dash with your pen) and your work stands thus:  
Then set your 4 under the 5, and say how many times 4 can you have from 5? Once. Set down 1 in the Quotient, and multiply it by 4, saying, once 4 is 4, from 5, and the remainder is 1: set 1 over the 5, cancelling your 4 and 5 by a dash with your pen (as you speak:) now because there is no more figures, your sum is done, and stands thus: The 1 which remains is ¼ of a unit (because the Dividend is units) for whatsoever the Dividend is, the Quotient and the Remainder shall be a part of it: it hath its Denomination or Name from the Divisor 4.  
Quest. 2. A man had an Estate of 78998 l. now he oweth to 23 several men 27821 l. apiece. I demand what each man shall have of the Estate, so that none of them may be wronged?  
You will conclude because they were owed alike, they should have alike of that which was left, therefore divide the Estate by 23, and the Quotient with the Remainder will be each man's due.  
Each will have 3434 l. 16/23 of a pound for his Debt.  
This is Division of two figures when the Divisor is two figures, and it is something harder than the other Example: The working of it is thus, first set your Sum in order thus: 78998 (23 Then I say, how many times 2 can you have from 7? Three times: 3 times 2 is 6, from 7 and there remains 1 (set 3 in the Quotient) and say again 3 times 3 is 9, from 18, and there remains 9, than your Sum stands thus:  
You might have said 3 times 3 is 9, from 8 I cannot, but 9 from 18, and the Remainder is 9, and 1 (for your borrowed one ten) from 1, and there remains 0, which is most easy for memory.  
Then set your Divisor on place further towards the right hand in this form:  
How many times 2 can you have from 9? Four times: then 4 times 2 is 8, from 9, and there remains 1, and 4 times 3 is 12, from 19 and the Remainder is 7: and 1 (for the ten) from 1, and the Remainder is 0; and then your Sum stands thus:  
Then remove your Divisor to the next place, and see how many times 2 you can have out of 7, which is 3 times: 3 times 2 is 6, from 7 and the remainder is 1: and 3 times 3 is 9, from 9 and the remainder is 0; and your Sum stands thus:  
Then remove your Divisor to the next place, and see how many times 2 you can have from 10? which is 5 times: now 5 times 2 is 10, from 10 and there remains 0, and 5 times 3 is 15, from 8 I cannot (you see 8 is all that is left, so that you have nothing to horrow from) therefore you could not take 5 times 23 from 108, try for 4 times, 4 times 2 is 8, from 10, and the remainder is 2, and 4 times 3 is 12, from 28 I can, you see you may take 4: set 4 in the Quotient, and say 4 times 2 is 8, from 0 I cannot, but 8 from 10, and the remainder is 2, and 1 that I borrowed from 1, and the remainder is 0, (cancel as you speak) then 4 times 3 is 12: now 12 from 28, and the remainder is 16; but it is better to use yourself to this method, namely, to take your units from the place of units, and your ten from the place of ten, as here 2 is in the place of units, and 8 above, therefore say 2 from 8, and there remains 6, and 1 (for the ten) from 2 and the remainder is 1: Now if this had fallen out so that the upper number had been 21, and the other 12, you know that 12 will be taken out of 21, but 2 would not be taken out of 1; now in such a case borrow one 10, and say as in Subtraction, 2 from 1 I cannot, but 2 from 11 and the remainder is 9, one 10 that I borrowed and 1 is 2, from 2 and there remains 0, and thus it is done at once: The reason why it is done this way is, because the Sum you are to take from may be big, so that a man cannot tell readily what the remainder will be without Subtraction; for this differs nothing from Subtraction, as it is showed before, but only that you carry one number in your head, whereas there both the numbers are before you.  
The hardest Sum in Division hath but these Difficulties in them:  
First, Be sure take no more to set in the Quotient, than your Divisor multiplied by it) will come under the Number you are to take it from, or equal to it.  
Secondly, Be sure you take the one as many times as you can from the other.  
Thirdly, Be careful in takeing one number from another, to use this way of Subtraction.  
This Sum is done, and stands thus: And it signifies that the 23. part of 78998 is 3434 units and 16/23 of a unit.  
Here following is a Sum of four figures done, and the way to prove any Sum.  
34771262 (2345  
This Sum is set down in its several Operations. From what hath been said already, I suppose you may be able to examine it; the like is to be understood of any Sum else in Division, therefore I shall say no more, only show how to prove them when they are thus wrought.  
The best way to prove Division is by Multiplication, for if you multiply your Quotient by the Divifor, and add in the Remainder, that product will be equal to the Dividend, if the Sum be right, and it stands to reason it should be so, for if the Quotient with the remainder be in this Sum the 2345th. part of the Dividend, then 'tis evident that 2345 times that Quotient must be the same that the Dividend is (with the remainder which is added in.)  
Example. Quotient 14827 Divisor 2345  74135  59308 Remainder 7  44481 4  26654 9   1 Divid. 34771262  
Here you see the Example of it, and the Product the same that the Dividend of the Sum divided was, and this I say is the best way for young Practitioners to prove Division, because this makes them perfect in Multiplication.  
There is a shorter way to prove any Sum in Division, and that is this: Cast up your Quotient, and cast away all the nine from it, (as you do in Multiplication) and what remains set down upon one side of a Cross thus:  
Then cast away your nine from your Divisor in like manner, and there remains 5, set it down on the other side against 4 thus:  
Multiply them one by the other, and they produce 20, cast away all the nine from 20, and the remainder is 2, keep this 2 in mind, and cast up the remainder of the Division, and cast all the nine from it: which if it be done in this Sum, afterwards there will remain 3, which add to the 2 that was in your mind, and it makes 5, which set down thus:  
Then cast away all the nine from your Dividend, and the remainder will answer to the last number you set down in the Cross, if the Sum be right, which here it is   

Reduction.  
REduction showeth how to bring gross or great Denominations into small, or small into great.  
Quest. 1. Suppose a Man had been at Sea twenty five Years, and for every Minute of that time he was to have on Farthing, I demand how many farthings is due to him?  
Here I see how many minutes is in that quantity of time, and so many minutes so many farthings there is due, which is, 13140000 farthings. Now to find this, I first multiply the years by the days contained in one year, which is 365, and that produceth 9125 days, the number of days in that number of years: this multiplied by 24 gives for its Product 219000 the number of hours in the whole time, because 24 hours is a day. That Product multiplied by 60, is the number of minutes in the whole time (because every hour is 60 minutes) which is the last Product, or the thing required, which I set down as the Answer to my Question thus, Facit 13140000 farthings.  
Quest. 2. Pray tell me what these farthings come to in Pounds, Shillings, and Pence, 1740120 Farthings.  
 
Four farthings make a penny, therefore I divide the whole Sum of Farthings by 4, and the Quotient is ½ of the number of farthings, or the farthings reduced into pence, 435030 d. of which Sum I take the 1/12 by dividing it by 12, which is the next Quotient (the number of shillings that is the farthings) which is 36252 s. 6 d. Twenty shillings make a pound, therefore I take the twentieth part of this last Product, by dividing it by 20, and that Quotient shall be pounds, and the remainder of the Division shillings: 1812 l. 12 s. The Answer of your desire I will set down thus: 1812 l.— 12 s.— 6 d.  
Quest. 3. Some say that in a Mile there is 1760 Paces, three Feet being one Pace, twelve Inches one Foot, three Barley-corns being one Inch: I pray tell me how many Barley-corns will reach three Mi.  
 
Because there is 3 miles, 3 times 1760 is the number of paces contained in 3 miles: then because 3 foot is 1 pace, multiply 5280 by 3, and it produceth the number of feet that is in 3 miles; then because 12 inches is a foot, multiply your feet by 12, and it produceth 190080, the number of inches contained in 3 miles: and because 3 Barley-Corns make an inch, multiply your inches by 3, and it gives you the number of barly-corns that are in 3 miles, and the same is to be understood for any other number of miles.  
Suppose I would know how many Miles is in the same number of Barley-corns, namely, 570240 Barley-corns.  
As before you multiplied to bring Miles into Barley-corns, so now divide to bring things of small Denomination into great, only begin at the lowest, where before you ended, for the demand of this is the beginning of the other.  
Example.  
 
Here the Quotient of the first Division is Inches, because three Barly-corns make an Inch, namely, 190080 Inches.  
The Quotient of the second Division is the number of Feet contained in those Inches, (and therefore the Inches were divided by 12) namely, 15840 Feet.  
The Quotient of the third Division is Paces, and therefore it was divided by 3, the number is 5280 Paces.  
The Quotient of the last Division is the number of Miles contained in either 570240 Barly-corns, or 190080 Inches, or 15840 Feet, or 5280 Paces, which is 3 Miles, the thing required: And after the same manner you may reduce minutes into degrees, or seconds into minutes and degrees, or degrees into hours, pounds into shillings, shillings into pence, and pence into farthings, or the contrary, by knowing how many of the one makes one of the other.   

The Proof of Reduction.  
The Proof of Reduction is the same thing which we did last, I framed it so for brevity's sake. When you would prove any Sum that is brought into a small Denomination (by Multiplication,) divide it backwards, beginning with the Product of your last Multiplication, and compare them in each as you do it, and if the Sum be right they will agree.  
When you would prove any Sum that is brought into a greater Denomination (by Division) then as before you divided, so now multiply; for I have showed that Multiplication and Division is the Proof one of the other.  
Quest. 4. I will reduce 28 degrees 20′ 40″ into seconds.  
First Multiply your degrees by 60, and it brings them into minutes, add your odd 20 minutes to that Product, and it makes the number of minutes in that given number of degrees and minutes: multiply that by 60, and it produceth the number of seconds contained in that given number of degrees and minutes: to that Product add your odd 40 seconds, and it gives the number of seconds contained in the degrees, minutes, and seconds, as here in this Example: and the reason is, because 60 minutes is a degree, and 60 seconds is a minute: The same is to be understood of the reducing of any number of several Denominations into one. 82 Degrees, 60 4920 Minutes, 20 added. 4940′ in all. 60 296400 Seconds. 40 added. 296440 in all. 
Facit 296440 seconds in 28 degrees, 20 minutes, 40 seconds.   
Thus much for what Mariners have need to know in Reduction.  
Note, That the Remainder of any Division either in Reduction, the rule of Three, or in any Division, is some part of a Unit, namely thus: If the Quotient be shillings, the remainder is some part of a shilling: If the Quotient be pounds, the remainder is some part of a pound. The remainder is the Numerator, and the Divisor is the Denominator of the Fraction; so that if the Quotient were pounds, and the remainder 12, the Divisor is 20, and the Fraction is the 12/20 of a pound, which is 12 shillings; the like for any thing else.   

The Golden Rule, or Rule of Three.  
THis Rule well understood, is the Sum of the rest that follow it; for they are all (but as it were this Rule) it is called The Golden Rule, because it is the Foundation of the rest: It is called The Rule of Three, because three numbers are given to find a fourth.  
Now there is the Rule of Three direct, and the Back Rule of Three: the Rule of Three direct, is for such questions as are so stated, that the third number and second multiplied together, and that Product divided by the first gives the fourth.  
The Back Rule of Three is so stated, that the first number and the second are multiplied together, and that Product divided by the third, finds the fourth.  
And this is all the distinction that I can make between the forward and backward Rule of Three, which is all the Rule of Three: I will not make therefore two distinct Rules of them, (but one) therefore pray observe this difference:  
In all Questions in the Rule of Three there is three things given to find a fourth, which number will have alike proportion to the third, as the second hath to the first.  
In stating your Question, be sure let it be so that you may set your sum in order with the first and third number, speaking both of one thing: I mean if the first be els let the third be els; if the first be Degrees, let the third be degrees, or something that Degrees can be reduced to.  
If your Sum be set with the numbers in several Denominations, before you bring your Sum in order, you must reduce your numbers into the least Denomination: As suppose your Sum was this, If 5 pound 10 ounces of Tobacco, cost 10 shillings 9 pence, what shall 510 pounds cost? Reduce your 5 pounds 10 ounces into ounces, and your 510 pounds into ounces: likewise your shillings into pence, taking in the odd pence and odd ounces, as I have showed how to do it in page 16 and 17.  
Your Sum being reduced (as I have showed) into the least Denominations, set it in order thus:  
If 90 Ounces cost 129 pence, what costs 81600 ounces?  
Your Sum being in order, consider whether your third number require more or less: if it require more, multiply the second number by the greatest of the two extremes, and divide that Product by the lesser, it produceth the fourth number in the same Denomination that the middle number was.  
If your third number require less, multiply the middle number by the least of the two extremes, and divide that product by the greatest, and it gives your fourth number in the same denomination that your second number was; then if you desire to reduce it into greater Denomination, do it as I have showed already.  
For your better understanding of this Rule of Three, I have here following done an Example of it in right Lines, and have found it upon this Question.  

Quest. 1. If 10 pounds of Chocolato cost 15 shillings the first penny, what shall 28 pounds cost?  

Arithmetically.  
If 10 l. cost 16 s. what will 28 l. cost?  

Facit 42 s. or 2 l. 2 s.   
Here is no need of bringing things into lower Denominations, because the numbers are in one Denomination already.   

Geometrically.  
Draw two lines that may touch one the other at pleasure, as the lines A S and A E touch one an other in A, then upon the line A E, set 10 l. fixing one foot of your Compasses in A, (I take the 10 from a scale of equal parts) from the same scale I take 15 s. and fix one foot  depiction of geometrical figure of my Compasses in A, and set it off upon the line A S, and draw a line from 15 to 10, then burelakorkupon the line A E, I set 28 (from A) and from 28 I draw a line parallel to the line 10 15, than it is evident that if A 10 giveth A 15, A 28 will give A 42, because 42 28 is drawn parallel to 15 10, and the same holds in proportion: I marked it 42, because it is 42 from A.  
A line parallel to another, is when two lines go so evenly one by anoiher, that they be in all places equally distant.    

Quest. 2. If 27 men rig a Ship in 8 days, how many days shall 20 men be doing it?  
 
Here you see the last number required is more than the middle most, for 20 men must be longer about the work than 27 men, therefore I multiplied the middle number by the first, because it is bigger than the third.  
If in 10 days 16/20 of a day 20 men rig a Ship, how many shall do it in 8 days?  
Here the number of days must both be brought into 20 of days (that is such parts as 20 makes 1) which is done by multiplying the days by 20, now the reason is because the first number is of several Denominations, and must be brought into the lowest; which is the Denomination here mentioned.  
 
This is the other Sum set backwards as it were, and you see it produceth the same number that was given to do it in 8 days before, which assures you that your other work is right, and this is the way the Rule of Three is proved.   

Quest. 3. If in four hours the Sun goeth 60 degrees, how far shall he go in 24 hours?  
If 4 hours give 60 degrees, what shall 24 hours give?  
Facit 360 degrees.  
I need not prove this Sum, or any other, but leave it to your own practice to set it backwards and do it.  
If 22 deg. 30 min. of the Horizon be two points of the Mariner's Compass, how many points is 360 deg.  
 
360 deg. is the Circumference of the Horizon or of any Circle, and the Mariners have 32 points, that they have names to in the Compass, and no more; then from hence it is evident that 11 deg. 15 min. is one point, 22 deg. 30 min. is two. 33 d. 45 m. three, and so forth.  
I might be large in handling Arithmetic, but it is a thing that hath been very well handled by Mr. Record, and lately by Mr. Hodder, whose Works are cheap and very admirable: I should have said nothing of this Subject, if I had not found the want that Seamen have of it, by daily experience in them which I have taught: and seeing I have persisted thus far; I will but show how every Rule that followeth is worked from this Rule of Three.  
The Backward Rule you see is it.  
The Rule of Practice is the Rule of Three abbreviated.  
The Rule of Proportion, sometimes called the Rule of five Numbers, or by some the double Rule from its double working, is nothing but the Rule of Three  
The Rule of Interest is the Rule of Three, as also Interest upon Interest.  
The Rule of Fellowship is the Rule of Three, I'll do Examples of no more, but all Rules depend upon it: I'll begin with the Rule of five numbers.   

Quest. 4. If 100 l. in 12 months' gains 6 l. what shall 900 l. gain in 24 Months?  
If 12 months' gain 6 l. what shall 24 months' gain? I work the proportion , and I find that 100 l. will gain 12 l. in 24 Months.  
Then I say if 100 l. gains 12 l. what shall 900 l. gain in the same time? If you work it you will find it will gain 108 l. and this is the Rule of Three, for first you see what 100 l. will gain in the time that is set to 900 l. (provided it get 6 l. in 12 Months) and then you see what 900 l. will gain in the same time, (provided that 100 l. gains such a proportion as you find it to do.)    

The Rule of Interest, and Interest upon Interest.  
THis is the Rule of Three, for it runs thus:  
Quest. 1. I let 478 l. for two years, what shall the Use of it come to, after the rate of 6 l. a year, for the lent of 100 l. which is the common Use of it in a year.  
I'll set my Question in order thus:  
If 100 l. gives 6 l. what shall the principal 478 give?  
Facit 28 l.  
Here we have found what the Principle hath gained for a twelvemonth, next consider what proportion 12 Months is to the time your money was lent, which was 24 Months: now if it had been three times as long a time as 12 Months, you must have made it 3 times as much as you find it for 12 Months, but being but twice as much, you must make the Use to to be twice 28 68/100 which is 57 l. 36/100 of a pound; the same is to be understood of any number else; but in stead of seeing what proportion the time limited for the lent bears with 12 Months, you may work it as you do the Rule of five Numbers, by saying (after you have abbreviated your Fraction.)  
If 12 Months give 28 l. 68/100, what shall 24 months' gain? But I say the other way is best, after your other manner; If it had been one fourth of the time, it would have been ¼ of the use; if one fifth of the time, ⅕ of the use. For it stands to reason, that if 100 l. or any other quantity gains 6 l. in twelve Months. it gains four times as much in four times the time, or five times as much in five times twelve Months, and so forth.  
I might say something of Interest upon Interest, but that differs but little from this: At the years' end, add in your simple Interest to your Principal, and make it one entire Sum till the next year, then add that Interest to the Principal the year before, and receive the Interest of it: this is all it differs from the former.   

The Rule of Fellowship.  
THis Rule is the Rule of Three, done several times in one Sum or Question: for here are several Stocks, and several men that own them; now if the principal, or all their Stocks gain so much, how much shall each man gain according to the Stock he put in? I'll say, if the whole Stock gained so much, what shall the first man's Stock gain? and so for the rest.  
Quest. 1. Six men make a Stock, the first puts in 30 l. the second 40. l. the third 52 l. the fourth 58 l. the fifth 60 l. and the sixth 78 l. If the whole Stock together gains 200 l. what shall each man have, so that there may be no wrong?  
First, find the whole Stock by adding every man's Stock together, which is 318 l.  
Then say, If the whole Stock 318 l. gains 200 l. what shall the first man's Stock gain?  
 l. First 30 Second 40 Third 52 Fourth 58 Fifth 60 Sixth 78  318  
For the Second.  
If the whole Stock 318 l. gains 200 l. what shall 40 l. gain, which is the second Stock?  
The like is to be understood of the rest: I forbear to work them, as being out of my intentions: I leave their operation to your genius who never learned them.  
Quest. 2. There where five men made a Stock, the first put in 200 l. the second 59 l. the third 180 l. the fourth 78 l. the fifth 240 l. they lost at the return of the Ship 120 l. I demand each man's Loss proportionable to his Venture?  
First find the whole Stock, by adding every man's Venture together: this done, say for the first man's Loss, if the whole Stock lost 120 l. what shall the first man's Stock 200 l. lose?  
For the second, If the whole Stock lose 120 l. what shall the second man lose which put in 59 l?  
And so for the rest: The way to prove one of these Questions is less trouble than the Rule of Three, for after you have done, add all your Facits together, and it will make the same that was lost in the whole Stock, or gained in the whole Stock, (if the Question be for Gain) and this stands, to good reason; for every man's Loss or Gain must together be equal to the whole, or else they do not contribute to the Loss, or enjoy the Gain that they have lost or gained.  
There is more intricacy in this Rule than I have here cited, but the Foundation-work lies on the Rule of Three, and the intricacy that I speak of, will be understood with some consideration: I had thoughts to have showed it, but it is showed as well as can be (I think) by Mr. Record and Mr. Hodder: besides it would take up much paper, and I should digress from my intentions to Navigation.  
What I have showed since I treated of the Rule of Three, is only a show how all Rules depend upon this Golden Rule. Blame me not for being so large in Arithmetic; but both that, and all faults else, season with the Salt of a charitable Construction, remembering that Navigation is imperfect without it, I end.    

GEOMETRY.  
THat which I shall handle in Geometry, will be only that part of it which is used in Navigation: He that will treat of it at large, had best to put it in a Treatise by itself, and he that will learn to be a good Geometrician, let him apply himself to Euclid's Works. There is no Art but hath a dependency on it, and Navigation depends much on it, which enforceth me to treat upon some principles of it.  

Geometrical Definitions.  
A Point or Prick is this. (.) and is void of length, breadth or thickness.  
A Line is length without breadth or thickness, and is properly called the nearest distance (if a straight line) between two places: but if not straight but circular, it is termed an Archippus   depiction of geometrical figure 
A Triangle is when three lines meet, making three angular points; now there be two sorts, right lined, and spherical: spherical being all the sides Arches of great Circles. There be three sorts of Angles, namely, obtuse, acute, and right angles.   depiction of geometrical figure  depiction of geometrical figure  depiction of geometrical figure  depiction of geometrical figure 
If divers Circles be described having all one Centre, they be called Concentrics, but if they have divers Centres, they be called Excentrics.  
The Circumference of any Circle consists of 360 degrees, every degree being 60 minutes.  
The Diameter of a Circle is a line drawn from one side of the Circle to the other through the Centre, As A C.  
The Semidiameter of a Circle is half the Diameter, or the distance from the Circumference to the Centre.   

PROP. I. To raise a Perpendicular from the middle of a Line given  
THe line given is A B, divide it  depiction of geometrical figure into two equal parts at F, then open your Compasses to any convenient distance (above half the length of the given line) and fix one foot in B, and with the other describe the Arch C (with the same distance) fix one foot of your Compasses in A, and cross the other Arch by the Arch D (mind where they intersect) and from that place (which is ⊙) draw a line to the middle of your given line A B, namely to F, then is F ⊙ a true perpendicular from the middle of the line A B, for it is so directly from it, that it leans no way.   

PROP. II. To divide a Line into two equal Parts by a Perpendicular.  
THe given line is A B, fix  depiction of geometrical figure one foot of your Compasses in A, and setting them at any convenient distance, above half the given line, describe the Arches D and f, carry your Compasses to B (with the same distance) and describe the Arches C and e, so as they may cross the other Arches as they do in ⊙, mind it, and by those intersections lay your scale, and draw the line ⊙ ⊙, which will cut the given line in the midst, and be a perpendicular to it, which was required.   

PROP. III. To raise a Perpendicular to the end of a Line given  
HEre your given line is  depiction of geometrical figure A B, fix one foot of your Compasses in that end of your given Line, which you would raise your Perpendicular from, which is B, and extend your Compasses to any convenient distance, (as in the example, from B to R) and describe the Arch R c D, (continue the same distance in your Compasses) and fix one foot in R, extend the other upwards to c, then making c the Centre, describe the Arch 1, 2, 3, and beginning at R, set off that distance three times upon that Arch, as 1, 2, 3; from the place where your Compasses fall the third time, (which is 3) draw a line to the place where you began your work, which is B, and it is a Perpendicular to your given line.  

Another way.  
After you have described the Arch R c D, (with the same distance which described it) extend your Compasses from R to c: from c (with the same distance) describe the Arch u: which is part of the Arch R 1 2 3: then extend the same distance again which falls at D, and cross the Arch before described as N, and from the place of their intersection, draw your line to the end of your given line, and it shall be a Perpendicular: you see it is the same the other was.    

PROP. IU. To let fall a Perpendicular from a Point to a given Line.  
LEt the given line be D A,  depiction of geometrical figure the point from whence the Perpendicular is to be let fall, be at C.  
From the point C, draw a white line to the given line (by guess) as C A, divide it into two equal parts, which is done at B, then continuing ½ the line C A, which is A B or C B in your Compasses, and your Compasses fixed one foot at B, describe the Arch C D, and where it cuts the given line, there will your Perpendicular fall from the given point, for C D is Perpendicular to the given line D A.   

PROP. V To draw a Line parallel to a Line given.  
LEt the given line be A S: It is required to draw a line so, that the two lines may run (at both ends) one by the other and never meet, which is parallel one to the other. Open your Compasses to that extent as you would have the two lines asunder, and go towards one end of the given line as at S, and  depiction of geometrical figure describe the Arch u, and with the same distance come towards the other end as at A, and describe another Arch which is N, and by the top of these two Arches draw the line R O, which is parallel to A S.   

PROP. VI To draw a Line parallel to a given Line, from any Point assigned.  
LEt the given line be S L, the point assigned be A, take the distance from S to A, and carry it towards the other end of the given line as at L, describe the Arch n, then take the distance from L to S, and fixing one foot of your Compasses in  depiction of geometrical figure the given point A, cross the Arch n with the arch o, and by the place of their intersection, and the point assigned draw a line, which shall be parallel to S L.   

PROP. VII. To make a Square of a Line given.  
LEt the Line given be A, equal to which draw the side of the Square B E, and from one end of it raise a Perpendicular, and by it set off the length of A, as here from E the Perpendicular was raised, and the length  depiction of geometrical figure of A set upon it, which is the side of the Square E D, continue the same distance in your Compasses, and go to D, and describe the Arch 8: carry the same distance to B, and cross the Arch 8 with f, and from the intersection of those two Arches, draw the sides C D and C B, which makes the Square BEDC, & this is a true square.   

PROP. VIII. To make a Square, whose Length and Breadth is given.  
THese sorts of Squares are called Geometrical Squares, when but two sides are equal: namely, the two longest sides, or the two shortest sides: The Angles are all equal, namely right Angles.  
Suppose the Length of the Square be A, the breadth B, I desire to make it: first draw a line equal to A (for the length of it) as S V, then from any end of that Line raise a Perpendicular (as here from V) and set off the line B for the breadth of it (upon it) which falls in the Perpendicular line at L, then take the length of A, and describe the Arch n, (fixing your Compasses one foot in L) then take B the breadth, and cross that Arch by another, fixing your  depiction of geometrical figure Compasses (one foot) in S, and draw L K, and K S from the place of their intersection, as you did in the other: Thus the two opposite sides in this Square are equal, and the Angles in both all equal, for they are right Angles.   

PROP. IX. To make a Diamond Figure of a Line, and an Angle given.  
A Diamond Figure is a Figure of four equal sides, but the Angles are two of them acute, and two of them obtuse, the acute Angles are equal,  depiction of geometrical figure and the obtuse Angles are equal one to another.  
Let A B be the given line, C B the measure of the given Angle, (A being the angular point) first take the line AB, and draw a line of its length, for one side of the figure (namely 8 0) then take the Semidiameter of the Arch C B which is C A, and fixing one foot of your Compasses in 8, describe the Arch S 0, and take  depiction of geometrical figure the Arch B C, and set it off from O to S, then draw the line S 8 equal to 8 O, this done, keep the length of O 8 in your Compasses, and from S and O describe the Arches n and t, and draw the sides O R and S R, as you did in the other figures, and thus S R is equal to 8 O or R O is equal to 8 S, and the opposite Angles also equal. I forbear to show the reasons of their being equal, because it hath been handled by others, and indeed it is so plain, that with a little consideration you may know it.   

PROP. X. To make a Rhomboiades of two given sides, and an Angle included.  
A Rhomboiades is a figure whose opposite sides, and opposite Angles are equal (as a Geometrical Square is) but in this they differ: a Rhomboiades hath never a right Angle, but two obtuse, and two acute, whereas the other hath all right Angles; it differs from the Diamond figure also; for in one the sides are all equal, and in this but two equal sides: I need not show the working of it, because it differs not from a Diamond figure only in taking the two sides apart, to describe the Arches at L;  depiction of geometrical figure I suppose you may conceive how it is made, by seeing this which is here made: the given sides and Angle is s O N, I have not set down the Arch to measure the Angle at O. I suppose from what hath been said, you will conceive how that is.   

