THE DESCRIPTION and USE OF AN INSTRUMENT, Called the DOUBLE SCALE OF PROPORTION. By which Instrument, all Questions in 

Arithmetic,  
Geometry,  
Trigonometry,  
Astronomy,  
Geography,  
Navigation,  
Fortification,  
Gunnery,  
Gauging Vessels,  
Dialling,   May be most accurately and speedily performed, without the assistance of either Pen or Compasses.  
By SETH PARTRIDGE.  
LONDON, Printed by William Leybourn for William Wright under Grace-Church in Grace-Church-Street, and at the Hand and Bible in Ducklane. 1661.   

To the Right Worshipful, Sir RICHARD COMBE Knight, The Author presents these, with his other best Services.  
SIR,  
THe transcendency of your knowledge in all noble and learned Sciences, hath heightened my ambition presumptuously to affix on you this Dedication, and to this boldness it's your love and favour that invites me, and prevents me too, of all Apologies for my thus doing. For truly I should not so boldly have adventured to shroud this unworthy Piece under the wings of your worthy Patronage, were I not assured, that as in judgement you are able to discern of it; so in your noble disposition willing favourably to accept it. The poor Widows two mites being the utmost of her ability, was accepted of by him, who was greater than the greatest of Mortals, far above the great gifts of the Rich: which emboldens me (without doubt of obtaining) humbly, to desire your Worship, to vouchsafe me both pardon for my presumption, and acceptance of my little mite, which having obtained, I shall think myself and work, to have a sufficient protection against all Cynical and illaffected Detractors, a thing which all men's works (how excellent soever) are subject unto, (much more mine) and therefore into the wide Ocean I dare not commit my weak vessel, without some able Pilot, such as your Noble Self, whose sound judgement and great learning is so well known, that your name prefixed before my Book, will be not only a defence against the carping of Zoylist's, but also to beget a good affection in the Readers. Not ta trouble your Worship more with words, my hope is, that this my boldness will be imputed to a will, rather to do you service than otherwise, might your Worship be pleased thereof so to make construction, I have attained my desired end, who am  
Your Worships ever to serve you, Seth Partridge.   

To the Reader.  
FRiendly Readers, you have here presented to your view those most excellent Soles, or Lines of Numbers, Sins, and Tangents doubled, by means of the moving whereof, the use of Compasses is wholly avoided, and the question resolved, by applying the Scales one to another. Some men (I know) have laid aside the using of the single Scales, because with an ordinary pair of Compasses on a large Scale, they could not work many examples they desired. But by these double Scales, how large soever, he shall never be troubled with any Compasses, nor the work never to outrun the Scale. And besides, upon the Instrument may be inserted any other Scales or Lines, as are for measuration, or otherwise, such as each man's calling and occasions do most require, and so make the Instrument of general use. To plead for my Book, I will not, the Subject whereon it treats, will do that better than I can, I am sure here is a good Subject, a good piece of cloth, if the garment be not marred in the making, if it be, the fault is in the botching Tailor, not in the stuff. The Ingenious (I know) will wink at faults, for they know themselves subject to them, and faults declare men to be but men. As for curiosity in the method, here is none, nor was any intended, my dishes being dressed, not as at a Feast, but as at an Ordinary, nor placed in so methodical an order as they might have been, I taking ●hings not orderly, but at an adventure, and as it happened to come into mind: I know my Subject, nor method, will please all, yet I doubt not, but it will please some. What the ignorant say of it, I care not, they are like the Fox, that despised the grapes, because they grew so high he could not reach them. And what the spiteful speak of it, I pass not, they are like the filthy Fly, that seeks all over the body for a sore, and when it cannot find one, it makes one. It is the censure of the impartial, judicious and solid Judgement which I respect, to whom I will stand, and to which only I will submit, resting Servant to all such, while I am  
Seth Partridge.   

THE Double Scale of Proportion.  
The Description and Use of an Instrument consisting of Doubled Scales.  
Whereby all Conclusions Mathematical may be performed only by Application of Scales.  

CHAP. I. The Instrument described.  
THe Instrument whereof I treat in this Book, I call the Double Scale, for that the Scales, or Lines thereupon, are doubled; so and in such manner, that by applying the one to the other, they will of themselves resolve any question Mathematical, that may be done by the Pen, or by Tables of Sines, Tangents, and Logarithms. The Scales principally inscribed thereupon, are those most admirable Lines of Numbers, of Sines, and of Tangents, whose use hath been heretofore set forth, only to be wrought upon with a pair of Compasses, and no otherwise. But I have so contrived them double, upon an Instrument to move, or slide along one by the other, in such manner, that upon the same, without any assistance of Compasses, I can work all Conclusions Mathematical, that can be wrought by the single Lines, with the help of Compasses, both in Arithmetic, Geometry, Trigonometry, Astronomy, Geography, Navigation, Gauging of Vessels, Fortification, Gunnery, Dialling. Yea, and the Usurer too, may hereby also compute the true interest of his money. In all which, I shall give you some examples for your instruction, in the use of my Double Scale.  
The body, or matter whereof the Instrument is made, may be either of Brass, or of very good and well seasoned Box. It consisteth of three pieces, or Rulers, each one about half an inch in breadth, and about a quarter of an inch in thickness, more or less, as the Maker and user of them pleaseth; and for their length, they may be made to what length you will, either one foot, two foot, three foot, or more or less, for they are not limited to any length, only the longer they are, the larger and more will be the divisions of the Scales, and so consequently the more exact in operation. These three Rulers, or pieces of this Scale, are to be all of one even length and thickness, and the edges so evenly jointed, that they may justly slide along close one by the other, having at each end a little plate of Brass, or wood fitted to hold them close together, & so fastened to the two outside pieces, that they may be kept steady, and the middle Ruler to slide to and fro between them.  
The Lines, or Scales engraven on this Instrument, are the ordinary lines of Numbers, of Sines, and of Tangents, only they are set on double, that is, once upon one Ruler, and once upon the other, upon one and the same joint; as the line of Numbers is set both upon one of the outside pieces, and on the middle piece, that is, on both sides the joint, and numbered on both, and so set to the very edges of both Rusers, that both lines being joined together, may appear to be but as one line of Numbers, and this line of Numbers is (as it were) twice repeated, or doubled in the length of the Ruler, that is, beginning with 1 at one end of the Ruler, which I call the lower end, and continued to 1 or 10 at the middle, and from thence begin again, and continued to 10 or 100 at the upper end. Also upon the other side of the Rulers, upon the same joint, is in like manner set a like line of Numbers, and this line of Numbers is fittest to be used with the line of Tangents, as that on the other side is with the lines of Sines, without any turning of the Instrument. But you may omit on the line of Numbers upon one side, only observing to turn the Instrument, when the work requireth to be done on several lines, as in the sequel it will appear.  
The Scales, or Lines of Sines, are in like manner set on twice, that is, once upon the other edge of the middle piece, and also on the inside of the other outside piece, and they are to be set on, upon both sides the joint, that they may appear as one line of Sines, being laid close together, and nnumbers set to the Divisions on both parts, and is in the lines of Numbers.  
The lines of Tangents is set upon the other side of the Rulers, opposite to the lines of Sins upon the same joint with them, and likewise so set upon both parts of the Instrument, that the parts being laid close together, the lines may appear to be both as one line of Tangents, and numbered on both parts to 45 at the upper end, against 90 on the lines of Sines, and from 45 back again to 89 at the lower end, as is usually done in the lines of Tangents.  
This Instrument having those Scales, or (as it pleased their first Inventor to call them) Lines, thus engraven or set on it, will work all conclusions, as may be wrought by Mr. Gunter's Lines, or Mr. Wingates, by only applying the Lines one to another, without the use of Compasses, which must be always had and used with theirs; yet you may use Compasses with these Scales, if you please, and so try and examine your work by both ways, and when you have made such trial, use which way you like best, and that (for aught I know) may be my Double Scales.  
To show the making of the Scales of Numbers, Sins and Tangents, is a thing altogether needless, the making of them being already so sufficiently set forth by others, that for me to do it again, were but labour lost, both to me in writing, and to the Reader in reading. And those that are makers of Mathematical Instruments, do already well understand the making of them, and for a man to make one for his own use, is but vain, for that he may buy one at a cheaper rate than make it, I shall therefore proceed to the use: Wherein note, that for distinction of the sides of the Scales, or Lines on the Rulers, I use the Terms of First and Second, as being as proper for the purpose as any other could have been: For evermore, that side of any Line, whereon the first term in the rule of Proportion is taken, I call the first side; and the other side of the Line, whereon the second term in the Rule is taken, I call the second side. And then for the third term in the Rule, it is always taken on the same side that the first term is taken on: and for the fourth term, which is the term sought, it is evermore found on that same second side, whereon the second term is taken. As if the first term be taken on any Scale, upon the outside Ruler, than the second term is on the middle Ruler, and if the first term be taken on the middle Ruler, than the second term is on the same Scale upon an outside Ruler: And when the work is by several Lines, than the two outside Rulers both bear the name of First, or Second.  
Other Scales may be added to this Instrument, and set on the sides and edges thereof, as a Scale of equal parts, or a Line of inches, a Meridian-line, a Gage-line, a line of Chords, the lines of Board and Timber-measure, or any others, such as your calling and occasions have most use for.   

CHAP. II. The Use and application of the double Scale of Numbers upon the Instrument, in the principal Rules of Arithmetic.  

PROBLEM I. Of Multiplication.  

Two numbers being given to be multiplied together, to find their Product, by the double lines.  
IN Multiplication the Analogy is this; 

As 1 is to one of the numbers given, to be multiplied together:  
So is the other of them, to the Product.    
Wherefore it may be said, 

As 1 is to the Multiplicator;  
So is the Multiplicand to the Product.   Or, 

As 1 is to the Multiplicand;  
So is the Multiplicator, to the Product.    
When two numbers are to be multiplied together, the greater of them is usually counted for Multiplicand, and the lesser for Multiplicator.  
To multiply two numbers by the double lines, the manner of working is thus: Place 1 on the first side, to the Multiplicator on the second side; And then against the Multiplicand on the first side, is the Product on the second side.  
Or else, place 1 on the first, to the Multiplicand on the second, and then right against the Multiplicator found on the first, is the Product on the second.  
Example 1. Let 8 and 4 be two numbers given to be multiplied together, to find their Product, do thus: upon any one of the sides, look out 1. (This side whereon I take the 1, I call the fitst side) I set this 1 to 4 on the other side (which I call the second side,) and then right against 8 on the same first side, whereon I did take 1; is 32 on the second side, whereon the 4 was taken: This 32 is the Product of 8 multiplied by 4, the thing required. Or otherwise thus.  
Set 1 on the first, to 8 on the second, and then right against 4 on that first, is 23 on the second, as before.  
Example 2. Let 25 be a Multiplicator, and 30 the Multiplicand, and the Product of them multiplied together, required.  
Set 1 on the first, to 25 on the second, and then right against 30 on the first, is 750 on the second. This 750 is the Product of 30 multiplied by 25, the thing required. In like manner, If you set 1 to 30, then against 25 taken on the same side the 1 was, is 730 on the other side whereon the 30 is.  
Example 3. Let 45 and 25 be two numbers given to be multiplied together, and their Product required. To resolve this by the double lines, set 1 on the first, to 25 on the second; and against the other number given 45 on the first, is 1125 on the second. This 1125 is the Product of 45, multiplied by 25, which was desired. Or,  
Set 1 on the first, to 45 on the second, and then right against 25 on the first, is 1125 on the second, as before.  
Example 4. Let 8 75/100 be given to be multiplied by 6 45/100. To perform this work, set 1 on the first, to 6 45/100 on the second, and then against 8 75/100 found on the same side that the 1 is on, is 56 43/100 and something more: therefore 56 43/100 is the Product of 8 75/100, multiplied by 6 45/100, the thing sought.  
Or thus, Set 1 or 10 at the upper end of the line on the first, to 8 75/100 on the second, and then against 6 45/100 on the first, is 56 43/100 on the second, as before. The like practice is to be observed in multiplying any other numbers.  
Note, that in working by the double lines, it will be all one, whether you work from 1 at the beginning of the line upwards, or from 10, at the upper end of the lines downwards. As if you set 10 at the upper end on the first, to 4 on the second, and then against 8 on that first, you shall have 32 on the second.   

How to square any number, or to multiply a number by itself, as also to cube any number.  
Set 1 on the first, to the number to be squared on the second, and then against that given number on the first, is its square on the second. As if 12 be a number to be squared, then set 1 on the first, to 12 on the second, and then against 12 on the same first, is 144 on the second. Therefore 144 is the square of 12. And then again, against 144 on the first, is 1728 on the second, which is the Cube of 12.    

PROBLEM II. Of Division.  
Any number being given, to be divided by another number, to find the Quotient.  
IN Division, the Analogy is thus: 

As the Divisor, is to an Unite:  
So is the Dividend, to the Quotient.    
Wherefore the work by the double lines is thus; Look out the Divisor on the first, and set it to 1 on the second, and then right against the Dividend on the first, is the Quotient on the second.  
Example 1. Let 273 be a number given to be divided by 13. Look 13 the Divisor on any one side, which we call the first side, and set it to 1 on the second side. And then right against the Dividend 273 upon that first, is 21 on the secondd. Wherefore 21 is the Quotient of 273 divided by 13, which was required.  
Example 2. Again, if 1728 be given to be divided by 12, than set 12 the Divisor on the first, to 1 on the second; and when that is done, you shall right against 1728, the Dividend on that first, see 144 on the second, which 144 is the Quotient of 1728, divided by 12, and so of any other.  
In Division note this, that so many times as the Divisor may be placed under the Dividend, so many places of figures shall be in the Quotient. As if 34785, be to be divided by 75, the Quotient shall consist of three figures only, and no more; because 75 can be placed only three times under 34785: And under 1728 the Dividend given, the Divisor 12 can three times be placed, and therefore three figures in the Quotient. And if you will divide 144 by 12, than set 12 on the first, to 1 on the second; which done, against 144 on the first, you have 12 on the second.  
Example 3. Let 46 75/100 be given, to be divided by 8 5/10, and the quotient required. In this case, as in all other: Set the Divisor 8 5/10 on the first, to 1 on the second, and then against the Dividend 46 75/100 on that first, is 5 5/10 on the second: therefore 5 5/10 is the quotient required, very speedily and exactly found.   

PROBLEM III. Of Reduction of Fractions.  
To reduce any vulgar Fraction into a decimal Fraction.  
LEt 63/84 be a vulgar Fraction propounded, to be reduced into a decimal Fraction. Now it is to be noted, that vulgar Fractions are reduced into decimals by the rule of proportion, in this manner: 

As the Denominator of the Fraction given,  
Is to the Numerator thereof:  
So is an Unite with cyphers, one, two, or three  
To a new Numerator, which Numerator will have so many places of figures in it, as the new Denominator hath cyphers, and therefore the Fraction given is thus reduced into a Decimal.    
As the Denominator 84, Is to the Numerator 63.  
So an Unite with two cyphers, thus 100, to 75, which 75 is the new Numerator to that new Denominator 100 Therefore 75/100 is a decimal Fraction, equal in value to 63/84. To work this kind of reduction by the double lines, do thus:  
Set 84 (the Denominator of the vulgar fraction given) on the first, to 1 on the second, & then right against 63, the Numerator on the first, is 75 on the second, which 75 is the new Numerator of a decimal Fraction, whose Denominator is an Unite with two cyphers. Thus is 63/84 changed into 75/100, which new Fraction is equal in value with the former.  
Again, if 8 12/40 be a number given to be reduced into a decimal; set 40, the Denominator of the fractional part, to 1, and then right against 12, on that part as you took the 40, is 2 on that second side whereon the one was, and so the Fraction is become 3/10, which joined to the 8, makes the mixed number that before was 8 22/40, to be 8 3/10, and of equal value.  
Note that the point 1 representeth also 10, 100, or 1000, and therefore the new Denominator may be taken either 10, or 100, or 1000, as in your own judgement it shall seem most fitting. As here in this last Example, we found 3 to stand right against 12, and therefore we put but one cipher to the unite, and made the Fraction 3/10. And in the last Example before, we found 75 to be against the 63, which because it consisteth of two places of figures, we put two cyphers to the unite for Denominator of the new Fraction, and so made it 75/100, and so of all other.  
Also be it remembered, throughout the residue of this work, that all Fraction figures are set right in the line, like whole numbers, without any Denominator under them, being separated or distinguished from the whole part of that number by a Comma, as all such Fractions as these 46 75/100, 6 45/100, 8 5/10, and 5 5/10, are thus expressed 46,75. 6,45. 8,5. and 5,5. which manner of writing Fractions being well observed, is a readier way than that other, of one number above, and another below, with a line between. Also note, that the Fractions so expressed are all Decimal Fractions, whose Denominator is a Unite with so many Ciphers as there be Fraction figures, as the Denominator of the Fraction figures of the 75 belonging to the whole number 46 is 100, being a unite with two cyphers, because two figures and the Denominator to 5 is 10, being a unite with one cipher, because but one figure. And when a Fraction is to be expressed alone without a whole number, than the Numerator is first expressed, and after it the Denominator right on in the line, with a Comma betwixt, as 75/100, and 5/10, are thus expressed 75,100. 5,10. and so of other the like.   