PROP. XI. To find the Centre of a Circle.   depiction of geometrical figure 
Draw a line from side to side of the Circle at a venture, as A C, and divide that line into two equal parts by a Perpendicular (as was showed before) that Perpendicular line draw through the Circle from side to side, as is u S, and it shall be the Diameter of the Circle, the half of which is the Semidiameter or very Centre ⊙.  
It is possible to find the Centre of a Triangle after the same manner: Suppose the Triangle, whose Centre you would find were A C n, divide any side into two equal parts by a Perpendicular, and it will go through the Centre of the Triangle, as the side A C is divided into two equal parts by the Perpendicular B S; then, I say, B S goeth through the Centre of this Triangle; but to find whereabouts in this Perpendicular the Centre of the Triangle is, I know by no other means, but by removing your Compasses in this line, from place to place till you find it, which is here found to be at u.  
But this is but a botchingly way, and with a little more labour you may find it at once, therefore mind this Geometrical Example.   

PROP. XII. To find the Centre of a Circle, of a Triangle, or any three Pricks that be not in a straight Line.   depiction of geometrical figure 
Here might be several Questions deduced from this Demonstration, but I'll content myself with one, and leave others that might be framed to your consideration.  
Quest. I have a Piece of Ground that hath three conveniences in it, which be these; the first is a Garden, now the Gate of this Garden is distant from a Gate that leads into an Orchard 300 Yards, that Gate is distant from the head of a curious River 350 Yards: Now if the head of this River be distant from the Garden-Gate 500 Yards, I demand where I may build me a House that may stand of a like distance from these Places, and how far it shall be from them?  
I consider that these places cannot be in a straight line, because then the House will not be of an equal distance from those Conveniences, and therefore if lines be drawn from one place from another, it will make a Triangle.  
The Operation.   depiction of geometrical figure 
Make a prick with your Compasses for your first station, which is G, then from a Scale of equal parts, take off the distance between the Garden-gate, and the Orchard-gate 300, (counting every equal part 100) and fixing one foot of your Compasses in G, set them upwards to O, which is the distance between the Garden and the Orchard, then take the distance between the Orchard-gate O and the River, which is 350 yards (or three equal parts and an half) and fix one foot of your Compasses at the Orchard-gate O, and with the other describe the Arch n t (which Arch will sweep the head of the River) lastly take the distance between the first place and the third, (which is the distance between the Garden and the River) 500 yards or 5 equal parts, and fix one foot in G the Garden-gate, and cross the Arch n t by the Arch u s, and draw the sides O K 350 yards, and G K 500 yards to the place of their intersection.  
Now because this House is to be built of an equal distance from every of these places, therefore the Centre of these three Conveniences (which are represented by the three pricks O K and G) must be the place where the House must be built, which is at L, and the reason is, because the distance between the Centre and any part of the Circumference is alike: I have showed already how to find the Centre.   

PROP. XIII. To divide the Circumference of a Circle into 2, 4, 8, 16, 32 equal parts.  
THe Circle to be divided is A B C D; first draw a line parallel (if you will) to the bottom of your Book as A C, through the Centre ⊙, which divides it into two Semicircles, divide that line into two equal parts by a Perpendicular (which Perpendicular will fall through the Centre) it is B D, then  depiction of geometrical figure is the Circle divided into four equal parts, divide them four equal parts (each) into two equal parts, and the Circle is divided into eight equal parts; divide them eight parts into two, and the Circle is divided into sixteen equal parts, and them sixteen (each) into two equal parts, and the Circle is divided into thirty two equal parts, which is the thing desired. This is sufficient in Geometry for our use in Navigation.   


The Projection of the Mariner's Compass.   depiction of mariner's compass  
Describe your Circles one within another, as you see here (making one centre for them all when you have so done, divide it into 32 equal parts, as I have before shown, and draw every other point to the inner Circle, the rest to the fourth Circle) thus the Compass is made. If you have a desire to divide it into half points, divide the whole ones into two equal parts round about, and draw them to the third Circle; if into quarters, divide your halfs into two equal parts, and draw them to the second Circle: draw none but the North and South, East and West, Northeast and Southwest, Northwest and Southeast points through the centre. For drawing the points, it is but observing to make the North point a whole point, and the other seven great points will fall by course (if you observe to make a great point of every other from the North point.) It is common for a Circle without all this to be divided into 360 deg. and then you may see how 11 deg. 15′ is a point of the Compass. But I have proved that in Arithmetic, which saves me a labour here to do it.   


The Projection of the Plain Scale.   depiction of compass and plain scale  
Describe a Semicircle upon the line A C (if you are minded to have your Scale large, let your Circle be large, for the Scale will be according to the Circle) and divide it into two equal parts by the Perpendicular B 0: the one of these Quadrants divide into 8 points of the Compass, the other Quadrant, namely B A, divide into degrees, which is done by dividing it into 9 equal parts, and every one of them parts into 10 equal parts, which is 90 degrees for the Points of the Compass, they are showed sufficiently before. Suppose your Scale were gauged fit for your turn, namely, a place to set the line of Chords, a place to set the line of Sines, and a place to set the Points of the Compasses: (there are Lines of Longitude, and Tangent lines, but we shall meddle with neither of them at this time, because without them these will be sufficient with a Scale of equal parts.)  

For the Line of Chords.  
THe Chord of an Arch is that quantity of a Circle that is between the two ends of a line that is drawn from one part of the Circumference to the other, as the line A 1 is the Chord of 75 degrees (for every pricked Division is a fifth) and so a straight line drawn from A to 10, is the length of the Chord of 10, a line drawn from A, to 20 degrees is the Chord of 20 degrees, and so of the rest. Now your Circle being thus divided, take of every degree, as I have, and apply it to your Scale.   

For the Points of the Compass.  
THey are taken off after the manner of a Chord: first C 1 for one Point, then C 2 for two points, and so of the rest; but you must be sure to take your Line of Chords and Points of the Compass from one and the same Arch (namely, the uttermost) else they are not proportionable (which they must be) take off every quarter in the manner as you took the degrees of Chords.   

For the Lines of Sines.  
A Sine of an Arch is always half the Chord of twice that Arch, as my Father showeth in his Description of a Sine, (p. 2. of his Trigonometry, Book 1.) It is the nearest Distance between the place you have a desire to measure, and a line from the other place that goeth through the Centre of the same Arch (which is a Perpendicular let fall to it) namely 460 is the measure of 60 degrees (the Arch A 60). For your better understanding, observe this following Example for both a Chord and a Sine.   depiction of geometrical figure 
Or if one of then be extended to the Radius (as is ⊙ I) and the other so far that the perpendicular is let fall upon it (as ⊙ u) that perpendicular is the Sine of the Arch included between these sides.  
The Sine of an Arch is half the Chord of twice that Arch, as I u is the Sine of the Arch I s, and is half I B, the Chord of the Arch I s B, which is twice I s.    


The Projection of the Sphere in Plain Lat. 50 deg.   depiction of geometrical figure  
There be many that do not attain to the Doctrine of Spherical Triangles, and are loath (indeed or cannot) spend time to study them: I have thought good to show the way to measure any thing that is useful for Seamen by the plain Scale (before made:) and though this cannot be so exact as to come to a minute of a degree, yet it is exact enough for our use at Sea: For in taking an Azimuth, or Amplitude, or such like, half a degree of the Compass will not make your Course the worse (after you have allowed for Variation) it being not full the 22 t●. part of a point, which is no sensible error in steering: or for finding the length of the day (for ordinary uses) it will breed but little error, 30 minutes being but 2 minutes of time; but for other things that require exactness to a minute or less, of a degree, this way is of no considerabie use; the way by the Tables being exact, and so better. (I shall do both) The reason I handle this here, is in order to Navigation, that you may understand the Circles of the Sphere. This by way of Advertisement.  
BEfore I show how to project the Sphere, it will be necessary to understand the Circles of the Sphere.  
A Sphere is a Scheme or Figure which represents the Heavens, and therefore is round exactly, though upon a Plain it doth not seem so; from whence it is evident, that the lines here drawn in it cannot be straight lines, but Circles: Now there are two sorts of Circles, namely, the Greater Circles, and the Lesser; the Greater Circles are such as go round the very Body of the Globe, and so cut through two opposite points, dividing it into two Hemispheres, and are six in number, which are these: 

1. The Axis of the World. (From South to North Pole.)  
2. The Horizon.  
3. The Aequinoctial.  
4. The Ecliptic.  
5. The East and West Azimuth.  
6. The Meridian.    
The smaller Circles of the Sphere, are all such Circles as do not divide the Sphere of Heaven into two Hemispheres, and so are less than those that do; and they be four, namely, The two Tropics, and the two Polar Circles, which to distinguish the Southermost from the Northermost, we call the Southermost Tropic, Capricorn; the Northermost, The Tropic of Cancer; the Southermost Polar Circle, The Antarctic; and the Northermost, The Arctic Circle.  
All Circles that cut through the two Poles of the World, are called Meridian's; and are also Great Circles, because they divide the Sphere into two Hemispheres, cutting in two opposite points: it is certain, if a Circle goeth through two opposite points, it is as great a Circle as that globous Body can bear, and must divide that Body into two Hemispheres, as doth the Meridian A B C D.  
The Poles are two opposite places in the Heavens, and are the ends (as it where) of that Line called the Axletree, which the Heavens may be imagined to turn upon: here that Line is called the Axis of the World.  
The Aequinoctial is a Great Circle of the Sphere, which lieth between the Pole so equally, that its distance from either Pole is 90 degrees (the half of the distance between the Poles) and because it is so equal between them, there gins Latitude; so that whatsoever Latitude you are in, so many degrees and minutes the Aequinoctial is below that part of the Heavens which is right over your head, and is called the Zenith; or from being right under you, which is called the Nadir. All Meridian's cut this Aequinoctial at right Angles.  
The East and West Azimuth is a Great Circle of the Sphere that cuts through two opposite points of the Heavens, namely, the Zenith and Nadir: it also cuts the Horizon at right Angles in the points of East and West; and therefore is called the East and West Azimuth: all Azimuths are great Circles and cut the Horizon at right Angles.  
The Horizon is a Great Circle of the Sphere, and divides that part of the Heavens which we do not see, from that part which we do see: or it is the furthest part of the Heavens which we can see for the Body of the Sea; it is only to be seen at Sea, or upon the shore where there is no land between you and it, and so you have divers Orisons according to your motion, for as you raise one part of the World, you lay the other.  
The Zodiac is a great Circle, it is the bounds of the twelve Signs: now in the middle of it is the Ecliptic Line, in which Line the Centre of the Sun goeth, and passeth every day its motion of Declination, till it comes to its utmost bounds, which is to the Tropics: it cuts the Aequinoctial in two opposite points, and makes from these points an Angle of 23 deg. 30 min. which is the Angle S R t, so the Sun is never out of it. The twelve Signs divide this Ecliptic into twelve equal parts, and to every Sign there is a Name, and a Character for that Name, and a Month to that Sign; which for your better knowledge I have here following inserted.  

The Names and Characters of the twelve Signs, with the Months they belong to.  March Aries ♈ These be the Northern Signs April Taurus ♉ May Gemini ♊ June Cancer ♋ July Leo ♌ August Virgo ♍ September Libra ♎ These be the Southern Signs October Scorpio ♏ November Sagittarius ♐ December Capricornus ♑ January Aquarius ♒ February Pisces ♓  
Six of these be Northern Signs, and are in the North part of the Zodiac; and six of them are Southern Signs, because they be in the South part of the Zodiac. The twelve Signs are twelve Constellations, the Months answering to them are agreeable. Thus much for the great Circles of the Sphere.  
The smaller Circles are the Polar Circles and the Tropics.  
The Polar Circles are distant 23 deg. 30 min. from the Poles, and between them and the Poles is counted the Frozen Zones.  
The Tropics are the bounds of the Sun's Declination, and they go parallel to the Aequinoctial (and 23 deg. 30 min. from it) the Tropic of Cancer being 23 deg. 30 min. to the Northwards, the Tropic of Capricorn being as far to the Southwards, which is the Sun's furthest distance from the Aequator at any time.  

How to Project a Sphere.  
THis Sphere or Scheme is projected on this wise: First, take an Arch of 60 deg. from your Scale of Chords, and describe the Meridian M A E R. Now the reason you take just 60 deg. is because that number of degrees taken from the Scale is the Semidiameter of the Circle the Scale was made by: so than whatsoever you take from that Circle, and apply to the Scale, it will be the quantity of it in degrees and minutes; or from the Scale if you take any distance, and apply it to the Circle, it will be the same quantity upon the Circle, that it was upon the Scale.  
The Meridian being thus described, draw the line A R through the Centre, which represents the Horizon: Cross that at right Angles, and draw the line E M, which is the East and West Azimuth.  
Then because the Sphere is made for the Latitude of 50 deg. take 50 deg. from your Scale of Chords, and fix one foot in the Horizon at A, and set the other foot upwards to B, (make a prick or mark) which is to signify that the North Pole is elevated or raised above the Horizon so many degrees as the Latitude is.  
Keep the same distance in your Compasses, and go to E with one foot, and let fall the other upon the Meridian on the left hand, which will be at G, and will represent that point of the Aeqinoctial that cuts the Meridian of that place which is the nearest point of the Aequinoctial to that Zenith, and is as much below the Zenith, as the Pole is above the Horizon.  
Latitude 50 deg. 00 min. Declination. Northerly 13 deg. 15 min.   depiction of geometrical figure 
Keep the same distance, and carry your Compasses to the Horizon at R, and let fall the other point to K in the Meridian, which is the South Pole; for as much as the North Pole is elevated above the North part of the Horizon (at A) so much is the South Pole depressed below the South part of the Horizon at K.  
Continue the same distance between your Compasses, and fix one foot in the Nadir at M, and set the other foot upon the Meridian upwards to O, which is the other point of the Aequinoctials intersection in the Meridian opposite to the former; and it stands to good reason, that as much as the Aequinoctial is below the Zenith, so much it ought to be above the Nadir: These marks being set off, draw the Axis of the World B K, and the Aequinoctial G O. And because (as I said before) the outmost bounds of the Sun's Declination is 23 deg. 30 min. (either to the Northwards or Southwards of the Aequinoctial, which bounds are expressed by the Tropics) take 23 deg. 30 min. from your Scale of Chords, and fix your Compasses at O in the Aequinoctial, and set it off on both sides to P and N, carry the same distance to G, and set it off in like manner from G to F and H, then draw F P and H N, which are the Tropics: also draw the Ecliptic from F to N, which will fall through the Centre. Always set the same distance from both Poles on both sides upon the Meridian, and draw D C the Arctic, and I L the Antarctic Circles. And thus you may project a Sphere by a Plain Scale: The like for any other.  
Before I proceed to the Questions, it will be necessary to give you the meaning of things that will be spoke to.    

The Meaning of the Terms used in the following Work.  
THe Latitude of any place is the Distance between the Zenith of that place, and the Aequinoctial in degrees and minutes.  
The Elevation of the Pole is always equal to it, and is the Poles height above the Horizon in degrees and minutes.  
Altitude signifies the Height of a thing: If it be meant of a Star, or of the Sun (not being upon the Meridian) it is meant the Distance between the Sun or Star and the Horizon.  
Meridian Altitude is the Sun's height above the Horizon, when he is upon the Meridian of that place, which is at 12 of the Clock; or at any other time when a Star is upon the Meridian either North or South.  
The Sun's true Amplitude of rising or setting signifies the number of degrees and minutes that the Sun riseth from the East, or sets from the West points of the Compass in the Horizon, as it is found in the Sphere.  
The Sun's true Azimuth of rising or setting signifies the distance (in degrees and minutes) of the Suns rising or setting (in the Horizon) from the North or South points in the Compass, as it is found in the Sphere.  
There is the true Amplitude of the Suns rising and setting, and the Magnetical Amplitude; the Magnetical Amplitude is the degree of the Compass that he is observed to rise or set upon, and is Called Magnetical Amplitude from the Loadstone: The Magnetical Azimuth is Called so for the same reason, and is the Suns rising from the South or North.  
Longitude is the Easting or Westing between two places in degrees and minutes, if in sailing.  
Departure from the Meridian is the Easting or Westing between two places in miles or leagues.  
Latitude, as I have said, is the Distance of the Aequinoctial from the Zenith of that place in degrees and minutes, or the Elevation of the Pole; but Difference of Latitude is the Northing or Southing that a Ship makes in any Distance run, or between two places, either in degrees, leagues, or miles.  
Declination (if it be meant by the Sun, or any Star that is near the Aequinoctial) is the nearest distance between the Sun or Star and the Aequinoctial.  
Stars that are nearer the Poles, their Declination are their nearest Distance to the Poles.  
The Sun's place in the Ecliptic is his nearest Distance from the place where the Sun's Parallel of Declination cuts the Ecliptic to the next Aequinoctial point, which is ♈ or ♎.  
There be seven Planets which I have here inserted, with their Characters under their Names.  
Saturn Jupiter Mars Sol Venus Mercury Luna. ♄ ♃ ♂ ☉ ♀ ☿ ☽  
I have noted already in pag. 3 of this Book, the quantity of degrees that make an hour in time, of minutes that make an hour, the quantity of degrees and minutes that are in a point of the Compass and the like: if you want to know, look there.  
I'll now proceed to the Work of the Sphere in Plano, and I will handle no more than is necessary for our use in Navigation.  
I had thoughts to have made a Table of the hard Terms of Art used (which our vulgar Capacities do not ordinarily comprehend) with their meaning, but then I considered what my own Method would be; and so I would not trouble myself with that at the Beginning of my Book: There are no intricate Terms used, but the meaning of them is commonly showed in the Beginning of the Book.  
Latitude and Declination given, to find the things following in the several Questions.  
Latitude 50 deg. 00 min. Northerly.  
Declination 13 deg. 15 min. Northerly.   depiction of geometrical figure 
Rule. All distances of the Sphere that you desire to know, must be extended to the Arch of a great Circle, for by great Circles is the Sphere measured.   


QUESTION I. To find the Meridian Altitude of the Sun.  
IN this or any other Sphere, M G E is that part of the Heavens that is visible, the other half invisible to us; for it is parted from our sight by the Earth and Sea, and the furthest part of it, which seems to mix as it were with the Heavens, we call the Horizon, which is the great Circle M E (for it is a great Circle, though here we are forced to represent it by a straight Line) M is the South point of the Horizon: Now the Sun's Meridian Altitude is his distance between that point and the place he cuts the Meridian that day, which is M I; fix your Compasses in M, and extend the other foot to I, and apply it to your Scale of Chords, and as many degrees as you find it there, so many degrees is the Sun high when he is upon the Meridian that day, which is the thing required.  
Note that at O the Sun riseth, at P it is 6 of the Clock, and P I is equal to ♈ K (when it is extended to a great Circle) and both the Sine of 90 deg. which is extended to a Chord, must be the Chord of as many degrees, which is 6 hours in time. The time between 6 a Clock and 12, which proves that M I is the Meridian Altitude of the Sun: and this measuring any Distance from the Meridian is called The first Way of Measuring.   

QUESTION II. To find the Sun's Amplitude of Rising and Setting.  
FIx one foot of your Compasses in the place of the Suns Rising, which is O, and extend the other foot to the Centre of the Sphere (which is termed the East point of the Horizon) and this Distance apply to your Line of Sines (if you have any) but if you have no line of Sines, extend it to the Chord of the same Arch, thus: Fix your Compasses with that distance, one foot in the Meridian, so that the other may just sweep a line that goeth through the Centre of the Sphere, then, say I, the Arch between that foot of your Compasses that stands in the Circle, and the place where this line you sweep cuts the Circle, is the Chord of the thing required, and will be the same number of degrees upon the line of Chords, as O R would be upon the line of Sines: and after this manner is the Chord of any distance (taken from a line that goeth through the Centre of the Sphere) found: and this is called The second Way of Measuring.  
I find that the Sun riseth 20 deg. 53 min. to the Northwards of the East (for she hath North Declination, and the Latitude is Northerly) or sets so much to the Northwards of the West.   depiction of geometrical figure  

QUESTION III. To find the Sun's Azimuth at six of the Clock.  
THe Sun's Azimuth at 6 of the Clock, is the nearest distance between the Sun at 6 of the Clock, and the East and West Azimuth, which is P z: Now if you mind it, P z is taken off from a Circle, which is not so great as the Horizon, and yet is parallel to it, as the line n w is parallel to the line M E; and as many degrees as the Sun is from the nearest part of the East and West Azimuth in the little Circle, so many he is from it (if an Azimuth where drawn) in the great Circle, for P z is as many degrees in the little Circle, as n ♈ is in the great one: Now because your Scale is made by the great Circle, therefore extend the distance taken from the lesser Circle to the measure of the greater, which is done thus:  
Take half the length of the pricked line which is z w, and fixing one foot of your Compasses in the Centre of the Sphere, describe an Arch from some line that goeth through the Centre of the Sphere, as the Arch M l n, then set the distance z P as a Sine upon that Arch (for it is a Sine upon that Arch as well as o ♈ was a Sine upon the Meridian) I find the Sine of it is the Sine of the Arch l M, lay your Scale from the Centre of the Sphere by l, and draw the line l q than shall q M be the same quantity of degrees upon the great Circle, as l M is upon the little one: therefore take M q and apply it to the Scale of Chords, and it answers your desire. And thus is the Sun's Azimuth at 6 of the Clock, P z found to be 8 deg. 36 min. to the Northwards of the East, and this is called The third way of Measuring, to measure any distance from a line that doth not go through the Centre (which must represent a small Circle) And thus you may find the Sun's Azimuth at any time of the day.   

QUESTION iv To find the Sun's height at six of the Clock.  
THe Sun is at P at 6 of the Clock: fix one foot of your Compasses in P, and extend the other to sweep the Horizon, which is the same as though you let fall the Perpendicular P n, set it off by The second way of Measuring, as was showed in Quest. 2. and apply it to your line of Chords; and the reason is, because P n represents a part of a great Circle, and so is to be understood to be of the nature of those lines that go through the Centre of the Sphere, for all Azimuths pass through the Zenith and Nadir, which are two opposite points.  
I find the Sun is 10 deg. 7 min high at 6 of the Clock.   depiction of geometrical figure  

QUESTION V To find the Sun's height being due East or West.  
Where the Sun's parallel of Declination cuts the East and West Azimuth, is the place the Sun is in, when he is due East in the morning, for you see he is then over the point of East in the Horizon, which is ♈ (or the Centre) therefore take the distance between that place and the Centre (which is S ♈) and apply it to the line of Sines; but if you have no line of Sines, extend it to a Chord after the manner of the second Question, which I call the The second way of Measuring, for it is the distance of a line which goeth through the Centre.  
I find the Sun's height being due East or West, is 17 deg. 25 m.  
Note, that the same height that the Sun is being over the East point, so high he is being over the West point in the afternoon.   

QUESTION VI To find the Difference of Ascension.  
THe Difference of Ascension, is the portion of time that is between the Sun's Rising and six of the Clock: If the days be longer than the nights, the Sun riseth before 6; but if shorter, after 6; but whether it be before or after 6 that he riseth, so many hours and minutes as it is from 6, so much is the half day or night longer or shorter than 6 hours, from whence it is evident, that if the Sun riseth due East, he riseth at 6 of the Clock, and so there is no Difference of Ascension, for he is then in the Aequinoctial, which cuts the Horizon in the two opposite points of East and West.  
In this Question the Difference of Ascension is O P, and is a distance upon a line that goeth not through the Centre, therefore take half the length of the parallel I D, and proceed as you did in the third Question, and after you have found it in degrees and minutes, convert it into hours and minutes of time, and set it down.  
I find it to be here 16 deg. 18 min. which is 1 hour 5′ 3/15.  
By the Difference of Ascension thus found, you may find the length of the day or night, the hour of Suns Rising, or the hour of uns Setting.   

QUESTION VII. For the time of the Suns Rising or Setting in this Example.  
I Consider as much as the Sun riseth before 6, so much he sets after 6: here he riseth before 6 of the Clock 1 hour 5′, subtract that from 6 hours, and you have the hour of the Suns Rising 4 h. 55′: add it to 6 hours, and you have the time the Sun sets 7 h. 5′. double that, and it is the length of the whole day, which is here 14 deg. 10 min. Subtract the length of the  depiction of geometrical figure day from 24 hours, and it leaves the length of the night 9 h. 50′. I omitted the Fractions.  
But if the Sun riseth after 6 of the Clock, and you have a desire to find these things (as he doth when he is South Declination) add the Difference of Ascension to 6 hours, and it gives the time of the Suns Rising; subtract it from 6, and it gives the time of Suns Setting; and that doubled is the length of the whole day. Again, that subtracted from 24 hours, is the length of the night.   

QUESTION VIII. To find the hour of the Suns being due East or West.  
THe Sun is due East (in this Example or any other) when he goeth by the East point of the Horizon; or West, when he goeth by the West point. In this Example, the Sun's parallel of Declination cuts the East and West Azimuth in S, which is later in the morning than 6 of the Clock, by the distance S P: therefore see what S P is by The third way of Measuring, and convert it into time, and add it to 6 hours, it shall give the hour of the Suns being due East.  
I find it in this Example to be 6 h. 46′.  
Subtract this from 12 hours, and it gives the time of the Suns being due West that day: for as many hours and minutes as the Sun is due East before 12 of the Clock, so many hours and minutes must it be due West after 12: According to this Example the Sun is due West at 5 of the Clock 14 min.  
If the Sun hath South Declination, he passeth the point of East before he riseth, and is set as long before he comes to the point of West, (provided the Latitude be Northerly, as this is) but if the Latitude and Declination be both one way, the Sun is always up before he cometh to the point of East: and the work is as I have showed.   

QUESTION IX. To find the time of Day breaking, and Twilight ending.  
IT is an ancient Observation and concluded Opinion, That the Sun makes some show of Day when he is 17 degrees under the Horizon; therefore take 17 deg. from your Scale of Chords and set it from both ends of the Horizon downwards, and draw the line T r, then fix your Compasses in r (the place where the Sun's parallel of Declination intersects that line) and extend them to 6 of the Clock; set it off by The third way of Measuring, and convert it into time: Subtract that from 6 of the Clock, and it gives the time of Day breaking.  
I find in this Example Day breaks at 3 of the Clock 6 min. ●8/15.  
Add it to 6 hours, and it gives the time of Twilight ending 8 of the Clock 3 53 14/13.  
It may happen so sometimes, that it may be past 6 of the Clock before the Day breaks, in such a case you must add (in the morning) for break of Day, and subtract for Night from 6 hours, which is the contrary: These things your own Reason will give you, after you are used to it, which makes me forbear to give any more reasons of it.   

QUESTION X. To find the Continuance of Twilight.  
THe Continuance of Twilight is the time between the Day breaking and Sun rising, which is r o; take it off by the Third way of Measuring, and convert it into time. I find it to be 1 hour 48 min.   

QUESTION XI. To find the Length of the longest Day in that Latitude.  
When the Sun is nearest the Zenith (in any Latitude) that day must be the longest: now in places near the Aequinoctial, as between the Tropics, there is but little difference all the year long; but in places nearer the Poles, there is more.  
The Sun is nearest the Zenith in this Latitude, when he is in the Tropic of Cancer, so that then must be the longest Day. Imagine the Tripick of Cancer to be the parallel of the Sun's Declination (as indeed it is that day) take the distance between Y and R, which is between Suns Rising and 6 of the Clock that day, and set it off by the Third way of Measuring, the number of degrees and minutes convert into time, and add it to 6 hours, which makes the length of the Forenoon, and that doubled is the length of the whole day (as in Quest. 7.) which is equal to the longest Night in that Latitude.  
I find it here to be 16 hours 10 minutes.  
Subtract the length of this longest day from 24 h. 00 min.  
And it leaves the length of the shortest night 7 h. 50 min.  
Equal to which is the length of the shortest day 7 h. 50 min.  
But you may measure the length of the shortest day, and subtract that from 24 hours, and it will be the length of the longest night, which is equal to the longest day.  
Now the shortest day is when the Sun is in the Tropic of Capricorn, for than it is latest before he riseth: Now to measure it, take the distance from 6 of the Clock, which is T, and the Suns Rising which is U, and set it off by the Third way of Measuring, convert the degrees into time, and subtract that time from 6 hours, it leaves the length of the Forenoon, which doubled is the length of the shortest day; (as in Quest. 7.) I have spoke to that purpose. I need say no more, that 7 th'. Question is light sufficient.  
I might do an Example of what is before done, in a South Latitude, but Reason gives, that the same which the South Pole or Southern parts of the Heavens are depressed in a Northern Latitude, the same will the Northern parts of the Heavens be depressed in a Latitude as far Southerly, so that there will be no difference in the work at all: And indeed my desire to be brief, makes me omit things that I think may be understood without treating of them.  
I'll only touch upon an Example in a Sphere which hath South Declination.  

Latitude Northerly 50 deg. 00 min. Declination 13 deg. 15 min. Southerly.   depiction of geometrical figure 
Also, if you subtract this Amplitude V ♈ from ♈ H 90 deg. the remainder is the Sun's Azimuth from the South V H, or if you add V ♈ to ♈ A (90 deg.) it is V A the Sun's Azimuth of rising from the North; forasmuch as V A is as much above 90 deg. as V H wants of it: The same is to be understood in any Sphere, that the Amplitude and Azimuth of the Sun are Compliments one of an other, to 90 deg. in that Quarter.  
When the Sun's Declination is Southerly (in any Northern Latitude) the days are not so long as the nights, so that the Sun cannot be up at 6 of the Clock, which is at R; therefore the Sun's Azimuth at 6 of the Clock is of no use.  
Neither is he up, when he passeth the East point of the Horizon, which is at P, so that you cannot take his height; but you may measure how far he is under the Horizon at either of those times, or you may find his true Azimuth at 6, which is of no value to us that are Seamen, because we cannot have a Magnetical Azimuth; they are measured as before.  
The Difference of Ascension is here after 6 of the Clock, as much as it was before 6, in the last Example: And the reason is, because the Latitude is the same, and the Declination of the Sun is just as far Southerly, as before it was Northerly, so that the Sun will now be just as long before he riseth after 6, as before he risen sooner: it is V R, take it off by the third way of Measuring, as you did before, and set it down; I find it to be 1 hour 5 min. 3/15, it serves for the same uses that it did in the other Example.  
For the length of the day, subtract the Difference of Ascension V R, from 6 hours, which is R f, and the remainder is f V, the length of the Forenoon, which doubled is the length of the whole day; that subtracted from 24 hours is the length of the night.  
For the hour of the Suns being due East, it is P R, which is from the place where the Sun cuts the East and West Azimuth to 6 of the Clock, and set it off by the third way of Measuring, convert your degrees and minutes into time, and subtract that time from 6 hours, and it gives the due time of the Morning that the Sun cuts the East point of the Compass; the reason why you subtract is, because the Sun is due East so long before 6 of the Clock.  
Subtract the hour of the Suns being due East from 12 hours, and it will give the true time of the Suns being due West.  
For the time of day breaking, draw the line 17 ⊙, parallel to the Horizon, and 17 deg. under it, as was showed before, and measure from P to R, that is, from the place where the Sun's Parallel of Declination cuts that Circle of 17 deg. (under the Horizon) to 6 of the Clock, convert it into time, and subtract it from 6 of the Clock, and it leaves the time of Day breaking, namely, the time in the morning that the Sun is in P; you subtract, because the day breaks so long before 6 of the Clock: If it where so that the Sun were in the Tropic of Capricorn, you must take the Distance S q. You see the Sun is passed the hour of 6 before it is break of day; now in this case you must add, whereas before you subtracted.  
For the continuance of Twilight, it is measured as before, from the place, where the Parallel of the Sun's Declination cuts the Circle of 17 deg. to the place where it cuts the Horizon; convert it into time, and set it down. In this Example it is P V.  
What I have not mentioned here, I have given sufficient instructions about in the other Example before this.   

Latitude 51 deg. 30 min. Northerly, Declination 11 deg. 10 min. Northerly, I demand the Sun's place and right Ascension.  
BEfore we can find the Sun's place in the Ecliptic, you are to consider that the Ecliptic is divided into 12 equal parts, by the 12 Signs; six of these Signs divide that half of the Ecliptic, which is to the Northwards of the Aequinoctial, and are called Northern Signs; and the other 6 divide the half that is to the Southwaads of the Aequinoctial, and are called Southern Signs, as I have set them down in this Book before, with their Names and Months to their Characters, and have set the Northern Signs by themselves, and the Southern Signs by themselves.   depiction of geometrical figure 
This plain Superficies can show but one part of the Globous Body, but you must know it is round, which makes the Ecliptic to be divided; on both sides from ♈ to ♋ is 90 deg. from ♋ to ♎ is 90 deg. from ♎ to ♑ is 90 deg. and from ♑ to ♈ is 90 deg. every one of these quarters containing 3 Signs.   

The Ecliptic being thus divided into Signs; we may find the Sun's place, and right Ascension.  
THe Sun's place in the Ecliptic is the nearest Distance between the next Aequinoctial point, and the Sun in the Ecliptic, (by Aequinoctial points is meant ♈ and ♎, the two points of intersection that the Aequinoctial and the Ecliptic make) Now, in this case, the Sun must be nearest the Equinoctial point at ♈, because the Month is belonging to a Sign nearer ♈ than ♎.  
Therefore, fix your Compasses in the point of ♈, and extend the other foot to the Sun in the Ecliptic, which is I, and apply it to the line of Sines; or if you have no line of Sines, convert it to a Chord (by the second way of Measuring) and as many degrees as it is, so far is the Sun distant from the nearest Aequinoctial point; if it exceed 30 deg. the Sun must be in that degree of ♉ that is above 30 deg. if it exceed 60 deg. the Sun must be in that degree of ♊ above 60; but here I find it in less than 30, namely, 29 deg. 3 min. therefore I conclude the Sun is in 20 deg. 3 min. of ♈.  
But suppose it were time of year that the Sun were returning from the Tropic towards the Aequinoctial, and were the same Declination, then would the Sun's place be the same from entering into ♎, that it now is in ♈, namely, in 57 min. of ♍, which wants 29 deg. 3 min. of entering into ♎. Thus much for the Sun's place.  
The Sun's right Ascension is measured from the place where the Sun's parallel of Declination cuts the Ecliptic to 6 of the Clock, which is I 6; set it off by the third way of Measuring to extend it to a Great Circle: for the Sine of I 6 is equal to the Sine of S ♈; when you have measured it, convert it into time, and set it down; I find it to be 1 hour 44 min.     

The Meridian Altitude of the Sun being given, and the Latitude, to find the Sun's Declination, as also all the other things before found.  
Latitude 51 deg. Northerly, Meridian Altitude 52 deg. 30 min. I demand the Sun's Declination, or any other thing that I found before.  
FOR the doing of this, project your Sphere (as before was showed) that done, I consider the Meridian Altitude of the Sun is given to be 52 deg. 30 min. I will take 52 deg. 30 min. from the Scale of Chords, and fix one foot of my Compasses in the South point of the Horizon at A, and extend the other foot upwards upon the Meridian, which falls at B; then I conclude the Sun is upon the Meridian, in the point B: Take the Distance between B and the Aequinoctial, which is B C, and apply it to the Scale of Chords, and so much as it is, so much is the Declination of the Sun; and you see it is Northerly, for the Sun is to the Northwards of the Aequinoctial. If you have a desire to find any thing else in this Sphere, draw the Parallel of the Sun's Declination (thus found) and work with it as before in other Examples; also draw the Ecliptic and Tropic.   depiction of geometrical figure 
Many other things might be done, but for Brevity's sake I omit them, judging this to be as much as we have occasion for in Navigation, which is the thing I aim at. If I have committed a fault in being too tedious in these things, pardon me, and remember it was my Beginnings; riper years (which may bring more ability, it is probable) will bring forth something more new and admirable. Be charitable.  

Terms of Art used in Navigation.  
LAtitude (as I have cited before) is the Distance between the Aequinoctial and the Zenith; what is meant by Longitude and Difference of Latitude, I have showed before; also the difference between Longitude and Departure from the Meridian, in this Book before.  
Now in every Distance that is sailed upon any Course that makes an Angle from the Meridian, either of North or South, or the Parallels of East and West, that is to say, any Distance that is sailed upon a point that is between those Courses; we have these things considerable, namely, The Distance run, (which must be the longest side) the Difference of Latitude, and Difference of Longitude, or Departure from the Meridian, and the Course; the Course is the Point of the Compass or Rumb you steer on.  
A Dead Reckoning is when a man cannot observe, but is forced to judge of the Course and Distance, to find the Difference of Latitude and Departure.  
A Reckoning is an Account of the Difference of Latitude and Departure from the Meridian, that is made in the time a Ship hath been out.  
Read Tan. for Tangent, Sine come. for Sine Compliment, Tan. come. for Tangent Compliment, comp arith. for Compliment Arithmetical.     

PROPOSITIONS of SAILING BY THE PLAIN SCALE.  

Difference of Latitude and Course given, to find the Distance run, and the Departure from the Meridian.  
QUESTION I.  
Admit I set from the Lizard, lying in the Latitude of 50 deg. and sail S W b W, till I altar my Latitude 1 degree or 60 miles: I demand my Distance run, and Departure from the Meridian.  
IN this Question we have the Course 5 points from the Meridian towards the West, which is 5 times 11 deg. 15 min. or 56 deg. 15 min. for 5 times 15 min. is 75 min. or 1 deg. 15 min. and 5 times 11 deg. is 55 deg. to which add the 1 deg. and 15 min. and the whole is 56 deg. 15 min.  
To protract this Question or any other of this nature, I wish you to observe the end of your Book, that lies from you to be the North part of your Book, then that to the right hand will be the East part, that to the left hand the West, and that right to you the South part: The Reason why I wish you to observe this Method, is, that so in any Travis you may set every Angle the right way, which must be observed then, and is good to be always used.   depiction of geometrical figure 
But besides this Geometrical way of resolving Triangles, there is an exacter way by the Tables of Sines, Tangents, and Logarithms, which is done thus,  

A General Rule.  
Note that in all right lined Triangles the Sine of every Angle is proportional to its opposite side, or every side is proportional to the Sine of its opposite Angle.  
The greatest side will have the greatest Angle opposite to it, the last side will have the least Angle opposite to it: As here in this Question, A B is the longest side, and opposite to it is the right Angle B D A, which is the greatest Angle, A D is the next, and hath A B D to it, D B is the least side, and hath D A B for its opposite Angle, which is the least Angle. Now as the Sine of A is to B D, so is the Sine of D to A B, or the Sine of B to A D, and the contrary.   

To find the Distance run.  
As Sine come. B 56 deg. 15 min. comp. arith. 0,25526 To B D 60 min.  1,77815 So is Radius or Sine D 90 deg.   To A B 108 miles  2,03341  
If Radius be not in the first place, you must take the Compliment arithmetical of the first number, as here I have done; but if Radius be in the first place, take no Compliment arithmetical at all: The Compliment arithmetical of a number is what every single figure wants of 9, take them in their due places, as here the Sine in the Tables answering to A 33 deg. 45 min. (which is the Sine Compliment of B 56 deg. 15 min.) is 9,74473, whereas I have set it down 0,25526, now if you compare them, the number that I have taken is what the figures in the other want of 9, as 9 wants 0, and 7 wants 2, and 4 wants 5, and so forward; after the Compliment arithmetical of the first number is taken, take the others as they be in the Tables, and add them together, casting away Radius always, and look for what comes out, either in the Sins, Tangents, or Logarithms, according as the order requires into which the proportion runs. You see here I took no notice of Radius, which is a casting of him away: in like manner if the things added up come to above Radius, always cast Radius away, and set down what is above Radius; as here the last came out to be 11, and I cast away the 10, and set down 1.   

For the Departure from the Meridian.  
As Sine comp. B 56 deg. 15 min. comp. arith. 0,25526 Is to B D 60 miles  1,77815 So is Sine B 56 deg. 15 mim.  9,91984 To A D almost 90 miles  1,95325  
But I have been often asked by them that I have learned, why I contradict the Rule, which saith, that the Sine of every Angle is proportional to its opposite side or the contrary. Now here you say, as Sine compl. B to B D, either you are false, or else the Sine comp. of B is the Sine of A, for A is the Angle opposite to B D, whose Sine the Rule order you to take.  
I answer that A is the Compliment of B, for this is certain, that the three Angles of any right lined Triangle consists of a Semicircle or 180 deg. and neither more nor less: Now if it be so, B A D and D B A must together be but 90 deg. because B D A is 90 deg. itself; then if B is 56 deg. 15 min. the Sine comp. of it must be the Sine of 33 deg. 45 min. which is the Sine of A, opposite to B D.  
But you will ask how you should know that the three Angles of a right lined Triangle is but 180 deg. Pray mind this Demonstration.   depiction of geometrical figure 
Set your Compass at any convenient distance, and from every angular point as a Centre, describe an Arch, as I O, U n, and s D; now I D is an Arch of 90 deg. 0. min. because it is included between two sides that are perpendicular one to the other; then U n and s O together are equal to I D; measure them, and set them upon the Arch I D, and they will be equal to it. From hence then it is evident, that s O and U n, or the two Angles beside the right Angle of any right angled Triangle, are Compliments one of the other to 90 deg. or that the three Angles of any right lined Triangle are 180 deg. or two right Angles or a Semicircle.    

Course and Difference of Longitude in miles given, to find the Difference of Latitude and Distance.  
QUESTION II.  
Admit I set from the Lizard, lying in the Latitude of 50 deg. and sail S W b W, till I depart from my Meridian 90 miles: I demand my Difference of Latitude and Distance run.  
LEt A represent the place your Ship is in, after she hath sailed; from whence draw the East line A D, then take 90 miles from your Scale of equal parts, and set off upon that line from A, which falls at D.   depiction of geometrical figure 
The Difference of Latitude is 60 miles.  
The Distance run is 108 miles.  

By the Tables. For the Distance run.  As Sine comp. A 33 deg. 45 min. comp. arith. 0,08015 To A D 90 miles 1,95424 So is Rodius  To A B almost 108 miles 2,03439  

For the Difference of Latitude.  As Radius to the Distance run 108 A B 2,03342 So is Sine comp. the Course 56 deg. 15 min. B 9,74473 To the Difference of Latitude D B 60 miles 1,77815  
Here to find the Distance run, I made use of the things that were given me, but for brevity's sake that I might show the taking of a Compliment arithmetical, I used Radius with the Distance before found, to find the Difference of Latitude.  
But if you would not do so, than the things given will do as well, and the Proportion holds thus.  
As Sine comp. A (which is Sine B) is to A D so is Sine A to B D.   

Course and Distance run given, to find the Difference of Latitude and Departure from the Meridian.  
QUESTION III.  
Admit I set from the Latitude of 50 deg. 0 min. and sail N E b N 108 miles: I demand the Difference of Latitude and Departure from the Meridian.  
LEt A D be a Meridian line, A the Latitude of the Lizard, D A B the Angle of the Course, set off by the Arch of 60 deg. N O s (N O being 3 points) A B a line drawn from the Latitude of the Lizard A by O, and the Distance that the Ship hath sailed set upon it, and no more, namely, 108 miles; then I conclude that the Ship is at B, from B let fall the Perpendicular B D, which cuts the Meridian in D, and leaves D A for the Difference of Latitude, and is itself the Departure from the Meridian; measure them as you have been showed before, and apply them.  
Difference of Latitude D A 90 miles.  
Departure D B 60 miles.   depiction of geometrical figure 

By Proportion.  
I have two things given besides the right Angle, namely, the side A B 108 miles, and the Angle B A D 33 deg. 45 min. therefore to find the Difference of Latitude A D.  
As Radius  To the Distance run 108 miles B A 2,03342 So is Sine comp. A 33 deg. 45 min. 9,91984 To A D the Difference of Latitude 90 miles 1,95326  

For the Departure from the Meridian.  As Radius  To the Distance run 108 miles B A 2,03342 So is Sine A 33 deg. 45 min. 9,74473 To D B the Departure from the Meridian 60 miles 1,77815  
In your putting things in proportion, follow these directions; always put the required thing in the last place, and the things you know, make use of, and put in proportion to find the things you know not.  
If it be a Side you are to find, begin with an Angle; if it be an Angle, begin with a Side  
Here it was a Side that was to be found (in both these Proportions) you see I began with an Angle, that I knew which was Radius; but you may ask why I did not begin with the given Angle at A or B as well, I answer, because neither of the things opposite to them were given, so that I could not proceed, if I had.    

Distance run and Difference of Latitude given, to find the Course and Departure from the Meridian.  
QUESTION IU.  
Admit I set from the Latitude of the Lizard 50 deg. 0 min. and sail upon some point between the North and the West 108 miles, till I altar my Latitude 1 deg. 30 min. or 90 miles: I demand the Departure from the Meridian and Course.   depiction of geometrical figure 
Take the Difference of Longitude D B, and apply it to your Scale as before.  
Take O N the measure of the Angle of the Course, and apply it to the Scale of Chords (if you are minded to see her Course from the Meridian in degrees and minutes) to the points of the Compass, if you desire to know it in points.  
Note that A is the Centre of the Arch O n 60 deg. It is plain, that if an Arch of 60 deg. described, and any number of points or degrees that is set upon it, be an Angle of those points set down (as it hath been in every Example before) than whatsoever you take from an Arch of 60 deg. is fit to apply to the Scale.  
I find the Course to be North 33 deg. 45 min. Westerly, or N W b N. I find the Departure from the Meridian to be 60 mil.  

By the Tables.  
Here I have given the Distance run A B, and the right Angle opposite to it, and the Difference of Latitude A D, whereby I may find the Angle opposite to it B, or the Compliment of it, which is A, the Angle of the Course.  

For the Course.  As the Distance run A B 108 mile's comp. arith. 7,966576 Is to Radius   So is the Difference of Latitude A D 90 miles 1,954242 To the Course or Rumb Sine comp. B 33 d. 33 m. 9,920818  

For the Departure.  As Radius  To the Distance run A B 108 miles 2,0334237 So is Sine the Course A 33 deg. 33 min. 9,9424616 To D B the Diff of Longi. here found to be 60 miles 1,7758853  
You may ask the reason why the Course is not 33 deg. 45 min. as it was first given to be: I answer, because of the part of an unite, that the Difference of Latitude was set down more than it should be; for if you observe, I said it was almost 90 miles, when I found it before: Now the error is not worth minding in Sailing, for the Difference that the Fraction causeth is but 12. min. of a degree of a point of the Compass, which is no more than the 56 th'. part of a point.  
If you mind, I have worked the Course thus found, and it doth not alter the Difference of Longitude or Departure (which should be 60 miles, if I had worked to the least Fraction) for the Logarithm here is nearest to the Logarithm of 60 in the Tables: Indeed I worked on purpose thus, because those that I have taught, when they could not find it come out right as the other was, could not tell where the fault lay; now this will direct them to know that some part of a unite missing in the finding of the Sides, may make some minute's Difference in the Course: But now, here in the following Examples we will work to a Fraction, namely, to the 10 th'. or 100 th'. part of a unite, (in the sides) to show how it is done; though in Sailing if you work to half an unite, it is near enough, or for any ordinary uses (and that the Tables give) and most commonly to less.    

Departure from the Meridian and Distance run given, to find the Course and Difference of Latitude.  
QUESTION V.  
Admit the Departure from the Meridian were the same that the Difference of Latitude was in the first Question 60 miles, the Distance run also 108 miles: I demand the Course and Difference of Latitude, let the Latitude you set from be 50 deg.  
HEre the Course from the Meridian will be the Compliment of the Course there, and the Difference of Latitude will be the Difference of Longitude, or Departure from the Meridian there.  
But instead of taking the Logarithm answering to the number here, take the Logarithm answering to ten times the number, and so you shall always have your sides come out to the tenth part of a unite; I say it will come out in tenths, for you must divide by 10, to bring it into miles after it is done.  
Now instead of multiplying your given sides by 10, do but set a Cipher more to the number and it is done, as now for 60 set down 600, which is ten times 60; for 108 set down 1080, which is ten times 108; and thus for any number that will be comprehended in the Radius of the Tables.  

Proportion for the Course  As Distance run given B A 1080 tenths comp. arith. 6,9665762 Is to Radius,  So is the Departure from the Meridian 600 tenths D B 2,7781512 To the Sine of the Course A 33 deg. 45 min. 9,7447274  

For the Difference of Latitude.  As Radius  To the Distance run 1080 tenths B A 3,0334327 So is the Sine comp. of A 33 deg. 45 m. 9,9198464 To the Diff. of Latitude in tenths D A 898 tenths 2,9533708  
Divide these ten by 10, and it gives the miles contained in the side D A the Difference of Latitude, which is 89 miles 8 tenths of a mile, whereas before it was 90 miles, whis is 2 tenths of a mile more: Now when you have the sides thus found in tenths or Centisms; work for the Angles, and you will find them to come out roundly alike: But if you desire more exactness, work then in Centisms, provided, that the numbers in Centisms do not outrun the Tables.  
A Centism is the hundredth part of an unite, or of any thing; and if I would put the number 60 into Centisms, I will set two cyphers behind it, and it is the same as before; I multiplied 60 by 100, for it is 6000, the like is to be understood of any other number.  
I would work to the nearest Fraction for the Angles here, but I conceive it to be no way beneficial, and therefore I'll refer it to the work of the Sphere, where it is of ●uch more use; this finds the Angles to a minute, which is near enough.  
It is like that some will be so curious, that this way of finding things to the tenth or hundredth part of unit will not suffice them, but they would have the real number answering to the Logarithm to the 1000 th'. part of a unite; because I find it of no real consideration or use in Navigation, I omit that here, and desire such to look in Book 1. Page 11. of my Father's Trigonometry, and there it is plainly showed.  

By the Plain Scale.  
This Question differs nothing from the other in its operation, only as you work in the other from a North and South line, here you work from an East and West line; The work is this.  
First, draw an East and West line, and from the west end of it, set off your Departure from the Meridian, which is from D to B 60 miles; from D raise the Perpendicular D A (which is a North and South line) and draw it at length; then take the Distance run 108 miles, and fix one foot of your Compasses in B, with the other cut the Meridian line D A, which it doth in A. Draw the side A B: then is B the place your Ship is at after her Sailing, A is the place of her setting out, A D is the Difference of Latitude and N O s is an Arch of 60 deg. from the Angle D A B (as a Centre) the measure of N O is the Course, and is performed as before in the last Example; and also of the Difference of Latitude as in other Examples.  
The Difference of Latitude is D B 90 miles.  
The Course is North 33 deg. 45 min. Easterly, D A B or N O.  
The Plain Scale of equal parts will not resolve to the least Fraction as the Tables will now to a tenth part of a mile, I confess a good Diagonal Scale will produce things to a small Fraction.  
If you would know what Latitude you are in, divide 90 by 60, and the Quotient will be degrees, namely, 1 deg. the Remainder minutes, namely 30 min. Now because your Latitude was Northerly, and you have gone to the Northwards, you have increased it 1 deg. 30 min. so that you must add the Difference of Latitude to 50 deg. and it makes 51 deg. 30 min. if you have gone to the Southwards, you must have subtracted.   depiction of geometrical figure   

Difference of Latitude and Departure from the Meridian given, to find the Course and Distance run  
QUESTION VI.  
Admit the Departure from the Meridian of the Lizard that I make be 89 miles 8 tenths of a mile, and I am in the Latitude of 51 deg. 0 min. I demand how the Lizard bears off me, and what distance I am from it, provided that I sail between the North and the West.   depiction of geometrical figure 
The Course I find to be N W b W.  
The Distance run 108 miles.  

By Proposition.  
The Rule that we have used for resolving all the Questions hitherto laid down was this.  
That the Sine of every Angle is proportional to its opposite Side, or every Side is proportional to the Sine of its opposite Angle: and this served where you had opposite things given, but here you have two sides, and an Angle between them given, so that you have nothing opposite to any of them, therefore that general Rule is of no value here in this case; this is a general Rule for this case, That, As the Sum of the two given Sides is to their Difference; So is the Tangent of the half Sum opposite Angles, To the Tangent of the Difference of either of them: That is, as the Sum of D B and D A together, is to their Difference, So is the Tangent of the half of A and B, to the Tangent of their Difference. Which Angle added to the half Sum makes the greatest, or subtracted from the half Sum it makes the least Angle. But this Rule though it be rare and true, yet in right angled Triangles we have a briefer way, which we will use, and leave this till we come to Obliqne Triangles, as a place fit for it: I only named it here, because I would give notice that my Father in saying all, he means 〈◊〉 for he saith (Axiom 3. Book 1.) All plain Triangles, etc. The reason why this Rule is not so good here, is this, if you subtract the Right Angle which is here always known from 180 deg. the Remainder is 90 deg. the Sum of the unknown Angles together, the half of which is 45 deg. now 45 deg. is the Radius of a Tangent, which is indifferent whether you take it or no, for it must be cast away in your work: and therefore if you only take the Logarithm answering to the Sum of the two Sides comp. arith. and the Logarithm of the Difference of them and add them together, they will produce this Angle of the Difference, which added to 45 deg. is the greater or subtracted is the lesser; but which is better, you may do it with less trouble this way. Say for the Course,  
As the Difference of Latitude 60 mile's comp. arith.  
Is to Radius  
So is the Departure from the Meridian A D 89 8 tenths  
To the Tangent of the Course A B.  
Or, As the Departure is to Radius; So is the Difference of Latitude to Tangent Compliment of the Course, Tangent A.  
We will work it in tenths.  
As Diff. of Latitude 600 tenths B D comp. arith. 7,2218487 Is to Radius  So is the Departure 898 tenths A D 2,9532763 To the Tang. of the Course, Tan. B 56 d. 15 m. 10,1751250  
Now the Angle comes out to a minute of a degree. And thus you have your desire at once; but this will be thus in none but right angled Triangles (whereas I said before the Radius of a Tangent 45 deg.) is always half the Sum of the required Angles.  
Note, that you always begin with the Side that is not opposite to the Angle you would have, and then you will put the Side that is opposite to the Angle in Proportion, with the Tangent of it; and thus for any other right angled Triangle of this nature:    

Difference of Latitude given alone, Difference of Longitude and Distance run in one entire Sum, I demand the Course and Sides several.  
QUESTION VII.  
Admit I set from the Latitude of 50 deg. and sail upon some Point between the South and the West, till I come into the Latitude of 49 deg. 0 min. (by Observation.) Now those that keep account of things, tell me, that the Distance that the Ship hath run, and the Departure from the Meridian, is together 197 8/10 miles: I demand them several and the Course.  
IN this case (because it would take up more room than my Book would hold) I'll make every 10 upon my Scale of equal parts 20, than every one of the small divisions will be 2. For the protracting of it Geometrically, first draw the North and South line B R at length, and upon it set off your Difference of Latitude, which is D B, make B the place of your Ships setting out, because it is the Northermost Latitude, and your Difference of Latitude is Southerly, raise a Perpendicular (or  depiction of geometrical figure East and West line) from D, and it must cut through the place where the Ship is; because she is in that Parallel or Latitude: upon this East and West line, from D set the Distance run, and Departure together, which is 197 miles 8/10, and it reacheth to E: by E and B draw the line E B, and divide it into two equal parts by a Perpendicular (you see it is a Perpendicular by the Arches h I and f g) draw it from P, till it cuts the Parallel E D, which it doth in A; from A then draw a line to the place you proposed to yourself the Ship set from, which is B; now the length of that line, namely, A B, shall be the Distance run, and A D the Departure from the Meridian.  

Demonstration.  
B E is divided into two equal parts in P, then E P and P B must be equal, P A is a Perpendicular to E P and B P, then must A B and A E be equal, because the Perpendicular P A serves for them both, and their Bases, or the sides E P and P B are equal, as was said before.  
It is evident then, that B A and A D are equal to E D, for if E A added to A D maketh E D, A B which is equal to A E, added to A D must be the same: and thus A B is the Distance run, and A D the Difference of Longitude or Departure from the Meridian, measure them as before; also from B as a Centre, describe the Arch of 60 deg. S N, and extend the Sides A B and B R to it, and measure the Angle of the Course A B D (as before) whose measure is N s, And thus much for the Geometrical way of working and demonstrating.  
The Distance run A B is 108 miles: The Departure from the Meridian A D is 90 miles: The Course is S W b W.   