PROBLEM iv Of Continual Proportionals.  
Two numbers being given, to find a third, a fourth, a fifth (or many numbers) in continual proportion Geometrical to them two.  
Example. LEt the two numbers given be 2 and 4, and it be required to find several numbers in continual Geometrical proportion to them two.  
Set 2 on the first, to 4 on the second, and then against 4 on that first, is 8 on the second, which is the third number in continual proportion Geometrical to them two; and then against 8 on the first, is 16 on the second, which 16 is the fourth number in continual proportion to them two; and against 16 on the first, is 32 on the second, the fifth continual proportional; and against 32 on the first, is 64 on the second, the sixth continual proportional; and against 64 on the first, is 128 on the second, the seventh continual proportional; and against 128 on the first, is 256 on the second, which is the eighth proportional to the two proposed numbers. Wherefore 8, 16, 32, 64, 128, and 256, are a rank of numbers in continual Geometrical proportion, to 2 and 4, the thing that was required.  
Example 2. Let it be required to find a rank of numbers in continual proportion, as 2 to 3. Here set 2 on the first to 3 on the second, and then without moving the Instrument, against 3 on the first, you have 4,5 on the second, and against the same 4,5, found on the first, is 6,75 on the second; and against 6,75 on the first, is 10,125 on the second. Therefore 4,5, 6,75, and 10,125, are a rank of numbers in continual proportion to 2 and 3, as is required.  
If it be required, to find such a rank of proportionals to the numbers 2 and 4, which may bear the same proportion to one another, as 2 bear's to 4: Set 4 to 2, and then against 2 on that first side whereon the 4 is, you have 1 on the second side, which is the third proportional to 4 and 2, bearing the same proportion to 2, as 2 doth to 4. And against 1 on the first, is 5,10 on the second, the fourth proportional; and against 5,10 on the first, is 25,100 on the second, which is the fifth number in continual proportion inverse, or backward.  
If the two numbers given be 10 and 9, and a rank of numbers to them in an inverse proportion Geometrical be required; set 10 on the first, to 9 on the second, and then against 9 on the first, is 8,1 on the second, which is the third proportional; and then against 8,1 is 7,29 the fourth proportional: wherefore 10,9 8,1 and 7,29 are numbers in a continual proportion. But if the numbers given be 1 and 9, and a third and fourth numbers in proportion to them, as 9 is to 1, be required, then must the numbers found be accounted 81 and 729, they being the third and fourth numbers in a Geometrical proportion to 1 and 9 In like manner, if the two numbers given be 10 and 12, then if you set 10 to 12, you shall see on the first against 12, 14,4; which is the third proportional: and against 14,4 on the first, is 17,28 on the second, which is the fourth proportional. But had the two numbers given been 1 and 12, then bring 1 on the first, to 12 on the second, and you shall have against 13 on the first, 144 on the second, for the third proportional, and the fourth will be 1728, and so of all other.   

PROBLEM V Of the Rule of Proportion direct.  
Three numbers being given, to find a fourth, the Analogy standeth thus.  


AS the first number, is to the second,  
So is the third number, to a fourth.    
Therefore work thus: Set the first number in the proportion on the first side, to the second number in the proportion on the second side: And then against the third number on the first, is the fourth number sought for on the second.  
Example 1. Let the Diameter of a known circle be 7, and its circumference 22, and it be required to know what the circumference of another circle is, whose Diameter is 14. To resolve this quaere: Set 7 on the first to 22 on the second, and then against 14, the Diameter of the other circle found on the first, is 44 on the second: This 44 is the circumference of that other circle, whose Diameter is 14.  
Example 2. If 45 yards of Stuff cost 30 pound, what will 84 yards of it cost?  
Set 45 (the first number in the Rule) on the first, to 30 (the second number) on the second; and then against 84 (the third number in the Rule) on the first, is 56 on the second, which 56 is the fourth number, and showeth that 84 yards will cost 56 pounds. Or,  
If 45 acres of land be worth 30 pounds a year, what will 84 acres be worth by the year? The answer is as before 56 pounds.  
And if 26 of any thing give 64, what will 36 of the same give? Set 26 to 64, and then against 36 on the first, is 88,615 the answer to the question demanded on the second.  
Note, that generally in the Rule of direct proportion; If the third number be greater than the first, then will the fourth number be greater than the second. But if the third number be less than the first, than the fourth number will be less than the second.  
Example. 3. If the circumference of a circle be 22, and its Diameter 7, what will the Diameter of another circle, whose circumference is 44?  
Here set 22 on the first, to 7 on the second; and then against 44 on the first, is 14 on the second, which 14 is the Diameter of that circle, whose circumference is 44, be their measures taken in inches, feet, or any other measure whatsoever.  
To make proof of the work, whether truly wrought, or not. Multiply the first term in the Rule, and the fourth term newly found, the one by the other, and likewise the second and third terms; and if the two Products be equal, the work is truly wrought, or else not.  
To prove the last question by the lines; Set 1 on the first, to 22 on the second, and then against 14 on the first, is 308 on the second: Next, set 1 to 7, and then against 44 on that first, is 308 on the second, here both Products being equal, proves the work to be truly wrought.   

PROBLEM VI The Rule of Proportion Inverse.  
Three numbers being given, to find a fourth in an Inversed Proportion.  
IT is to be noted, in this Inverse Rule of Proportion, that if the third number be greater than the first, then will the fourth number be less than the second. And contrariwise, if the third number be less than the first, than the fourth number is to be greater than the second.  
But in the Rule of Proportion direct; If the second number, or term, be more than the first, than the fourth term is also more than the third. And if the second term be less than the first, then is the fourth term less than the third.  
This Inverse Rule may be wrought two ways on our double lines. One way is thus:  
Set the first term, on the first, to the other term of the same denomination on the second: And then against the other term of contrary denomination, sought out on that second, is the fourth number sought for, on the first; Or else, set the third term on the first, to the first term on the second, and then against the second term on the first, is the answer on the second.  
Example 1. If 60 Pioners can make a trench in 45 hours; In how long time can 40 pioneers make it?  
Set 40 on the first, to 60 on the second (the two numbers of the same denomination:) and then against 45 on that first, (the number of contrary denomination) is 67,5 on the second, which 67,5 is the fourth term in reciprocal proportion to the other three, and of the same denomination with 45, viz. hours, and is the answer to the question demanded, showing that 40 men can do as much in 67,5 hours, as 60 men can do in 45 hours. Or set 60 on the first, to 40 on the second, and then against 45 on that second, is 67,5 on the first.  
Another way is thus: Set 40 the third term, on the first, to 45 on the second, (which is the term of contrary denomination to the other two) and then right against 60, on the first, (which is the number of the same denomination with the 40) is 67,5 on the second, the number sought. Thus one way of working proves the other.  
Example 2. If 45 men do a work in 30 days: In how many days will 270 men do it?  
Set 270 on the first, to 45 on the second, and then against 30 on the first, is 5 on the second. Or,  
Set 270 on the first, to 30 on the second, and then against 45 on the first, is 5 on the second: And therefore the fourth number sought for is 5, showing that 270 men will do as much work in 5 days, as 45 men can do in 30 days.  
This Rule is proved by multiplying together the first and second terms, and also the third and fourth: And if the two Products be equal, the work is truly wrought, or else not.  
By the lines it is thus proved: Set 1 on the first, to 30 on the second, and then against 45 on the first, is 1350 on the second. This done: Set 1 on the first to 5 on the second, and then against 270 on the first, is 1650 on the second; here both Products being equal, declareth the work to be truly wrought.   

PROBLEM VII. Of Duplicated Proportion.  
Three numbers being given, to find the fourth in a Duplicate Proportion.  
THis Rule chief concerns the proportion of Lines to Superficies, or of Superficies to Lines.  

1 Of the Proportion of Lines to Superficies.  
Example 1. If the diameter of a circle be 14 inches, and its Content 154 inches; What will the Content be of another circle, that is 28 inches in diameter?  
Set 14 on the first, to 28 on the second, (they being the terms of one denomination, viz. Lines) and then against 154 on the first, (the content of the circle given) is 308 on the second: This 308 seek on the first, and against it on the second is 616, which 616 is the content of that other circle of 28 inches diameter.  
Example 2. Let the diameter of one circle be 7 foot, and the Area of it 38,5 foot, and let it be demanded, What the superficial Area of another circle is, whose diameter is 18 foot.  
Because 7 and 18 be terms of one denomination, viz. Lines; Set 7, (the diameter of the the circle, whose Content is known) on the first, to 18 on the second, being the diameter of the other circle, whose Content is sought, and then against 38,5 (the Content known) on the first, is 99 on the second, and then seek this 99 on the first, and against it is 254,5 tenths on the second, which is the superficial Area, or Content in feet of that other circle which was demanded.  
Example 3. If a piece of land that is 20 pole square be worth 30 pounds; What is a piece of land of the same goodness worth, that is 35 pole square?  
Set 20 on the first, to 35 on the second, and then against 30 on the first, is 52,5 on the second; and lastly, against 52,5 on the first, is 91,8 on the second: that is 91 pounds, and eight tenths of a pound, or 16 shillings. So much is the worth of that piece of land of 35 pole square.  
Example 4. How many acres of land of our English measure of 16,5 foot to the pole, are contained in 30 Irish acres, of 21 foot to the pole.  
Place 16,5 on the first to 21 on the second, and then against 30 on the first, is 38,2 on the second; and against 38,2 on the first, is 48,6 on the second. So many English acres are in 30 Irish acres.   

2 Of the Proportion of Superficies to Lines.  
If the two terms of like denomination be of superficial Contents, and a diameter, or a line sought for.  
Example. Let two circles be given, the Content of the one being 154, and its diameter 14: the Area of the other circle is 616, and its diameter is required.  
Set the Area of the circle known, viz. 154 on the first, to its diameter 14 on the second, and then against 616 on the first, is 56 on the second. The half whereof, viz. 28, is the diameter of that other circle, whose Content is 616, which 28 is feet, or inches, or any other measure, such as the diameter of the other circle was measured by.    

PROBLEM VIII. Of Triplicate Proportion.  
Three numbers being given, to find a fourth in a Triplicated Proportion.  
THis Rule concerneth the proportion betwixt Lines and Solids. Example 1. There is a Bullet whose diameter is 4 inches, weigheth 9 pounds: What will another Bullet weigh, whose diameter is 8 inches, and of the same metal.  
Set 4 on the first, to 8 on the second, (rhat is the one diameter to the other:) And then against 9 on the first, (which is the weight of the Bullet of 4 inches diameter) is 18 on the second, and against 18 on the first, is 36 on the second; and thirdly, against 36 on the first, is 72 on the second. This third sum found, is the fourth proportional number, which was required, showing that the weight of that other Bullet of 8 inches diameter, is 72 pounds.  
Example 2. If a Gun of 5 inches diameter, require for her due charge 16 pound of powder; How much powder will a Gun of 4 inches diameter in the bore require, for her due charge, of the same powder?  
Place 5 on the first, to 4 on the second, and then against 16 on that first, is 12,8 on the second; and next against 12,8 on the first, is 10,24 on the second; and thirdly, against that 10,24 on the first, is 8,2 on the second, which third number 8,2 is the answer to the question: showing that 8 pounds, and 2 tenth parts of a pound of powder, is a due charge for a Gun of 4 inches bore.   

PROBLEM IX. A Company of men laying down several sums of money together into one stock, wherewith they trade and get gain; to find out how much each man's part of the gain must be, answerable to his part of money laid down in stock.  
LEt five men, whose names let be represented by these five letters, A, B, C, D, E, make a stock of 300 pounds, of which stock A put in 84 pounds, B put in 72 pounds, C put in 48 pounds, D put in 54 pounds, and E put in 42 pounds, which all together make the 300 pounds. Now at the end of a time, having traded therewith, they gained clearly 50 pounds: And let it be demanded, what portion of the gain each man must have, according to his proportion of money laid down in stock.  
The Rule to answer this demand is thus: 

As 300, the whole stock,  
To 50 pounds the whole gain;  
So is each man's portion of the stock,  
To his portion of the gain.   Therefore,  
Set 300, the stock on the first, to 50 the gain on the second; and then against each man's particular portion laid down, being sought out on the first, is his portion of the gain on the second. As against 84, the portion laid down by A on the first, is 14, his portion of the gain on the second; and against 72, the portion of B, is 12 pounds, his portion of the gain; against 48, the portion of C, is 8 pounds, his portion of the gain; against 54, the portion of D, is 9, his portion of the gain; and against 42 pounds on the first, the portion of E, is 7 pounds on the second, his portion of the 50 pounds' gain. Thus much is each man's several portion of the 50 pounds' gain; all which several portions of the gain added together, make up the whole gain of 50 pounds.  
Thus upon our lines can we work that rule of Arithmetic, called The Rule of Fellowship; which is, when divers men adventure a stock of money together, and therewith trade, and either gain or lose a certain sum of money, to find each man's portion of the gain or loss, answerable to his portion of money put into the stock.   

PROBLEM X. Of Interest and Annuities.  

To find the Interest of any sum of money, after any rate by the 100 propounded.  
WHat is the Interest of 65 pounds for a year, after the rate of 8 in the 100?  
By the rule of Proportion, the question is thus resolved: 

As 100, is to 108;  
So is 65, to the fourth term.    
Set 1 or 100 on the first, to 108 on the second, and then against 65 on the first, is 70,2 on the second, which is 70 pounds 4 shillings; so much doth the principal and interest arise unto in a year, that is to say, five pounds and four shillings: And without stirring the Instrument against any other sum of Principal on the first, is the Principal and Interest thereof on the second. As against 40 pounds, is 43 pounds 4 shillings; and against 80 pounds, is 86 pounds and 8 shillings. From whence it appeareth, that 3 pounds 4 shillings is the Interest of 40 pounds for a year; and 6 pounds 8 shillings, the Interest of 80 pounds for a year. In like manner, against 27 pounds 14 shillings Principal, or 27,7 is 29 pounds 18 shillings, and a little more. Or else work thus:  
Set 1 or 100 back to 8, and then against 65 on that first, is 5,2 on the second; and against 40 pounds on the first, is 3,2 on the second: So the Interest of the one is 5 pounds 4 shillings, and of the other 3 pounds 4 shillings.  
If the rate of the Interest proposed be 6 in the 100, than set 100 on the first, to 106 on the second; and then against 65 pounds on the first, is 68 pounds 18 shillings on the second, which is the Principal and Interest together of 65 pounds for a year. Or,  
Set 100 on the first, to 6 on the second, and then against 65 on the first, is 3,9 on the second; this 3,9 is 3 pounds 18 shillings, the Interest alone of 65 pounds for a year, and so of any other.   

Of Interest of money continued from year to year.  
The increase or Interest of money from year to year, for many years, is in continual proportion to the Principal, as 100 is to its Interest; as if 40 pounds were to be continued at Interest for many years, at the rate of 6 in the 100  
Set 100 on the first, to 106 on the second, and then against 40 on the first, is 42,4 on the second, that is 42 pounds 8 shillings, so much is the first years Principal and Interest. And now, if you look 42,4 on the first, you shall have right against it 44,9 and more, that is 44 pounds 18 shillings and more, for the Principal and Interest of two years; and against 44,9 on the first, is 47,6 and better, on the second, that is 47 pounds 12 shillings 8 pence, the Interest and Principal of 40 pounds in three years. Again, against 47,6 on the first, is 50,45 and better, that is 50 pounds 9 shillings 7 pence: So much is the Principal and Interest together of 40 pounds at the end of 4 years, and so forth to as many years as is required.   

Of Annuities.  
When lands are sold at certain years purchase, according to the yearly rent, to find what their value upon the purchase will be: Set 1 on the first, to the number of years purchase on the second; and then against their yearly rent on the first, is the value of the purchase on the second.  
Example. Let a house and land worth 16 pounds a year, be set to sale at 14 years' purchase, and demand made, how much money it will arise unto at that rate.  
Set 1 on the first, to 14, the number of years purchase on the second; and then against 16; the yearly rent on the first, is 224 on the second, which is 224 pounds: So much money doth the purchase arise unto, of 16 pounds a year bought at 14 years' purchase.  
If the price of the lands be given, and that it cost after 14 years' purchase, to find what yearly rend it was sold at. In this case, Set 14 on the first, to 1 on the second; and then against the sum of money paid 224 pound, is 16, which is 16 pounds: So much is the yearly rent sought.  
A man borrowed 666 pounds 13 shillings and 4 pence for 2 years, and covenanted that he would repay at the 12 years' end 1333 pounds 6 shillings and 8 pence. It is desired to know after what rate of Interest by the 100, he paid for his money borrowed.  
Set 1333● on the first, (the sum of money to be paid, to 666● on the second (the sum lent) and then against 12 on the second (the term of years it was lent) is 5,95, which is 5 pounds 19 shillings: So much by the 100 doth he pay for the money lent.   

A Sum of money being due at a certain time to come, to find what it is worth in present money, to take in.  
There is 402 pounds 2 shillings due at the end of 5 years to come, I would know what it is worth in ready money, abating Interest for the money received in, before due, after the rate of 8 in the 100  
First, set 108 on the first, to 100 on the second, and then against 402,1 on the first, is 37,1 on the second: Secondly, against 37,1 on the first, is 34,46 on the second: Thirdly, against 34,46 on the first, is 31,95 on the second: Fourthly, against 31,95 on the first, is 29,71 on the second: Fifthly, against 29,71 on the first, is 373 on the second. So much, namely, 373 pounds may be received in present money for the 402 pounds and 2 shillings due five years hence, as being the present worth thereof.  
Here take notice, that as the Principal and Interest of money forborn for many years, increaseth in a proportion direct; So in this case, where money is paid many years before due, it decreaseth in the like proportion.     

CHAP. III. The Use of the double Scale of Numbers in Superficial measure, as Board, Glass, Land, and the like.  

PROBLEM I. The length and breadth of any square, or long square Superficies being given, to find the Content thereof.  
IF the length and breadth be given in inches, Then, 

As 1 to the breadth in inches,  
So is the length in inches,  
To the Content in inches.    
Example. Let a plain Superficies, as a Board or Plank be given to be measured, the breadth is found 30 inches, and its length 183, and the Content required.  
Set 1 on the first, to 30 on the second, and then against 183 on the first, is 5490, the Content sought in inches.  
If the Superficies given, be a piece of land, 30 perches broad, and 183 long, the Content is 5490 perches.  
Again, let a piece of Wainscot be 2,5 foot in breadth, and 15,25 foot in length, the Content will be found 38,12 foot. For,  
Set 1 on the first, to 2,5 on the second, and then against 15,25 on the first, is 38,12 foot, the Content sought.   