By the Tables.  
I consider that the side D B (the Difference of Latitude) and D E the Distance run and Difference of Longitude together are given in the Question at first; whereby we may by the last Example, find the Angle at E, in the right angled Triangle B D E: Thus 
For the Angle at E and so for the Course.  As D E Dist. run, and Depar. 197 8/10 which is 1978 tenths comp. arith. 6,7037736 Is to Radius   So is the Difference of Latitude D B 600 tenths 2,7781512 To Tangent E 16 deg. 53 min.  9,4819248  
The Compliment of which to 90 deg. is E B D 73 deg. 7 min.  
Now because the Angle P E A and P B A is equal, subtract the Angle E B A from E B D, and the Remainder is A B D the Course  
Example.  
E B A 16 deg. 53 min. equal to A E P, subtracted from E B D 73 deg. 7 min. the Remainder is A B D 56 d. 15 m. the Course.  
The reason that I take 3 from 7, and set down 5 is, because if you look to the absolute minute, and part of a minute of the Arch that answers to this artificial Tangent 9,4819248, it will fall to such a Fraction, or to be the 15 part of a minute less than 3, as will bring it out; but it is needless to be so exact in cases or Questions of this nature.  
I will not show how to work that, till I come to a work that requires such exactness; here you see if you take it without allowing for the Fraction, it is but 1 min. less or more, which is but the 675 th'. part of a point of the Compass, but to go to the exactness of it, it is the same.  

For the Departure from the Meridian A D.  As Sine comp. the Course A B D 56 d. 15 m. co. are. 0,2552609 To the Difference of Latitude 600 tenths B D 2,7781512 So is the Sine of the Course 56 deg 15 min. A B D 2,9198464 To the Departure from the Meridian A D 898 tenths 9,9532585  
Which divided by 10, brings out 89 miles 8 tenths of a mile, as it was in the former Examples.  

To find the Distance run A B.  As Sine come. A B D the Course 56 d. 15 m. co. are. 0,2552609 To the Difference of Latitude 600 tenths B D 2,7781512 So is Radius to the Distance A B 108 miles 3,0334121    

A TRAVIS.  
Admit I set from Sommars Islands lying in the Latitude of 32 deg. 20 m. North Lat. and sail S E 40 leagues; from thence S S E, till I altar my Latitude 1 d. 0 m. from thence upon some point between the North and the East 40 leagues, till I altar my Lat 1 d. 45 m. I demand what Lat. I am in, my Distance run upon a straight line, my Departure from my first Meridian, and my Course made good.  
FOr the doing this Geometrically, I have the Distance run, and the Course given; to lay down the Triangle A C B, make A the place you set from, and lay it down as was showed in an Example of that nature before, Quest. 3.   depiction of geometrical figure 
The departure from the first Meridian is F R 56 leagues.  
The Course is the measure of the Arch s e O, which is E S E 9 deg. 8 min. Easterly.  
The Distance that I am from the Port I set is A F 57 leagues.  
Diff. of Lat. made since I set out is R A 13 leagues or 00 d. 39 m. Which subtracted from the Latitude I set from 32 20 The Remainder is the Latitude you are now in 31 41  
And thus you see you sailed from A to C, and then altered your Course, and sailed from C to e, and then from E to F; the like is to be understood of any Travis else. And thus if you had a Travis of 24 hours, you might find what Distance your Ship is from the place she was the day before, and he other things, as Difference of Latitude, the Latitude you are in, the Departure, the Course made good upon a straight distance; so that you need not set down every thing to every alteration in your Reckoning which is endless and fruitless.  

By the Tables.  
To give an answer to the demands in the Travis, you need find nothing but the Difference of Latitude, and Departure that is made in every Angle and note which way it is, either North or South for Latitude; East or West for Longitude, and taking this notice after you have done, see which is greatest, (in the Difference of Latitude) either the Northing or Southing, or for Longitude the Easting or Westing, and that which is greatest, that way it is, and so much as it is above the other: We will show an Example of this, when we have wrought the Travis.  
I mind the Travis, and I find I have the Course and Distance run in the first Triangle, to find the Difference of Latitude and Departure.  

For the Departure.  As Radius  To the Distance run A C 40 leagues 400 tenths 2,6020600 So is the Course B A C 45 deg. 9,8494800 To the Depar. BC 283 tenths or 28 leag. 3 tenths 28 leagues 3 tenths East Departure. 2,4515450  
For the Difference of Latitude.  
Forasmuch as the Angle opposite to the Difference of Latitude, is equal to the Angle opposite to the Departure, namely, both 45 deg. the Sides will be also equal, therefore is A B the Difference of Latitude South 28 leagues 3 tenths.  
For the Second Triangle.  
Here I have given the Difference of Latitude and Course to find the Departure from the Meridian, which is all I need.  
As Sine come. the Course D C E 22 d. 30 m. co. are. 0,0343846 Is to the Diff. of Latitude 20 leagues or 291 tenths 2,3010300 So is the Sine of the Course D C E 22 deg. 0 min. 9,5828307 To the Departure D F 8 8/10 leagues or 88 tenths 1,9182543  
The Departure is 8 leagues 8 tenths East, because the Course is Easterly.  
For the third Triangle.  
In the third Triangle I have given the Difference of Latitude, and Distance run, to find the Departure.  

For the Course.  As the Distance run G F 40 leagues comp arith. 7,3979399 Is to Radius  So is the Difference of Latitude G e 35 leagues 3,5440680 To the Sine comp. of the Course s G F e, the comp. of which is the Course G e F 28 deg. 58 min. 9,9420079  

For the Departure  As Radius  To the Distance 40 leagues 2,6020600 So is the Course 28 deg. 58 min. 9,6851151 To the Departure 19 4/10 leagues 2,2871751  
The Departure is East, because the Course is Easterly.  
Thus now in every Angle you know the Difference of Longitude and Difference of Latitude in leagues, and also which way either North or South, East or West, set them down apart thus: Lon. East. Lon. West. Northing. Southing. 28 3/10 00 00 28 3/10 8 3/10 00 00 20 19 4/10 00 35 00 56 1/10 00 35 48 3/10  
Here I have added them up, and I find no West Longitude, so that the Departure from the Meridian is East 56 leagues 1/10, the Difference of Latitude North I find to be 35 leagues, the Difference of Latitude Southerly is 48 3/10, subtract the Northing which is least from the Southing, and the remainder is the Difference of Latitude in the whole run 13 leagues 3/10; if you had West Longitude, you must have subtracted thus likewise, and that which was greatest the remainder belongs to.     

OF A RECKONING.  
FRom what hath been already showed, you understand, that if the Latitude of two places be known, and the Difference of Longitude between them, you may find the bearing of them one from the other, their distance asunder, or any thing you desire; and this is all so plain, that if I set from a place, and am bound to a place, whose Latitude is known, and the difference of Longitude also known between them, and keep a reckoning of the departure both East and West that I make, setting them down in two distinct Columns; it is but subtracting one from the other, and the remainder will be the East or West Longitude that I have made, (that of the two which was most,) also, if I know what Latitude I am in, and what Latitude the place lies in, it is but subtracting the lesser from the greater, and the remainder will be the difference of Latitude between your ship and that place (provided the Latitudes be both one way.) Also, subtract the Difference of Longitude that you have made, (provided that it be that way which shortens your Longitude between the place you set from, and are bound to,) from the whole difference of Longitude between the two places, and the remainder will be what is still between you and the place you are bound to: but if your departure which you have made, hath increased the difference of Longitude between the place you set from, and are bound to, add the remainder (after your subtracting the Columns one from the other) to the whole difference of Longitude between the places; the like for Latitude.  

Example in the Reckoning following.  
Suppose the Island of Ditiatha lies so, that it hath been found to have 950 Leagues departure West, from the Meridian of the Lizard (according to the Courses that be ordinarily sailed by Plano) the Latitude of it is 16 deg. 16 min. North Latitude, the Lizard lies in the Latitude of 50 deg. 00 min. Now upon some Occasion that fell in the term of our Voyage, namely, the 29th. of January on Tuesday, the Captain demands of me, what Distance Ditiatha was from me by Plano, how it bears, what Leagues of Departure I have to it, and the like: I look in my Reckoning, and I find in the West Column 434 Leagues, I look in the East Column and see 60 Leagues; I subtract 60 (my whole East Column) from 434 (the whole West Column) and the Remainder is 374 Leagues; which signifieth the Ship hath departed from the Meridian of the Lizard 374 Leagues West; for the West Column was greatest.  
Subtract this West departure 374 Leagues, from 950 the whole Departure, and I find there is still wanting 576 Leagues; This 576 Leagues is the Departure that is from the Meridian that my Ship is in, and Ditiatha.  
Subtract also the Latitudes one from the other, (namely, the Latitude your Ship is in, and the Latitude of Ditiatha) and the Remainder is the Difference of Latitude in degrees and minutes between the Ship and the Place you are bound to, which in this Example I find to be 6 degrees 24 minutes. Take the Latitude out of the Column of Latitude for that day in the Reckoning, and you will find it so, if you subtract 16 deg 16 min. from it.  
Thus you have the Difference of Latitude between you, that day, and your Port, and the Departure that the Port hath from the Meridian that you are in that day, given you to find the things demanded; I will not work it, for it hath been showed sufficiently already.  
I have here following set down a Reckoning, and afterwards how I work it, and lastly the reason that makes me use this way of keeping my whole Reckoning in the last line, so that I am not troubled to add it up, when I give account of it.  

January the 3d Anno Dom. 1655, we departed from the Lizard, lying in the Latitude of 50 d. 00 m. being bound for Ditiatha, which lies in the Latitude of 16 d. 16 m. it having West Departure from the Meridian of the Lizard 950 Leagues (according to the courses which we then steered) at 4 of the clock, the Lizard bore from us N N E, about 6 leagues distance by estimation.  
EVery day at Sea you set down at noon that 24 hours' work: Now I advise you to set it down as it is here: In the first Column on the left hand, I set the day of the Month, in the next the day of the Week, in the third the Latitude I am in by dead Reckoning or by Observation, only I make a distinction between them by the letter E, which stands for Estimation, and is set down only when I do not observe: that so you may know where to begin to correct. In the 4 th'. Column I set down the Easting, in the fifth column the Westing.  
The manner that I set my Easting and Westing down is thus. Every day there is the Ships course minded, and an observation, whereby you have the difference of Latitude, and course to find the departure; or else if you cannot observe, you have the distance run, guessed at, and the course; to help your D. M. Week days. Latitude Dep. E. in leag. Dep. W. in leag. deg. min. 4 Friday 49 00 000 021 5 Saturday 47 55 000 059 6 Sunday 46 20 000 085 7 Monday 45 00 000 088 8 Tuesday 43 E 54 000 108 9 Wednesday 42 4 000 147 10 Thursday 41 E 19 000 262 11 Friday 41 10 000 264 12 Saturday 40 46 000 274 13 Sunday 40 45 000 178 14 Monday 39 E 49 000 179 15 Tuesday 37 E 24 000 200 16 Wednesday 35 59 000 229 17 Thursday 34 42 000 249 18 Friday 33 15 000 278 19 Saturday 31 15 000 322 20 Sunday 29 5 000 364 21 Monday 28 E 20 000 400 22 Tuesday 27 5 012 400 23 Wednesday 27 30 022 400 24 Thursday 27 22 031 400 25 Friday 26 E 49 049 400 26 Saturday 25 36 060 400 27 Sunday 24 31 060 406 28 Monday 23 28 060 406 29 Tuesday 22 E 40 060 434 30 Wednesday 22 08 060 434 31 Thursday 21 53 060 444 February. 1 Friday 21 40 060 450 2 Saturday 20 47 060 456 3 Sunday 19 22 060 460 4 Monday 19 53 060 472 5 Tuesday 18 23 060 485 D. M. Week days. Latitude Dep. E. in leag. Dep. W. in leag. deg. min. 6 Wednesday 17 19 060 518 7 Thursday 16 47 060 556 8 Friday 16 54 060 606 9 Saturday 16 29 060 643 10 Sunday 16 19 060 678 11 Monday 16 21 060 726 12 Tuesday 16 28 060 784 13 Wednesday 16 35 060 840 14 Thursday 16 27 060 889 15 Friday 16 24 060 929 16 Saturday 16 2 060 973 17 Sunday 16 44 060 1003 guess is the Log-line every 2 hours; your course and distance is set down upon the Logboard every 2 hours, whereby you may see it, and work it every 24 hours, to find the departure and difference of Latitude: Now whatsoever I find the departure to be, I add it from the first day all along thus: the departure when I set the Lizard (which was N N E 6 Leagues off) is 2 League's West; I keep mind of that till the next day at noon, and I find what departure I have made from the Meridian that I was in the day before at noon (by the things I have given me) and here I found it was 19 Leagues, I add it to my departure from the Meridian West yesterday 2 Leagues, and it makes my whole departure West 21 Leagues.  
Again, the next day at noon, I work my Ships Travis, as hath been showed, and I find that she hath departed from the Meridian she was in yesterday 38 Leagues West, which I add to what I had before, which was 21, and it makes 59 Leagues; and thus I go forwards, adding my last days departure in its true Column to all the rest, (which is in one sum) and by this means the last line is the whole Reckoning. 21 38 59  
In the East Column I set cyphers to fill it up, till I have some East departure to put in it, and as that increaseth, so I add it (as the other) every day.  
Also, I carry the West departure along in its full number of Leagues in the same line, without increase or decrease, till such time as I have more to increase it, and then I increase that, and carry the whole Easting with it in the same line: and thus you have need to look upon nothing, but the last line, to resolve you any thing that you desire, either of your course made good since you set out, your distance upon a strait line that you have failed, your Ports bearing from you, what departure is yet between you and your Port, your distance to it upon a strait line; for there you have your whole Westing, and your whole Easting, and the Latitude you are in, as also the Latitude of the place you are bound to; whereby (as I have formerly showed) you may find them.  
If there be Longitude but one way, you will see nothing but cyphers in the other Column: this way you see is done without a Plat. I hold a Plate to be necessary only to show what dangers lie in your way, that so you may shape a course clear by it, and I should use a Plate for nothing else, except it be for Coasting, where it is really useful.  
I commonly set down the Easting or Westing between the place I set and the place I am bound to at the beginning of my reckoning, as also the Latitude of each place, as I have done here. You may ask why I do not set down the course that I made good every day: indeed that is not unnecessary, but the reason why I omit it is, First, because I find it of little use in my Reckoning; and secondly, I find it can be better expressed in a Journal which is kept with my Reckoning; and indeed there I set it with the Winds, and the reasons for steering upon such Courses: but I leave every one to their own Judgement for that, as also for Distances and Winds, for Difference and Variation.  
It would have been necessary here to have set down some Tables to have worked Triangles by, for (as I have said) sometimes in 24 hours you have a Travis of 4 or 5 several courses, and to work them this way may seem tedious. I commend you to my Father's Practice, where there is as good Tables as can be, to every degree of the Compass; its use is easy, and works to the tenth part of a Mile or League.     

CONCERNING the VARIATION OF THE COMPASS.  
THere is always two things given to find the Variation of the Compass; that is, the true Amplitude of the Suns rising, and the Magnetical Amplitude of the Suns rising or setting: the true Amplitude of the Suns rising, is, (as I have said before in another place) the true and absolute quantity of degrees, that the Sun riseth from the East, (either Northwards or Southwards) or sets from the West, and is found (as I have before shown) in the use of the Sphere.  
The Magnetical Amplitude of the Sun, is what the Sun riseth from the East, or sets from the West by the Compass: Now because the first gives the Truth how far the Sun riseth from the East, or sets from the West; therefore whatsoever difference there is between them, so much is the variation.  
As now, Suppose I find in a certain Latitude such a day of the Month (by the Sphere) that the Sun riseth to the Northwards of the East 17 degrees; that is, the true Amplitude: but I observe the Sun at her Rising, with an Azimuth Compass (which is made for that purpose) and find that she riseth but East 10 degrees Northerly; then I conclude the Variation is 7 deg. 0 min. or the Compass is false so much; so that whereas if I direct my Course East 10 degrees Northerly by the Compass, I do not go on that Course, but I go East and by North 5 deg. 45 min. Northerly, which is just 7 d. from my expectation, or East 17 deg. Northerly, the true Amplitude. Now that is a gross error, and in long runs may deceive a man much, and perhaps be a means to lose a Ship, when one little thinks of it; and therefore it ought to be looked to.  
An Azimuth Compass is no other in effect, but a Compass fitted for the exact taking of the Sun at her Rising or Setting, or upon other certain times of the day, as you may have occasion.  
In like manner you may find the Variation of the Compass at other times, by taking the Sun's Azimuth at any time of the day.  

Example.  
Suppose I were in the Latitude of 33 deg. 20 min. Northerly, and upon the 8 th'. day of November the Sun's Declination is Southerly 19 deg. 20 min. I demand the Sun's Azimuth at 8 of the Clock.  
The Sun's Azimuth at her Rising (as I have showed) is the Compliment of the Sun's Amplitude; but after the Sun is up, it may be also the Distance of the Sun from the East and West Azimuth: Now in this Proposition, you desire to know how many Degrees the Sun is from the East and West Azimuth, (which is that Part of the Heavens that is distant from the Sun) parallel to the Horizon over the East Point of it; or if it had been at 4 of the Clock in the Afternoon, it would have been required from the West Point.  
For the resolving this Question project a Sphere (as hath been taught) for the Latitude proposed, with the Parallel of the Sun's Declination drawn as it is given: Divide this Parallel of Declination into the hours of the day from 6 to 12, or which serves from 12 to 6 in the Afternoon; the way to divide it is thus; From 6 of the Clock set off 15 degrees by the third way of measuring, which will be 7, than (because 30 degrees is two hours in time) next set off 30 degrees by the third way of Measuring from 6, and it makes 8 a Clock; then take 45 deg. which is three hours, and it makes 9 a Clock, than 60 deg. for 10 a Clock, than 75 deg. for 11 a Clock, and then 90 deg. which will just fall in the place where the Sun comes to the Meridian for 12 a Clock. Thus you divide the Parallel of the Sun's Declination into hours, as you may see in the Sphere here following: it being thus divided, the Sun's Azimuth at 8 of the Clock is 8 s, taken off by the third way of Measuring, I find it to be 33 deg. 0 min. (by the Scale) so it is certain that the Sun should be just 33 deg. 0 min. to the Southwards of the East at 8 of the Clock, than I will see how high he should be at that time, and I find him to be 13 deg. 50 min. equal to s ♈, take off s ♈ by the second way of Measuring. I will observe the Sun till I find him at that height, and then I know it is 8 of the Clock; see by your Azimuth Compass whether he be 33 deg. 0 min. from the East or no, and if not, what it differs, so much is the Variation of the Compass.  
I have wrought this Example no nearer than the plain Scale works it, which possibly may be 15 min. or half a degree of the Compass out of the way, but that is no considerable error in a Course, which is the thing we here use it for.  
The same may be done at any time of the day, as Latitude 32 deg. 20 min. North, Declination 19 deg. 20 min. South.  
Suppose it were cloudy, but at some time the Sun breaks out, so that I have an opportunity to take his height with my Quadrant, and also his magnetical Azimuth with my Azimuth Compass: Now I find his height to be r t, equal to N P, and the Azimuth at that time is N t, it being drawn parallel to the Horizon; this N t set off by the third way of Measuring, and so much as it comes to, so much is the true Azimuth of the Sun from the East towards the South: Take your magnetical Azimuth at the same time, and as much as they differ, so much is the Variation: And thus much for the Variation of the Compass.   depiction of geometrical figure 
When a man observes the Sun with a Quadrant (which is our usual Instrument) he takes the upper edge, which is to the Northwards in this Latitude. Suppose then I set from the Lizard, and am bound to Barbadoes, I make observations of the Sun as often as I can all the way, till I bring the Sun to the Zenith: after I pass the Zenith, the edge of the Sun that was highest is lowest: Now he that doth not consider this, loseth the breadth or Diameter of the Sun, so that your observations may differ from your expectation 30 miles, which is enough to miss an Island; therefore I advise you always to allow the Sun's Diameter (which some count 30 min.) to every observation you make in places where the Sun is to the Northwards of you; that is, make your Meridian Altitude 30 min. less than it is, which is the Altitude of the lower edge, which was your upper edge before you crossed the Zenith, where the Sun was; or if you work your observations by the Compliment of his height, make that 30 min. more.  
I confess the best way would be at all times to subtract the Semidiameter of the Sun, which is 15 min. from your Meridian Altitude: But because the Latitude of Places are not set down with this consideration, it is more safe to do as before, unless you know otherwise by your own experience. This only by the by to give notice.   

Example, for the laying open of this Error.  
Suppose I observe the Sun at Barbadoes the 11th. day of December, and I find that the Meridian Altitude of the Sun's upper edge is 53 deg. 18 min. to which I add the Sun's Declination for that day, which is Southerly 23 deg. 30 min. and it gives the height of the Aequinoctial above the Horizon of the Compliment of the Latitude to 90 degrees, which is 76 deg. 48 min. which subtracted from 90 degrees, leaves the Latitude or Aequinoctial's Distance from the Zenith, 13 deg. 12 min.  
Upon the 11th. of June, I observe the Sun again in the same place, and I find the Meridian Altitude of the upper edge of the Sun is 80 deg. 12 min. (Note that this upper edge is opposite to the edge before, for the Sun's Declination is 23 deg. 30 min. Northerly, which makes that the Sun gives a South Horizon) I add the Sun's Declination to the Meridian Altitude, and it gives 103 deg. 42 min. the Distance between the Aequinoctial and the North Horizon, from which subtract 90 deg. and the remainder is the Aequinoctial's distance from the Zenith, or the Latitude which is 13 deg. 42 min. Now this Latitude should agree with the other, for the place stands still; but for want of this subtracting the Semidiameter of the Sun, it differs 30 min. and when men meet with things thus at Sea, for want of minding of this they run 30 min. more Southerly than they should do, which I suppose may be the greatest reason of their missing this Island, and others that lie near the Aequinoctial; but for those places that lie far from the bounds of the Sun, it is not to be minded: Yet it were good (methinks) if all Places had been laid down with allowance for the Sun's Semidiameter; but because they are not, you ought to be careful, when at any time you cross the Zenith of the Sun: I have not been myself in a time of year to cross the Sun's Zenith, and therefore I only set this down by way of Caution to those that may.    

The USE of a PLAIN SEA-CHART.  
MOst Plaits have all the Points of the Compass drawn out from several Centres through the Plate, and because they are common, and the way to work by them so ordinarily known, I have thought it necessary to draw the manner of a Plate, and to show its use that hath only Meridian's and Parallels drawn in it, which is the Plate following.  
Suppose then that I have a Plate that hath the Meridian's and Parallels drawn to every fourth degree, as this following is. You see this Plate is divided into equal Squares throughout, by the Meridian's and Parallels so drawn, also one of these Squares is divided into eight Points of the Compass, which is the Square E C D O, (the manner of dividing it we will show anon.) Now if A where the Lizard, and B were the Island Madera, I demand their distance asunder. Let B A represent the edge of a Scale lying between them, take the Distance A B, and apply it to the Meridian, and see how many degrees it reacheth, and convert them into miles, and it answers your desire. If you cannot take the distance at once, take it at twice, and add the several Sums together, and convert them into miles: Then  

For their bearing one from the other.  
Imagine (as I said before) that A B were the edge of a Scale cutting upon the two places. Look for that Corner that is nearest without the edge (which is S) fix one foot of your Compasses in S, and with the other sweep the edge of your Scale, which will be S G, keep fast your Scale, and from the Corner S run that foot of your Compasses (keeping it at the same Distance from the Scale by the other foot running by the edge of the Scale) till you come to the next side with it (which in this example fails at F) then take the Distance E L, and apply it to the graduated Square, either from D or E, and as many points as it is there, so many Points doth the Lizard bear to the Eastwards of the North, from the Maderas, which is here two points and almost a quarter. The same will be done from a Corner farther from the Scale, and this is evident; for by the Flower de luys you see that S L is a North line from S, and a line drawn from S to F is parallel to B A, or runs upon the same point to the Eastwards of the North from S. Therefore whatsoever line S F is, the same is B A, the like is to be understood of any other example, as, suppose we had not found the bearing of the Lizard from the Maderas, and would find the bearing of the Maderas from the Lizard. Because my Scale runs no farther than B, I cannot use the Corner at S (which is nearest) and the rest are upon the graduated Meridian, therefore I will use the Corner at L, as I did the Corner at S; but because it is the contrary thing, I'll run the contrary way, namely from L to H, then is L H and F S parallel, and because S L is the Meridian to them both, therefore are H S and L F equal, so that if S H be applied to the graduated Square, it is as many points to the Westwards of the South, as L F is to the Eastwards of the North, therefore the Maderas bear from the Lizard South South-west almost a quarter of a point Westerly: if it had been from any other Corner, it would have been the same, the like for any other.   

For the Departure from the Meridian between two Places.  
Suppose X were one Place, and (‖) the other, and I would know the Departure or Difference of Longitude between them.  
Fix one foot of your Compasses in (‖), and extend the other to sweep the Meridian nearest short of the other place, which is the Meridian V, that Distance apply to the graduated Meridian, and see how much it is, then take the Distance from the other place (which is X) to sweep the nearest part of the same Meridian, and apply that to the graduated Meridian, see how much it is, and add it to the former, and it makes the whole Departure between the two places, X and (‖), convert it into miles or leagues, according as you desire; in most Plaits you have a Scale or Leagues, which is necessary for small Distances.   

For the Difference of Latitude between two Places.  
The Difference of Latitude between two places is taken after the same manner from a Parallel, as the Departure is from a Meridian Line; as suppose I would take the Difference of Latitude between X and (‖), first, take the nearest Distance between X and the Parallel that T is in, and see how much it is. Then take the nearest Distance between (‖) and the same Parallel, and see how much it is, and add them together, and that convert into miles or leagues according as you desire.  
To find what Latitude any place lies in, take the nearest Distance between that Place and the next Parallel with your Compasses, and look where that Parallel cuts the graduated Meridian, and carry your Compasses with that distance there (one foot) and observe to set the other that way as the place is, either to the Northwards or Southwards of that Meridian, and see upon what degree and minute of a degree it falls, and that is the Latitude of the place.  
As, suppose I would know the Latitude of the North side of the Maderas, I will fix my Compasses, one foot in B, and extend the other to sweep the Parallel that S is in, that Distance I will carry to the place were the Parallel cuts the graduated Meridian which is at 34, and fixing one foot there, I will see where the other falls to the Southwards towards L (because it is the Southwards of that Parallel) and I find it falls in the Latitude of 32 deg. 30 min. from whence I conclude, that the Northermost part of the Maderas lies in the Latitude of 32 deg. 30 min.  
The way that the Square E C D is divided into Points, is on this manner: First, I draw the line O C, then the other lines for half-points, and quarter-points, than I describe the Circles, making the outmost come to the innermost of the lines (but whether it be or no is no great matter) the Circles being divided into Points, half-points, and quarters, I lay my Scale by the Centre of these Circles, which is O, and the divisions on the Circles, & where my Scale cuts upon the outer line, from thence I draw my Points, half-points, and quarter-points, as you see here in the following Blank; and the Points upon the sides of the Square thus set off are the Tangents of the Points upon the Arch: for when from the outermost edge of an Arch a Perpendicular is raised from a side that bounds that Arch (as the side N 4 is,) and the other side that bounds the same Arch be extended without the Arch to intersect it (as the line O C intersects N 4 at 4,) that Perpendicular is the measure of that Arch, either in degrees or points; as here N W is 45 degrees or 4 points, and so is N 4. And thus you may divide a Tangent line for degrees or points, from an Arch divided into degrees or points.   

To set off any Course from a Place assigned upon a Blank.  
Let the Place assigned be X, from which I would draw a N W b N line.  
Take three points from the graduated Square (making D or E the place you take it from) and fix one foot of your Compass in the next convenient Corner to X (which is at M) and set off the three points which fall at K; then from the next Corner upon the same Meridian that M is, (and to the Southwards of it) which is at P, lay your Scale to K, than your Scale lieth N W b N from P, and a line Parallel to the Scale from X must be a N W b N line, the like for any Course else, observing the quarter of the Compass, it is to be set off in, as also whether it be nearer the East or West than the North or South; for if it be nearer the East or West than the North or South, you must do the same from a Parallel, as here you did from a Meridian; for the side of a Square is but the measure of four points, which is but half the points between a Meridian and a Parallel. This way of working may seem hard and tedious at first, but you will soon find that it is free from mistakes, and both exact and easy, if you practise it.  

Place this between Page 112 and 113.   depiction of a map with Lizard and Madera    

Of OBLIQNE TRIANGLES.  

Two Sides with an Angle opposite to one of them given, to find the other Angles and Side.  
QUESTION I.  
Two Ships set sail from the Rock of Lisbon, one sailed W S W, the other sailed N W b W 38 leagues, and at the end of their sailing, they were 58 leagues asunder, I demand the Southermost Ships Distance run, and how the Ships bear one from the other?  
FIrst draw a North and South line white, and then from that set off the Northermost Ships Course, make the Rock or Place setting out the Place in the North and South line that you draw that Course from, which is C, upon this Course set off 38 leagues (because the Question saith you sailed 38 leagues upon it) and extend the side C A to the Arch of 60 deg. at t, and set off five points from t to S, and draw S C a white line, which is W S W Course: (for it is five points between the W S W and the N W b W) this done, take 58 leagues from the Scale of equal parts, and fix one foot of your Compasses in A, and where the other intersects the Course B C (which it doth in B) there is the other Ship, then to measure the Angle of the Ships bearing one from the other it is B, and B C is a E N E line; extend B A to the Arch of 60 deg. whose Centre is at B, and see how many degrees or points it is more Northerly than an E N E line, and so the other Ship (namely, the Ship at A) bears from the Ship at B;  depiction of geometrical figure then take the length of B C, and apply it to your Scale, and see how many leagues or miles it is. I have wrought this in leagues, but I will work the rest in miles, because it is more exact. I find that the Ship at A bears from B, N E b N 45 min. Easterly, the Distance run of the Southermost Ship is 70 leagues.  

By the Tables.  
The proportion of this and all others of this kind, is the same that holds in right angled Triangles; namely, that the Sine of every Angle is proportional to its opposite side, or every side is proportional to the Sine of its opposite Angle.  
Here we have given the side C A 38 Leagues, the side A B 58 Leagues, and the Angle at C which is 5 points or 56 deg. 15 min.   

Say then for the Angle at B.  
As A B 58 leagues, 580 tem. comp. arith 7,2365719 Is to A C B 56 deg. 15 min. Sine  9,9198464 So is A C 38 leagues, 380 tem.  2,5797836 To A B C Sine 33 deg. 0 min.  9,7362019  
Which subtract from 6 points or 67 d. 30 m.  33 00 Remainder is the Course from North 34 d. 30 m.  
Which the Ship A bears from the Ship B, which is N E b N 45 min. Easterly.   

For the Distance run of the Ship at B, the Side B C.  
As Sine C 56 deg. 15 min. comp. arith 0,080153 Is to A B 58 leagues  2,763428 So is Sine B A C 90 deg. 45 min. Take the Sine of the acute Angle B A t 89 d. 15 m. 9,999961 To B C 68 8/10  2,843542  
The reason why you take the acute Angle is, because the Tables go no further than 90 deg. neither indeed is any Sine beyond 90 deg. but (as my Father saith in his Trigonometry, p. 2) the Sine of an Arch less than a Quadrant is also the right Sine of an Arch as much greater than a Quadrant: So than the right Sine of the Arch 90 deg. 45 min. which is 45 min. greater than a Quadrant, must be the right Sine of the Arch 89 deg. 15 min. which is as much less (namely, 45 min.) the Geometrical Demonstration of it is there laid down.  
You may ask how I came to know the Angle B A C, which was thus: I found one of the other two Angles, namely B, and the other I had given me, I added them both together, and that Sum I subtracted from 180 deg. the Remainder must then be the Angle at A, because 180 deg. is the Sum of the three Angles of any right lined Triangle: And if I subtract two of them from three of them, there will remain one of them, which was 90 deg. 45 min.    

The three Angles of a Triangle given, with one of the Sides, to find the other two sides.  
QUESTION II.  
Admit I set from a Head-land lying in the Latitude of 50 deg. 00 min. North Latitude, and sail W S W 38 Leagues, and then meet with a Ship that came from a Place which lies S S W from the Head-land. Now this Ship hath Sailed N W. I demand the Distance of that Place from the other, and also the Distance the Ship hath sailed that came from the Southermost Place.   depiction of geometrical figure 
The distance between the two places is A B 
58 miles.   
The distance that the Ship sailed is A O 
44 4/10 miles.   
If you have occasion to find the Latitude the place at A lies in, let fall the Perpendicular A N upon the South line that comes from B, and it cuts it in N, and leaves the Difference of Latitude B N, see how many miles it is, and subtract it from 50 deg. and you have your desire, measure A N, and you have the Westing that A lies from the Head-land B.  

By the Tables.  
Here the Angle at B is an Angle of 4 points or 45 d. 00 m. The Angle at A is an Angle of 6 points or 67 d. 30 m. The Angle at O is an Angle of 6 points or 67 d. 30 m. The side B O is 58 miles.    

First for the Side A B.  
The general Rule saith, That the Sine of every Angle is proportional to its opposite side. Then from this I conclude, that B A should be equal to B O, because the Angel's opposite to them are equal, and so you will find them For as Sine A 67 deg. 30 min. comp. arith. 0,034384 Is to O B 58 miles  1,763428 So is Sine O 67 deg. 30 min.  9,965615 To A B 58 miles, the two places distance 1,763428  
You see it is so exactly, for these three numbers added together, and Radius cast away, produceth the same Logarithm that 58 (taken out of the Book) did, and this Question I do on purpose to show the truth of the general Rule. The same way other Questions of this nature are wrought.   

For the Distance that the Southermost Ship sailed, A O.  
As A O B Sine 67 deg. 30 min. comp. arith. 0,034384 Is to A B 58 miles  2,763428 So is Sine A B O 45 deg. 00 min.  9,849485 To A O 44 4/10 miles  2,647297  
If you desire to find the Difference of Latitude between the two places B and A, you have the Distance A B, and the Course A B n given to find it or the Longitude.  
If you desire to find the Latitude the Ships are in when they meet, let fall a Perpendicular from O upon the South line (which is B N) and you have the distance O B given, 58 miles, and the Angle O B R 67 deg. 30 min. to find it, or the Departure between the Ships at their meeting, and the Head-land O r.  
The like is to be understood of any other Question of this nature.     

Two Sides and their contained Angle being given to find the third Side and the other Angles.  
QUESTION III.  
Two Ships set from one Port, and make an Angle of 58 deg. one sailed N b E, and the other sailed N W 1 deg. 45 min. Westerly, the Eastermost Ship sailed 70 miles and came to her Port: The Westermost Ship sailed 89 miles, and came to her Port: I demand the Distance of these Ports asunder, and how they bear one from the other?  
FIrst draw the North and South line A S. Let A B represent the Port you set from: Then from A as a Centre, describe the Arch of 60 deg. F S D: From S to the Eastwards set off one point: for N b E the Eastermost Ships Course draw the line A D at length, and set the first Ships distance 70 miles upon it, which is A B, then from A draw a N W line 4 deg. 45 min. Westerly, and set off the Westermost Ships distance run upon that, which is 89 miles, and it reacheth from A to C, draw the line C B from the ends of the other two sides, which is the Distance between the Ports (for the ends of each Ships run must be the Ports) then is A C B the Angle of the bearing of the Eastermost Port from the Westermost, or A B C the bearing of the Westermost Port from the Eastermost, C A is a S E line, 1 deg. 45 min. Easterly from C, see how many points are contained between the C B and C A, by the way that is showed in the Question before, or because B A is N b E line (from B) you may find the Point of the Compass that B C runs upon from B, you need find it but one of these ways, and the opposite point must be the other, as I have showed before, then take the length of the side C B, which is the distance of the Ports asunder, I find their distance and bearing to be as followeth.   depiction of geometrical figure 
The Distance between the two Ports is C B 79 miles.  
The Bearing of them is East 5 deg. 35 min. Northerly, or West 5 deg. 35 min. Southerly.  

By the Tables.  
Here is given the Angle at A, 58 deg. 00 min. the side A B 70 miles, the side A C 89 miles, which is two sides, and their contained Angle, and here is required A C B or A B C, and the side B C.  

First For the Angles.  
For all Questions of this nature, this Rule is beneficial, and will work your Question. Note,  
That in all plain Triangles, the Sum of two Sides are in such proportion to their Difference, as the Tangent of the half Sum of their opposite Angles, is to the Tangent of an Angle, which Angle shall be the Difference between the half Sum, and the Angles: so that if it be added to the half Sum, it shall make the greater Angle; if it be subtracted, it shall make the lesser.  
I have said something of this Rule before, which makes me demonstrate it no farther here.   

Proportion.  
As the Sum of A B and A C 159 mile's comp. arith. 6,79860 Is to their Difference 19 miles  2,27875 So is the Tan. of the half Sum of B and C 61 d. om. 10,25624 To Tangent of an Angle 12 10 9,33359  
This added to the half Sum makes the Angle at B 73 d. 10 m. subtracted, it makes the Angle at C 48 deg. 50 min. which is 4 points, and 3 deg. 50 min. then say I, if C A be a S E Course, 1 deg. 45 min. Easterly from C, 'tis certain that C B must be 4 points and 38 deg. 50 min. from it, which is East 5 deg. 35 min. Northerly, the bearing of the Eastermost Port from the Westermost.  
To find the half Sum of the Angles, Note that the three Angles are 180 d. therefore if one of them be known, subtract that from 180 d. and the Remainder must be the other two.   

Now for their Distance asunder, C B.  
As Sine A C B 48 deg. 50 min. comp. arith. 0,123321 Is to B A 70 mil.  2,845098 So is Sine B A C 58 deg. co min.  9,928420 To the Distance of the Ports asunder C B 78 9/10 2,896839     

The three Sides of a Triangle given, to find the Angles.  
QUESTION IU.  
Two Ships set from one Place, one sails N b E 70 miles, the other sails between the North and the West 89 miles, and at the end of their sailing they are distant asunder 78 miles 9/10 of a mile, I demand what Course the Westermost Ship sailed, and how the two Ships bear one from the other?  
FIrst draw the North and South line A e, make A the place of the Ships setting out, set off a N b E Course upon the Arch of 60 deg. I e, which is e D (because the Eastermost Ship sails N b E) likewise set off her Distance which she sailed upon that Course, which is A B 70 miles, then take the other Ships  depiction of geometrical figure Distance, which is 89 miles, and fixing one foot of your Compasses in A, with the other describe the Arch at C: Lastly, take the Distance of the two Ships asunder 78 miles 9/10, and fixing one foot of your Compasses in B, across that Arch, and from the place of intersection draw the sides of the Triangle C A and C B: then measure the Angles from an Arch of 60 deg. as hath been showed before; and to know them, consider that A B is a N b E line, so that A F will be upon a Course so many points, or degrees, and minutes from a N b E, as there are points, or degrees and minutes in the Arch D F. Secondly, for the bearing of the Ships, which is the Angle at B: consider that B A is a S b W line from B, and then the degrees and minutes, or points contained in the Arch O R, is so many degrees and minutes, or points from the S b W.  

I have resolved the Demands, which are as followeth.  
The Course that the Westermost Ship steered was N W 2 d. Westerly.  
The bearing of the Eastermost Ship from the Westermost is W b S 5 deg. 30 min. West.  
I set the Angles by the plain Scale a little differing, because the Scale cannot be guessed to a minute, or 2 or 5, nor the miles to a single tenth.  

By Proportion.  
Here the three sides are given us to find the Angles; the side A B is 70 miles, the side B C is 78 9/10 miles, the side C A is 89 miles.   

A General Rule in Obliqne Triangles.  
As the true Base  
Is to the Sum of the other two sides:  
So is the Difference of the sides  
To the alternate Base.  
You may make which side you will the true Base. If you make either of the shortest sides the true Base, and let fall a Perpendicular from the Angle opposite to it, the side must be extended in some cases, for the Perpendicular will fall without the Triangle, and the extent of the alternate Base will fall as far without the Perpendicular, as the Perpendicular is without the length of the true Base: but if you make the longest side the true Base, the Perpendicular will fall within the Triangle in any Question whatsoever, so that you will have no occasion to extend the side, which makes me use the longest side for the true Base, though in this Question take any side and it would fall within.  
Here the side C A is longest, and therefore we will make that the true Base: take the length of the shortest side, which is A B, and fixing one foot of your Compasses in B, (the Angle opposite to the true Base) cut the true Base with that side B A in the point S, then draw the line S B which is equal to A B, the Perpendicular N B will fall in the middle between S and A, (which is in the point N) then shall S N, and N A be equals C S is the alternate Base: say, As C A the true Base 89 mile's comp. arith 7,0506699 Is to C B and A B 148 9/10 miles  3,1728947 So is their Difference 8 9/10 miles  1,9493900 To the alternate Base S C 14 9/10 miles 2,1728946  
Subtract S C the alternate Base thus found 14 9/10 89 74 1/10 37 ½ tenth  
From C A the true Base 89 miles  
And the Remainder is S A 74 1/10 miles  
The half of which is N A 37 miles or  
N A being thus found, I have A B besides it in the rightangled Triangle A N B to find the Angle at A:  
Say then, As A B 70 mile's comp. arith 7,154901 Is to Radius   So is A N 37 miles  2,568201 To Sine comp. N B A 58 deg. 6 min. 9,723102  
Which is Sine B A C, or the quantity of the Arch D F.  
Consider then what Course the side A B runs upon, and 58 deg. 6 min. from that Course is the Course that the side A C runs upon.    

Example.  
A B runs upon a N b E Course from A: the side A C runs upon a Course 58 deg. 6 min. from it to the Northwards, and so to the Westwards, 58 deg. 6 min. is 5 points and 1 deg. 49 min. which is N W 1 deg. 45 min. Westerly.  
What this is above a Northwest 1 deg. 45 min. Westerly, is by the neglect of taking the absolute Logarithm answering to 2,1728946; and also the neglect of the half tenth in dividing: you see it is but 6 min. of a degree of the Compass from the other, in the Question before this, which is no sensible error in sailing, and indeed I neglect working so near, to let any Learner see what a necessary thing it is to be exact: the Angle at A being thus found, you may find the other Angles at pleasure, and also know what point the side B C runs upon, because you know what point B A runs upon from B.  

For the Angle C B A.  
As C B 78 9/10 mile's comp. arith. 7,102922 To Sine B A C 58 d. 6 m.  9,928893 So is C A 89 miles  2,949390 To Sine C B A 73 d. 16 m.  9,981205  
That is 6 points, and 5 deg. 46 min. more; then is B C a line that runs from B, W b S 5 deg. 46 min. Westerly, or the Ship at C bears from the Ship at B, W b S 5 deg. 46 min. Westerly.  
Then must B bear from C, E b N 5 deg. 46 min. Easterly; and thus your desire is known.  
But you may think much, that a Fraction should put you out so much as 6 min. (and if you know no better) you may be afraid it will cause some difference in the sides; but if you examine this Example, you will find the sides will not differ for this a full tenth of a mile.    

Example.  
By the Angles to find the sides to agree with the sides in the other Question, within a tenth part of a mile.  
As Sine B A C 58 deg. 6 min. comp. arith. 0,07110 To C B 78 9/10  2,89707 So is Sine A B C 73 deg. 16 min.  9,98120 To C A 89 0/10 miles  2,94938  
Here it is evident, that all this Difference doth not alter the sides the tenth part of a mile.  
If you desire to see farther proof of the Rule, by which this Question is wrought, you have a clear Geometrical Demonstration for it in page 19 and 20 of my Father's first Book of Trigonometry, which makes me forbear the demonstrating of it by any Geometrical way.    

QUESTION V.  
Admit I be sailing along the Shore, and see two Headlands, the Eastermost bears N N E off me, the Westermost bears N W from me, at the same time I see an Island that bears South from me; but I sail West 8 miles, and find the Eastermost Head-land to bear N E b E 5 deg. Easterly from me, the Westermost N N E, and the Island E S E: I demand how these Places bear one from the other, and their Distance asunder?  
SUch Questions as these are used in situating of places that are in sight one of the other; or, as a man sails by a land, he may take the Headlands bearing one from the other, and also the Imbays as well as the Headlands, if he be not too far out.  
This Question is wrought as the second Question, for here you have the bearing of the places you note in both Stations, and also the Course that you steer from one Station to the other, with the Distance; which is two Angles and a Side, to find the other two Sides, which is your Distance at each Station from the places.  
For the protracting of this Question; from your first Station draw a line upon those several Courses, as the Question saith the places bear; As, suppose the first Station be A, from A draw A I a N N E line, and (mind) that line must be cut upon the Eastermost Head-land, because the Question saith it bears so.  
Next draw a N W line from A, which is A C, and cuts the the Westermost Head-land.  
Lastly, draw out a South line from A, which is A D, and cuts the Island.   depiction of geometrical figure 
Lastly, A s and R s are the Courses that lead from the two Stations to the Island; and therefore s is the Island, for there they cut.  
Draw a line from every one of them as from s to O, and from O to I, from I to s, and take the distance of them asunder; then for their bearing one from another, draw a North and South line from each place, and by an Arch of 60 deg. measure it, as hath been showed.  

By the Tables  
First, I consider that I have the Distance between the two Stations, R A (which is a side of the triangles A R I, A R O, and A R s) and in every of these Triangles, the Question gives me all the Angles, so that I can find my Distance from my Station to the Headlands, after the form of this Example, which I shall do from each Station to the Eastermost.  

For A I.  
As Sine R I A 38 deg. 45 min. comp. arith. 0,20347 To A R 8 miles  1,90309 So is Sine A R I 28 deg. 45 min.  9,68213 To A I 6 miles 2/10  1,78869   

For R I  
As R I A Sine 38 deg. 45 min. comp. arith. 0,20347 To A R 8 miles  1,90309 So is R A I Sine (its comp. to 180 d.) 67 d. 30 m. 9,96561 To R I 11 miles 8/10  2,07217  
Here I have found the Distance from each Station to the Eastermost Head-land, next find the distance from the Westermost Head-land to the first Station A O.   

For A O.  
As Sine R O A 6 points or 67 d. 30 m. co. are. 0,034384 Is to R A 8 miles  1,903090 So is O R A 67 deg. 30 min. Sine  9,965615 To A O 8 miles  1,903090  
Thus you might find R O, R s, and A s, for you have the same things given, namely, R A and the Angles: Here we have found the sides A I and A O, and because A I is upon a N N E Course, and A O upon a N W Course, the Angles contained between O A I must be 6 points or 67 deg. 30 min. so that we have two sides and their contained Angle, to find the Angles of the bearing of these Headlands A I O or A O I, and the Distance of them asunder O I: this is wrought as the third Question is. Thus, as this is found, so you may find the bearing of the rest of the places one from the other, if there were twenty of them, for you have as much given in the other Triangles; but I leave it to your own practice also to make such Experiments, or to frame such Questions.     

Two Sides and a contained Angle being given, to find the third Side and the Course (that each Side runs upon) provided that no Course be named, only the half of the Compass that you sail in (or the quarter which is most commonly) and the Difference of Latitude that is made between the extent of the two Sides.  
QUESTION VI.  
Two Ships make an Angle of 50 deg. the one sails between the South and the West 40 leagues, the other sails between the South and the East 50 leagues, and at the end of their sailing, the Eastermost Ship is more Southerly than the Westermost, so that they differ their Latitude 15 leagues; I demand each Ships Course, and their Distance asunder.  
THe place the Ship set from, I make to be B, take off 60 deg. from your Scale, and describe an Arch, making B your Centre, as the Arch T R: then I take 50 deg. from my Scale, and set off somewhere upon that Arch, namely, from T to R, and from B, upon the Course that leads to T, I set off 40 leagues, my Westermost Ships Distance, and so conclude my Westermost Ship is at C: then from B upon the Course that leads to R, I set off my my Eastermost Ships Distance run, which is 50 leagues from B to A. Lastly, draw A C, which is the Distance of the Ships at the end of their sailing.  
Then because the Question saith the Eastermost Ship was 15 leagues more Southerly in Latitude, than the Westermost; I will take 15 leagues from the Scale of equal parts, and fix one foot of my Compasses in A, and with the other describe the Arch I S N, and draw the line from C, by the upper edge of the Circle at length, which must be an East and West line, because A is just 15 leagues to the Southwards of the nearest part of it: then let fall a Perpendicular from B upon C N, and extend it as far as the Arch of 60 deg. which you first described T R, and it cuts in O, this must be a South line from B, namely B S; then measure upon the Arch of 60 deg. from O to R, for the Eastermost Ships Course, and from O to T for the Westermost Ships Course, as you do in any other Question. I find them here as followeth,  
The Eastermost Ship sailed S b E 4. d. 40 m. Easterly.  
The Westermost Ship sailed S W 11 deg. Southerly.  
The Distance asunder is C A 39 leagues.   depiction of geometrical figure 

By Proportion.  
Here are the two sides B A 50 leagues, and B C 40, with the contained Angle C B A given, to find the Angles A C B and B A C; therefore for the Angles, As B A and B C 90 leagues comp. arith. 7,045757 Is to their Difference 10 leagues  2,000000 So is Tan. half the Angles B A C and B C A 65 d. 0 m. 10,331327 To Tang. their Difference 13 deg. 24 min. 9,377084  
Which added to the half sum 65 deg. is 78 deg. 24 min. the Angle A C B, subtracted from 65 deg. it makes 51 deg. 36 min. the Angle C A B.   

Then for the Ships Distance asunder, C A.  
As Sine A C B 78 deg. 24 min. comp. arith. 0,008962 Is to B A 50 leagues  2,698970 So is Sine C B A 50 deg. 0 min.  9,884254 To C A 39 deg. 1/10 leagues.  2,592186  
Here we have found the required Angles, and the required side, but we know not what point of the Compass either of the sides that bounds the Triangles runs upon: but this which we have found makes way to it, for now we know the side A C to be 39 1/10 leagues; and if you let fall a Perpendicular from A, upon the East and West line C N, it cuts it in S, than A S is known to be the Semidiameter of the Arch I N S u 15 leagues, so that in the Triangle A S C (right angled at S) we have A S 15 leagues given, and A C 39 1/10 leagues, to find A C S or C A S, we will find A C S.  
As A C 39 1/10 leagues 7,407823 Is to Radius or A S C 90 deg.  So is A S 15 leagues 2,176091 To Sine A C S 22 deg. 34 min. 9,583914  
Here it is evident, the place A bears from the place C East 22 deg. 34 min. Southerly, which is E S E 4 minutes Southerly, because C S is an East and West line.  
Also, if you subtract A C S 22 deg. 34. min. from A C B 78 deg. 24 min. the remainder will be B C S 55 deg. 50 min. the Angle that the Westermost Ship hath made from a West Course Southerly (for that T C V and B C S are equal) so that I say, the Course that the Westermost Ship hath sailed upon, is S W 10 deg. 50 min. Southerly, or the Course that leads to the place she set from B, is N E 10 deg. 50 min. Northerly.  
Then for the Eastermost Ships Course, consider that B C is a S W Course 10 deg. 50 min. Southerly, and B A must be to the Southwards, and so to the Eastwards 50 deg. from it (which is 4 points and 5 deg.) that is, S b E 4 deg, 35 min. Easterly.    

Two Sides of an obliqne Triangle being given in one Sum, and the other Side alone, with an Angle opposite to one of the Sides, to find the Sides several, and the other Angles.  
QUESTION VII.  
Two Ships set from two several Ports, these Ports were both in one Parallel, and distant 92 miles, the Westermost Ship sails N E b E, the Eastermost Ship sails upon some point between the North and the West, and they both meet, if the Distance run of these Ships together be 159 miles: I demand them several, and the Eastermost Ships Course.  
LEt B A be the Parallel line the Ports be in. Let A be the Eastermost Port, B the Westermost, then must A B be 92 miles: from B set off a N E b E line, which is B C O, set off both the Ships runs upon it, which is 159 miles B C: then draw C A, which is the other side of the Triangle B C A, and divide it into two equal parts by a Perpendicular (as you see the Perpendicular D S doth, in the point R) draw this Perpendicular from one side of the Triangle to the other, as R I; then from I draw a line to A.  
Then forasmuch as A R is the half of A C or equal to R C, and I R is a Perpendicular to A C from R, I A and I C must be equal, and if B I C be the length of the two sides 159, B I and I A are equal to them, so that B I is the Distance that the Westermost Ship sailed, and I A is the Eastermost Ships run, measure them, and set them down; also I A B is the Angle from the West toward the North, that the Eastermost Ship hath steered, I have found them to be as followeth.  
The Eastermost Ship sailed North 7 d. 20 m. Westerly.  
The Westermost Ship's Distance run is 102 miles.  
The Eastermost Ship sailed 57 miles.   depiction of geometrical figure 

By the Tables.  
In the Triangle B A C I have the side B C given, 159 miles, the side B A 92 miles, and the contained Angle at B 23 deg. 30 min. whereby I may find the Angle at C, which we will do first.  
As the Sum of B A and B C 251 miles comp. arith. 6,600326 Is to their Difference 67 miles 2,816074 So is Tan. half B A C and B C A 73 d. 7 m. 10,517833 To Tan. their Difference 41 d. 20 m. 9,944  
Which added to the half Sum, it makes B A C 114 deg. 27 min. the greater Angle: subtracted from the half Sum, it makes B C A 31 deg. 47 min. (which is equal to I A C, for if A I and I C, as also A R and R C be equal (as was proved before) and I R a Perpendicular to R A and R C, of necessity, the Angles in the Triangles I A R and I C R must be equal.  
So that then, if from B A C 114 d. 27 m. You subtract I A C equal to B C A 31 47 The Remainder is I A B 82 40  
So that because A B is a West line (from A) of necessity A I is a line that runs upon a Course 82 deg. 40 min. to the Northwards of it, which is North 7 deg. 20 min. Westerly.  
And thus we have the Angles and the side B A of the Triangle B I A to find the sides A I and B I  

First then for B I the Eastermost Ship's run.  
As Sine B I A 63 deg. 35 min. comp. arith. 0,047894 To B A 92 miles  2,963787 So is Sine B A I 82 deg. 40 min.  9,996433 To B I the Eastermost Ship's run 101 9/10 miles 3,008114  
If you question how the Angle B I A is known, it is easily resolved; for if B A I be found, A B I was given, add them together, and subtract that from 180 deg. and the remainder must be B I A, because the three Angles of any right lined Triangle is but 180 deg.   

Lastly, for I A the Westermost Ships run.  
As Sine B I A 63 deg. 35 min. comp. arith. 0,047894 To B A 92 miles  2,963787 So is Sine A B I 33 deg. 45 min.  9,744739 To I A the Westermost Ships Distance run 57 d. ●1/10 mil. 2,756420  
I handled a Question of this nature before, by which you may be directed: for though that be a right angled Triangle, and this an obliqne angled Triangle, there is no difference in the demonstration.     

The three Sides of a Triangle being given in one entire Sum, and the Angles apart, to find the Sides apart.  
QUESTION VIII.  
Two Ships set from one place, one sails S W, the other sails S b W, and at the end of their sailing, they arrive at two several Ports; the Westermost of these Ports bears from the Eastermost N W and by N: Now if the Ships Distances which they sail, with the Distance between the Ports be 148 leagues, (which is the three sides of the Triangle) I demand the Distance that each Ship sailed, and how far the Ports are asunder?  
FOr the resolving of this Question, I will suppose the side that the Westermost Ship sails upon to be 30 leagues, and so imagine you have the three Angles and a Side given, as you have in Quest. 2. We will protract this after the directions there given, it being so laid down: I find by the Plain Scale, that these sides are as followeth:  
The side C N is given to be 30 leagues.  
The side N B is found to be 42 leagues.  
The side C B is found to be 24 leagues.  

By the Tables the Proportion is,  
As Sine B 45 deg. 0,15051 To C N 30 leagues 2,47712 So Sine N 33 d. 45 m. 9,74473 To C B 2 6/10 leagues 2,37236  

For N B.  
As Sine B 45 deg. 0,15051 To C N 30 leagues 2,47712 So Sine C (take its Comp. to 180 d.) 78 d. 45 m. 9,99157 To B N 41 6/10 leagues 2,61921  
These sides with the side C N (added together) makes 95 leagues 2/10.  
Then it stands to good reason, that being the Angles in this Triangle are the same with the Angles in the Question; therefore, as the sum of these sides are in proportion to the Sum of the sides there, so is the sum of the side 30 here, to the sum of the side that is correspondent to it there, or runs upon the same Course.   