PROBLEM II. The breadth and length of any Superficies being given in one kind of measure, to find the Content in another kind of measure.  
LEt the length and breadth be given in inches, and the Content required in feet. The Rule is thus: 

As 144 to the breadth in inches,  
So is the length in inches,  
To the Content in feet.    
Example. Let the breadth be 30 inches, and the length 183 inches, and the Content in feet required.  
In this case, because 144 inches make a foot of superficial measure, Set 144 on the first, to 30 the breadth on the second, and then against 183, the length on that first, is 38 foot, and a Fraction of a foot, being a little more than one tenth part of a foot: So many foot are in that Board, or what other platform it be, that is given to be so measured.  
If the platform were a piece of land 30 perches broad, and 183 perches long, than the Analogy is thus: As 160, (the perches making an acre) to the breadth in perches; So is the length in perches, To the Content in acres.  
And therefore in this case of land measure, Set 160 on the first, to 30 the breadth on the second; and then against 183, the length on that first, is 34,31 on the second: So many acres of land are contained in that ground.  
If the place, whose Content is to be cast up, be a Triangle, a Trapezia, or of any other form whatsoever, the Analogy in general is this.  
As 144, or 160, etc. Is to one of the numbers given to be multiplied together; So is the other of them, to the Content in Feet or Acres, etc.  
There is a piece of Wainscot, that is 3,5 foot broad, and 21 foot long: How many yards is in it?  
Seeing that in a yard are contained 9 foot; Therefore, Set 9 on the first, to 3,5 on the second; and then against 21 on the first, is 8,16 on the second: So many yards is in that piece.   

PROBLEM III. The breadth of a Superficies being given in one kind of measure, and the length in another, to find the Content in the greater measure.  
LEt the breadth of a Superficies given be in inches, and the length in feet, and the Content in feet required:  
Here the Analogy is: 

As 12 to the breadth in inches;  
So the length in feet, to the Content in feet:    
So that if the breadth be 30 inches, and the length 15,25 foot, the Content will be 38,12 foot: For,  
Set 12 on the first, to 30 on the second; and then against 15,25 on the first, is 38,12 on the second, being the Content sought for: Or else,  
Set 12 to 15,25, and then against 30 on that first, is 38,12 on the second.  
By this rule also: If the breadth of a plot of land be given in perches, and the length in chains (being measured by a chain of 4 perches long) the Content in Acres is readily had.  
Example. Let a piece of land be in breadth 30 perches, and in length 15,25 chains, measured by a chain of 4 perches in length.  
In this case the Analogy is thus: 

As 4 is to the breadth in poles,  
So is the length in chains to the Content in Acres.   Therefore, Set 4 to 30, and then against 15,25 on that first side whereon the 4 is, you shall have 11,4 on the second side: So much is the Content in Acres of that piece of land.  
Again, let a piece of land be 36 poles broad, and the length 23 chains and an half; to find the Content.  
Set 4 on the first, to 36 on the second, and then against 23,5 on that first, is 21,1 and better on the second: the Content sought.   

PROBLEM IU. The length and breadth of a Superficies being given in feet, to find the Content in yards.  
TAke this for a general Rule: As 9 is to the breadth in feet; So is the length in feet, To the Content in yards.  
Example. Let the breadth of a pane of Wainscot be 4 foot, and the length 12 foot, and the Content in yards be sought for.  
Here is no more to be done, but to set 9 on the first, to 4 on the second; and then against 12 on the first, is 5,35 on the second: The Content in yards of that pane. which is almost 5 yards, a quarter and an half, or rather 5 yards and one third part of a yard.   

PROBLEM V The breadth of any Superficies being given in inches or feet, to find how much in length will make a superficial foot.  
The Rule is thus. 

AS the breadth in inches, to 144;  
So is 1 to the length in inches, to make a foot.    
Example. Let the breadth given be 30 inches, and the length to make a foot at that breadth be required.  
Set 30 on the first, to 144 on the second, and then against 1 on the first, is 4,8 on the second: So much in length makes a foot at 30 inches broad.  
But if the breadth be given in feet; Then, 

As the breadth in feet is to 1,  
So is 1 to the length to make a foot:   Therefore,  
Set 2,5 on the first, to 1 on the second, and then against 1 on the first side, is 4 on the second, which 4 signifieth four tenth parts of a foot: So much in length makes a foot at that breadth.   

PROBLEM VI The length and breadth of a plot of land being given in chains, to find the Content in Acres.  
HAving a chain of 4 perches long, divided into 100 links, that is, 25 in the perch; measure the length and breadth of the land to be measured in chains and links: And then cast up the Content into Acres thus: 

As 10 to the breadth in chains;  
So is the length in chains, to the Content in Acres.    
Example. Let the breadth given be 7 chains, 50 links, and the length 45 chains, 75 links, and the Content in Acres sought.  
Set 10 on the first, to 7,5 on the second, and then against 45,75 on the first, is 34,31 on the second, which is the Content in Acres of that plat of land. Or,  
Had the breadth been 15 chains, 25 links, and the breadth 22 chains, 50 links; then set 10 or 1 on the first, to 15,5 on the second, and then against 22,50 on the first, is 34,31 on the second, the Content in Acres sought for: which Fraction 31 above the 34 Acres, contains 1 rood 10 perches. So that the true Content is 34 Acres, 1 rood and 10 pole.  
If the plot of land be of a Triangle form, or any other figure whatsoever, the Analogy is this: 

As 10 is to one of the two numbers of chains, that are to be multiplied together:  
So is the other of them, to the Content in Acres.    
As if the half perpendicular of a Triangle be 3,75 chains, and the Base 45,75, the Content will be found to be 17,15 Acres.  
For set 10 on the first, to 3,75 on the second, and then against 45; 75 on that first, is 17,15, the content.  
Or else, having the whole Base, and whole perpendicular, say thus: As 20 to the whole perpendicular 7,50; So is the whole Base 45,75; To 17,15 the content in Acres, as before: For,  
Set 20 on the first, to 7,50 on the second, and then against 45,75 on the first, is 17,15 on the second.  
Or if the Base of a Triangle be 36,83 chains, and the perpendicular 17,59 chains, the content will be found by either way of working, to be 32,39, which is 32 acres, 1 rood 22 perches.   

PROBLEM VII. The Content of a piece of land being measured by one kind of perch, to find the Content thereof, after another kind of perch.  
THese kind of proportions are wrought by the Rule of Three reverse, after a duplicated proportion; and the Analogy is thus: 

As the length of the second perch, is to the length of the first perch:  
So the content in acres given, to a fourth number: and so is that fourth number to a fifth number, which is the content sought.    
Suppose a piece of land measured by the 16½ foot-pole, do contain 34,3 acres, and it be demanded; How much it would contain, if it were measured by an 18 foot-pole.  
Note that I call the 16½ foot-perch, the first perch, because by it the land was measured; and the 18 foot-perch I call the second perch, because according to it the content is sought for. Wherefore,  
Set the second perch 18 on the first, to 16,5 on the second, (the perch first used) and then against 34,3, the content in acres given, on the first, is 31,45 on the second; and then against that 31,45 on that first, is 28,8 on the second: So much is the content in acres by the 18 foot-pole, which was demanded.  
In like manner, were the content given 5 acres, 2 roods, 20 pole, or 5,62: Set 18 on the first, to 16,5 on the second, and then against 5,62 on the first, is 5,15 on the second. And lastly, against that 5,15 on that first, is near about 4,7 on the second: So much is the content of that close by the 18 foot-pole.   

PROBLEM. VIII. The one side of any piece of land being given, to find how much in breadth the other way will make an acre of land.  
LEt the side of a close be 20 pole, and it be required; How much in breadth will make an acre of land at that length.  
Set the breadth given, 20 on the first, to 160 on the second; and then against 1 or 10 on that first, is 8 on the second. So much in breadth makes an acre at 20 pole long.  
Or if the side measured be 25, than set 25 to 160, and against 1 on that first, is 6,4 on the second, the breadth sought for. And if the length be 32 pole, the breadth to make an acre will be 5 pole.   

PROBLEM. IX. A plot of land being laid down, and cast up by any Scale, to find how much it will contain by any other Scale, either greater or lesser.  
SUppose a plot of land being laid down, and cast up by a Scale of 10, in the inch, does contain 28,5 acres; and it is required to know how many acres it will contain, should it be cast up by a Scale of 12 in the inch.  
Here because 12, the Scale to be used, is lesser than 10, the Scale formerly used: And so by consequence, the content of the given plot by the Scale of 12, will be more acres than it is by the Scale of 10 in the inch. Therefore,  
Set the Scale used 10 on the first, to the Scale to be used 12 on the second, and then against the content known 28,5 on the first, is 34,2 acres, or near thereabouts on the second: So much is the content by the Scale of 12 in the inch.  
But if the plot had been laid down, and cast up by a Scale of 12, and the content required by a Scale of 10 in the inch, which is the greater Scale, and therefore the content is the lesser. Then,  
Set 12 on the first, to 10 on the second, which done, right against the content in acres by the Scale of 12 on the first, is the content by the Scale of 10 on the second: As if the content by the Scale of 12 be 34,2 acres, than the content by the Scale of 10, will be found to be 28,5 acres near. For,  
Set 12 on the first, to 10 on the second, and then against 34,2 on the first, is 28,5 on the second.   

PROBLEM. X. The Diameter of a circle being given, to find the Circumference.  
The Analogy stands thus: 

AS 1 is to the Diameter;  
So is 3,142 to the Circumference.   Or,  
As 7 to 22; So is the Diameter to the Circumference. If the Diameter be 15 inches, what is the Circumference? Set 1 on the first, to 15 on the second, and then right against 3,142 on the first, is 47,13 on the second: So much is the Circumference of that circle. Or,  
Set 7 on the first, to 22 on the second, and then against 15 on the first, is 47,13 on the second, as before.   

PROBLEM XI. The Circumference of a circle being given, to find the Diameter.  


AS 22 is to 7,  
So is the Circumference to the Diameter.   Or, 

As 3,142 to 1,  
So is the Circumference to the Diameter.   So if the Circumference be 47,13, what is the Diameter?  
Set 22 on the first, to 7 on the second, and then against 47,13 on the first, is 15 on the second: So much is the Diameter of that circle. Or,  
Set 3,142 to 1, and then against 47,13 is 15, as before.   

PROBLEM XII. The Diameter of a circle being given, to find the side of a Square equal to it.  
THe Diameter of a circle is 15 inches, what is the side of the square equal to it in Content?  
Set 1 on the first, to 15 on the second, and then right against this number 8862 on the first, is 13,29 on the second: So much Is the side of a square, that is equal in Content to that circle.   

PROBLEM XIII. The Circumference of a circle being given, to find the side of a square, equal in Content to that circle.  
LEt the Circumference of a circle be 47,13, and the side of a square equal to it be required.  
Set 1 on the first, to 47,13 on the second, and then always against this number 2821 on the first, is 13,29 on the second: So many inches is the side of a square, that is equal in Content to the circle given.   

PROBLEM. XIV. The Diameter of a circle being given, to find the side of a square, that may be inscribed within it.  
LEt the Diameter of a circle be 15 inches, and it be required, what the side of that square will be, that may justly be inscribed within it?  
Set 1 on the first, to that 15 on the second, & then against this number 7071 on the first, is near about 10,6 on the second: Therefore 10,6 inches, is the side of a square that may be inscribed within a circle of 15 inches Diameter. Or thus it may be found:  
Double the square of the Semidiameter, the Root square of that Product is the side of the square inscribed; but this is by the By.   

PROBLEM. XV. The Circumference of a circle being given, to find the side of a square that may be inscribed within it.  
LEt the Circumference given be 47,13, and the side of a square that may be inscribed within it, be required.  
Set 1 on the first, to the Circumference 47,13 on the second; and then against this general number 2251 on the first, is near about 10,6 on the second: So many inches is the side of that square, that can be inscribed within the circle given.   

PROBLEM. XVI. The Diameter or Circumference of a circle, either of them being given, to find the side of an Equilater triangle, to be inscribed within that circle.  
FIrst, having the Diameter given, which suppose 14 inches and it be required, to find the side of the Equilater Triangle that may be inscribed in that circle.  
Set 1 on the first, to 14 on the second, (that is, to the Diameter given) and then against this general number 8,65 on the first, is 12,1 on the second: Therefore conclude, that 12 inches and one tenth part of an inch, is the length of the side of that Equilater Triangle, which may be inscribed within a circle of 14 inches Diameter.  
Or you may set 1 to the general number 865, and then against 14 is 12,1, as before.  
Having the Circumference given, which let it be supposed to be 44, then to find the side of that Equilater Triangle, the work is thus.  
Set 1 on the first, to the Circumference given 44 on the second, and then against this general number 272,6 on the first, is 12,1 on the second, the side of the Equilater Triangle sought for.  
Or, Set 1 to that general number 272,6, and then against 44, is 12,1, and so of any other.   

PROBLEM. XVII. The Diameter or Circumference of any circle being given, to find the superficial Content.  
LEt the Diameter of a circle proposed be 15 inches, and the superficial Content of it demanded.  
Set 1 on the first, to 15 on the second, & then then against this general number 7854 on the first, is 11,83 on the second; and then again against that 11,83 on the first, is 176,74 on the second: This 176,74 is the Content of that circle proposed.  
But if the Circumference of any circle only be given, as if it were given 47,13, and the Content required. Then,  
Set 1 on the first, to the Circumference given, 47,13 on the second; and then against this general number 7958 on the first, is near 37,8 on the second: And lastly, against this 37,8 on the first, is 176,74 on the second, the Content required.   

PROBLEM. XVIII. The Diameter, with the superficial Content of any circle given, to find the Content of any other circle that is twice the Diameter of the first.  
LEt the Diameter of a circle given be 7 inches, and the Content 38,5 inches, and the demand be, to know what another circle is, that is double in Diameter to the former.  
Set 7 on the first, to 14 on the second, and then against the known Content 38,5 on the first, is 77 on the second; and then again, against that 77 found on the first, is 154 on the second: So much is the Content of that other circle, whose Diameter is double to the Diameter of the circle given, that is to say, of 14 inches Diameter.   

PROBLEM. XIX. The Content of a circle being known, to find the Diameter and Circumference.  
LEt the Content of a circle be known to be 176,74 inches, and it be required to know what the Diameter of that circle is, as also what the Circumference is.  
For the Diameter, Set 1 on the first, to 1,273 on the second, and then against the Content known, 176,74 on the first, is 225 on the second; the square root of this 225 is 15: So many inches is the Diameter of that circle.  
For the Circumference, Set 1 on the first, to this general number 12,57, and then against 176,74 on the first, is near 2221, whereof the square root is 47,13: So many inches is the Circumference of that circle.  
The Root square of the Content of any circle, is the side of a square equal to it.    

CHAP. IU. The Use of the double scales in solid measure, such as Timber, Stone, etc.  

PROBLEM. I. The side of a square solid being given in inches, or feet, to find how much in length will make a foot solid in inches or feet.  
SVppose the side of a squre log be 25,45 inches, and demand is made, How many in length will make a foot.  
Set the breadth given, 25,45 on the first, to this general number 41,57 on the second; and then against 1 on the first, is 1,63 on the second, and against this 1,63 found on the first, is 2,67 on the second: So much in length makes a foot of Timber, viz. 2 inches, and 67 parts of 100  
If the breadth be given in parts of a foot (the foot being divided into 100 or 1000 equal parts) to find what parts of a foot in length will make a foot of Timber. As if the breadth of a piece of Timber be 2 foot and 120 parts of 1000 of a foot; then to find how much in length will make a foot solid.  
Set 2,120 on the first, to 1 on the second, and then against 1 on the first, is 471 on the second; and lastly, against that 471 on the first, is 222: So many parts of a foot (it being divided into 1000) do make a foot of Timber at that breadth.   

PROBLEM. II. The breadth and depth of an unequal square solid being given in parts of a foot, or in inches, to find how much in length will make a foot.  
LEt the breadth of a piece of Timber be 2,5 foot, and the depth 1,8 foot. Here first, by Multiplication find out the Content of the Base or end, which you shall find to be 4,5, which being had, get the length of a foot solid thus.  
Set the Content of the head found, 4,5 on the first, to 1 on the second; and then against 1 on the first, is 222 on the second: So many parts of a foot divided into 1000, do make a foot of Timber.  
But if the breadth be given 30 inches, and the depth 21,6, than the Content of the Base will be 648, and the work thus wrought.  
Set 648 on the first, to 1728 on the second, and then against 1 on the first, is 11,27 on the second; and lastly, look this 11,27 on the first, and right against it on the second, is 2,67 inches, the length of a foot of Timber in that log, whose breadth is 30 inches, and depth 21,6. Or you may work thus:  
Set 12 on the first, to 21,6 on the second, and then against 30 on the first, is 54 on the second; This done, Set the new found number 54 on the first, to 144 on the second; and then against 1 on the first, is 2,67 on the second, the length to make a foot, as before.  
Such an unequal squared piece of timber, as is here mentioned, may be reduced into a perfect square, and wrought as in the first Problem. And so the perfect square of this unequal squared log, whose breadth is 30 inches, and depth 21,6, will be found to be 2,120 foot. Or 25,40 inches, and now you may proceed herewith, as in the first Problem you do with square Timber.   

PROBLEM III. The breadth of a square solid, or the side of a square equal to the Base of any unequal square solid, and the length of the same solid being given in inches or feet, to find the Content in feet.  
FIrst, suppose the breadth and length be given in inches, as let the breadth be 25 inches, and 45 parts of an inch, and the length 183 inches, and the Content demanded.  
Set this general number 41,57 on the first, to the side of the square given 25,45 on the second, and then against the length in inches 183 on the first, is 112 on the second: And again, against that 112 on the first, is 68,62 on the second: Therefore the Content of the piece of Timber proposed is 68 foot, and 62 parts of a foot, into 100 divided.  
But if the breadth be given in feet, and parts of a foot; As let the breadth of the solid be 2 foot, and 12 parts of a foot, such as the whole foot is divided into 100, and the length 15 foot, and 25 parts. Then,  
Set 1 on the first, to 2,12 on the second, that is to the breadth given; and then against 15,25 on the first, is 32,35 on the second: And lastly, against that 32,35 on the first, is 68,62 on the second: So much is the Content in feet of that solid.  
If an unequal squared solid be to be measured, it is as good a way as any, to reduce it to a perfect square, and then compute the Content as here before, and this reduction is easily made; for multiply the breadth by the thickness; and the Product is the Content of the Base, whereof the square root is the side of a perfect equal square; to that unequal square given.   