To find the Ships Distance that sailed S W, correspondent to C N 30 leagues.  
As sum of the sides of the Triangle C N B 95 le. co. are. 7,021363 Is to the sum of the sides in the Quest. 148 leagues 3,170261 So is the side C N 30 leagues 2,477121 To the Ships Distance that sailed S W 46 leagues 6/19 2,668745   depiction of geometrical figure 
This side being thus found, we (to abbreviate the work) will lay it down in the Triangle before, whose Angles are equal to to the Angles here: set 46 6/10 leagues, for the length of the side, the Westermost Ships run, which is N R.  
Draw R s parallel to C B, and extend N B till it intersect R s, which it doth in s.  
This is as much as protracting of it anew, for the Angles at R, at s, and at N, are the same that they were given to be in the Question, and the side R N 47 leagues, as it was found.  
You may measure the sides unknown by the Plain Scale, and set them down if you will, or you may work by the Rule of Three, using this Proportion.  
As N C 30 leagues is to N R 46 6/10 leag. So is N B 41 6/10 to N S.  
Multiply and divide, and it maketh 64 6/10 leagues, 14/57 of a Tenth.   

By the Tables for N s.  
As N C 30 leagues comp. arith. 7,52287 To N R 46 6/10 leagues  2,66839 So is N B 41 6/10  2,61909 To N s. 64 6/10 leagues  2,81035  
We might have done this by the Tables, putting the Angles and Sides in proportion, as we have done all along in the other Triangles. But I suppose your own reason will give you, that this hath the same given in it that the Angle C N B had; and is wrought so.  
The Fractions here, and the other agree, only this is not so true altogether, because here we do it but to tenths, there in smaller Fractions, but this is within a small part of one tenth of a unite.   


For the Side R s.  As Sine N s R 45 deg. 00 min. comp. arith. 0,15051 Is to R N 46 leagues 6/10  2,66839 So is Sine N 33 deg. 45 min.  9,74473 To R s. 36 6/10 leagues  2,56363  
If you add the sides now found all together, it will come to 2/10 of a unite less than the given sides together, in the Question is; now that ariseth by neglecting the taking of the absolute number, answering to the Logarithm that comes out; but this is sufficient in cases of this kind. If you had set it in paces, and wrought to the tenth part, you would have been out but 2/10 of a pace, therefore use your mind in such cases. If you had wrought to Centisms, you would have been nearer: but this I count near enough for sailing, and I am sure is more proper to be used, than such small Fractions, because no long distances can be guessed to a mile or a league. You might have wrought your last proportion by the Rule of Three, as you did the other.     

Three Sides of a Triangle given, to find the Centre.  
QUESTION IX.  
There be three Ships bound to one place, the Eastermost is distant from the middlemost 40 leagues, and bears S E b E, the middlemost is distant from the Westermost 50 leagues, and bears N E; now they every one know as much as I have writ, they also know they are of a like distance from this Port. I desire to know what distance the Port is from them, and how it bears from each Ship?  
I Have applied this Question to sailing here, but before in this Book I have used it in a thing which is very proper for it; and indeed it may be of use many times, which makes me give it place here. I shall say nothing of this Question by the Plain Scale, for it is done, as before you see: but we will do it by the Tables of Tangents and Logarithms.  
Here I have given the side M er 40 leagues, the side M W 50 leagues, and the Angle at M, namely, its Compliment to 180 deg. R M o 78 deg. 45 min. to find the Angles M ere W and M W oer: As W M add M o 90 leagues comp. arith. 7,04575 Is to their Difference 10 leagues  2, So is the Tan. of ½ M er W and M W oer 39 d. 22 m. 9,91404 To Tan. of their Difference 5 deg. 12 min. 8,95980  
Subtracted from 39 deg. 22 min. and the Remainder is M W oer 34 deg. 50 min. added to it makes M oer W 44 deg. 34 min.  

For the Side er W.  As Sine M o W 44 deg. 34 min. comp. arith. 0,153824 To W M 50 leagues  2,698970 So is Sine R M o 78 d. 45 m. its comp. log. 180 d. 9,991573 To W oer 69 9/10 leagues  2,844367  
But now you will demand of me, how we shall find any thing in the Triangle W F er, being there is but one thing known in it, namely, the Distance of the Eastermost Ship from the Westermost er W, I'll tell you.   depiction of geometrical figure 

Mr. Euclid proves this, which makes me forbear it.  Mr. Euclid (Book 3. Prop. 20.) saith, That in a Circle an Angle at the Centre is double to an Angle at the Circumference, (provided) that both the Angles have to their Base one and the same part of the Circumference. The Centre is F here, the Triangle W F oer and W M ere have W oer to both their Bases: then must B F A be double to I M O, or C B A double to D M I (that is to the Arch I O D) so that if you double D O I, 101 deg. 15 min. it makes C B A 202 deg. 30 min. that subtracted from a Circle 360 deg. leaves the Triangle W F oer: It is evident than that W F ere is double to R M I because R M I is what I O D wants of a Semicircle, as W F ere is what C B A wants of a whole Circle, which is double to it.  

Example.  
 deg. min. W M ere is 101 deg. 30 min. doubled it is C B A 202 30 That subtracted from a Circle 360 00 The Remainder is W F oer 157 30 And this subtracted from 180 00 The Remainder is F W oer and F oer W (for the Angles of a right angled Triangle are 180 d.) 22 30 The half of which is F W oer or F ere W 11 15  
For they must be equal, because the sides opposite to them are equal, namely, F oer and F W.  
And thus we have the three Angles of the Triangle W F er, and the side W oer to find F oer or F W or F M, the distance of any of the Ships from the Port.  

For the Distance of the Ships from the Port.  
As Sine W F oer 22 deg. 30 min. comp. arith 0,417160 Is to W oer 69 9/10 leagues  2,844477 So is Sine F W oer 11 deg. 15 min.  9,290235 To any of the Ships Distances from the Port which I find to be 35 6/10 leagues 2,551872  
I need not count the Triangle (for the points) for I have showed that sufficiently: So that you, I suppose, can tell how to find the Ships bearing from the Port, for it differs nothing from the way you count in other Questions.     

QUESTION X.  
There was two Ships set from one Port, one sailed N W b N 6 leagues a watch, the other sailed N E 8 leagues a watch, these Ships arrived at two several Ports in one instant of time, the Ports were 12 leagues asunder: I demand the Distance run of each Ship, and the Ports bearing one from the other?  
FOr the doing of this Question, First draw a N W b N line, and set off 6 leagues upon it which is A I; then draw a N E line, and set 8 leagues off upon it, which is A D; draw the side I D, then consider that as A I is to A D, so is the Westermost Ships run to the Eastermost; so that the sides A D and A I extended so far that a line of 12 leagues long, drawn parallel to D I, will just touch the extended sides which must cut them in the places were the Ports are, namely, A B in B, and A I in C: measure the sides C A and A B, and also the Angles of the Ports bearing A B C, or A C B and count them as hath been showed before.   depiction of geometrical figure 
I find the side A B the Eastermost Ships run to be 10 leag. 7/10. I find A C the Westermost Ships run to be 8 leagues. The Westermost port bears from the Eastermost West, 4 deg. 14 min. Southerly.  

By the Tables.  
First for the Ports bearing one from the other A D I equal to A B C, or A I D equal to A C B: they are equal, because I D and C B are parallel, and I A D is an Angle opposite to them both.  
As A D and A I 14 leagues comp. arith. 6,853871 Is to their Difference 2 leagues  2,301030 So is Tan. ½ A I D and A D I 50 deg. 37 min. 30 sec. 10,085698 To Tan. their Diff. 9 deg. 51 min.  9,240599  
Which added to the half Sum is A I D 60 deg. 28 min. A C B equal to A D I or A B C the bearing, subtracted from the half sum is 40 deg. 46 min.  

For I D.  As Sine D 40 deg. 46 min. 0,185100 To I A 6 leagues 0,778151 So is Sine I A D 78 deg. 45 min. 9,991573 To D I 9 Leagues 0,954824  

For C A.  As I D 9 leag. co. are. 8,04575 Is to C B 12 leagues 2,07918 So is I A 6 leagues 1,77815 To A C 8 leagues 1,90309 
For A B.  As I D 9 leag. co. are. 8,04575 Is to C B 12 leagues 2,07928 So is A D 8 leagues 1,90309 To A B 10 7/10 leagues. 2,02802  
The reason that the proportion stands thus, is because the Angles are equal in each Triangle, and therefore the sides must be proportional.  
Thus the Eastermost Ship sailed 10 7/10 leagues, the Westermost 8 leagues, and the Eastermost Port bears from the Westermost East 4 deg. 14 min. Northerly.  
There is no Distance that a man sails, but if he returns the same way that he went out, he shall find the same Distance backwards as he did out.  
But now, the World is round, and if a man goeth out one way, and returns another (as it is common) he will find a great deal of difference in long runs. For to demonstrate the reason of this I shall not, because my Father and others have done that sufficiently, besides, it would take up too much room in this little Book.  
These Questions of Sailing may be understood in such cases as this is, namely, when a man sails the same way home as he doth out: But because the Winds will not suffer a man to do so, it is necessary to know some better way to sail by. Let a man return never so contrary to what he sailed outwards, yet at all times to know how near or how far he may be off.  
Now there is no way better than Mercator's way of Sailing, it is very excellent, and hath been treated of by many, and that makes me presume to think that I may treat of it as well as others. And first here are the Tables of Meridional Parts, by which you are to work.     

A TABLE OF MERIDIONAL PARTS, To every Third MINUTE.  
Latitud. Mer. Parts D. M. 0 3 3  6 6  9 9  12 12  15 15  18 18  21 21  24 24  27 27  30 30  33 33  36 36  39 39  42 42  45 45  48 48  51 51  54 54  57 57 1 0 60  3 63  6 66  9 69  12 72  15 75  18 78  21 81  24 84  27 87  30 90  33 93  36 96  39 99  42 102  45 105  48 108  51 111  54 114  57 117 2 0 120  3 123  6 126  9 129  12 132  15 135  18 138  21 141  24 144  27 147  30 150  33 153  36 156  39 159  42 162  45 165  48 168  51 171  54 174  57 177 3 0 180  3 183  6 186  9 189  12 192  15 195  18 198  21 201  24 204  27 207  30 210  33 213  36 216  39 219  42 222  45 225  48 228  51 231  54 234  57 237 4 0 240  3 243  6 246  9 249  12 252  15 255  18 258  21 262  24 264  27 267  30 270  33 273  36 276  39 279  42 282  45 285  48 288  51 291  54 294  57 297 5 0 300  3 303  6 306  9 309  12 312  15 315  18 318  21 321  24 324  27 327  30 330  33 333  36 337  39 340  42 343  45 346  48 349  51 353  54 355  57 358 6 0 361  3 364  6 367  9 370  12 373  15 376  18 379  21 382  24 385  27 388  30 391  33 394  36 397  39 400  42 403  45 406  48 409  51 412  54 415  57 418 7 0 421  3 424  6 427  9 430  12 433  15 436  18 439  21 442  24 445  27 448  30 451  33 454  36 457  39 460  42 463  45 466  48 469  51 472  54 475  57 478 8 0 482  3 485  6 488  9 491  12 494  15 497  18 500  21 503  24 506  27 509  30 512  33 515  36 518  39 521  42 524  45 527  48 530  51 533  54 536  57 539 9 0 542  3 545  6 548  9 551  12 554  15 557  18 560  21 563  24 567  27 569  30 573  33 576  36 579  39 582  42 585  45 588  48 591  51 594  54 597  57 600 10 0 603  3 606  6 609  9 612  12 615  15 618  18 621  21 624  24 627  27 630  30 634  33 637  36 640  39 643  42 646  45 649  48 652  51 655  54 658  57 661 11 0 664  3 667  6 670  9 673  12 676  15 679  18 682  21 685  24 689  27 692  30 695  33 698  36 701  39 704  42 707  45 710  48 713  51 716  54 719  57 722 12 0 725  3 728  6 731  9 734  12 738  15 741  18 744  21 747  24 750  27 753  30 756  33 759  36 762  39 765  42 768  45 771  48 774  51 777  54 782  57 784 13 0 787  3 790  6 793  9 796  12 799  15 802  18 805  21 808  24 811  27 814  30 818  33 821  36 824  39 827  42 830  45 833  48 836  51 839  54 842  57 845 14 0 848  3 851  6 855  9 857  12 861  15 864  18 867  21 870  24 873  27 876  30 879  33 882  30 886  39 889  42 892  45 895  48 898  51 901  54 9●4  57 9●7 15 0 910  3 9TH  6 97  9 920  12 923  15 926  18 929  21 932  24 935  27 938  30 942  33 945  36 948  39 951  42 954  45 957  48 960  51 963  54 966  57 969 16 0 973  3 976  6 979  9 982  12 985  15 988  18 99●  21 994  24 998  27 1●●1  30 1004  33 1007  36 1010  39 1013  42 1016  45 1019  48 1023  51 1026  54 1029  57 1032 17 0 1035  3 1038  6 1042  9 1045  12 1048  15 1051  18 1054  21 1057  24 1060  27 1063  30 1067  33 1070  36 1073  39 1076  42 1079  45 1082  48 1086  51 1089  54 1092  57 1095 18 0 1098  3 1101  6 1104  9 1107  12 1111  15 1114  18 1117  21 1120  24 1123  27 1126  30 1130  33 1133  36 1136  39 1139  42 1142  45 1145  48 1149  51 1152  54 1155  57 1158 19 0 1161  3 1164  6 1168  9 117●  ●2 1174  15 1177  18 1181  21 1184  24 1187  27 1190  30 1193  33 1196  36 1200  39 1203  42 1206  45 1209  48 1212  51 1215  54 1219  57 1222 20 0 1225  3 1228  6 1232  9 1235  12 1238  15 1241  18 1244  21 1247  24 1251  27 1254  30 1257  33 1260  36 1264  39 1267  42 1270  45 1273  48 1276  51 1279  54 1283  57 1286 21 0 1289  3 1292  6 1296  9 1299  12 1302  15 1305  18 1308  21 ●31●  24 1315  27 1318  30 1321  33 1324  36 1328  39 1331  42 334  45 1337  48 1341  51 ●344  54 1347  57 1350 22 0 1354  3 1357  6 1360  9 1363  12 1367  15 1370  18 1373  21 1376  24 1380  27 1383  30 1386  33 1389  36 1393  39 1396  42 1399  45 1402  48 1406  51 1409  54 1412  57 1415 23 0 1419  3 1422  6 1425  9 1428  12 1431  15 1435  18 1438  21 1441  24 1445  27 1448  30 1451  33 1454  36 1458  39 1461  42 1464  45 1467  48 1471  51 1474  54 1477  57 1480 24 0 1484  3 1487  6 1491  9 1494  12 1497  15 1500  18 1504  21 1507  24 1510  27 1513  30 1517  33 1520  36 1524  39 1527  42 1530  45 1533  48 1537  51 1540  54 1543  57 1546 25 0 1550  3 1553  6 1557  9 1560  12 1563  15 1566  18 1570  21 1573  24 1577  27 1580  30 1583  33 1586  36 1590.  39 1593.  42 1596  45 1599  48 1603  51 1606  54 1610  57 1613 26 0 1616  3 1619  6 1623.  9 1626.  12 1630  15 1633  18 1637  21 1640  24 1643  27 1647  30 1650  33 1653  36 1657  39 1660  42 1663.  45 1666  48 1670  51 1673  54 1677  57 1680 27 0 1684  3 1687  6 1690  9 1693  12 1697  15 1700  18 1704  21 1707  24 1710  27 1713  30 1717  33 1720  36 1724  39 1727  42 1731  45 1734  48 1738  51 1741  54 1744  57 1747 28 0 1751  3 1754  6 1758  9 1761  12 1765  15 1768  18 1772  21 1775  24 1778  27 1781  30 1785  33 1788  36 1792  39 1795  42 1799  45 1802  48 1806  51 1809  54 1813  57 1816 29 0 1819  3 1822  6 1826  9 ●829  12 1833  15 1836  18 1840  21 1843  24 1847  27 1850  30 1854  33 1857  36 1861  39 1864  42 1868  45 1871  48 1875  51 1878  54 1881  57 1884 30 0 1888  3 1891  6 1895  9 1898  12 1902  15 1905  18 1909  21 1912  24 1916  27 1919  30 1923  33 1926  36 1930  39 1933  42 1937  45 1940  48 1944  51 1947  54 1951  57 1954 31 0 1958  3 1961  6 1965  9 1968  12 1972  15 1975  18 1979  21 1982  24 1986  27 1989  30 1993  33 1996  36 2000  39 2003  42 2007  45 2010  48 2014  51 2017  54 2021  57 2024 32 0 2028  3 2031  6 2035  9 2039  12 2043  15 2046  18 2050  21 2053  24 2057  27 2060  30 2064  33 2067  36 2071  39 2074  42 2078  45 2081  48 2085  51 2088  54 2092  57 2096 33 0 2100  3 2103  6 2107  9 2110  12 2114  15 2117  18 2121  21 2124  24 2128  27 2131  30 2135  33 2139  36 2143  39 2146  42 2150  45 2153  48 2157  51 2160  54 2164  57 2167 34 0 2171  3 2175  6 2179  9 2182  12 2186  15 2189  18 2193  21 2197  24 2201  27 2204  30 2208  33 2211  36 2215  39 2218  42 2222  45 2226  48 2230  51 2233  54 2237  57 2240 35 0 2244  3 2248  6 2252  9 2255  12 2259  15 2262  18 2266  21 2270  24 2274  27 2277  30 2281  33 2284  36 2288  39 2292  42 2296  45 2299  48 2303  51 2307  54 2311  57 2314 36 0 2318  3 2321  6 2325  9 2329  12 2333  15 2336  18 2340  21 2344  24 2348  27 2351  30 2355  33 2359  36 2363  39 2366  42 2370  45 2374  48 2378  51 2381  54 2385  57 2389 37 0 2393  3 2396  6 2400  9 2404  12 2408  15 2411  18 2415  21 2419  24 2423  27 2426  30 2430  33 2434  36 2438  39 2442  42 2446  45 2449  48 2453  51 2457  54 2461  57 2464 38 0 2468  3 2472  6 2476  9 2480  12 2484  15 2487  18 2491  21 2495  24 2499  27 2503  30 2507  33 2510  36 2514  39 2518  42 2522  45 2526  48 2530  51 2533  54 2537  57 2541 39 0 2545  3 2549  6 2553  9 2556  12 2560  15 2564  18 2568  21 2572  24 2576  27 2580  30 2584  33 2588  36 2592  39 2595  42 2599  45 2603  48 2607  51 2611  54 2615  57 2619 40 0 2623  3 2627  6 2631  9 2634  12 2638  15 2642  18 2646  21 2650  24 2654  27 2658  30 2662  33 2666  36 2670  39 2674  42 2678  45 2682  48 2686  51 2690  54 2694  57 2698 41 0 2700  3 2706  6 2710  9 2714  12 2718  15 2722  18 2726  21 2730  24 2734  27 2738  30 2742  33 2746  36 2750  39 2754  42 2758  45 2762  48 2766  51 2770  54 2774  57 2778 42 0 2782  3 2786  6 2790  9 2794  12 2798  15 2802  18 2806  21 2810  24 2814  27 2818  30 2822  33 2826  36 2830  39 2834  42 2839  45 2843  48 2847  51 2851  54 2855  57 2859 43 0 2863  3 2867  6 2871  9 2875  12 2880  15 2884  18 2888  21 2892  24 2896  27 2900  30 2904  33 2908  36 2913  39 2917  42 2921  45 2925  48 2929  51 2933  54 2938  57 2942 44 0 2946  3 2950  6 2954  9 2958  12 2963  15 2967  18 2971  21 2975  24 2979  27 2983  30 2988  33 2992  36 2996  39 3000  42 3005  45 3009  48 3013  51 3017  54 3022  57 3026 45 0 3030  3 3034  6 3039  9 3043  12 3047  15 3051  18 3056  21 3060  24 3064  27 3068  30 3073  33 3077  36 3081  39 3085  42 3090  45 3094  48 3098  51 3102  54 3107  57 3111 46 0 3116  3 3120  6 3124  9 3128  12 3133  15 3137  18 3142  21 3146  24 3150  27 3154  30 3159  33 3163  36 3168  39 3172  42 3176  45 3180  48 3185  51 3189  54 3194  57 3198 47 0 3203  3 3207  6 3212  9 3216  12 3220  15 3224  18 3229  21 3233  24 3238  27 3242  30 3247  33 3251  36 3256  39 3260  42 3265  45 3269  48 3274  51 3278  54 3280  57 3287 48 0 3292  3 3296  6 3301  9 3305  12 3310  15 3314  18 3319  21 3323  24 3328  27 3332  30 3337  33 3341  36 3346  39 3350  42 3355  45 3359  48 3364  51 3368  54 3373  57 3377 49 0 3382  3 3386  6 3391  9 3396  12 3401  15 3405  18 3410  21 3414  24 3419  27 3423  30 3428  33 3432  36 3437  39 3442  42 3447  45 3451  48 3456  51 3460  54 3465  57 3470 50 0 3475  3 3479  6 3484  9 3488  12 3493  15 3498  18 3503  21 3507  24 3512  27 3517  30 3522  33 3526  36 3531  39 3535  42 3540  45 3545  48 3550  51 3554  54 3559  57 3564 51 0 3569  3 3573  6 3578  9 3583  12 3588  15 3593  18 3598  21 3602  24 3607  27 3612  30 3617  33 3622  36 3627  39 3631  42 3636  45 3641  48 3646  51 3651  54 3656  57 3660 52 0 3665  3 3670  6 3675  9 3680  12 3685  15 3690  18 3695  21 3700  24 3705  27 3709  30 3714  33 3719  36 3724  39 3729  42 3734  45 3739  48 3744  51 3749  54 3754  57 3759 53 0 3764  3 3769  6 3774  9 3779  12 3784  15 3789  18 3794  21 3799  24 3804  27 3809  30 3814  33 3819  36 3824  39 3829  42 3834  45 3839  48 3844  51 3849  54 3855  57 3860 54 0 3865  3 3870  6 3875  9 3880  12 3885  15 3890  18 3896  21 3901  24 3906  27 3911  30 3916  33 3921  36 3927  39 3932  42 3937  45 3942  48 3947  51 3952  54 3958  57 3963 55 0 3968  3 3973  6 3979  9 3984  12 3989  15 3994  18 4000  21 4005  24 4010  27 4015  30 4021  33 4026  36 4031  39 4036  42 4042  45 4047  48 4053  51 4058  54 4063  57 4068 56 0 4074  3 4079  6 4085  9 4090  12 4096  15 4101  18 4106  21 4111  24 4117  27 4122  30 4128  33 4133  36 4139  39 4144  42 4150  45 4155  48 4161  51 4166  54 4172  57 4177 57 0 4183  3 4188  6 4194  9 4199  12 4205  15 4210  18 4216  21 4221  24 4227  27 4232  30 4238  33 4244  36 4250  39 4255  42 4261  45 4266  48 4272  51 4277  54 4283  57 4289 58 0 4295  3 4300  6 4306  9 4311  12 4317  15 4323  18 4329  21 4334  24 4340  27 4346  30 4352  33 4357  36 4363  39 4369  42 4375  45 4380  48 4386  51 4392  54 4398  57 4403 59 0 4409  3 4415  6 4421  9 4427  12 4433  15 4439  18 4445  21 4450  24 4456  27 4462  30 4468  33 4474  36 4480  39 4486  42 4492  45 4498  48 4504  51 4510  54 4516  57 4522 60 0 4528  

A Declaration of the Table of MERIDIONAL PARTS.  
THese Tables are in my Father's Works to every six minutes, but here to every three minutes; so that if a man light not upon the very number that he desireth, he may the easier take it, for it will be but one third or two thirds of the Difference between the two nearest to be added or subtracted; I suppose it will be the less troublesome; I have gone no farther than sixty degrees, because I would not take up too much room, and I suppose there is but little Trading to the Northwards of that Latitude, so that it would have been seldom in use, These Tables were calculated by my Father's Tables.  
He that sails according to Mercator's Projection, notes the true Quantity of Longitude in degrees and minutes, namely, such degrees as in that Parallel, 360 makes the Circumference. But they which sail by Plano, measure the Sea, which is round, as though it were a flat or plain, now it stands to reason that there must be great Difference: as thus,  
All North and South Courses (which we call Meridian's) intersect in the Poles: Now suppose there be two Ships under the Aequinoctial, 5 degrees or 100 leagues asunder, and these Ships sail both North, they will meet in the North Pole of the World, but by Plane they count them as far off there, as they were at first, for all North and South Lines are parallel one to another, as you may see in the Work all along.  

Illustration.  
All Meridian Lines (as I said before) cut through the Poles of the World, and all Parallel Lines go round about the Poles, continuing the same distance from them. The Aequinoctial is one of these Parallel Lines, drawn equally distant from the Poles, and so is upon the body of the Earth; and there a degree of Longitude is 60 miles, and 360 times 60 miles is the Circumference of the Earth, which is 360 degrees.  
But now if there be a drawn Parallel, 50 degrees nearer to one of the Poles, our own reason will give that that Circle will not contain so many miles round as the other did, because it must be less; yet he that sails by Plano counts that it doth, and though he hath Winds that carry him no better than due North from the Aequinoctial to the Parallel or Latitude of 50 degrees, he counts he hath as many Leagues from the Meridian of the Place he is bound to, as he had when he was under the Aequinoctial. Now tell me how it is possible, (if Meridian's intersect in the Poles, as they do) that these Meridian's should be as far asunder upon a Parallel of 50 degrees nearer the Poles, as they were under the Aequinoctial: I hope you understand me.  
It is the work of M●rcator to show (therefore) the Quantity of a degree of Longitude in every Latitude, (for in these Parallels there is 3●0 degrees in the Circumference, but these degrees cannot contain so many miles as they do nearer the Aequator) It showeth how you may sail upon any Course from one Latitude to another, and reckon your Longitude according to the truth of it in degrees and minutes: So that having two things given besides the right Angle, you may find your desire at Sea according to Mercator's Direction, which is the surest way of Sailing that I know, except Sailing by the Arch of a great Circle, and that and this will agree in degrees of Longitude, but that showeth the nearest way to sail from place to place, which is all that it differs from this, (they being both true) this showeth a general way of sailing from place to place, that a particular way of sailing the nearest way from place to place: It is but seldom used, because it is not very commonly known, but it is an excellent thing, I thought to have touched it, but I considered that my Father hath demonstrated it as evidently as can be. For Mercator use the Tables here inserted.   

The Use of the foregoing Tables of Meridional Parts.  
When you work this way, find first what Latitude you are in (if you cannot observe) do it by the things given that 24 hours: Then take notice what Latitude you were in the day before, and look in the Tables what answers to each Latitude, (setting them down) subtract the lesser from the greater, and the remainder are the Meridional Parts contained between the two Latitudes, which number make for the length of the side, which contains the Difference of Latitude, and so work according to the Directions following: I will be brief.   

Example. To find the Meridional Parts answering to any Latitude.  
Let the one Latitude be 50 deg. 9′ in the Tables 3488 3274 214 answers it.  
Let the other Lat. be 47 deg. 48′ in the Tables answers it.  
Subtract the lesser from the greater, and the Remainder is the Meridional Parts contained in that Difference of Latitude.  
The same is to be understood of any Latitude else.    

QUESTION I.  
The Latitude of two Places being given, and the Rumb, to find the Distance and Difference of Longitude.  
SUppose I set sail from a Place in the Latitude of 39 deg. 12 min. North Latitude, and am bound to a Place in the Latitude of 13 deg. 12 min. North Latitude, which bears S W by W from the other; I demand the Distance between those Places, and Difference of Longitude.  
Let A represent the Northermost place, C the Southermost, then is A and B the Latitudes of each place, C B the Difference of Longitude, B A C is the Course from the Northermost to the Southermost place.  

For the Distance between the Places A C.  As Sine comp. A 56 deg. 15 min. comp. arith 0,25526 To A B the Difference of Latitude in miles 1560 3,19312 So is Radius  To the Distance between them C A 2808 miles 3,44838  

For the Difference of Longitude.  The Meridional parts answering to 39 deg. 12 min. are 2560 The Meridional parts answering to 13 deg. 12 min. are 0799 The Meridional parts contained between the two Lat. 1761   depiction of geometrical figure 
As Radius  To B A 1761 parts 3,245759 So is Tan. the Rumb A 56 deg. 15 min. 10,175107 To C B the Diff. of Longitude in minutes 2635 m. 3,420866  
The Quotient is the number of degrees and the Remainder is the number of minutes contained in the Difference of Longitude 43 deg. 55 min.   

QUESTION II.  
Course and Distance run, with one Latitude given, to find the other Latitude, and the Difference of Longitude.  
SUppose I set from the Latitude of 49 deg. 9 min. North, and sail S W b W 2808 miles. I demand the Latitude I shall then be in, and the Difference of Longitude.  
Let A be the Latitude of 49 deg. 9 min. let A C be the Distance run 2808 miles, then is A B the Difference of Latitude, and C B the Difference of Longitude.  

For the Difference of Latitude A B.  
To A B 1560 miles  
Which being divided by 60  
The Quotient and the Remainder is the Difference of Latitude in degrees and minutes, which is 26 d. 0 m.  
That subtracted from 49 d. 9 m.  
Leaves the Latitude you are now in 23 d. 9 m.  
The reason why there is no inequality between degrees of Latitude, is because all Meridian's are great Circles, so that 60 miles is a degree of them always.   

For the Difference of Longitude.  
The Meridional parts contained between these two Latitudes I find to be 3396 1428 1968 pts.  
As Radius  To A B in parts 1968 3,294025 So is Tang. A the Rumb 56 deg. 15 min. 10,175107 To the Difference of Longitude 2945 m. 3,469133  
Now you may ask a reason why this Difference of Longitude should differ from the other in the last Question, seeing the Difference of Latitude, Course and Distance in each Question are alike: the reason is, because this Question runs between two less parallels than the other did, and therefore the same things must give a greater part of this Parallel than it did of the other, though the number of miles of Departure be alike.  
Diwide 2945 min. by 60, and you have the Difference of Longitude in degrees and minutes 49 deg. 5 min.    

QUESTION III. The Latitudes of two places, and the Distance between them given, to find the Course and Difference of Longitude.  
LEt one place (namely A) be in the Latitude of 49 deg. 9 min. North, let the other place (namely C) be in the Latitude of 23 deg. 0 min. North, and let the Course be between the South and the East, the Distance between them is A C 2808 miles.   depiction of geometrical figure 

For the Course A.  As A C Dist. 2808 mil. 6,551602 Is to Radius  So is the Diff. of Lat. A B reduced into miles, 1560 miles 3,193124 To the Sine C, whose Comp. is A 9,744726  
Which is 56 deg. 15 min. to the Eastwards of the South.  

For the Difference of Longitude B C.  As Radius  To B A in parts 1968 3,294025 So is Tangent the Course A 56 deg. 15 min. 10,175107 To the Difference of Longitude in minutes 2945 3,469133  
Which is as it was in the last Question 49 deg. 5 min.   

QUESTION IU.  
The Latitude of two places, and their Difference of Longitude given, to find the Course and Distance, provided the Course be between the North and the West.  
LEt the Latitudes be as before, and the Difference of Longitude 49 deg. 5 min. or 2945 parts or minutes.  

For the Course.  As the Diff. of Latitude in parts A B 1968 co. are. 6,70597 Is to Radius  So is the Diff. of Longitude in minutes or parts B C 2945 3,46908 To the Tangent of the Course, A 56 deg. 15 min. 10,17505  
To the Westwards of the North.  
As Sine C 33 deg. 45 min. comp. arith. 0,25526 To the Difference of Latitude in miles A B 1560 3,19312 So is Radius  To the Distance run C A 2808 miles 3,44839   depiction of geometrical figure  

QUESTION V  
The Difference of Longitude, Course, and one Latitude given, to find the other Latitude and the Distance in the Triangle CBA.  
LEt C B be the Difference of Longitude 49 deg. 5 min. or 2945 min. A the Latitude of 23 deg. 9 min. North Latitude, C A B the Angle of the Course North-westerly 56 deg. 15 min. let it be required to find A B the Difference of Latitude, (which added to the Latitude of A 23 deg. 9 min. will be the Latitude of the other place B) and A C the Distance between them.  
Here you must find your Difference of Latitude in parts first, which will be reduced into miles (after the Latitude of the place at B is found) and then find your Distance 49 deg. 5 min. reduced into minutes (as I said in the Question) is 2945 min. the Difference of Longitude given.  

For the Difference of Latitude in parts.  
As Radius  To the Difference of Longitude in parts 2945 parts 3,469085 So is Tan. comp. the Course A 56 deg. 15 min. 9,824897 To the Difference of Latitude in parts, 1968 parts 3,293977  
This done I look in the Tables against the Latitude that is given me 23 deg. 9 min. and the Meridional part answering to it, is 1428 To which I add the parts here found, 1968 because the Latitude is increased,  And there comes forth 3396  
I look for that in the Tables, and I find answering to it the Latitude of 49 deg. 9 min. which is the Latitude of the unknown place B.   

For the Distance between these Places.  
I subtract the Latitude of A 23 deg. 9 min. from the Latitude of B (now found) 49 deg. 9 min. and the Remainder is the Difference of Latitude between them, 26 deg. 0 min. which converted into miles is 1560 miles: then say, As Sine comp. 56 deg. 15 min. A comp. arith. 0,255260 Is to A B 1560 miles 3,193124 So is Radius  To the Distance run A C 2808 miles 3,448384  
I had thoughts to have showed the Difference between Plano and Mercator, in an Example, but my Father hath done it in page 170 of his Book. If you are desirous to know Great Circle Sailing, you may learn it from his Book, for he hath treated of it largely; and indeed it is the most necessary way of Sailing that is, for a man cuts his way much shorter (in many runs) than the usual way that men go, doth: If you desire to know in any Latitude the number of miles contained in any number of degrees of Longitude, convert your degrees and minutes of Longitude into minutes, and say,  
As Radius  
To Sine comp. that Latitude;  
So is the Difference of Longitude in minutes  
To the Difference of Longitude in miles.  
If you desire to convert miles of Departure (in any Parallel) into degrees, say,  
As Sine comp. the Latitude or the Parallels dist. from the Pole  
Is to Radius;  
So is the Departure in miles  
To the Difference of Longitude in minutes.  
Divide that by 60, and the Quotient with the Remainder is the degrees and minutes contained in that Departure.    

Of the Longitude and Latitude of Places.  
BEfore I come to show how to keep a Reckoning by Mercator's way, I will set down a Catalogue of the Latitude and Longitude of some Places in degrees and minutes, the Longitude taken from Flowers and Corves.  
Places Names. Latitude, Longitude. D. M. D. M. The Tassel North 53 00 35 10 The Maze North 52 00 34 29 The Willing 51 30 33 53 Isle of Wight 50 24 29 16 Dover 51 05 32 04 Portland 50 24 28 16 The Start 50 07 27 07 The Lizard 50 00 25 20 Silly 50 04 24 00 Vshin 48 30 25 30 Cape Finistre 43 08 20 30 The Rock of Lisbon 38 52 20 47 Cape St. Vincent 37 00 21 29 Gilbraltar 36 01 24 39 Cape Blanco 20 32 13 09 Cape de Verde 14 31 13 10 Serrat Lion 08 00 17 53 South side of St. Ann 06 42 17 32 Cape the Palmose 04 00 24 30 Cape Terras Pantas 04 06 30 30 Cape Formosa 04 04 39 15 River Comrans 03 19 42 32 Nethermost of the Isle Farmandosa 03 26 41 56 Middle of the Isle of Thoma 00 00 40 00 Cape de Lopez 10 00 42 01 Cape Negro South 16 00 4 521 Cape Bonesperanza South 34 25 52 30 North end of Martinass 23 02 17 16 Isle of Picos 23 01 17 16 Isle Terrestinam 37 15 23 31 Y Digon Aluarum 38 54 26 45 Ascension 08 21 20 16 St. Ellna 16 00 28 21 Cape Agulhas 35 00 54 00 Pomte Primero 32 25 63 06 Cape Corentas 24 00 69 50 Mosambique 15 10 75 32 Pemba 04 41 77 02 Cape Derfue 10 20 87 40 Cape Guardafa 12 01 87 41 Cape de Raffas 22 10 97 29 Serrat 21 01 108 35 Goa 15 40 109 22 Cape Cormoram 07 53 112 20 Point Gada 06 02 115 00 Northwest Point of Sumatra 05 30 129 02 The Flat Point 05 51 137 26 Crequeta in Sinda 06 02 137 21 East end of St. Brandon 17 15 102 05 East end of Diego Roderigo's 19 24 99 22 Southeast Harbour of Maurice 20 11 92 54 Southeast end of Mastorbos' 20 16 90 22 Cape de Roman 25 02 83 24 Isle St. Paulo 38 30 108 40 Isle St. Maria Dogusto 19 02 13 00 Assemaon 20 01 02 42 East end of the Foul Ground 18 21 00 00 Isle de Lobas at Rio de Plato 35 25 343 02 Cape St. Thoma 22 21 358 01 Bay Todas Sanctos 13 15 357 41 Cape St. Augustin 08 41 001 12 Isle Fernando 03 45 005 14 Vizia 03 30 003 01 Venedo St. Paulo 01 52 006 25 The West-Indies. Barbadoss North 13 12 327 42 Ditiatha North 16 18 326 32 St. Christopher's 17 20 324 52 Tobago 11 16 327 41 East side of Trinidado 11 37 326 41 Mattinino 14 48 325 30 St. Vincent 13 10 326 00 New point of Portorica 19 02 319 22 Altanalla near Hispaniola 17 26 313 52 Cape Tibarune in Hispaniola 18 14 310 22 West end of Margareta 11 11 321 51 Cartagena 10 49 310 41 Island Guanatho 16 51 297 11 West end of Cuba 22 00 298 44 Cape Florida 25 20 302 35 Bermuda Island the body of it 32 20 325 15 Island of Sable 43 40 334 02 Cape Race at Newfoundland 46 32 339 32 Cape Sable 43 41 327 21 Cape Cod 41 20 322 01 Cape Hattarass 35 40 314 11 Cape Henry 37 00 344 00 Cape Charles 37 20 313 55 West side of CORVES 40 00 000 00 East side of FLOWERS 37 20 000 00 The Road of Fial 38 50 002 17 West end of Fial 38 41 002 48 West end of St. George 39 00 003 00 Gratiosa 39 15 003 20 West end of Tercera North 39 00 03 40 East end of Michael North 38 05 06 10 East end of Maria 37 00 06 01 East point of Porto Santo 33 05 15 26 West end of Madera 32 20 13 48 East side of Palma 28 45 12 48 North point of Gomera 28 10 13 20 North side of Ferro 27 40 12 30 North-east side of Gammest 28 36 14 31 North-east point of grand Canaria 28 20 15 10 East end of Fortaventure 20 21 17 18 East end of Lausareta 28 53 17 21 East end of Gratiosa 29 00 17 29 East end of St. Antonio 17 21 04 20 East end of St. Vincent 17 12 05 00 East point of St. Lucia 16 54 05 25 East point of Sal 16 54 07 30 East side of Bonavista 16 10 08 00 East side of the Isle of May 15 00 07 30 East side of St. Jago 15 10 07 02 Fogo 14 36 06 00 Bramma Island 14 24 05 38  
The Longitudes and Latitudess of these places here inserted, are some of them my own Experiences, and the rest Accounts of able men.  

Concerning the foregoing Tables.  
HE that is bound to any place, the first thing that he ought to consider is the Latitude of each place, and the Difference of Longitude between them, which things may be taken from such a Catalogue as this is, after the manner following.  
Suppose I were bound from the Lizard to Barbadoes, I look in the Column of Latitude against the Lizard, and find 50 deg. 0 min. for its Latitude, and 25 deg. 20 min. for its Longitude, from Flowers and Corves.  
I look in the Column of Latitude against Barbadoes, and find 13 deg. 12 min. for its Latitude, and 327 deg. 42 min. for its Longitude from Flowers and Corves.  
Here you have the Latitude of each place; but to find the Difference of Longitude between these places, I consider that Barbadoes hath 327 deg. 42 min. of East Longitude from these Islands; but if I subtract that from 360 deg. (the whole Circumference) there must remain the West Longitude that is between them, which I find to be 32 d 18 m.  
Circumf. 360 00 East Long. 327 42  32 18  
To it add the longitude between the Lizard and Flowers and Corves 25 d. 20 m.  
That Sum is 57 d. 38 m.  
The Difference of Longitude between the Lizard and Barbadoes (the Meridian of Barbadoes being to the Westwards of the Lizard so many degrees.)  
And thus any Difference of Longitude between two places is taken; provided, that one of them is nearer than 19 deg. to the Westwards of the place you begin your Longitude from, and the other nearer than 90 deg. to the Eastwards.  
And the reason of this is, because the Longitude of all places is set down from Flowers and Corves to the Eastwards; so that if a place were but 1 deg. to the Westwards of Flowers and Corves, it would be set down in the Catalogue to lie in the 359 deg. of Longitude.  
For to know the Difference of Longitude between any two places, consider which way they are nearest, and that take: Suppose there were two places, one lieth in 20 deg. 0 min. of Longitude, the other lieth in 195 deg. of Longitude: I consider that the difference between them is not 180 deg. and therefore I will subtract 20 deg. 0 min. from 195 deg. and the Remainder is the Difference of Longitude between them 175 deg. but if the Difference between two places be above 180 deg. as they are laid down in the Catalogue: Subtract that difference from 360 deg. and the Remainder is the nearest Difference of Longitude: Your own reason will guide you in this, and therefore I will proceed.    

How to keep a Reckoning of the Longitude and Latitude a Ship makes at Sea.  
HAving found the Difference of Longitude between the places I set from, and am bound to, as also the Latitude of each place. I will begin my Reckoning thus:  
Imagine it were between the Lizard and Barbadoes, and I set from the Lizard the 3 d. of January 1658.  

From the Lizard we departed January 3, 1658 being bound to Barbadoes, lying in Latitude North 13 d. 12 m. Longitude, being West from the Meridian of the Lizard 57 d. 38 m.   Mo. Day Week Days. Latitude. East Long. West Long. D. M. D. M. D. M. January 4 Friday 49 00 00 00 03 00  5 Saturday 47 24 00 00 06 27  6 Sunday 46 09 00 46 06 27  7 Monday 45 00 01 05 06 27  8 Tuesday 43 01 01 05 06 27  9 Wednesday 41 10 01 05 09 00  10 Thursday 40 00 01 05 11 02  11 Friday 38 11 01 05 13 20  12 Saturday 36 12 01 05 14 40  13 Sunday 34 10 01 05 16 00  14 Monday 32 20 01 05 18 47  15 Tuesday 30 11 01 05 20 00  16 Wednesday 27 00 01 05 21 40  17 Thursday 25 01 01 05 22 30  18 Friday 23 10 01 05 24 00  19 Saturday 21 45 01 05 26 10  20 Sunday 20 45 01 05 28 44  21 Monday 19 00 01 05 31 00  22 Tuesday 18 30 01 05 33 32  23 Wednesday 18 00 01 05 36 00  24 Thursday 17 27 01 05 38 20  25 Friday 16 49 01 05 40 40  26 Saturday 16 10 01 05 42 50  27 Sunday 15 40 01 05 44 59  28 Monday 15 00 01 05 47 00  29 Tuesday 14 30 01 05 49 02  30 Wednesday 13 45 01 05 50 57  31 Thursday 13 12 01 05 52 00 February 1 Friday 13 14 01 05 54 30  2 Saturday 13 12 01 05 54 40  3 Sunday 13 12 01 05 54 40  4 Monday 13 12 01 05 55 30  5 Tuesday 13 12 01 05 57 30  6 Wednesday 13 12 01 05 58 35  
Here in this Reckoning I have set down the Longitude in degrees and minutes, as before I set down the Departure in leagues, carrying all in the last line: the last line of my Reckoning saith, that they were in the Latitude of 13 deg. 12 min. and the whole East Longitude that we have made since we set from the Lizard is 1 deg. 5 min. the whole West Longitude is 58 deg. 35 min. subtract the Easting from the Westing, and the Remainder is 57 deg. 30 min. which lacks but 8 min. of 57 deg. 38 min. the Meridian of the Place.  
Also the same line saith, that we are in the Latitude of 13 deg. 12 min. which is the Latitude of the place: therefore I conclude I am but 8 min. from it, so that I expect to see Land.  
There is no Difference in the setting down of this Reckoning from that before, but only this is degrees and minutes, that leagues or miles; neither is there any other difference in their casting up, for there your Easting from your Westing in leagues, here in degrees.  

That there be nothing wanting to make me be understood, I will here following do the three first days works in three Examples.  

SUppose the first day I set out, I sailed S W b W 6 deg. 25 min. Westerly, till I come into the Latitude of 49 deg. 0 min. which was the next day at noon.  
I consider that I have the two Latitudes given me, and the Course to find the Difference of Longitude.  
The Meridional part answering to 50 deg. 0 min. is 3475 The Meridional part answering to 49 deg. 0 min. is 3382 The Meridional parts contained between them is 93  

For the Difference of Longitude.  As Radius  To the Difference of Latitude in parts 93 1,96848 So is the Tangent of the Course 62 d. 40 m. 10,28661 To the Difference of Longitude in minutes 180 2,25510  
This 180 divided by 60 produceth 3 deg. 0 min. the Difference of Longitude made that 24 hours; set it down in your Reckoning with the Latitude you were in, filling up the East Column with cyphers: I need not show how to find the Distance run, because I suppose by what hath been said before, you can do it.   

For the second days Work.  
Suppose till the next day at noon we sail S W 10 deg. 11 min. Westerly, 169 miles by Estimation (for we could not observe.)  
I consider here that I have given the Course and Distance run and one Latitude, namely, the Latitude I was in yesterday, 49 deg. 0 min. to find the Latitude I am now in, and my Difference of Longitude.  

For the Latitude I am now in.  As Radius  To the Distance run 169 miles 2,227886 So is Sine comp. the Course 55 d. 11 m. 9,756599 To the Difference of Latitude 96 miles 1,984486  
This 96 miles is 1 deg. 36 min. which subtracted from the Latitude I was in yesterday (because the Course is Southerly) leaves the Latitude I am now in 47 deg. 24 min. (set it down in your Column of Latitude in your Reckoning) Thus I have both Latitudes and the Course, to find the Difference of Longitude.  
I take the Meridional part answering to each Latitude, and subtract the one from the other, and there remains 144 parts.  

For the Difference of Longitude.  As Radius  To the Difference of Latitude in parts, 144 parts 2,158362 So is Tangent the Course 55 deg. 11 min. 10,157734 To the Difference of Longitude in minutes 207 2,316097  
Which divided by 60, produceth 3 deg. 27 min. this I add to my west Longitude (because the Course is Westerly) 3 deg. 00 min. and it makes 6 deg. 27 min.  
This 6 deg. 27 min. I set down in the West Column, and thus you may keep your whole Longitude in the last line.   

The third days Work.  
Suppose the Winds be cross, and that we are forced to tack, and we make our way good S S E, and I observe and find myself in the Latitude of 46 deg. 9 min. I demand the Difference of Longitude.  
I find the Meridional part answering to my Latitude yesterday 47 deg. 24 min. is 3238 The Meridional part answering my Latitude gives this day by observation 46 deg. 9 min. is 3128 The Meridional parts contained between these two Latitudes is 110  
As Radius  To 110 2,04139 So is Tangent 22 deg. 30 min. 9,61722 To the Difference of Longitude in minutes 46 m. 1,65862  
This 46 min. is East Longitude, and must be set down in the East Column: and let the West Column hold its Sum still, as was done in the plain Reckoning before, and when you come to increase your Westing again, carry your whole East Longitude along, and thus you have your whole Reckoning in the last line: The setting down of this differs nothing from the other; therefore to that I refer you.  
Sometimes your Latitudes may be such, that the Tables do not answer them to a minute, (for they are but to every 3 min.) as suppose the one Latitude were 13 deg. 20 min. the other 15 deg. 16 min. and I would take the Meridional part contained between them.  
I look for 15 deg. 16 min. and cannot find it, but the nearest less than it, is 15 deg. 15 min. against which I find 926 I look for the nearest above it in the Tables, which is against 15 deg 18 min. and find 929 I subtract the lesser from the greater, the remainder is 003  
Then I consider that the difference between 15 deg. 15 min. and 15 deg. 18 min. is but 3, and that the portion that I want is but ⅓ of this Difference, 1; therefore add 1 to the Meridional part answering to 15 deg. 15 min. which is 926, and it makes 927, the Meridional parts answering to 15 deg. 16 min. required.  
The same I do for 13 deg. 20 min. and find the Meridional parts answering it to be 807, subtract them one from the other, and the remainder is the Meridional part contained between them; the like for any other, As   

Suppose one demands of me the 20 th'. day of January, how the Island of Barbadoes bears off me by my account in this Reckoning.  
I look upon my reckoning, and I find that day I was in the Latitude of 20 d. 45 m. and I have made West Longitude 28 d. 44 min. East Longitude 1 deg. 5 min. I subtract the East Longitude 1 deg. 5 min. from the West 28 deg. 44 min. there remains 27 deg. 39 min. the West Longitude that I have made since I set from the Lizard; subtract this 27 deg. 39 min. from the Difference of Longitude between the Lizard and Barbadoes 57 deg. 38 min. and the remainder is 29 deg. 57 d. 38 m. 27 39 29 59 59 min. the Difference of Longitude which I have yet to make: I look at the top of my Reckoning for the Latitude of the place I am bound to, and find it to be 13 deg. 12 min.  
Thus I have the two Latitudes, and the Difference of Longitude, to find the bearing of the place I am bound to: You may find it as is showed before, Quest. 4. by bringing your Difference of Longitude into minutes, and taking the Meridional parts contained between the Latitude you are now in, and the Latitude of Barbadoes; and using them as the sides to work by.  
If you have a mind to know your Distance (upon the Course) to Barbadoes, see what Difference of Latitude is between the Latitude you are in, and the Latitude of Barbadoes, reduce it into miles, and say, 
As Sine comp. the Course  
To the Difference of Latitude,  
So is Radius  
To the Distance.   
And thus in brief you are able to give an account of your whole Reckoning at any time, or of any part of your Reckoning, if the day be given. The like you may do for the Course that was steered from day to day; but that (as I shown before) ought to be taken notice of in a Journal, or if you like it better make a Column for it in your Reckoning. The reason why I refer that and the Winds and Variation, is, because it can better be expressed with the reasons of it than here.  
Those days that you observe, correct your Reckoning by your Observations, as my Father showeth in his practice after the Tables.  
I had thoughts to have projected a Chart after the manner of Mercator, but because few of them are made true, I forbear, and advise none to trust them, except they are able to examine them.  
If I have been too tedious either in this, or any thing else, remember this is but the buds of my beginnings, and hereafter I may learn to comprise a great matter (more plainly) under fewer words, which is most docible.     


Stars near the Aequinoctial, or declining 52 deg. their Names and Declinations, with their Seasons.    D. M. Season In the Girdle of Andromeda North 33 52 July In the Ram's horn the first  17 36 Aug. In the South foot of Andromeda  40 41  In the Rams Head  21 51  Perseus right Shoulder  52 09  Medusa's Head  39 37  Perseus right side  48 33  Bulls eye  15 46 Sept. Orion's left foot South 08 38  Wagoners right North 20 15  Second in Orion's Girdle South 01 28 Oct. Wagoners right Shoulder North 44 51  The great Dog in his mouth South 16 15  In the upper head of the Twins North 32 34 Nou. The lesser Dog  06 03  Hydra's Heart South 07 14  Lions Heart North 13 35 Dec. Lion's Back  22 23  Lions Neck  21 31  Virgins Spike South 09 21 Jan. Arcturus North 21 19 Feb. South Balance South 14 34  North Balance  08 04  The brightest in the Crown North 27 54  Scorpions Heart South 25 34 Mar. In Ophiuch's right foot  20 30  In the Harp the brightest North 38 32 Apr. eagle's heart alias Vultures  08 03 May. Dolphin's Tail  10 13  Fomahant South 31 20 June In Pegasus leg North 26 16  The Head of Andromeda  27 15 July  

Stars near the North Pole, their Declination from it, with their Seasons.    D. M. Season North Star  02 29 July In the hip of Cassiopeia  31 06  In Cassiopeia's knee  31 33  In Perseus right Shoulder  37 51 Aug. In the great Bear's side  31 49 Dec. In the great Bear's thigh  34 25 Jan. In the great Bear's rump  31 06  First in the great Bear's tail  32 10  Middlemost in her tail  33 17  In the end of her tail  38 57  In the bending of Dragon's tail  24 01 Feb. The foremost Guard  14 22  The hindmost Guard  16 42  In Dragons head foremost  37 25 Apr. In Cepheus Girdle  20 52 June In the back of Cassiopeia's Chair  32 42 July  

A Table showing ho● much the North Star is above or beneath the Pole, for every several Position of the former or greater Guard, Anno 1660 complete.   The nether part of the Meridian accounted North. The upper part of the Meridian accounted North.  Lat. 40 Lat. 60  d. m. d. m. If the Guard bear from the North Star S W b W N W b W Then the North Star is 0 13 0 16 Beneath the Pole. S W N W 0 41 0 44 S W b S N W b N 1 08 1 10 S S W N N W 1 33 1 34 S b W N b W 1 54 1 55 South North 2 10 2 11 S b E N b E 2 22 2 22 S S E N N E 2 2● 2 28 S E b S N E b N 2 29 2 29 S E N E 2 24 2 24 S E b E N E b E 2 15 2 15 E S E E N E 1 59 1 00 E b S E b N 1 39 1 41 East East 1 16 1 18 E b N E b S 0 50 0 53 E N E E S E 0 22 0 24 If the Guard bear from the North Star N E b E S E b E Than the North Star is 0 08 0 05 Above the Pole. N E S E 0 36 0 34 N E b N S E b S 1 05 1 02 N N E S S E 1 31 1 28 N b E S b E 1 51 1 50 North South 2 08 2 07 N b W S b W 2 21 2 20 N N W S S W 2 28 2 28 N W b N S W b S 2 29 2 29 N W S W 2 24 2 24 N W b W S W b W 2 13 2 13 W N W W S W 1 57 1 56 W b N W b S 1 36 1 35 West West 1 12 1 09 W b S W b N 0 44 0 42 W S W W N W 0 16 0 13   

Rules to find the Latitude, or Poles Elevation by the Meridian Altitude of the Sun or Stars, having the Tables of their Declination.  
FIrst, if the Sun or Star be on the Meridian to the Southwards, and have South Declination, add the Sun's Declination to the Meridian Altitude, and it gives the Aequinoctial's height above the Horizon, that total subtracted from 90 deg. gives the Latitude Northerly, which is the Aequinoctials Distance below the Zenith or Poles Elevation.  
Example.  
Suppose upon the 11 th'. of February, 1663., I find the Sun's Declination to be Southerly 10 d. 22 m. Sun's Meridian Altitude I observe to be 30 10 That Total is 40 32 Which being subtracted from 89 60 The Remainder is the Poles Elevat. or Lat. North 49 28  
If when you have added the Sun's Declination to the Meridian Altitude, it exceed 90 deg. then cast away 90 deg. and the Remainder is the Latitude Southerly.  
Secondly, if the Sun or Star be on the Meridian to the Southwards, and have North Declination, subtract the Sun's Declination from the Meridian Altitude, and it leaves the height of the Aequinoctial above the Horizon, that subtracted from 90 deg. leaves the Latitude Northerly.  
Example.  
Admit upon the 19 th'. of May 1663. the Sun's Declination is Northerly 13 d. 17 m. The Sun's Meridian Altitude is 69 58 Suns Declination being subtracted the Remainder is 56 41 This subtracted from 89 60 Leaves the Latitude Northerly 33 19  
Thirdly, if the Sun or Star be on the Meridian to the Northwards, and have North Declination, add the Sun's Declination to the Meridian Altitude, and it gives the height of the Aequinoctial above the Horizon, that being taken from 90 deg. leaves the Latitude Southerly.  
But if it exceed 90 deg. after the Declination is added, subtract 90 deg. from it, and the remainder is the Latitude: See in the first Rule an Example, in its operation as this is.  
Fourthly, if the Sun or Star be to the Northwards and have South Declination, subtract the Sun's Declination from the Meridian Altitude, and it gives the Aequinoctial's height above the Horizon; subtract that from 90 deg. and it leaves the Latitude Southerly.  
Fifthly, if you observe when the Sun or Star hath no Declination, than the Compliment of his Meridian Altitude is the Latitude.  
Sixthly, if the Sun or Star be observed in the Zenith, the Declination is the Latitude.  
If you have a desire to work by the Compliment of the Sun or Stars Meridian Altitude (which is used for its brevity by Seamen) do as followeth:  
If the Sun hath South Declination, and be to the Southward of you, subtract the Sun's Declination from the Compliment of the Sun or Stars Meridian Altitude, and the Remainder is the Latitude Northerly: But if the Sun's Declination be more than the Compliment of the Sun's Meridian Altitude, subtract the Compliment of the Sun's Altitude from the Declination, and the remainder is the Latitude Southerly.  
Secondly, if the Sun or Star be on the Meridian to the Southwards, and have North Declination, add the Compliment of its Altitude to the Declination, and the remainder is the Latitude.  
If the Sun or Star be on the Meridian to the Northwards, and have North Declination, subtract the Declination from the Compliment of the Meridian Altitude, and the remainder is the Latitude Southerly: If the Sun's Declination be more than the Compliment of the Sun or Stars Meridian Altitude, subtract the Altitude from the Declination, and the Remainder is the Latitude Northerly.  
If the Sun or Star be to the Northwards, and have South Declination, add the Declination to the Compliment of the Meridian Altitude, and the remainder is the Latitude Southerly.  
If you observe the Sun when he hath no Declination, the Compliment of his Meridian Altitude is the Latitude.  
If you observe the Sun or Star in the Zenith, the Declination is the Latitude. And thus much for the finding the Latitude by the Sun's Meridian Altitude, or the Compliment thereof; also for such Stars as have their Declination from the Aequinoctial.  
For those Stars whose Declination is their distance from the Pole, if you observe them under the Pole, add the Declination to the Meridian Altitude; if above the Pole, subtract their Declination, and the Remainder is the Latitude. And thus much for finding the Latitude by the Declination, and Meridian Altitude of the Sun or Stars, or the Compliment of the Meridian Altitude of the Sun, or such Stars whose Declination is their distance from the Aequator.    