PROBLEM IU. The side of a squared solid given in inches, and the length in feet, to find the Content in feet.  
LEt the side of a perfect square solid be 25 inches, and 45 parts of an inch, such as the whole inch is divided into 100; and the length 15 foot, and 25 parts of a foot, Such as the whole foot is divided into 100, and the Content in feet required.  
Set always 12 on the first, to the breadth in inches 25,45 on the second; and then against 15,25 the length in feet, is 32,35 on the second: This 32,35 found on the first, and right against it on the second is 68,62: So many foot of Timber, and parts of a foot, are in that log.  
If an unequal square solid be to be measured, reduce the Base of that unequal square solid into a perfect square, and then find the Content, as here is taught.   

PROBLEM V The length, breadth and depth of a square solid being given in inches, to find the Content in feet.  
BY the breadth and depth, get the Content of the Base; as let the length of a piece of Timber be 183 inches, the breadth 20 inches, and the depth 13 inches, and the Content in feet desired: Here 20 and 13 multiplied together, make 260 for the Content of the Base.  
Now, Set 1728 on the first, to 260, the Content of the Base on the second; and then against 138, the length on the first, is 27, ● near on the second: Wherefore you may conclude, that in the piece of Timber proposed, is contained 27 foot and an half, with a little more, which is inconsiderable. Or else you may work it thus:  
Set 12 on the first, to 13 the depth on the second, and then against 20 the breadth on the first, is a fourth number, viz. 21,70 on the second. This done, Set 144 on the first, to this fourth number new found 21,70 on the second, and then against 183, the length of the piece on the first, is 27,5 on the second: So that the Content of that piece of Timber, is 27 foot and an half, as before.  
Or you might have set 12 on the first, to 20 on the second, and then against 13 on the first, you should have had 21,70 on the second, for the fourth number, as afore; to which set 144 on the first, and then against 183, is 27 foot and an half.   

PROBLEM VI The Base of any square solid being given in inches, and the length in feet, to find the Content in feet.  
LEt the Base of a piece of Timber contain 260 inches, and the length 15 foot and a quarter, and the Content thereof in feet desired.  
Set 144 on the first, to the Content of the Base 260 on the second; and then against 15,25 on the first, is 27,5 on the second. So then 27 foot and an half is the Content desired.  
If both the length and content of the Base be given in feet; then, Set 1 on the first, to the Content of the Base in feet on the second, and then against the length in feet on the first, is the Content in feet on the second: As if the length of a piece of Timber be 15 foot and a quarter, and the Content of the Base 4 foot and an half.  
Set 1 on the first, to 4 and an half on the second, and then against 15 and a quarter on the first, is 68,62 on the second; that is 68 foot and an half, and almost half a quarter: So much is the Content of that piece.   

PROBLEM VII. The Diameter of a Cylender given in inches or feet, to find the length of a foot, according to that Diameter.  
LEt the Diameter of a Cylender, or round piece of Timber or Stone be 15 inches, and it be demanded, How much in length makes a foot solid?  
Set the Diameter 15 on the first, to this general number 46,90 on the second, and then against 1 on the first, is 3,13 on the second; and then against that 3,13 found on the first, is 9,78 on the second: So much in length makes a foot of Timber, that is 9 inches, and 78 parts of 100, of an inch.  
But if the Diameter be taken in feet, and parts of feet, than set the Diameter 1,25 on the first, to this general number 1128 on the second, and then against 1 on the first, is 9027 on the second, and against this 9027 found on the first, you have 8,15 on the second, that is 8 parts of a foot, such as the whole foot is divided into 10 parts, and 15 parts of 100 of one tenth over: So much in length makes a foot.   

PROBLEM VIII. The circumference of a Cylinder given in inches, or tenth parts of feet, to find the length to make a solid foot.  
LEt the circumference of a round log be 47 inches, and 13 hundred parts of an inch, and the length to make a foot, desired.  
Set the circumference 47,13 on the first, to 147,36 on the second, and then against 1 on the first, is 31,2 on the second, and against that 31,2 on the first, is 9,78 on the second: So many inches and parts in length make a foot solid.  
But if the circumference be given in feet, and parts of feet, as let the forementioned Cylinder be measured by feet, and parts of feet, and be found to contain 3 foot and 93 parts of a foot. Then,  
Set the circumference 3,93 on the first, to this number 3545 on the second, and then against 1 on the first, is 903 on the second; and then against 903 found on the first, is near 8,15 on the second: So much is the length in foot measure to make a foot.   

PROBLEM. IX. The Diameter and length of a Cylinder given in inches or feet, to find the Content in inches, or tenth parts of feet.  
LEt the Diameter of a Cylinder be 15 inches, and the length 105 inches, and the Content in inches sought for.  
Set this number 1128 on the first, to 15 on the second, the Diameter known; and then against the length 105 on the first, is 1396 on the second; and again, against this 1396 on the first, is 18553,5 on the second: The whole Content in inches, which divided by 1728, giveth 10 foot, and almost three quarters.  
But if the Diameter and length be taken by foot-measure, as let the Diameter be 1,25 foot, and the length 8,75 foot. Then,  
Set the number 1128 on the first, to 1,25 (the Diameter in foot-measure) on the second; and then against 8,75, the length on the first, is 9,69 on the second. And lastly, against 9,69 on the first, is 10,737 foot, or 10 foot, and almost three quarters: the Content sought for, as before.   

PROBLEM. X. The Diameter and length of a Cylinder given in inches, to find the Content in feet.  
LEt the Diameter of a circle be 15 inches, and the length 105 inches, and the Content in feet desired.  
Set this number 4690 on the first, to 15 the Diameter on the second, and then against 105, the length on the first, is 33,58 on the second; and again, against this 33,58 on the first, is 10,737 on the second: So then 10 foot, and almost three quarters, is contained in that Cylender.   

PROBLEM. XI. The Diameter of a Cylinder given inches, and the length in feet, to find the content in feet.  
LEt a Cylinder be given to be measured, whose Diameter is 15 inches, and the length 8 foot, and 75 parts of 100 of a foot, and let it be required by these measures only, to give the Content in feet.  
Set this general number 1354 on the first, to 15 the Diameter on the second; and then against 8,75 the length in feet on the first, is 969 on the second: And lastly, against this 969 on the first, is 10,74 on the second: So much doth that Cylinder contain, viz. 10 foot and almost three quarters.   

PROBLEM. XII. The circumference and length of a Cylinder given in inches, to find the content in inches.  
LEt a Cylinder be given to be measured, whose length is 105 inches, and its circumference 47,13 inches, and let it be required only by these measures, to give the Content in inches, which to do:  
Set this general number 3545 on the first, to 47,13 on the second (that is the circumference) and then against 105, the length on the first, is 1396 on the second. Now again, against this 1396 on the first, is 18555: So many inches are contained in that Cylinder.   

PROBLEM XIII. The circumference and length of a Cylinder given in inches, to find the content in feet.  
LEt the forementioned Cylinder be proposed, whose length is 105 inches, and the circumference 47,13 inches, and the content in feet required by this measure only. In this case:  
Set this general number 14736 on the first, to 47,13 on the second, and then against 105, the length in inches, is 33,58 on the second, and against this 3358 on the first, is 10,74 on the second, which is 10 feet and 74 parts of 100 of a foot.  
But if the circumference and length be given in foot-measure, as if the circumference be 3,927 foot, and the length 8,75 foot, and the Content in feet required. Then,  
Set this general number 3545 on the first, to 3,927 the circumference on the second, and then against 8,75 the length on the first, is 969 on the second; and again, against that 969 on the first, is 10,74 on the second: So much is the Content in feet required.   

PROBLEM. XIV. The circumference of a Cylinder taken in inches, and the length in feet, to find the content in feet.  
IN the Cylinder afore proposed, having the circumference in inches 47,13, and the length in feet 8,75. I thereby to find the content in feet, do work thus:  
Set this general number 4254 on the first, to the circumference in inches 47,13 on the second, and then against 8,75 the length in feet on the first, is 969 on the second; and lastly, against this 969 on the first, is 10,74 on the second: So much is the content in feet.   

PROBLEM. XV. The Diameters of any vessel at the head, and at the bung, with the length in inches had, to find the content thereof, first in inches, and then in gallons, either of Wine or Beer.  
LEt the length of a vessel be 40 inches, the Diameter at the head 18 inches, and the Diameter at the bung 32 inches, and the content of the vessel in inches and gallons is sought for. Which to find,  
You must first get two third parts of the Content of a circle answerable to the Diameter at the bung; and one third part of the Content of a circle answerable to the Diameter at the head, and add them two numbers together, and their total multiply by the length of the vessel, and the Product is the Content in inches.  
Now to get the two third parts of the circle at the bung; Set 1 on the first, to this general number 5236 on the second, and then against 1024, the square of 32, the Diameter at the bung is 536,166, which is two third parts of the Content of that circle at the bung.  
For the one third part of the circle at the head; Set 1 on the first, to this general number 2618 on the second, and then against 324, the square of 18 the diameter at the head on the first, is 84,823 on the second, which is one third part of the content of the circle at the head.  
These two numbers 536,166 and 84,823 added together, make 620,989. This 620,989 multiply by 40, and it giveth 24839,56: So many inches are contained in it.  
And now to know how many Wine-gallons are in this vessel, divide the content in inches 24839,56 by 231, which is the number of inches contained in a Wine-gallon, and the quotient is 107,53: So many Wine-gallons are in that vessel.  
To know how many Ale or Beer-gallons are in it, divide 24839,56 the content in inches. By 272,25, which is the number of inches in a Beer-gallon, the quotient is the content of the vessel in Beer-gallons. How to multiply and divide is showed before, and therefore I need not show it here again.    

CHAP. V The Use of the double scales of Numbers in Spherical Bodies, such as Globes, Bullets, etc.  

PROBLEM. I. The Diameter of any Spherical Body being known, to find the circumference.  
LEt the Diameter of a Bullet be 9 inches, and the circumference sought for.  
Set 7 on the first, to 22 on the second, and then against 9 on the first, is 28,28 on the second: So many inches is the circumference of a Bullet of 9 inches diameter.   

PROBLEM. II. The circumference of any Spherical Body being known, to find the Diameter.  
LEt the circumference of a Bullet be 28,28 inches, and the diameter sought for.  
Set 22 on the first, to 7 on the second, and then against 28,28 on the first, is 9 on the second, that is, 9 inches is the diameter of that Bullet, whose circumference is 28 inches, and 28 hundred parts of an inch.   

PROBLEM. III. The Diameter and circumference of any spherical Body being known, to find the superficial content.  
LEt the diameter of a Globe be 9 inches, and the circumference 28,28 inches, and it is demanded, How many square inches, the Superficies of that Globe doth contain? Then,  
Set 1 on the first, to 9 the diameter on the second, and then against 28,28 on the first, is very near 254,5 on the second: So then 254,5 inches and an half, are contained in the Superficies of that Globe.  
Or else by knowing only the diameter, work thus; Set 1 on the first, to this number 3,1416 on the second; and then against 81, the square of the diameter on the first, is 254,5 on the second, as before; here you see are two ways of working, and both agreeing: So that one proveth the other.   

PROBLEM. iv The Axis or Diameter of a Globe being known, to find the solid content.  
IF the diameter of a Globe be 9 inches, what is the solid content in square cubicle inches?  
For resolving this and the like questions, the Rule is this; As the diameter is to the Cube of itself, So is 11 to the solid Content, The Cube of 9 the Diameter known is 729, which Cube, if you know it not, is thus found.  
Set 1 on the first, to 9 on the second, and then against 9 on the first, is 81, the square of 9, on the second; and lastly, against this 81 on the first, is 729 on the second, which is the Cube of 9, the Cube being had.  
Set 9 on the first, to 729 on the second, and then against 11 on the first, is 891 on the second: Therefore 891 cubick inches are contained in the solid body of that Globe. Or,  
Set 1 on the first, to 9 on the second, and then against 11 on the first, is 99 on the second, and next against this 99 on the first, is 891 on the second, as before.  
If the Diameter be 21 inches, than set 21 on the first, to 9261, its Cube on the second; and then against 11 on the first, is 4851 on the second: So many cubick inches are in a Globe of 11 inches diameter. Or,  
Set 1 to 21, and then against 11 is 235, and against that 235 is 4851, as before.   

PROBLEM V The Diameter of a Bullet being given, with the weight; to find the weight of another Bullet of the same metal, but of another Diameter, either greater or lesser.  
LEt there be propounded an Iron Bullet of 6 inches diameter, weighing 30 pound, and let the question be put, What another Bullet of the same metal will weigh, that is of 7 inches diameter.  
Set 6 on the first, to 7 on the second, and then against 30, the weight on the first, is 35 on the second; Secondly, against 35 on the first, is 40; 8 on the second; Thirdly, against 40,8 on the first, is 47,7 on the second: So than the conclusion is, that 47 pounds, and 7 tenth parts of a pound, is the weight of that other Bullet of 7 inches diameter, which was sought for.  
If a Bullet of 6 inches diameter weigh 32 pounds, what will a Bullet of the same metal weigh, that is of 3 inches diameter?  
Set 6 on the first, to 3 on the second, and then against 32, the weight on the first, is 16 on the second; Secondly, against 16 on the first, is 8 on the second; and lastly, against 8 on the first, is 4 on the second: So then 4 pounds is the weight of that Bullet of 3 inches diameter.   

PROBLEM VI Having the side of a cubick body of silver with the worth thereof, to find the worth of another cubick body of silver, whose side is greater or lesser than that of the body given.  
IF a Cube body of silver being 4 inches square be worth 12 pounds, What will another Cube body of the same metal be worth that is 5 inches square?  
Set 4 (the side of the body whose worth is known) on the first, to 5 (the side of that body whose worth is sought) on the second; and then against 12, the worth known on the first, is 15 on the second; next against 15 on the first, is 18,57 on the second; and thirdly, against 18,57 on the first, is 23,4 on the second, which is 23 pounds and four tenths of a pound; that is 8 shillings: So than the worth of that cubick body of silver of 5 inches square is 23 pounds and 8 shillings.   

PROBLEM VII. Having the weight of a Bullet of one kind of metal, to find the weight of a Bullet of another kind of metal, being equal in magnitude.  
BEfore I proceed to resolve this Problem, I will show the proportions between some several metals used for this purpose, as of Brass, Iron, Led and Stone, according to the best approved Authors, as followeth.  
The proportion between Lead and Iron is as 2 to 3; So that a Leaden Bullet of 3 pounds' weight, is equal in diameter with an Iron Bullet of 2 pounds' weight.  
The proportion between Iron and Stone is as 3 to 8: Therefore a Stone of 3 pounds' weight is equal in bigness to a piece of Iron of 8 pounds' weight, and a Stone body of 30 pounds' weight, is equal in magnitude to an Iron body of 80 pounds' weight, of the same form.  
The proportion between Lead and Stone is as 4 to 1: So that a Bullet of Lead of 4 pounds' weight, and a Stone Bullet of one pound weight, are equal in diameter; and a Leaden Bullet of 40 pounds' weight, and a Stone Bullet of 10 pounds' weight, are equal in diameter or magnitude.  
The proportion between Iron and Brass, is as 16 to 18; and the proportion between Lead and Brass, is as 24 to 19  
Remember that some Stone is heavier than other, and so likewise of metals, the finer they are, the heavier they be, being of the same magnitude: but we speak of the ordinary sort in use. And now for resolving the Problem.  
Having the weight of a Bullet of Lead, to find the weight of a Bullet of Marble of the same bigness.  
If a Bullet of Lead weigh 106 pounds, What will a Bullet of Marble weigh?  
By the former Rule it is found, that a Bullet of Lead to the like Bullet of Marble, beareth such proportion as 4 doth to 1. Therefore,  
Set 4 on the first, to 1 on the second, and then against 106 on the first, is 26,5 on the second: So much is the weight of a Stone Bullet, that is equal in bigness to that Leaden one of 106 pounds' weight.  
On the contrary, having the weight of a Stone Bullet, to find the weight of a Leaden Bullet, of the same magnitude.  
Set 1 on the first, to 4 on the second, and then against 26,5; the weight of the Stone Bullet on the first, is 106 on the second, the weight of the Iron Bullet.  
There is a Bullet of Iron weigheth 72 pounds, What will a Bullet of Lead weigh, that is equal to it in bigness?  
Set 2 on the first, to 3 on the second, and then against 72, the weight known on the first, is 108, the weight sought for on the second.  
But if the weight of the Leaden Bullet be given 108, then to get the weight of the Iron Bullet, Set 3 on the first, to 2 on the second, and then against 108 on the first, is 72 on the second: So much is the weight of the Iron Bullet. What is said of a Bullet, is to be understood also of all round bodies.    

CHAP. VI The Use of the double Scales in the Measuration of Concave Cylinders, such as great Ordnance.  

PROBLEM I. The Diameter and weight of any one Cylinder, or Piece of great Ordnance being known, to find the weight of any other Piece of the same metal and shape, either greater or lesser, its Diameter being only known.  
IF a Brass Saker, whose Diameter is 11,5 inches, do weigh 1900 pounds, What will another Piece weigh, whose Diameter is 8,75 inches? For answer:  
Set 11,5 on the first, to 8,75 on the second, and then against 1900 on the first, is 1440 on the second; Secondly, against that 1440 on the first, is 1100 on the second; Thirdly, against the 1100 on the first, is 837 on the second: So much is the weight of that other Piece of Ordnance of 8,75 inches Diameter, it being of the same metal and shape as the other.  
If a Piece of Ordnance of 4 inches Diameter weigh 1600 pounds, What will another Piece weigh that is 6 inches Diameter, being of the same metal and shape?  
Set 4 on the first, to 6 on the second, and then against 1600, the weight known, is 2400 on the second; Next against 2400 on the first, is 3600 on the second; And lastly, against 3600 on the first, is 5400 on the second: The weight sought for of that other Piece of 6 inches Diameter.   

PROBLEM. II. Having the Diameter and weight of any Piece of great Ordnance, of one metal; to find, the weight of another Piece of Ordnance of another metal, that retaineth the same shape.  
A Piece of Ordnance being of another sort of metal, there will be required a double work to find out its weight; As let there be a Brass Piece of Ordnance given, of 11,5 inches Diameter, weighing 1900 pounds; and let the question be, to find the weight of an Iron Piece of Ordnance of the same shape, that is 8,75 inches Diameter.  
In this and the like cases, you must in the first place, by the former Problem, find the weight of that Piece of 8,75 inches Diameter, as if it were a Brass Piece, and having found that the weight of it, had it been Brass, would have been 837 pounds: You must next seek the proportional numbers of the two metals, which by the seventh Problem of the last Chapter, you found was 16 and 18: Such is the proportion between Brass and Iron, Brass being the heavier metal. Therefore,  
Set 18 on the first, to 16 on the second, and then against 837 on the first, (the weight the Piece would have been of, had it been Brass) is 744 on the second, the weight of that Piece it being Iron.   