OF ASTRONOMY.  
I Had thoughts to have handled the Doctrine of Spherical Triangles fully, but if I should, I might do that which my Father hath done before, better than I can; the thing which induced me to this thought was that I might make the Application of it to Great Circle sailing; but because that way of sailing (though the best) is seldom used, I have not set it in my Book; if you desire to know it, you may see the way of it in my Father's works, as plainly as (I think) it can possibly be demonstrated by any man.  
I have thought good to make use of his fundamental Axiom, to show how to resolve those Questions in the Sphere, which I have wrought in this Book by the Plain Scale, though I confess (for Amplitudes of Rising and Setting, and Azimuths at certain given hours, with divers other things that are chief to be minded at Sea, for the Variation of the Compass, and other necessary uses) I hold the Plain Scale to be true enough, but that I conceive not to be so satisfactory to all men, and therefore I have done this, the Axiom is this: The Sine of a middle part with Radius, is equal to the Tangents of the Extremes adjacent, or to the Sine Compliment of the opposite Extremes.  
Now for Brevity's sake it is common to use these Characters, for this Character + signifieth with Radius or more, or one Side or Angle with an another, for equal =, for less −; so that the general Rule or fundamental Observation which we purpose to use, may be expressed thus:  
The Sine of the middle part + Radius is = to the Tangents of the Extremes adjacent, or to the Sine compliment of the opposite Extremes.  
This Rule being rightly understood gives such full directions, that a man may resolve any Question in the Sphere by it, which makes me judge those culpable of a fault, that accuse him of Prolixity, that made it in so brief a manner. I will define a spherical right angled Triangle, and so come to the use of this Axiom.   depiction of geometrical figure 
If you have two things given beside the right Angle, to find a third (as you always have) the first thing that you are to consider, is, whether it be an adjacent or an opposite Extreme.  
That is an adjacent Extreme, when the given parts and the required part join.  

For Example in Adjacent Extremes.  
Suppose the given things were A C and A C B, and the required thing were C B: Here you see that A C is joining to C, and so is C B, therefore they all join, and C is the middle part, for it is between B C and C A the Extremes.  
That is an opposite Extreme that hath two things lying together, and the third thing lieth alone, and that part which lieth alone is the middle part.  
Example.  
Suppose the three things (namely the two given things, and the required thing) were B C, the Angle at A, and C A; then I conclude that C B is the middle part, and that it is an opposite Extreme, because it is alone, and the other thing, namely, C A and A are together: If you ask how this can be, thus:  
A C B  between C B and C A, and A B is between C B and the Angle at A, which proves that B C is alone, and A C and the Angle at A joins: Then is A C and the Angle at A also the two Extremes.  
Pray observe this for a Rule, that the right Angle never parts any thing, for if C B, the Side B A, and the Angle at A be the three things, it is nevertheless an adjacent Extreme; for though the right Angle be between the Sides B A, and B C, yet do not count that it parts them, (as another Angle would) so as to make it an opposite Extreme.  
Or if the given and required things were the three Sides, you might judge B A the middle part, because it is alone, for the Angle at A parts it from A C, and the right Angle at B is between B C and it: Now never count that the right Angle at B parts B C from B A, but that they lie together, and A C lies alone, being parted by the Angle C from C B, and by the Angle A from A B, and so is the middle part: These be the first considerations (after you have a Question set you) that you ought to mind, without regarding which of the three things is given or required.  
We now come to apply this general Axiom. Suppose I have the Side B C given, the Angle at C, and that the Side C A be required.  
From what hath been said I know C to be the middle part and that it is an adjacent Extreme; I will apply the Rule thus, it saith the Sine of the middle part, which intimates that the middle part must be always so called, as you may take the Sine of it; now because it's noted to be a Compliment, you must call it Sine Compliment, to find the Sine of it. This done, I will go to one of the Extremes, and consider what the Axiom saith for that, (admit it be the Extreme C B) It saith the Sine of the middle part with Radius, is equal to the Tangents of the Extremes adjacent; from whence I conclude, that I must call C B so, as that I may have the Tangent of it; Now C B is not noted to be a Compliment, and therefore I call it Tangent C B. last, for the other Extreme C A, for the same reason I must call it so, as that I may take the Tangent of it: and because it is noted to be a Compliment, therefore I call it Tangent Compliment, to have the Tangent of it: Now as I go along thus in my thoughts, I will set it down as I consider it, and it frames this Rule:  
Sine come. C + Radius is = to Tang. C B + Tang. comp. C A.  
My Question being brought into this order, I will consider which is given and which is required: those which are given, I will set down first, the thing required undermost, and look for them in the Tables by the directions which I have framed from the Axiom.  
Look for Sine comp. C For Tangent B A Add them together, casting away Radius, and the Sum look for in the Tangents take Tan. come. C A it produceth your desire.   
And thus this Axiom sets your business in order; but now that you may know when to take a Compliment arithmetical, or let it a loan; and also, that you may know which thing it is that you must take the Compliment arithmetical of, observe this Rule, either in adjacent or opposite Extremes.  
That if the middle part and an extreme be given, take the Compliment arithmetical of the given Extreme; but if the two Extremes be given, to find the middle part, take no Compliment arithmetical at all: This may be sufficient for all Questions in right angled Triangles, will fall out one of these two ways; this Compliment arithmetical is what every figure in the Tables wants of 9: it was first invented by my Father, it saves a labour of Subtracting, and so abbreviates the work much.  
To many which I have taught, it hath seemed strange that a thing is called Sine comp. or Tan. comp. (when one is to find the Sine or Tangent of it) but the reason of it is this; if a thing be a Compliment already (as three things in every right angled spherical Triangle is noted to be) the comp. of that Compliment must be the Sine or Tangent of that Compliment.  
But pray mind me; I have noted that there are five things in every right angled Triangle considerable, besides the right Angle.  
Now observe this for a Rule, (when you are framing your Rule from the general Axiom to work by) that if you call any of these five things by their names, you shall not have the Compliments of them, but call any of them out of their names, and you shall; I mean thus. Suppose you have a mind to take the Sine of A B, it is noted not to be a Compliment, therefore call it Sine A B, to have the Sine of it, or Tangent A B to have the Tangent of it, the like for C B.  
Again, if you have a mind to take the Sine of C A, I consider that C A is noted by the name of a Compliment, call it Sine comp. so shall you have the Sine of it, or Tangent Compliment, and you shall have the Tangent of it; the like for the two Angles contrariwise: If you have occasion to take the Compliments of them, call them out of their names.   

Example in opposite Extremes.  
Suppose A B be given, and B C, and the Side A C be required.  
From what hath be said, I consider this is an opposite Extreme, and that A C is the middle part; (for that is alone) call A C Sine Compliment to find the Sine of it: and because the Axiom saith, that the Sine of the middle part with Radius, is = to the Sins Compliment of the opposite Extremes (for the former Reasons) call A B Sine comp. and B C Sine comp. which is out of their names, and you have this Rule:  
Sine come. A C + Radius is = to Sine come. A B + Sine come. B C.  
This done, set the things down by the order Sine come. C B Sine come. A B Sine come. A C of this Rule, with the required thing under most thus:  
Take the figures answering the number of degrees and minutes, which the given things may contain, (out of the Tables) add them together, casting away Radius, (if it comes to be above it) and look for the Remainder amongst the Sins, and take the Compliment of that Sine that answers it, to answer your demand.  
Suppose it were A B and A C that were given, to find the Angle at C; this is opposite Extreme, and A B is the middle part, call A B (the middle part) Sine, (which is by its name) to find the Sine of it, and A C, and the Angle at A out of their names (which is Sine A C, and Sine A) to find the Sine Compliment of them, and thus for any else.  
Be sure to remember the Rule that I propounded about the Compliment arithmetical, that so you may know when to take it, and when to let it alone. This is in brief.  
My Father hath showed the proportions that holds, and the reasons of them, which I purposely omit, knowing that this Rule works those Proportions: We will now come to the work itself; and I shall do only those Questions which I wrought by the Plain Scale, as being most necessary for our use that go to Sea.   

Latitude 50 deg. 00 min. Northerly, Declination 13 deg. 15 m. Northerly, I demand the Meridian Altitude of the Sun.  
LEt M G n R represent the Meridian, M n the Horizon, K y the Equinoctial, I ♈ the Parallel of the Sun's Declination, G the Zenith, R the Nadir, subtract the Latitude G K 50 deg. 0 min. (= to n f) from G K M 90 deg. and the Remainder is K M, the height of the Aequinoctial above the Horizon, 40 deg. 0 min. to which add the Sun's Declination K I 13 deg. 15 min. and the Sum is I M the thing required, 53 deg. 15 min. as you may see in this Example.   depiction of geometrical figure 
This, if the Sun's Declination be 50 00 90 00 40 00 13 15 53 15  
Northerly  
But if the Sun's Declination were Southerly, namely, so as that her Parallel of Declination cut the Meridian in s; subtract the Sun's Declination K s from the height of the Aequinoctial above the Horizon K M, and the Remainder is s M, the Sun's Meridian Altitude.  
In any Latitude, subtract the Latitude from 90 deg. and so you have the Aequinoctials height above the Horizon; then see whether the Meridian Altitude be greater or lesser than that; if greater, add the Sun's Declination to the Aequinoctials height above the Horizon, if lesser subtract it, and the Remainder is the Sun's Meridian Altitude: and this is evident, because the Sun's Declination is his Distance from the Aequator, and so he can be but as much higher or lower than the Aequator, as his Declination is.   

Latitude 50 deg. 0 min. Declination 13 deg. 15 min. Northerly, I demand the Sun's Amplitude of Rising and Setting.  
NOte, that the sides of a Spherical Triangle are three Arches of Great Circles, every Arch being less than a Semicircle; and therefore the Parallels or other lesser Circles of the Sphere must not be taken as the Sides of a Triangle.  
F O P is an Arch of the Meridian, cutting the Centre of the Sun at his Rising, which is at O (you have been told that all Meridian's cut the Aequinoctial at right Angles) then must of ♈ be a right Angle, and in it I have given s 0 the Sun's Declination (= to t c) and s ♈ 0 = 40 deg. 0 min. the measure of f ♈ 0, is E K = to t A the Compliment of t s to 90 deg. (t s being the Latitude 50 deg. 0 min.) and I am to find ♈ 0, the Sun's Amplitude of Rising; I consider it is an opposite Extreme, and that s 0 is the middle part.  
Sine f 0 + Radius is = to Sine ♈ + Sine ♈ 0.   depiction of geometrical figure 
Sine f 0 the Sun's Declination 13 deg. 15 min. 9,36021 Sine f ♈ 0 the come. of the Poles Elev. 40 d. co. ar. 9,19193 Sine ♈ 0 the Sun's Amplitude of Rising or Setting 20 deg. 53 min. 24 sec. 9,55214  
Use this way to find the Sun's true Amplitude of Rising, upon occasion of finding the Variation of the Compass.  
The Sun's place of Rising from the East towards the North, namely, East 20 deg. 53 min. 24 sec. Northerly; O ♈ is the Compliment of O E, the Sun's Azimuth of Rising from the North: so that if you subtract it from 90 deg. ♈ E the remainder is 69 deg. 6 min. 36 sec. O E; but if you have a desire to find the Sun's Azimuth by the things you have given, here it is in the Triangle F E O, and is done as in this following Question.   

Latitude 50 deg. Declination 13 deg. 15 min. I demand the Sun's Azimuth of Rising or Setting.  
LEt F O f P be an Arch of the Meridian, cutting the Sun at his Rising (which is at O) f O is the Sun's Declination 23 deg. 15 min. which subtracted from f F 90 deg. the remainder is O F 76 deg. 45 min. the Sun's distance from the North Pole, and F E is the Poles Elevation 50 deg. 0 min. and thus you have two sides given in the right angled Triangle O E F, to find the third side O F; it's an opposite Extreme, and F O is the middle part: The general Axiom produceth this: Sine come. F O + Radius is = to Sine come. F E + Sine come. O E. Sine come. O F 76 deg. 45 min. 9,360215 Sine come. F E 50 deg. 0 min. comp. arith. 0,191932 Sine come. O E 69 deg. 6 min. 36 sec. 9,552147  
Thus I find the Sun riseth to the Eastwards of the North 69 deg. 6 min. 36 sec. which is E b N 9 deg. 38 min. 24 sec. Northerly; the other Question shows the same, when you find the Sun's Amplitude.   

Latitude 50 deg. Declination 13 deg. 15 min. Northerly, I demand the Sun's height at six of the Clock.  
LEt G P m N be an Azimuth passing through the hour of 6: Now in the Triangle ♈ m P, right angled at m, you have given P ♈ the Sun's Declination 13 deg. 15 min. and the Angle at ♈, namely P ♈ m = to the Arch f O (for f O is the measure of it) it is the Poles Elevation 50 deg. 0 min. to find P m the Sun's height at six of the Clock: I consider it is an opposite Extreme, and the required thing P m is the middle part.  
Sine P m + Radius is = to Sine P ♈ + Sine P ♈ m.   depiction of geometrical figure 
Sine P ♈ the Sun's Declination 13 deg. 15 min. 9,360215 Sine P ♈ m the Poles Elevation 50 deg. 0 min. 9,884254 Sine P m the Sun's height at six of the Clock 10 deg. 6 min. 51 see. 9,244469   

The same things given, to find the Sun's Azimuth at six of the Clock.  
THe Sun's Azimuth at 6 of the Clock is m ♈, and the reason is, for that P u in the Heavens (being parallel to the Horizon) is as many degrees in that Circle, as ♈ m is in the Horizon; for ♈ is under m, and m under P.  
So that this Question will fall in the same Triangle as the other did, and we will use the same things to find it.  
The Angle P ♈ m is 50 deg. 0 min. the Poles Elevation.  
P ♈ is the Sun's Declination 13 d. 15 m. and ♈ m is required.  
I consider that this is an adjacent Extreme, and that P ♈ m is the middle part; The general Axiom produceth.  
Sine come. P ♈ m + Radius, is = to Tang. ♈ m + Tang. come. P  
Tan. come. P ♈ the Sun's Decl. 13 d. 15 m. come. arith. 9,371933 Sine come. P ♈ m the Poles Elevation 50 d. 0 m. 9,808067 Tang. ♈ m the Sun's Azimuth 8 deg. 36 min 24 sec. 9,180000   

Latitude 50 deg. 0 min. Declination 13 deg. 15 min. to find the Sun's height being due East or West.  
LEt O L I u be an Arch of the Meridian, cutting the Sun in the East and West Azimuth, and it helps to make the right angled Triangle L I ♈, in which lieth our business: For L ♈ is the Sun's height being due East; and to find it we have I L the Sun's Declination 13 deg. 15 min. and L ♈ I the Angle of the Latitude, besides the right Angle; I consider it it is an opposite Extreme and I L is the middle part, there will be this inference produced from the general Axiom.  
Sine I L + Radius, is = to Sine I ♈ L + Sine L ♈.   depiction of geometrical figure 
Sine I ♈ L the Latitude 50 deg. comp. arith. 0,1157459 Sine I L the Sun's Declination 13 deg. 15 min. 9,3602154 Sine L ♈ Suns height being due East 17 d 24′ 34″ 9,4759613  
The same height that the Sun is being due East in the morning, the same height he is when he is due West in the afternoon.   

The Latitude being 50 deg. 0 min. Declination 13 deg. 15 min. I desire to know the Difference of Ascension.  
I Have showed what the Difference of Ascension means in this Book before; and if you look in the Question to find the Amplitude of the Suns rising before, there is the same Scheme that this is, but for your better vuderstanding I have here set it.  
In the Triangle B D ♈ right angled at D, you have given D B the Sun's Declination 13 deg. 15 min. the Angle at ♈ which is the Compliment of the Arch I Z too 90 deg. or equal to O A the Compliment of the Latitude 40 deg. 0 min. to find D ♈, which is as many degrees and minutes as B 6 is: It is an adjacent Extreme and ♈ D is the middle part: The conclusion from the general Axiom is  
Sine Y D + Radius, is = to Tan. D B + Tan. come. D Y B.   depiction of geometrical figure 
Tangent D B Suns Declination 13 deg. 15 min. 9,371933 Tan. come. D ♈ B come. Poles Elevat. 40 d. 0 m. 10,076186 Sine ♈ D the Difference of Ascension 16 d. 17′ 40″ 9,448119  
Which converted into time is 1 hour 5′ 2/15 and ⅔ of 1/15 of a minute.   

The same things given, to find the time of Sun Rising.  
FOr the doing of this, first find the Difference of Ascension (as hath been showed in the last Question) and convert it into time; which done, subtract it from 6 hours, and you have your desire (in this case where the Declination and Latitude is both one way) but in any case take this in general, that if the Difference of Ascension be before 6 of the Clock, subtract it; if after 6, add it to 6 hours, and you have the time of Sun Rising.  
This stands to good reason, forasmuch as the Difference of Ascension is the portion of time that the Sun riseth before or after 6 of the Clock.  
In this Example you see the Difference of Ascension is before 6 of the Clock 1 hour 5′ 3/15 (we will omit the part of a part of a minute) I would know the time of the Suns Rising.  
Example.  
Subtract the Difference of Ascension 1 h. 5 m. 3/15 From 6 hours 5 60 The remainder is the time of Sun Rising 4 54 12/15 or ⅘  
Thus I conclude the Sun riseth at 4 a Clock 54 min. ⅘  
If you have a desire to find the time of Sun setting, subtract the time of Sun rising from 12 hours, and you have it; for as many hours and minutes as the Sun riseth before 12, so many hours and minutes he sets after 12.  
Example.  
Here the Sun riseth at 4 of the Clock, 54 m. ⅘ which we express thus 4 h. 54 m. ⅘ This subtracted from 11 60 Leaves the time of Sun setting 7 05 ⅕  
Which is 5 min. ⅕ past 7 of the Clock in the afternoon.  
This doubled is the length of the whole day, which is 14 hours 10 min ⅖.  
This subtracted from 24 hours, is the length of the night, 9 hours 49 min. 8/5.   

To find the length of the longest Day in that Latitude before proposed.  
When the days are at the longest in any North Latitude, the Sun is in the Tropic of Cancer; in a Southern Latitude, in the Tropic of Capricorn: This is a Northern Latitude, therefore make the Tropic of Cancer the Parallel of the Sun's Declination as here, O R is the Parallel of the Sun's Declination, then must I 6 be the Difference of Ascension, which is = to F ♈. In the right Angled Triangle I F ♈ right angled at F, you have I ♈ F the Compliment of the Latitude 40 deg. given, and I F the Sun's Declination 23 deg. 30 min. to find F ♈; find it as you was showed to find the Difference of Ascension before, convert it into time, and find the length of the day (as was showed before) This note, that Sine s I F L is an Arch of the Meridian, cutting the Horizon in that place of it where the Sun riseth.  
Sine ♈ F + Radius is = to Tang. I F + Tang. comp. I ♈ F.   depiction of geometrical figure 
Tang. I F Suns Declination 23 deg. 30 min. 9,638301 Tang. come. I ♈ F Poles Elevation 40 deg. 0 min. 10,076186 Sine F ♈ Difference of Ascension 31 d. 12 m. 39 sec. 9,714487  
But you may ask all this while how this Fraction is found, it is thus found: Admit I would find the sine of the Arch answering to 9714487, I look in the Sins, and find the nearest less than it to be  9,714352 I take the next greater than it and find it to be 9,714560 I subtract the lesser from the greater, the remainder is 208 Then from the figures that came forth 9,714487 I subtract the nearest in the Tables less 9,714352 And the Remainder is 135  
Then say, As the Difference between the nearest less and the nearest greater 280 comp. arith. 7,681936 Is to the Difference between the same and that less 135 2,130333 So is 60 min. 1,778151 To 39 sec. 1,590420  
The like is to be understood of any Fraction else; if you have occasion for the artificial Sine or Tangent of an Arch that hath a Fraction to it: Say,  
As 60 seconds Is to the seconds in the Fraction, which is here 39 So is the Difference between the nearest lesser and the nearest greater 208 To the Fraction 135 And this added to the lesser, makes the artificial Sine of the Arch required 9,714352 9,714487 The like is to be understood of a Tangent.   
To find the hour of the Suns being due East or West.   

Latitude 50 deg. 0 min. Declination 13 deg. 15 min. Northerly, I demand the time of the Suns being due East or West.   depiction of geometrical figure 
And for the time of the Suns being due West, you are to subtract it from 6 hours, (in this case) and the remainder is your desire: The reason is, because as many hours as the Sun is due East after 6 of the Clock in the morning, so long time is he due West before 6 of the Clock in the Afternoon. I have left the working of this to your own practice, only I have set down the Resolution of it:  
The Sun is due East 45 min. 9/15 past 6 of the Clock.  
The Sun is due West 45 min. 9/15 before 6 of the Clock.  
Which is at 5 of the Clock 14 min. 6/15.  
Note, That if the Sun's Declination be Southerly, then will he be East before 6; so that whereas here you add, there you subtract from 6 hours, to find the hour of the Suns being due East, or add to 6 hours for the time of its being due West; but your own Reason (if you look well on the Scheme) will guide you to know this; and also to know that, it is useless in such cases, for than he is not above the Horizon.   

Latitude 50 deg. 0 min. Declination 13 deg. 15 min. I demand the Continuance of Twilight.  
AS I have noted before, the Sun is accounted 17 deg. under the Horizon when the day breaks; therefore in the following Scheme let 17 I, be an Arch parallel to the Horizon, 17 deg. under it: Let D B C A be an Azimuth cutting the Aequinoctial, and the line of 17 deg. in the place of their intersection, which is at C: then in the Triangle r B C (right angled at B) you have given B C 17 deg. the Angle B r C the Compliment of the Poles Elevation 40 deg. to find C r the continuance of Twilight (for C r and B u are equal.) I leave it to your own Practice: I find the continuance of Twilight to be 27 deg. 4 min. which converted into time is 1 h. 48 min. 4/13 nearest.   depiction of geometrical figure 
So I conclude that between the Day breaking and Sun rising, it is 1 h. 48 min. 4/15, which is u B: Now if you add this to the Difference of Ascension (before found) B 6, which was 1 h. 5 min. 3/15 of a minute (I omit the smaller Fraction) you have the time between break of day, and 6 of the Clock, which may be termed u B 6 2 h. 53 min. 7/15.  
And because the day breaks so much before 6 of the Clock, if you subtract it from 6 hours, you have the hour and minute of day breaking, which is at 3 of the Clock 6 min. ●8/15.  
Add it to 6 of the Clock, and you have the time of Twilight ending, which is at 8 h. 53 min. 7/15.   

To find the Sun's Place and Right Ascension, provided, the Latitude and Declination be given.  Latitude 50 deg. 0 min. Declination 13 deg. 15 min, I demand the Sun's Place in the Ecliptic and right Ascension. 
RIght Ascension is an Arch of the Aequinoctial between a Meridian, cutting the Aequinoctial and the point of Aries or Libra; the Meridian must also cut the Sun or Star in the Ecliptic as here in this Scheme: A B C is an Arch of the Meridian, cutting the Sun at B in the Ecliptic, and the Aequinoctial in C; then C ♈ is the Sun's right Ascension.  
The Sun's Place (as I shown formerly) is the Sun's distance in the Ecliptic, from the nearest Aequinoctial point here B Y.  
To find the Sun's place ♈ B, you have given the Angle B ♈ C 23 deg. 30 min. and C B the Sun's Declination 13 deg. 15 min. in the right angled Triangle B C ♈; it is an opposite Extreme, and C B is the middle part, I find it to be 35 deg. 5 min. if you propose the month and the day of the month that the Sun hath this Declination, you may tell what Sign the Sun is in, and what degree of that Sign.   depiction of geometrical figure 
As suppose the Sun hath this Declination here given upon the 14th. of April, I consider what Sign belongeth to that Month, and I conclude it is ♉, then say I, as much as the Sun is distant from the next Aequinoctial point above 30 deg. (which is the Sine of ♈) so much he is entered into ♉, which is 5 deg. 5 min. I omit the Fraction.  

For the Sun's Right Ascension.  
You have the same things given in the same Triangle, to find the Sun's right Ascension C ♈; but this is an adjacent Extreme, and the required thing is the middle part.  
I find it to be 32 deg. 47 min. some part of a minute more, (but I omit the Fraction.)  
This must be converted into time, and then is your demand fully answered, it is 2 h. 11 min 2/15.  
These be things necessary and very useful in Navigation for several occasions, and that hath made me insist upon them: If you are unacquainted with this way of working, my Father hath insisted upon it largely, and in his works, commonly called The Doctrine of Triangles; you may be satisfied both in the Doctrine of right lined Triangles, and Spherical. This I judge sufficient for the two kinds of Sailing.  
I would have handled something in obliqne Spherical Triangles, and so have proceeded to that incomparable way of Sailing by the Arch of a Great Circle; but the goodness of it is not known because it is not practised: If you love that way of Sailing, (Reader) I commend thee to my Father's Doctrine of Triangles in large, where it is largely treated of.    
FINIS.   

ADVERTISEMENT.  
THere is newly reprinted the Mariner's Magazine, Stored with these Mathematical Arts: the Rudiments of Navigation and Geometry, the making and Use of divers Mathematical Instruments; the Doctrine of Triangles, Plain and Spherical; the Art of Navigation, by the Plain-Chart, Mercator's-Chart, and the Arch of a Great Circle; the Art of Surveying, Gauging, and Measuring; Gunnery and Artificial Fire-Works; the Rudiments of Astronomy; the Art of Dialling. Also with Tables of Logarithms, and Tables of the Sun's Declination; of the Latitude and Longitude, Right Ascension, and Declination of the most notable Fixed Stars; of the Latitude and Longitude of Places; of Meridional Parts; whereunto are annexed, an Abridgement of the Penalties and Forfeitures, by Acts of Parliament, relating to the Customs, and to Navigation; and a Compendium of Fortification; by Capt. Samuel Sturmy, the Second Edition, diligently Revised and carefully Corrected, by John Colson, Teacher of the Mathematics in London, in Folio.  
Also,  
A Mathematical Manual; Containing Tables of Logarithms, for Numbers, Sins, and Tangents; with the manifold Use thereof briefly Explained and Applied in Arithmetic, Geometry, Astronomy, Geography, Surveying, Navigation, Dialling, Gunnery, and Gauging; by Henry Phillippes. in Octavo.