PROBLEM. III. To find the superficial Content of the concave Superficies of any Piece of Ordnance, and also the solid Content of the Concavity thereof.  
SVppose the Circumference of the Concavity be 22 inches, and the length of it 12 foot, or 144 inches, & the question put, What is the superficial Content of the concave Face, and what the solid Content of the concave Boar. For the concave Superficies.  
Set 1 on the first, to 22, the Circumference of the Concavity on the second, and then against 144, the length in inches on the first, is 3168 on the second: So many square inches are in the Superficies of the concave face of the same Piece, which in feet makes 22. For the solid Content.  
First get the Semidiameter, which in this example is 3,5 inches; and also the Semicircumference, which here is 11: These being had,  
Set 1 on the first, to 3,5 the Semidiameter on the second, and then against 11, the Semicircumference on the first, is 38,5 on the second: So many square inches are contained in the Base or Plane of the Concavity of the mouth. This Base had:  
Set 1 on the first, to 38,5 the Content of the Base on the second; and then against 144, the length in inches, is 5544 on the second: So many cubick inches are in the solid Content of the Concavity of that Cylinder, which is 3 foot and 360 parts of 1728 of a foot.   

PROBLEM iv To know how much of every kind of metal is conned in any Brass Piece of Ordnance.  
IT is said to be an usual thing with Gun-founders, that for every 100 pounds of Copper, to put in 10 pounds of Latin, & 8 pounds of pure Tin. Now supposing this mixture to be true, let it be demanded, How much of every sort of these metals is in a Gun of 5600 pounds' weight?  
For answer to this & the like questions; First, join all the several mixtures together, that is 100, 10, & 8, and this must be the first number in the rule of proportion; the weight of the Piece the second number, which here is 5600; and the third number is each several sort of metal in the mixture, which here be 100, 10, and 8: wherefore by our double lines, this quaere is very easily resolved. For,  
Set 118, the total of the common mixture on the first, to 5600, the weight of the Piece on the second, and then against 100 on the first, is 4745,7 on the second; against 10 on the first, is 474,6 on the second, being one place less than the former, because 10 is one place less than 100; so much Latin is in that Piece: and against 8 on the first, is 379,7 on the second, so much Tin is in the Gun.  
Now if all these three sums thus found be added together, they make the just weight of the Piece propounded, as here it doth appear, 4745,7 474,6 379,7 5600,0 and proveth the work truly wrought.   

PROBLEM. V By knowing what quantity of powder will load some one Piece of Ordnance, to find how much of the same powder will load any other Piece of Ordnance, greater or lesser.  
IF a Saker of 3,75 inches Diameter in the Boar, require four pounds of powder for its load, What will a Demi-Cannon of 6,5 inches Diameter in the Boar require?  
Set 3,75 on the first, to 6,5 on the second, and then against 4, the weight known on the first, is 6,93 on the second, and then against that 6,93 on the first, is 22 on the second; and lastly, against that 12 on the first, is 20,8 on the second, very near; that is, 20 pounds of powder, and 8 tenth parts of a pound: So much doth the Demi-Cannon require.  
But note, that it is here understood, that the Demi-Cannon ought to be as well fortified as the Saker is, that is, it should bear the same proportion to the Saker in weight and thickness of metal, that the bore thereof beareth to the bore of the Saker, as in this example; The bore of the Saker is 3 inches and 3 quarters, the Cube whereof is 52,73, and its weight is 1600 pounds; The bore of the Demi-Cannon is 6 inches and an half, the Cube whereof is 274,62. Now the question is, How much the weight of that Demi-Cannon ought to be, that is proportional in its weight and thickness to the Saker, that so it may be able to bear a load of powder proportional to the Saker.  
The weight of such a Demi-Cannon is thus found; Set 52,37 the Cube of 3,75 on the first, to 274,62 the Cube of 6,5 on the second, and then against the weight of the Saker 1600 on that first, is 8351 on the second: The weight that the Demi-Cannon ought to be of, that is proportional in weight to the Saker, and able to carry a load of powder proportional to the Saker. Or you may work it thus without the Cubes.  
Set 3,75 on the first, to 6,5 on the second, and then, first against 1600 on the first, is 2766 on the second; secondly, against this 2766 on the first, is 4800 on the second; and thirdly, against that 4800 is 8351 on the second, as before: So the one work proveth the other.  
But if the Demi-Cannon be found to want of its proportional weight with the Saker, as if it weigh but 6000 pounds, then to find its due load in powder, answerable to its strength and weight of metal; Multiply the weight thereof 6000, by 20,8 the charge already calculated, & divide the Product by 8351, the weight it ought to have had, and the quotient is 14,8: Therefore 14 pounds, and 8 tenth parts of a pound, is a sufficient charge for such a Gun.  
By our lines, Set 1 on the first, to 20,8 on the second, and then against 6000, is 1248000, the Product. Now set 8351 the Divisor on the first, to 1 on the second, & then against 1248000 on the first, is 14,8 on the second, the quotient sought.  
But suppose there be another sort of powder brought to be used, that is stronger than the former, by such proportion as 5 is to 2. How much of this sort will serve to charge that Gun, which required 14,8 pounds of the other sort. For answer of this quaere.  
Set 5 on the first, to 2 on the second, and then against 14,8 on the first, is 5,9 and a little above: Therefore 5 pounds of powder and 8 parts of 10 of a pound, of this new sort of powder, will charge the Gun, having as much strength to carry forth the Bullet, as 14 pounds and 8 tenths of the other.   

PROBLEM. VI Knowing how far any Piece of Ordnance will carry her Bullet at point-blank, and at the best of her Random, to find how far any other Piece of Ordnance will carry her Bullet at her best Random, her level-range being known.  
A Saker at point-blank conveys her Bullet 200 paces, and at her best Random 900 paces. Now how far will a Cannon carry her Bullet at her best Random, that carrieth it at point-blank 360 paces.  
Set 200 on the first, to 900 on the second, and then against 360, the other Pieces point-blank on the first, is 1620 on the second: The number of paces it will carry at point-blank.  
If the best Random and point-blank of the one Piece be given, with the best Random of the other Piece, to find the point-blank thereof.  
Set 900, the best Random on the first, to 200 it's point-blank on the second, and then against 1620, the other Pieces best Random on the first, is 360, it's point-blank on the second.   

PROBLEM VII. By knowing how far any piece of Ordnance will carry a Bullet at the best of her Random, to find how far she will carry her Bullet at any other degree of Randon.  
IF a Piece at her best Random, which is 45 degrees of Mounture, carry her Bullet 1440 paces, How far will she carry it at 30 degrees of Randon?  
Set 45, the best Random on the first, to 30, the other Random on the second; and then against 1440, the paces of her carriage at her best Random on the first, is 960, the paces of her carriage at 30 degrees of Randon; and if you take 960, from 1440, the remain is 480: So much doth she shoot short of her best Random.  
If a Gun carry a Bullet 700 yards at her best Random, which as aforesaid is at 45 degrees, and the mark to shoot at be distant but 500 yards; To what degree must the Gun be mounted to make a good shoot?  
Set 500 on the first, to 700 on the second, and then against 45 on the first, is 63 on the second; or bring 500 to 45, and then against 700 is 63, which is 63 degrees: Therefore the Piece must be raised to 63 degrees of Mounture, to make a good shoot into the place.  
Note, this question is resolved by the backward Rule of Three.  
If a Gun at her best Random shoots 500 yards, How much will she shoot short of it, being elevated two degrees above her best Random?  
Set 45, the degrees of best Random on the first, to 2 the degrees she is elevated above it on the second, and then against 500, is 22,22: So many yards will she shoot short of her best Random, being elevated 2 degrees above it.  
How far is it to the place where a Bullet falleth, the Piece being mounted 15 degrees above her best Random, she abating 22,22 for 2 degrees elevation?  
Set 2 on the first, to 22,22 on the second, and then against 15 on the first, is 167 on the second; therefore the Gun being raised 15 degrees above her best Random, abateth 167 yards of the 500, her carriage at her best Random, wherefore 167 taken from 500, leaveth 333: So many yards is the place distant where the Bullet falleth.  
If a Gun shoot point-blank 240 paces, and being mounted to one degree, doth out-shoot the point-blank 30 paces, What will she out-shoot it at 20 degrees of Mounture?  
Set 1 on the first, to 30 on the second, and then against 20 on the first, is 600 on the second: So many paces will she out-shoot her point-blank, being mounted 20 degrees, in the whole shot 840 paces.   

PROBLEM VIII. To find out how much wide of the mark any Piece of Ordnance will shoot, by knowing how far it is to the mark shot at, and how wide the Pieces mouth lieth from the right line to the mark.  
A Gun of 10 foot, or 120 inches long, being to shoot at a mark 700 yards distant, her mouth lying one inch beside the right line to the mark, How far will she shoot her Bullet wide of the mark at that distance?  
For answer: First, bring 700 yards into inches, and they make 25200, which is done by Multiplication thus.  
Set 1 on the first, to 36 on the second, for that is the number of inches in a yard, and then against 700, the distance on the first, is 25200, the inches in 700 yards. And then,  
Set 120 on the first, the Guns length in inches, to 1; and then against 25200, the distance in inches, is 210 inches, or 17 foot and an half: So far the Bullet goeth wide of the mark.   

PROBLEM IX. Knowing the quantity of each sort of Ingredients for the making of Gun powder, to find how much of every sort is to be put into any wantity of powder that shall be required to be made.  
I Have read in some Authors, that for making of the best sort of ordinary Gunpowder, there is used to be taken 12 parts of Mercury, three parts of Cole, and two parts of Sulphur, in all 17 parts. Now a Gun-powder-maker is appointed to make 1000 pound weight of powder, How much must he take of each sort of these Ingredients, to make that quantity of powder?  
Set 17, the sum of the parts on the first, to 1000, the quantity to be made, on the second; and then without more ado, on the first, right against 12, is 706, and against 3 is 176,5; and lastly, against 2 is 117,5 on the second. These three sums being added together, make just 1000 Therefore you may conclude, that to make 1000 pounds of Gunpowder, he must take of Mercury 706 pounds, of Cole 176,5, and of Sulphur 117,5 pounds, the thing required. This Problem is grounded upon the rule of Proportion thus: 

As 17, the whole parts taken,  
Is to 1000, the quantity to be made:  
So is each several part of the mixture,  
To the quantity thereof to be taken towards the making of 1000 weight.    
I might have put down the Analogies of Proportion all along, as I did at the beginning; but I thought those things would have filled my book too fast, and every man that knows the rule of Proportion, cannot but see how every Problem is resolved thereby.    

CHAP. VII. The Use of the double Scales in Fortification.  

THe side of a Pentagonal Fort being propounded, with the perpendicular and all the other lines, with the Area: To find the sides and all other lines and Content of any other Pentagonal Fort, that shall be required to be made, either greater or lesser, according to any proportion assigned.  
Suppose there be a Pentagonal Fort made, such an one as is represented by this Diagram annexed, noted by the letters A B S T X K O G F E D N M L H P and C, whose sides AB, BS, SX, XR, and KA, are each 662 yards. And let it be required, to make another Fort that shall be a just fourth part in Content of ground of that propounded Fort. Now the Content of the propounded Pentagonal Poligon, described by the letters ABSXK, is 754680 yards, and the sides and lines of the Fort thereon made are as followeth.  

The side of the Pentagon AB, is 662 yards.  
The Perpendicular CI, 456 yards.  
The Semidiameter of the Pentagon CA, 564 yards.  
The Gorge-lines of the Fort AD, AO, NB, BP, each one of then 119 yards.  
The Flanks of the Bulwark DE, and OG, 100 yards.  
The Curtain DN, 424 yards.  
The Fronts lines FE, FG, and LM, LH, 264 yards.  
The Head-lines, or Capital OF, or BL, 224 yards.  
The Breadth of the Bulwark from G to E, or M to H, 330 yards.  
The Lines of Defence DL, and NE, 700 yards.  
The distance between the points of the Bulwark F and L, is 926 yards.  
The Line perpendicular to the Line FL, viz. CR, 637 yards.   
By these Lines and lengths known, it is required to make another Pentagonal Fort, whose quantity or Content in land shall be the just fourth part of the proposed Pentagonal Fort propounded: where note, that the half of the sides and lines of the Fort made will make another Fort, that will contain the fourth part of the ground in that Fort made: And that you may readily have half of all the lines of the Fort made, do thus:  
Set 2 on the first, to 1 on the second, and then against the number of any line on the first, is its half on the second: As against 662, the length of the side of the Pentagon given, is 331, the length of the Pentagons' side to be made. The Lines thus remaining without any moving, against 119 on the first, the length of the Gorge-line in the made Fort, is 59,5, the length of the Gorge-line in that Fort to be made: And against 424, the length of the Curtain in the made Fort, is 212 on the second, the length of the Curtain in that Fort to be made: And against 456, the length of the proposed Poligons perpendicular which is made, is 228, the length of the perpendicular in that Poligon which is to be made: And against the Semidiameter 564, is 282: And against 100, the length of the present Bulwarks flank, is 50, the length of the Flank in the Bulwark to be made. And against 224, the length of the Bulwarks head-line, is 112, the length of the head-line in the Fort to be made: And against 310, the distance of the shoulders GE, or MH in the Fort made, is 155, for the distance of the shoulders in that Fort which is to made: And in like manner, against the length of any Line in the proposed Fort is the length of that same Line in the Fort to be made.  
Thus very speedily is found out the true length of all the Lines in and belonging to such a Fort as is required to be made, whose lines and sides shall be just half the length of those in the Fort already made: And being drawn on the ground, and made up, will in all parts be like & proportional to the same proposed Fort. The like manner of working is to be used for any other proportion whatsoever, either greater or lesser.  
Suppose it were required to make a Pentagonal Fort, whose sides, and so all the lines thereof, should be the one third part in length of the sides and lines in that proposed Fort, which in the annexed Diagram is represented by the letters CABSTXKOGFEDNMLPH. To which demand, to give a ready answer, do thus:  
Because one third part is sought for, Set 3 on the first, to 1 on the second, and then against 662 on the first, the side of the Pentagon given, it 220⅔ on the second, which is one third part of 662, & showeth that the length of the side of the Pentagon to be made is 220⅔ yards; and against 119, the length of the Gorge-lines in the proposed Fort, is 39⅔, that is 39 yards 2 foot, which is the third part of 119, and is the just length of the Gorge-lines in the Fort to be made. In like manner, against any Line or distance in th' proposed Fort, is the length of the same Line in that Fort to be made. For to be short, Take any number on the first, and right against it on the second, is the one third part thereof.   

How to make a Fort greater than the Fort proposed.  
SVppose there be a Fort made, such an one as is that represented in the aforementioned Diagram, and another Fort is to be made like to it; so that in all the lines and sides thereof, it be double to those in the Fort that is already made.  
In this case and the like, we have no more to do, but to set 1 on the first, to 2 on the second, and then against the number of the length of any side or line on the first, is the double of it on the second. As if you set 1 on the first, to 2 on the second, then against 456 on the first, the length of the perpendicular in the Pentagonal Fort given, you shall have 912 on the second, the double of 465, which is the length of the perpendicular in that Fort to be made; and against 662, the length of the Curtain in the Fort made, is 1322, the length of the Curtain in the Fort to be made: And so of all the rest of the Lines.  
If it shall be required to make such a Fort, as in its sides and lines shall be one third part longer than those in the proposed Fort: Then take any one line, as for example, CI 456 yards, and thereof take the one third part, which is 152: This add to the same 456, and it makes 608. This done, Set 456 on the first, to 608 on the second, and then against any line of the Fort made on the first, is its match-line in the Fort to be made: As against DN, the Curtain in the made Fort, 424 yards, is 565●: So many yards will the Curtain be in that Fort to be made, and so of all other lines. And if a Fort be to be made, whose sides and lines must be half so much more in length, as are the sides and lines of the proposed Fort: Then take the half of any number, and add it to the same number, and so proceed as before.  
Note, that the lines in the Type or Diagram annexed, may be accounted in their lengths there set down, either so many foot, or so many yards, or so many perches, or what other measure you are to use: And so the side of the Pentagon 662, may be accounted 662 yards, or 662 foot, or 662 perches. Thus by supposing a Pentagonal Fort to have lines and sides in length, as those in the Diagram, you may by these double lines very easily discover all the lines and sides of any other Pentagonal Fort, either greater or lesser.   

 geometrical diagram of a pentagonal fort    

CHAP. VIII. The Use of the double Scale of Sines in Astronomy.  

PROBLEM. I. The Sun's greatest declination, with his place or distance from the next Equinoctial point being given, to find his present declination for the time given.  
THe Sun's greatest declination in this our age, is found to be 23 degrees 30 minutes. This is general, his place given on the day proposed, is 0 degr. of Taurus, which point is distant from the next Equinoctial point of Aries 30 degrees.  
The Analogy for Solution of this Proposition is thus: 

As the Radius, or Sine of 90 degrees,  
Is to the Sun's greatest declination 23 deg. 30 min.  
So is the distance of the Sun 30 deg. from the next Equinoctial point which is Aries,  
To the Sine of the declination 11 degr. 30 min. sought for.   Wherefore,  
Set 90 degrees on the first, to the Sun's greatest declination 23 deg. 30 min. on the second, and then against 30 deg. the Sun's distance from the next Equinoctial point on the first, is 11 deg. 30 min. on the second: So much is his declination sought for, his place being the very beginning of Taurus.  
Suppose the Sun be in the beginning of Leo, which is 60 degrees distant from the Equinoctial point of Libra; then set 90 deg. on the first, to 23 deg. 30 min. on the second, and then against 60 deg. on the first, is 20 deg. 10 min. on the second: The declination of the Sun, when he is entering into Leo.   

PROBLEM. II. The Sun's greatest declination, and his present declination for any time proposed being had, to find his distance from the next Equinoctial point, and thereby his place in the Ecliptic.  
SEt the Sun's greatest declination, 23 deg. 30 min. on the first, to 90 degrees on the second; and then against his present declination, being given 11 deg. 30 min. on the first, is 30 degrees on the second: So much is his distance from the next Equinoctial point, which is Aries; therefore his true place is just leaving Aries, and entering Taurus.  
Again, suppose the Sun's present declination be given 20 deg. 10 min. and his place sought: Set 23 deg. 30 min. on the first, to 90 deg. on the second, and then against 20 deg. 10 min. on the first, is 60 degrees on the second: So much is the Sun's distance from the next Equinoctial point.  
Now if the Sun be in the first quarter of the Ecliptic, between Aries and Cancer, than he is distant from the Equinoctial of Aries 60 de-degrees, and therefore in the beginning of Gemini: But if he be in the second quarter of the Ecliptic, as in this proposal he is, his distance is to be accounted from the Equinoctial of Libra, and then his place must be in the beginning of Leo; If he be in the third quarter of the Ecliptic, and 60 degrees from the next Equinoctial point, than he is in the beginning of Sagittarius: And if he be in the fourth quarter of the Ecliptic, and distant 60 degrees from the next Equinoctial point, than he must be in the beginning of Aquarius.   

PROBLEM. III. The Sun's declination, and the Latitude of the place being given, to find the Sun's Amplitude.  
LEt be given the Sun's declination, 11 deg. 30 min. and the Latitude of the place proposed 51 degrees 30 minutes, whose Compliment to 90 degrees, is 38 degrees, 30 minutes: And the Amplitude sought for.  
Set the Co-sine of the Latitude 38 deg. 30 min. on the first, to the Radius, or 90 degrees on the second, and then against the sine of the Sun's declination, 11 deg. 30 min. on the first, is 18 deg. 40 min. on the second: So much is the Sun's Amplitude, at such time as his declination is 11 deg. 30 minutes. Or,  
Suppose the Suns, or a Stars declination were known to be 20 degrees, then without altering the lines from the place they were first set to, right against 20 deg. on the first, is 33 deg. 10 min. on the second, the Amplitude sought: And so against any other declination on the first, is the Amplitude on the second.   

PROBLEM. iv The Latitude of the place, the Sun's greatest declination with his distance from the next Equinoctial point being had, to find his Amplitude.  
THe Latitude is 51 deg. 30 min. whose Compliment is 38 deg. 30 min. the Sun's distance from the next Equinoctial point is 18 deg. 40 min. and his greatest declination 23 deg. 30 min. and the Amplitude required.  
Set the Co-sine of the Latitude, 38 deg. 30 min. on the first, to 23 deg. 30 min. on the second, the sine of the greatest declination, and then against 30 degrees, the distance from the nearest Equinoctial point on the first, is 18 deg. 40 min. on the second: So much is the Sun's Amplitude, when his place is in the beginnings of Taurus, Virgo, Scorpio, or Pisces.   

PROBLEM. V The Latitude of the place, the Sun's place or distance from the next Equinoctial point, with his greatest declination being known; to find what altitude the Sun will have, when he is on the true East or West point.  
LEt the Latitude be 51 deg. 30 min. the Sun's place in 10 degrees of Taurus, which is distant from the next Equinoctial point 40 degrees, and the greatest declination 23 deg. 30 min. What is the Sun's altitude at the true East or West point?  
Set 51 deg. 30 min. on the first, (the Latitude of the place) to 23 deg. 30 min. the greatest declination on the second; and then against 40 degrees, the distance from the nearest Equinoctial point on the first, is 19 degr. 5 min. on the second: So much is the Sun's altitude, when he is on the true East or West point: And at his entrance into Taurus, his altitude at that point is 14 degrees 45 minutes. As this, so his altitude for any other place is found, without stirring the lines, after once set together.   

PROBLEM. VI The Latitude of the place, and Sun's declination being had; to find what altitude the Sun will be of, when he cometh upon the true East or West point.  
LEt the Latitude be 51 deg. 30 min. and the Sun's declination 11 deg. 30 min. and his height at his being on the due East or West point required.  
Set the sine of the Latitude, 51 deg. 30 min. on the first, to the Radius on the second, and then against the sine of the given declination, 11 deg. 30 min. on that first, is 14 degr. 45 min. on the second: So much is the Sun's altitude at the time of his being due East or West, when his declination is 11 deg. 30 min.   

PROBLEM VII. Having the Sun's greatest declination, and his distance from the next Equinoctial point; to find his right ascension.  
THe Sun's greatest declination is 23 degr. 30 min. and his place given is 0 degr. of Taurus. Therefore his distance from the nearest Equinoctial point Aries is 30 degrees: By these known, his right ascension is desired.  
Set 90 deg. on the first, to the Co-sine of the greatest declination, 66 deg. 30 min. on the second; and then on the backside, (being the lines of Tangents) against the Tangent of 30 degrees, the Sun's distance from the nearest Equinoctial point, on that first, is the Tangent 27 degr. 50 min. So much is the Sun's right ascension, at that time.   

PROBLEM VIII. The Sun's greatest declination, and his present declination, being given for a time; to find his right ascension for that time.  
THe Sun's greatest declination is 23 degr. 30 min. and his present declination for a time proposed, is 11 deg. 30 min. I demand his right ascension thereby.  
Set the Tangent of 23 deg. 30 min. on the first, to the Tangent of the present declination, 11 deg. 30 min. on the second; and then upon the other side, on the lines of sins, against the Radius on the first, is the sine of 27 deg. 50 min. on the second: So much is the Sun's right ascension demanded.   

PROBLEM. IX. The Latitude of the place, and the Sun's declination known; to find how long the Sun riseth and setteth, before or after the hour of six.  
THe Latitude of the place is 51 degr. 30 min. North, and the Sun's declination, 11 deg. 30 min. I demand the time of Sunrising before six.  
Set the Co-tangent of the Latitude, 38 deg. 30 min. on the first, to the Tangent of 11 deg. 30 min. the declination given on the second; & then on the other side, against the Radius on the first, is the sine 14 deg. 50 min. This 14 deg. 50 min. converted into time, by accounting four minutes of an hour to every degree, and one minute of an hour for ●… minutes of a degree, and it will amount to 59 minutes of an hour: So long the Sun riseth before six in the morning, and sets after six at night in the Summer season; and in the Winter season, or Southern signs, he riseth so long after six in the morning, and sets so long before six at night, where the Latitude is 51 deg. 30 min. and when the Sun's declination is 11 deg. 30 min. South.   

PROBLEM. X. The Sun being in the Equinoctial, by knowing his distance from the Meridian, and the latitude of the place; to find the Sun's altitude at that time.  
MArch the tenth, the Sun being in the Equinoctial at ten a clock before noon, or at two a clock after noon, he being at that hour distant from the Meridian 30 degrees, I would know his height in our Latitude of 51 deg. 30 minutes.  
Set the Radius on the first, to the Co-sine of the Latitude, 38 deg. 30 min. on the second, and then against the Co-sine of the Sun's distance from the Meridian, viz. against 60 degrees on the first, is 32 deg. 37 min. on the second: So much is the Sun's altitude on the tenth of March, at ten a clock in the forenoon, and at two a clock after noon.   

PROBLEM XI. The latitude of the place, and the Sun's declination Northwards being known; to find the Sun's altitude, at the hour of six.  
LEt the Latitude of the place be 51 deg. 30 min. the Sun's declination 11 deg. 50 min. North: And the Sun's height at six a clock is desired.  
Set the Radius on the first, to the sine of the declination, 11 deg. 50 min. on the second; and then against 51 deg. 30 min. the Latitude on the first, is 8 deg. 58 min. on the second: So much is the Sun's height at the hours of six, at morning and night. Or else thus:  
Set the Radius to 51 deg. 30 min. and then against 11 deg. 50 min. on the first, is 8 deg. 58 min. on the second: For note, that the Radius on the first may be set to either of the other two terms given on the second, and then against the other term on the first, is the answer on the second.   

PROBLEM. XII. The Latitude and declination of the Sun being known; to find the Sun's Azimuth at the hour of six, from the North part of the Meridian.  
IN the Latitude of 51 deg. 30 min. the Sun's declination given is 11 deg. 30 min. What is his Azimuth at the hour of six?  
Set the Radius on the first, to the Co-sine of the Latitude, 38 degr. 30 min. on the second; and then against the Co-tangent of the declination, 11 deg. 30 min. on the first, is the Tangent 82 deg. 47 min. on the second: So much is the Sun's Azimuth from the North towards the East, which was required.   

PROBLEM XIII. The Latitude of the place, and the Sun's altitude at the point of his being due East or West; to find the hour and minute when he will be so due East or West.  
THe Latitude of the place is 51 deg. 30 min. The Sun's height found at his being on the true East point, is 14 deg. 45 min. I demand the hour of the day.  
Set the Radius on the first, to the Co-sine of the Latitude, 38 deg. 30 min. on the second; and then on the Lines of Tangents, right against the Tangent of the Sun's altitude, at his being due East, viz. 14 deg. 45 min. on that first, is 9 deg, 18 min. on the second, the Tangent of time from the hour of six, either before or after, that the Sun will be due East, or due West. This 9 degr. 18 minutes reduced into time, by allowing to every degree four minutes of time, and for fifteen minutes of a degree, one minute of time, amounteth to 37 minutes of time: So than the Sun is due East on the day given, at 37 minutes after six a clock in the morning; and also will be due West at 37 minutes before six at night, which is at five a clock 23 minutes.   

PROBLEM XIV. The elevation of the Pole, the Suns (or a Stars) declination, and distance from the Meridian for any time being given; to find the Sun's altitude at that time.  
LEt the elevation of the Pole be 51 deg. 30 min. whose Compliment to 90, is 38 deg. 30 min. The Sun's declination 11 deg. 30 minutes, whose Compliment is 78 deg. 30 min. his distance from the Pole; his distance from the Meridian 30 degrees, or which is all one, two hours, that is, either at ten or two a clock, and the Compliment of this distance to 90 degrees, is 60 degrees: And let the Sun's Altitude for that time be required.  
1 Set the Radius on the first, to 60 deg. the Co-sine of the Sun's distance from Noon on the second; and then on the lines of Tangents against 38 deg. 30 minutes, the Co-tangent of the elevation on the first, is the Tangent 34 degr. 33 min. on the second; for the fourth term, which taken from the Sun's distance from the Pole, the Remainder is 43 degrees, 56 minutes.  
2 Set the Co-sine of that 34 deg. 33 min. the fourth term, viz. 55 deg. 27 min. on the first, to 46 deg. 4 min. the Co-sine of that Remainder, on the second: and then against the sine of the Elevation, 51 deg. 30 min. on the first, is the sine 43 deg. 18 min. on the second: So much is the altitude of the Sun at that time, the thing required.  
This is when the Sun hath Northern declination, but when he hath Southern declination, then add his declination to 90 deg. and it makes 101 deg. 30 min. the Sun's distance from the Pole, from which take the fourth term found, and the Remain is 66 deg. 27 min. This done, the second work would be thus:  
Set the Co-sine of the fourth term found on the first, viz. 55 deg. 27 min. to the Co-sine of that Remain, which is 23 deg. 33 min. on the second; and then against 51 deg. 50 min. the Elevation on the first, is 21 deg. 52 min. on the second. The Sun's height at ten or two a clock, when he hath 11 deg. 30 minutes of South declination.  
If the Sun be just in the Equinoctial, having no declination; then to find his height on that day at ten or two a clock, the work will be done at one working in this manner.  
Set the Radius on the first, to 60 degrees, the Co-sine of the distance from the Meridian, on the second; and then against 38 degr. 50 min. the Co-sine of the Elevation on the first, is the sine 32 deg. 62 min. So much is the Sun's altitude at the hours of ten and two a clock on the Equinoctial day.   

PROBLEM XV. Having the Poles elevation, and the Sun's declination; to find the ascensional difference.  
LEt the Poles elevation be 51 deg. 30 min. and the Sun's declination 11 deg. 30 min. And by them the ascensional difference required to be given.  
Set 38 deg. 30 min. the Co-tangent of the elevation on the first, to the Tangent of the declination 11 deg. 30 min. on the second; and then (on the Lines of Sines) against the Radius on that first, is the sine of 14 deg. 49 min. on the second: So much is the ascensional difference required at that time, as the Sun hath 11 deg. 30 min. declination.  
I could show by this ascensional difference converted into time, How to find the time of Sunrising and setting, with the length of day and night, etc. But these are things done without the Lines, and therefore I pass them by, having limited myself to show only the use of the double Lines.   

PROBLEM XVI. Having any Planet's declination and Latitude, with its distance from the next Equinoctial point; to find its right ascension.  
LEt a Planets declination be 26 degrees, its Latitude 4 degrees, and its distance from the nearest Equinoctial point 70 degrees, (it being in the 10 deg. of Gemini) I demand the right ascension thereof, which to attain:  
Set 64 deg. the Co-sine of the declination, to 20 deg. the Co-sine of the Planets distance from the next Equinoctial point; and then against 86 deg. on the first, the Co-sine of the Planets latitude, is 22 deg. 17 min. on the second, the Co-sine of the right ascension: Therefore 67 deg 43 min. (being the Compliment of 22 deg. 17 min.) is the right ascension of that Planet which was demanded.   

PROBLEM XVII. The Sun's greatest declination, with his distance from the next Equinoctial point, being had; to find the Meridian angle, that is, the intersection of the Meridian with the Ecliptic.  
LEt the Sun's greatest declination be 23 degr. 30 min. and his place just entering Taurus, which is distant from the nearest Equinoctial point 30 degrees, I would know the Meridian angle.  
Set the Radius on the first, to 60 degrees, the Co-sine of the Sun's distance from the nearest Equinoctial point on the second; and then on the lines of Tangents, against 23 deg. 30 min. the Tangent of the Sun's greatest declination on the first is 20 deg. 38 min. on the second, which is the Co-tangent of the angle sought: Wherefore the Meridian angle is 69 degr. 22 min. being the Compliment of 20 degr. 38 min. to 90 degrees.   

PROBLEM. XVIII. The Sun's declination and amplitude given; to find the height of the Pole.  
LEt the Sun's declination given be 14 degr. 51 min. and his amplitude 19 degr. 7 min. And the Poles elevation demanded.  
Set 19 deg. 7 min. the amplitude on the first, to 14 deg. 51 min. the declination on the second; and then against the Radius on the first, is 51 deg. 30 min. on the second, which is the sine of the latitude: Wherefore 51 degr. 30 min. is the Latitude sought.   

PROBLEM. XIX. The amplitude of the Sun, and time of his rising being known; to find thereby his declination.  
LEt the Sun's amplitude be 33 degr. 38 min. and the time of his rising be at 4 a clock 10 minutes; by which two things known, I would find the declination of the Sun. Here first convert the 4 hours 10 minutes into degrees and minutes, and they make 62 degr. 30 minutes.  
Now set this 62 degr. 30 min. the time of Sunrising converted into degrees on the first, to 56 deg. 22 min. the Co-sine of the amplitude on the second; and then against the Radius on the first, is 69 degr. 50 min. on the second, the Co-sine of the declination: Whereby it followeth, that 20 deg. 10 min. is the declination desired.   

PROBLEM. XX. Having the elevation of the Pole, and the Sun's amplitude; to find his declination.  
LEt the Elevation of the Pole be 51 degr. 30 min. and the Sun's amplitude 19 deg. 7 min. By which two things given, the declination is required.  
Set the Radius on the first, to 19 deg. 7 min. the Sun's amplitude on the second, and then against 51 deg. 30 min. the sine of the elevation on the first, is 14 deg. 51 min. the sine of the declination upon the second: The thing desired.   

PROBLEM. XXI. Having the hour of the day, the Sun's nltitude and declination; to find the Azimuth.  
LEt the Sun's altitude be 25 degr. 56 min. whose Compliment is 64 deg. 4 min. his declination 11 deg. 30 min. whereof the Compliment is 78 degr. 30 min. The hour of the day, 7 hours 56 minutes fore noon, which time is distant from noon 4 hours 4 minutes. This time converted into degrees, giveth 61 degrees. Now by these things known, the Azimuth is desired to be found.  
Set 64 degrees 4 minutes, the Co-sine of the Sun's altitude on the first, to 61 degrees, the Sun's distance from the Meridian on the second; and then against 78 degrees 30 minutes, the Co-sine of the declination on the first, is 72 degrees 22 minutes on the second: So much is the Sun's Azimuth from the South Eastwards.   

PROBLEM XXII. The Sun's declination, altitude, and azimuth being known; to find the hour of the day.  
SVppose the Sun's declination be 11 degr. 30 min. his altitude 25 deg. 56 min. and the Azimuth 72 deg. 22 min. Hereby to find the hour of the day.  
Set 78 deg. 30 min. on the first, (which is the Compliment of the declination) to 64 deg. 4 min. on the second, which is the Compliment of the Altitude; and then against 72 deg. 22 min. the Azimuth on the first, is 61 grease on the second. This 61 degrees converted into time, giveth 4 hours 4 minutes, which taken from 12 hours, the Remain is 7 hours 56 minutes before noon, the time of the day required.   

PROBLEM XXIII. The altitude of the Equator, and the Suns, or a Stars declination being given; to find the angle of the Meridian with the Horizon.  
LEt the altitude of the Equator be 51 degrees 30 minutes, and the declination 22 degrees. And the angle of the Meridian with the Horizon sought for.  
Set 68 degrees, the Co-sine of the declination on the first, to the Radius on the second; and then against 51 deg. 30 min. the Co-sine of the Equators altitude, on the first, is 57 degrees 34 minutes on the second: So much is the angle of the Meridian (or circle of Declination) with the Horizon.    

CHAP. IX. The Use of the double Scales in Geography.  

PROBLEM I. Two places lying without the Equinoctial, and having both one Latitude, differing only in Longitude, being propounded; to find their Distance.  
SVppose there be two places, lying both in the Latitude of 48 deg. 20 min. differing only in Longitude 16 degrees, as do Paris in France, and Vienna in Austria, whose distance is desired.  
Set the Radius on the first, to 41 degrees 40 minutes, the Co-sine of the Latitude on the second; and then against 16 degrees on the first, the difference of Longitude, is 10 degrees 36 minutes, which in miles makes 636: So much is the distance between the two places.   

PROBLEM. II. Two places being propounded, which differ in Longitude, the one lying under the Equinoctial, and the other having Latitude, to find their distance in miles.  
LEt two places be propounded, the one lying under the Equinoctial, having 313 degrees of Longitude, and the other lying in the Latitude of 32 deg. 42 min. and in the Longitude of 323 degrees, they differing in Longitude 10 degrees, I would know their distance.  
Set the Radius on the first, to 80 degrees on the second, the Co-sine of the difference of Longitude; and then against 57 deg. 18 min. the Co-sine of the one places Latitude on the first, is 55 deg. 58 min. the Co-sine of the distance in degrees; the Compliment of this 55 deg. 58 min. to 90, is 34 deg. 2 min. which converted into English miles, giveth 2042 miles: So much is the distance between the two places.   

PROBLEM. III. Two places having several Latitudes towards one Pole, and differing in Longitude; to find their distance.  
LEt London and Jerusalem be two places propounded, betwixt which the distance is required: The Latitude of London is 51 degrees 30 minutes, the Latitude of Jerusalem is 32 degrees; and their difference in Longitude is 47 degrees. Now to find their distance.  
1 Set the Radius on the first, to 43 degrees, the Co-sine of the difference of Longitude on the second; and then on the lines of Tangents, against the Co-tangent of the greater Latitude, 38 deg. 30 min. is 28 deg. 28 min. on the second; the fourth term. This fourth term, 28 deg. 28 min. taken out of the Compliment of the lesser Latitude, that is here, out of 58 deg. the Remain is 29 deg. 32 min. Then,  
2 Set 28 deg. 28 min. the fourth term found, to that Remain, 29 deg. 32 min. on the second; and then against the Co-sine of the greater Latitude, viz. 38 deg. 30 min is the sign 39 deg. 14 min. the distance sought for in degrees of a great Circle, which converted into miles, makes 2354 miles, the distance between London and Jerusalem.   

PROBLEM IV. Two places, one having North Latitude, and the other South Latitude, and differing in Longitude; to find their distance.  
SVppose the one place be in 50 degrees of North Latitude, and the other in 32 degr. 25 min. of South Latitude, and their difference in their Longitude 70 degrees, and betwixt these two such places, the distance is desired. To resolve this demand:  
1 Set the Radius on the first, to 20 degrees, the Co-sine of the difference of Longitude on the second; and then against 50 degrees, the Tangent of the greater Latitude, is the Tangent 16 degrees 1 minute, a fourth term. This fourth term, 16 degrees 1 minute, take from 57 degrees 35 minutes, the Compliment of the lesser Latitude, and the Remain is 41 degrees 34 minutes, for a fifth term. Now then:  
2 Set 73 degrees 59 minutes, the Compliment of 16 degrees 1 minute, the fourth term, to 48 degrees 26 minutes, the Compliment of that 41 degrees 34 minutes, the fifth term; and then against the sine, the greater Latitude 50 degrees on the first, is the sine 36 degrees 36 minutes, whose Co-sine 53 degrees 24 minutes is the distance in degrees by a great Circle. This 53 degr. 24 min. being turned into miles, by multiplying by 60, maketh 3204 miles, and in leagues 1068: So much is the distance sought for between the two places.    

CHAP. X. Of plain right lined Triangles.  

PROBLEM. I. In a plain right lined Triangle, right angled, the three angles being known, and one side; to find either of the other sides.  
IN the Triangle annexed ABDELLA, let the angle at A be a right angle, or 90 degrees, the angle at D 43 deg. 20 min, and the angle at B 46 deg. 40 min, and the side AB 230 yards; and let the side AD be the side sought.  
The Resolution of this Problem upon the Instrument is by this Analogy: 

As the Sine of the Angle opposite to the side known,  
Is to the Sine of the Angle opposite to the side required:  
So is the side known,  
to the side sought.   And therefore in this Example it is,  
As the sine 43 deg. 20 min. the angle at D opposite to AB the side known, is to the sine of the angle at B, 46 deg. 40 min. which is the angle opposite to the side sought AD: So is the length of the known side AB 230 yards, to 243 yards and very near three quarters of a yard, the length of the side AD, which was required. Wherefore,  
Upon the lines of sins, Set the sine 43 deg. 20 min. on the first, (the angle at D. opposite to the side given.) To the sine 46 deg. 40 min. on the second, (which is the sine of the angle opposite to the side required.) And then upon the lines of Numbers, against 230, the length in yards of the known side on that first, is 243 and three quarters very near: So than the length of the side AD, which is sought, is 243 yards, and very near three quarters.  
For further illustration hereof; Suppose you were standing at A, and would know how far it is from you to some notable mark, as to D. Here first erect some Mathematical Instrument, as the Theodilit, or some other at A, the place of your standing; and by help thereof, let a mark be set up sideways, (which way the ground will be best for the purpose) as at B, in a line making right angles with the visual line from you at A to D; this done, measure from A to your mark B, and find it 230 yards. Then place your Instrument at B, and observe the angle there as ABDELLA, which find 46 degr. 40 min. this angle had, and the other at B being a right angle, the angle at the mark D must needs be 43 deg. 20 min. being the Compliment of the angle observed at B to 90 degrees: These things had, the work by my Instrument is the same, as is afore showed. For,  
Set 43 deg. 20 min. on the first, to 46 deg. 40 min. on the second; and then upon the lines of Numbers, against the measured length 230 on the first, is 243,7 & better on the second: So than the distance from you to the mark is 243 yards, and something better than seven tenths of a yard, which is very near three quarters. If you please to work this Example by the Tables of Logarithms, you shall find the work of the Instrument to agree most notably with that of the Tables.  
I might have placed the Problems of right lined Triangles after those of Spherical; for that these require to be resolved by the lines of Sines, Tangents and Numbers together, whereas those of Spherical Triangles are resolved only by lines of Sines and Tangents alone, and therefore require the more variety: but the doctrine of plain Triangles being usually first taught, I have therefore first placed it; and the rather, because by these Problems of right lined Triangles, you may see the way to work with both the lines of Numbers, Sins, and Tangents altogether, in any case whatsoever.   

PROBLEM. II. In a plain right lined Triangle obliqne angled, the three angles and one side being known; to find either of the other sides.  
IN the Triangle CBD, let the obtuse angle at C be 112 deg. 0 min. the angle at D be 43 degr. 20 min. and the angle at B 14 degr. 40 min. and let the known side be CD 100 pole, and let the side CB be the side sought.  
The Analogy for resolution of the present Problem, is the same as in the former, wherefore it is; 

As 14 deg. 40 min. the angle opposite to the side known,  
Is to 43 deg. 20 min. the angle opposite to the side sought;  
So is 100, the length of the side known,  
To 271 the length of that side sought.    
Therefore, upon the lines of Sines; Set the sine 14 deg. 40 min. on the first, to 43 deg. 20 min. the sine of the angle opposite to the side required on the the second; and then upon the lines of Numbers, against 100 on the first, is 271 on the second: So much is the length of that side sought. For a further illustration:  
Suppose you were standing at C, and desired to know how far it is from you to B. Here by reason of some impediment, as a pond, a wall, a hedge, or some other thing, you cannot point out your second station at right angles with the visual line from C to B, but are enforced to work by an angle greater than a right angle; wherefore at C observe an angle, as BCD of 122 degrees, and measure along from C to D 100 pole, and at D observe the angle CDB 43 deg. 20 min. These two angles had, add them together, and they make 165 degr. 20 min. which taken from 180 degrees, leaveth 14 deg. 40 min. for the angle at B. (For in all Triangles, if any two of the angles be known, the third is also known, for it is evermore the Compliment of them both to 180 degrees.) These things thus had, the Analogy for resolution of the quaere by the lines, is the same as before. Therefore,  
Set the sine 14 deg. 40 min. on the first, (the sine of the angle opposite to the side known) to the sine 43 deg. 20 min. (the sine of the angle opposite to the side required) on the second; and then on the lines of Numbers, against 100, the length of the side known on the first, is 271 on the second: Therefore 271 pole is the length, or distance of B, the mark from your station C, which was desired.   

PROBLEM. III. If two sides of a right lined Triangle, and an Angle opposite to one of them sides being known; to find the angle opposite to the other side.  
HEre the Analogy standeth thus by the Lines. 

As that side opposite to the angle known,  
Is to the other known side;  
So is the known angle,  
To that angle sought.    
Example. In the Triangle BCD, let the one side known be the side CD, 100 yards; and the other side known be CB, 271 yards; and let the angle at B be known to be 14 degr. 40 min. and let the angle at D be sought.  
Upon the lines of Numbers; Set 100, the side CD opposite to the known angle B on the first, to 271 the side CB opposite to that angle at D sought for on the second; and then on the lines of Sines, against the sine of the known angle 14 deg. 40 min. on the first, is 43 deg. 20 min. on the second: So much is the quantity of the angle D, which was sought.   

PROBLEM. iv Two sides, and the angle included between them, being a right angle; to find the other two angles.  
IF you would find the lesser of the two unknown angles: Then the Analogy is, 

As the greater side is to the lesser,  
So is the Radius to the Tangent of the lesser angle.    
Example. In the Triangle ABC, let there be known the side AB 230 yards, and the side AC 143 and three quarters very near, and the included angle just 90 degrees; and let the angle B, being the lesser angle, be sought; the which to find,  
Upon the lines of Numbers, Set 230, the greater side known on the first, to 143 and three quarters, the lesser side on the second; and then upon the lines of Tangents, against the Radius on the first, is the Tangent 32 degr. 0 min. on the second: So much is the quantity of the lesser angle, which is the angle at B; and the Compliment of that to 90, is 58 degrees: So much is the quantity of the angle C, being the greater of the two angles sought.  
If you would find the greater angle first, of the two unknown angles, than the Analogy is thus: 

As the lesser side, is to the greater,  
So is the Radius, to the Tangent of the greater angle.   Therefore,  
Upon the lines of Numbers, Set the lesser side known AC, 143 and three quarters on the first, to the greater side AB 230 on the second; and then on the lines of Tangents, against the Radius on the first, is the Tangent 58 degrees on the second: So much is the quantity of the greater angle at C, and the Co-tangent thereof is 32 degrees: So much is the quantity of the lesser angle. For it is to be noted, that always the greater angle is opposite to the greater side, and the lesser angle opposite to the lesser side.   

PROBLEM. V In any obliqne angled Triangle, whether obtuse or acute, having two sides, and the angle included between them; to find the other angles.  
IN the Triangle BCD, let the angle known be the obtuse angle at C 122 degrees, and the two sides including that angle, be CB the greater side, containing 271; & CD the lesser side, containing 100, (which sides may be reckoned poles, yards, feet, inches, or what other measure you please) and let the two angles at B and D be sought.  
In this case the Analogy will be thus: 

As the sum of the two sides known,  
Is to the difference betwixt the same sides;  
So is the Tangent of the half sum of the two angles sought,  
To the Tangent of the half of their difference.    
Here the two sides including the known angle, are 100 and 271, which added together, their sum is 371, for the first number in the rule of Proportion. Next take the lesser side 100, from 271 the greater side, and the Remainder is 171, this is the difference between the sides, and is the second number in the rule: And then take the known angle 122 degrees from 180 degrees, or two right angles, and the Remainder is 58 degrees: So much is the quantity of both the other two angles at B and C, whereof the half is 29 degrees, this half sum is the third number in the rule of Proportion: These things had, then according to the Rule it is thus: 

As 371 is to 171,  
So is the Tangent 29 deg.  
To the Tangent 14 deg. 20 min.   Therefore,  
Upon the lines of Numbers, Set 371 the sum of the two containing sides on the first, to 171 the difference of the same sides on the second; and then upon the lines of Tangents, against 29 degrees, the half sum of the two angles sought on the first, is the Tangent 14 deg. 20 min. on the second. This 14 degr. 20 min. thus found, is the half of the difference of two angles sought; therefore, To 29 degr. the half sum of the angles sought, add this 14 deg. 20 min. and the sum is 43 degr. 20 min. So much is the quantity of the angle at D, the greater of the two angles sought; and this 14 deg. 20 min. taken out of that 29 degr. 0 min. the Remainder is 14 deg. 40 min. So much is the angle B, the lesser of the two angles sought.   

PROBLEM. VI In any right lined Triangle, whether right angled, or obliqne angled, any two sides b●ing known, with the angle included between them; to find the third side.  
IN the obliqne angled Triangle CBD, Suppose you have the two sides given CD 100 yards, and CB 271 yards, and the angle C included betwixt them 122 degr. and you require to know the side DB opposite to the obtuse angle C. Here first you must by the last Problem find out the two angles unknown, and then having the three angles and two sides known, you may by the angles and one of the sides find the third side, according to the second Problem: As suppose you would choose the side CD 100, and thereby work to find the side DB. Then the proportion will stand thus: 

As 14 deg. 40 min. the sine of the angle opposite to the chosen side,  
Is to 58 deg. the Compliment of 122 deg. the angle C opposite to the side sought:  
So is 100, the length of the chosen side CD,  
To 335, the length of that side DB, which is required.    
Therefore upon the lines of Sines, Set 14 deg. 40 m. the sine of the angle opposite to the chosen side on the first, to the sine 58 deg. the Compliment of 122 deg. to 180; and then upon the lines of Numbers, against 100, the length of the side chosen on the first, is 335 on the second: So many yards is the length of the side DB, which was required.  
Again, Suppose in the right angled Triangle BAC, right angled at A you have the two sides AB 230, and the side AC 143 and near three quarters, and the included angle 90 degrees. Then to find the other two angles, work as in the last Problem thus: Add the two sides together, 230 and 143 and three quarters, and the sum is 373 and three quarters, for the first number. Then take the lesser side, 143 and three quarters, out of 320 the greater side, and the Remainder is 86, and one fourth for the second number, and the angle A is the third number in the rule of Proportion. Therefore,  
Set 373 and three fourth's, the sum of the sides on the first, to 86 and one fourth, their difference on the second; and then against the Tang. 45 deg. the half sum of the two angles sought on the first, is the Tang. 13 deg. on the second, these 13 deg. added to the 45 deg. and the sum is 58 deg. for the greatest of the two angles sought: And taken from 45 deg. leaveth 32 deg. for the lesser of them two angles.  
The angles thus had, to find the third side CB. Set 32 deg. on the first, to 90 on the second, and then against 143 and three fourth's on the first, (the side chosen to work by) is 271 on the second: So much is the length of the side CB, which is required.  
Or else, making use of the side AB, Set the sine 58 deg. to 90; and then on the lines of numbers, against 320 on the first, is 271 on the second: The length of the side sought, as before.   

PROBLEM VII. In a right lined Triangle, the three sides only being known; to find the Perpendicular, and thereby the three angles.  
LEt the greatest of the three sides of your Triangle be taken for the Base, for so the Perpendicular will fall within the Triangle; then take the sum of the two sides by adding them together, and also their difference, by substracting the lesser from the greater: As in the Triangle CBD, make DB the Base, being the longer side 335 foot, and the other two sides be CD 100, and CB 271, these two sides added together make 371, and the lesser being substracted from the greater, their difference is 171. These had, the Proportion is thus: 

As the Base DB 335, is to the sum of the two sides 371;  
So is the difference of the two sides 171, to a fourth sum.   Therefore,  
Upon the lines of Numbers only, Set 335, the length of the Base on the first, to 371 the sum of the sides on the second; and then against 171, the difference of the sides on the first, is 189,4 on the second; this 189,4 taken out from the Base 335,0, and the Remainder is 145,6, the half whereof is 72,8: So many foot measured in the line DB, from D towards B, (because the angle at D is the greatest of the two angles at the Base) will reach to E, the point where the Perpendicular CE will cut the Base DB. By which Perpendicular, the Triangle CDB, is divided into two right angled Triangles, DEC and CEB, which being done, all the other angles may be found by the third Problem of this Chapter: For here you have two sides DC 100, and DE 72,8, and the right angle DEC opposite to the side DC. Therefore by that third Problem, to find the angle ECD.  
Set 100, the side opposite to the angle known on the first, to 72,8 the side opposite to the angle sought on the second; and then on the lines of sins against 90 degr. the angle known is 46 degr. 40 min. So much is the quantity of the angle ECD, and its Compliment to 90 is 43 deg. 20 min. for the quantity of the angle CDE. Thus have you the three angles of the right angled Triangle DEC. And for the length of the Perpendicular, it is thus attained unto by the first Problem. For,  
Set 46 deg. 40 min. the quantity of the angle C, opposite to the known side ED on the first, to 43 deg. 20 min. the angle D opposite to the Perpendicular EC; and then on the lines of Numbers, against 72,8 the length of the side known on the first, is 68,5 and something more: So much is the length of the Perpendicular EC, which was required.  
In like manner, as the Angles and Perpendicular in this right angled Triangle are found, even so are found the Angles in that other right angled Triangle CEB on the other side of the Perpendicular; for you shall find the angle EBC 14 deg. 40 min. the angle BE 75 deg. 20 min. And in conclusion, you shall find that the two angles at C, made by the Perpendicular, namely ECD and ECB will make just 122 degrees, which was the quantity of the obtuse angle at C, before the Perpendicular was drawn.   

PROBLEM VIII. In a right angled Triangle, the Angles and the Hypothenusal being given, to find any one of the sides including the right angle.  
IN the right angled Triangle BAC, let the angles known be as afore, at C 58 deg. and at B 32 deg. and the angle at A a right angle, and the Hypothenusal CB be 271 yards: and you would hereby find the sides AC and AB, which for to do, the rule is thus by the lines on my Scale. 

As the Radius, is to the angle opposite to the side sought;  
So is the Hypothenusal or side opposite to the right angle, to that side sought.   Therefore to find the side AC.  
Set the Radius on the first, to 32 deg. on the second, which is the angle opposite to the side sought; and then, on the lines of Numbers, against 271 the side known on the first, is 143 and three quarters very near on the second: So many foot is the length of the side AC. And for the side AB.  
Set the Radius on the first, to 58 deg. on the second, (which is the angle opposite to AB, the side sought;) and then on the lines of Numbers, against 271 the Hypothenusal on the first, 230 on the second: So many foot is the length of the side AB.  
Now for your further instruction; Let it be supposed, that AB is the altitude of some body, as a Tree, a Turret, a Steeple, or the like, which you would know, and can come no nearer to it than at C; therefore at C take the angle of altitude BCA 58 degrees, and then the angle at B the top of the altitude, made by the visual line CB, and the altitude AB, will be 32 deg. being the Compliment of the observed angle to 90 deg. for the altitude is supposed to stand perpendicular, and to make a right angle at the base A. Then from C, measure backwards in a right line 100 yards to D, and there observe the angle of altitude BDA, and find it 43 deg. 20 min. These two angles thus observed keep, and take the first observed angle at C 58 deg. out of 180 deg. and the Remainder is 122 deg. which is the quantity of the obtuse angle BCD made by the stationary line CD, and the visual line CB. This angle of 122 deg. and the angle observed at D 43 deg. 20 min. add together, and they make 165 deg. 20 min. which taken from 180 degr. the Remainder is 14 degr. 40 min. So much is the angle at B, the top made by the two visual lines CB and DB. Thus having attained the three angles of a Triangle CBD, and one side being the stationary distance, you may find the side or Hypothenusal CB, by the second Problem, and with the Hypothenusal, the altitude by this Problem.    

CHAP. XI. Of Trigonometria, showing the use of the double Scales of Numbers, Sins and Tangents, in the resolution of Triangles, either plain or spherical.  


PROBLEM. I. The two sides of a rectangled Triangle being given, to find the Base, which is the side opposite to the right angle.  
IN the Triangle ABC of the annexed Diagram, Let the side AC be 27 deg. 54 min. and the side CB 11 deg. 30 min. be given, and the base AB required.  
For resolution of this quaere, say thus: 

As the Radius is to the Co-sine of one of the given sides;  
So is the Co-sine of the other given side to the Co-sine of the base.   Therefore,  
Set the Radius on the first, to 62 deg. 6 min. the Co-sine of the one side given, viz. of AC, 27 deg. 54 min. on the second; and then against the Co-sine of BC, 11 degr. 30 min. which is 78 deg. 30 min. on the first, is 60 deg. the Co-sine of the side or base required: So than the base AB is 30 degrees.   

PROBLEM II. The two sides being given, to find either of the obliqne angles.  
IN the Triangle ABC aforementioned; Let AC be given 27 deg. 54 min. and BC 11 deg. 30 min. and let the angle A, next the side AC, be sought. Say, 

As the sine of the side next the angle sought, is to the Radius;  
So is the Tangent of the side opposite to the angle sought, to the Tangent of the same angle.   Therefore,  
Set 27 deg. 54 min. the side adjacent to the angle sought, to the Radius on the second; and then against the Tangent of 11 deg. 30 min. the side opposite to the angle sought on the first, is the Tangent 23 deg. 30 min. on the second: So much is the angle at A that was required.  
If the angle sought for do appear to be above 45 deg. as doth the angle B, and the sides less than 45 deg. than the best way to find B, is by the next Problem to find the base, and then having the base to find the angle B.  
But if the side opposite to the angle sought be above 45 degrees, as suppose the sides AC were 61 deg. 53 min. and CB 54 deg. 28 min. and the angle A sought for. Then work thus.  
Set the Radius on the first, to the sine of the side adjacent to the angle sought, viz. to the sine of AC 61 degr. 53 min. on the second; and then against the tangent 54 degr. 28 min. the side opposite to the angle sought on that first, is 57 degr. 47 min. the Tangent of the angle A sought for. Here note, that in examples of this kind the angle found is greater than 45 degrees.   

PROBLEM III. One side, and the obliqne angle next unto it being given; to find the Base.  
LEt the side AC 27 degr. 54 min. and the angle A 23 deg. 30 min. be given, and the Base AB sought.  


As the Co-sine of the angle given, is to the Radius;  
So is the tangent of the side given, to the tangent of the base.   Therefore,  
Set 66 deg. 30 min. the Co-sine of the angle given, to the Radius, and then against the tangent 27 deg. 54 min. on the first, is the tangent 30 deg. on the second, the base sought.  
If the side given be above 45 degrees, as were the side AC, 61 deg. 53 min. and the angle A 57 deg. 47 min. then to find the base; Set the Radius to 32 degr. 13 min. the Co-sine of the angle given, and then against the tangent of the side 61 degr. 53 min. is the tangent 74 degr. 6 min. for the base AB.   

PROBLEM IU. The Base, and one of the obliqne angles being given; to find the other obliqne angle.  
IN the Triangle aforesaid; Let there be given the Base 30 degrees, and the obliqne angle A 23 deg. 30 min. and the angle B required. Say, 

As the Radius, is to the Co-sine of the base;  
So is the Tangent of the angle given, to the Co-tangent of the angle required.   Therefore,  
Set the Radius on the first, to 60 degr. the Co-sine of the Base; and then against 23 deg. 30 min. the tangent of the angle given, is the tangent 20 deg. 38 min. which is the Co-tangent of the angle B. Wherefore the said angle is 69 deg. 22 min.  
If the angle given be above 45 degrees; As if the angle given be the angle B, and the lesser angle A sought, then, Set the Co-sine of AB 60 degrees on the second to the Radius; and then against the tangent of the given angle 69 degrees 22 minutes on that first, is 23 degrees 30 minutes, the Tangent of the Angle A.   

PROBLEM. V The Base, and one of the obliqne angles given; to find the side next the same angle.  
LEt the Base AB 30 degr. and the angle A 23 deg. 30 min. be given, and let the side AC adjacent to the angle A be required. Say, 

As the Radius, to the Co-sine of the angle given;  
So is the tangent of the base, to the tangent of the side sought.   Therefore,  
Set the Radius on the first, to 66 degr. 30 min. the Co-sine of the angle given on the second; and then against the tangent of the base 30 degrees, is the tangent of the side sought, 27 deg. 54 min.  
If the base be above 45 degrees; as, Let the base AB be 74 deg. 6 min. and the angle A 57 deg. 47 min. and then to find the base.  
Set 33 degr. 13 min. the Co-sine of the angle given on the first, to the Radius on the second; and then against the tangent of the base 74 deg. 6 min. on the first, is 61 degr. 53 min. the side AC sought for.   

PROBLEM. VI The Base, and one of the obliqne angles given; to find the side opposite to the angle given.  
LEt the Base AB 30 degrees, and the angle A 23 degrees 30 minutes be given, and let the side CB opposite to the same angle A be sought.  


As the Radius, to the Sine of the Base;  
So is the Sine of the angle given, to the Sine of the side sought.   Therefore,  
Set the Radius on the first, to 30 degrees, the Sine of the Base on the second; and then against 23 deg. 30 min. on the first is 11 degr. 30 min. on the second, the side CB, which was sought for.   

PROBLEM VII. One side, and the obliqne angle next it being given; to find the other side.  
LEt the side AC, 27 degr. 54 min. and the angle A 23 deg. 30 min. be given, and the side BC required. Say, 

As the Radius, to the Sine of the side known;  
So is the tangent of the angle known, to the tangent of the side required.   Therefore,  
Set the Radius on the first, to 27 degr. 54 min. the sine of the side known; and then against the tangent of the angle A, 23 degrees 30 min. on the first, is the Tangent 11 degr. 30 minutes for the side CB, which was required.   

PROBLEM VIII. One side, and the obliqne angle next it being known; to find the other obliqne angle.  
LEt the side AC, 27 degr. 54 min. and the angle A 23 deg. 30 min. be given, and the angle B sought. Say, 

As the Radius, to the Co-sine of the side given;  
So is the Sine of the angle given, to the Co-sine of the angle sought.   Therefore,  
Set the Radius on the first, to 62 deg. 6 min. the Co-sine of 27 degr. 54 min. the side given; and then against the Sine of the angle given, 23 deg. 30 min. is the Sine 20 deg. 38 min. the Compliment of 69 degr. 22 min. the angle B, which was required.   

PROBLEM. IX. One side, and the angle opposite to it being known; to find the Base.  
LEt the side BC 11 deg. 30 min. and the angle A 23 deg. 30 min. be given, and let the base AB be sought for. Say, 

As the sine of the angle given, to the sine of the side given;  
So is the Radius, to the sine of the base.   Therefore,  
Set 23 deg. 30 min. the sine of the angle given on the first, to 11 deg. 30 min. the sine of the side given; and then against the Radius on the first, is the sine 30 deg. on the second. Therefore 30 deg. is the base sought for.   

PROBLEM. X. One side, and the angle opposite to it being given; to find the other side.  
LEt the side BC 11 deg. 30 min. and the angle A 23 deg. 30 min. be given, and the side AC sought for. Say, 

As the tangent of the angle given, is to the tangent of the side given;  
So is the Radius, to the sine of the other side.   Therefore,  
Set the tangent of the angle A, 23 degr. 30 min. on the first, to the tangent of 11 degr. 30 min. on the second; and then against the Radius on the first, is 27 deg. 54 min. The sine of the side AC, sought for.  
Sometime it so falleth out in this Problem, and some others, by reason of a great angle or side, that the resolution cannot be readily done by the Analogy set down; yet you shall see ready ways upon the lines to perform the same, or by other Problems to gain the thing sought.   

PROBLEM XI. One side, and the angle opposite to it being known; to find the other angle.  
LEt the side CB 11 degr. 30 min. and the angle A 23 deg. 30 min. be given, and the angle B sought. Say, 

As the Co-sine of the side given, to the Co-sine of the angle given;  
So is the Radius, to the sine of the angle required.   Therefore,  
Set 78 deg. 30 min. the Co-sine of the side CB given on the first, to 66 deg. 30 min. the Co-sine of A the angle given; and then against the Radius on the first, is 69 degr. 22 min. on the second. The sine of the angle sought.   

PROBLEM. XII. The Base, and one side known; to find the obliqne angle adjoining to the same side.  
LEt the side AC, 27 degr. 54 min. and the Base AB 30 degrees, be given, and the angle A adjoining to the side given, sought for. Say, 

As the tangent of the Base, to the tangent of the side given;  
So is the Radius, to the Co-sine of the angle required.   Therefore,  
Set the tangent of the Base 30 degr. on the first, to 27 degr. 54 min. the tangent of AC, the side given on the second; and then against the Radius on the first, is 66 degr. 30 min. on the second, whose Co-sine is 23 degr. 30 min. The sine of the angle A that was sought.  
When the Base and side be above 45 degr. as, Suppose the Base AB to be 74 deg. 6 min. and the side AC 61 deg. 53 min. than set the tangent of that side next the angle sought, viz. 61 deg. 53 min. on the first, to the tangent of the Base 74 degr. 6 min. on the second; and then against the Radius on the first, is 32 degr. 13 min. the Co-sine of the angle A. After this manner, by altering the placing of the terms given, one way or other, any quaere may be resolved.   

PROBLEM XIII. The Base, and one side being given; to find the angle opposite to the same side.  
LEt the side AC, 27 degr. 54 min. and the Base AB 30 deg. be given, and the angle B required. Say, 

As the Sine of the Base, to the Radius;  
So is the Sine of the side given, to the Sine of the angle required.    
Set 30 deg. the sine of the Base on the first, to the Radius on the second; and then against 27 deg. 54 min. the sine of AC the side given on the first, is 69 deg. 22 min. The sine of the angle B, which was required.   

PROBLEM. XIV. The Base, and one side being given; to find the other side.  
LEt the side AC 27 degr. 54 min. and the Base AB 30 degrees be given, and the side BC sought for. Say. 

As the Co-sine of the side given, to the Radius;  
So is the Co-sine of the Base, to the Co-sine of the side sought.   Therefore,  
Set 62 degr. 6 min. the Co-sine of the side given, to the Radius; and then against 60 degr. the Co-sine of the Base, is 78 deg. 30 min. the Co-sine of the side BC, 11 degr. 30 min. that was required.   

PROBLEM. XV. The two obliqne angles being given, to find the Base.  
LEt the angle A be 23 degr. 30 min. and the angle B 69 degr. 22 min. be given, and let the Base AB be sought.  


As the tangent of one of the angles, is to the Co-tangent of the other angle;  
So is the Radius, to the Co-sine of the Base.   Therefore,  
Set the tangent of the one angle, (as of that at A 23 degr. 30 min. on the first, to the Co-tangent of the other angle, viz. to 20 degr. 38 min. on the second; and then against the Radius on the first, is the sine 60 degrees on the second. Which is the Co-sine of 30 degrees, the Base AB, which was required.   

PROBLEM. XVI. The two obliqne angles being given; to find either of the sides.  
LEt the angle A be 23 degr. 30 min. and the angle B 69 deg. 22 min. and let the side AC be sought. Say, 

As the sine of one of the angles, is to the Co-sine of the other angle;  
So is the Radius, to the Co-sine of the side opposite to that angle, whose Co-sine was taken.   Therefore,  
Set the sine of the angle A 23 degr. 30 min. on the first, to 20 deg. 38 min. the Co-sine of the other angle B 69 degr. 22 min. And then against the Radius, is 62 degr. 6 min. which is the Co-sine of 27 deg. 54 min. the sine of the side AC sought. Having showed the use of the double lines, through all the varieties of right angled Spherical Triangles, I come to the Obliqne.    

In Obliqne angled spherical Triangles.  

PROBLEM XVII. Two angles, and a side opposite to one of them being given; to find the side opposite to the other.  
LEt the Triangle proposed be ABE, wherein let be given the angle E 38 deg. 15 min. and the side BASILIUS 30 degrees, and the angle A 23 degrees 30 minutes, and let the side BE be sought. Say, 

As the Sine of the angle opposite to the side known, is to the Sine of the same side;  
So is the Sine of the angle opposite to the side sought, to the Sine of that same side.   Therefore,  
Set 38 deg. 15 min. the sine of the angle E, to the Sine of the side given 30 deg. (viz. AB) And then against 23 degr. 30 min. the Sine of the other angle A on the first, is 18 degr. 47 min. The Sine of BE, the side sought.   

PROBLEM XVIII. Two sides, and an angle opposite to one of them being known; to find the angle opposite to the other of them.  
IN the Triangle ABE, let the side AB be 30 degr. the angle E 38 degr. 15 min. and the side BE 18 degr. 47 min. and let the angle A, which is the angle opposite to the side BE, be sought. Say, 

As the Sine of the side opposite to the angle given, is to the Sine of the same angle;  
So is the Sine of the side opposite to the angle sought, to the Sine of that same angle.   Therefore,  
Set 30 degrees, the side AB on the first, to 38 deg. 15 min. the Sine of its opposite angle on the second; and then against 18 deg: 47 min. the Sine of BE the other side, is 23 degr. 30 min. the Sine of A its opposite angle, which is the thing that was to be sought.   

PROBLEM. XIX. Two sides, and the angle included between them, being known; to find the other side opposite to that angle.  
IN the Triangle ABE, let the sides AB 30 deg. and A 42 deg. 51 min. with the angle A included between them, be given, and the side BE required.  
Here first a Perpendicular must be supposed to be let fall from the unknown angle B, to the side A opposite to that angle, and now to find on what part of the side A it will fall. Say, 

As the Radius, to the Co-sine of the included angle A;  
So is the Tangent of AB, to the Tangent of AC.   Therefore,  
Set the Radius on the first, to 66 deg. 30 min. the Co-sine of the included angle A; and then against the Tangent of AB 30 degrees the lesser side, is 27 degrees 54 minutes, the Tangent of the distance from A to C, on the side A; therefore C is the point whereon the Perpendicular will fall: This distance AC. 27 deg. 47 min. deduct out of the whole side A 42 deg. 51 min. And the Remainder is 14 deg. 57 min. Then say, 

As the Co-sine of AC, the distanee found,  
Is to the Co-sine of CE, the Remainder;  
So is the Co-sine of AB, the lesser side,  
to the Co-sine of BE, the side required.   Therefore,  
Set 62 deg. 6 min. the Co-sine of AC, the distance found, to 75 degr. 3 min. the Co-sine of CE, the Remainder; and then against 60 degrees, the Co-sine of the side AB. is 71 deg. 13 min. the Co-sine of BE, 18 degr. 47 min. the side required.  
It is to be observed, that in obliqne angled Triangles, when the terms propounded are two sides and one angle, or two angles and one side, and yet not resolveable by the two last Problems before this: That then a Perpendicular is to be let fall, (or supposed to be let fall) from one of the unknown angles, to the side opposite thereto; and so of the obliqne angled Triangle given, to make two right angled Triangles. Which being done, all the parts of such a Spherical Triangle, may be found out by the former Problems of right angled Triangles. This Perpendicular will sometime fall within the Triangle, and that is, when the angles at the ends of the side whereon the Perpendicular falls, are both of one kind, that is to say, both acute, or both obtuse, as in the Triangle ABE. And sometime this Perpendicular falls without the Triangle, and that is, when the angles at the ends of the side whereon it falls, are of differing kinds; that is, one acute, and the other obtuse, as in the Triangle ADB, where the Perpendicular BC falls on the side AD, it being prolonged, as it must always be in those cases.   

PROBLEM XX. Two sides, with the angle included between them, being given; to find either of the other angles.  
LEt the sides AB 30 degr. and A 42 degr. 51 min. with the angle A 23 degr. 30 min. be given, and the angle E sought.  
As in the last Problem, so here a Perpendicular must be let fall on the given side A, from its opposite angle B, and so of the obliqne angled Triangle, make two rectangled Triangles. This Perpendicular, as afore was showed, is thus found: Set the Radius to 6● degr. 30 min. the Co-sine of the included angle; and then against the Tangent of ●0 deg. the further side, is the Tangent 27 degr. 54 min. So much is the part of the side A, betwixt A and the point C, where the Perpendicular falleth. Now if 27 degr. 54 min. be taken from 42 deg. 51 min. the Remain is 14 deg. 57 min. for the side of the rectangled Triangle BE. Now to find the angle E, the Analogy is thus. 

As the sine of the side CE, is to the sine of the side AC;  
So is the Tangent of the angle included, to the Tangent of the angle sought.   And therefore,  
Set the sine of CE, 14 deg. 57 min. on the first, to the sine of AC, 27 degr. 54 min. the part found; and then against the Tangent of A, 23 deg. 30 min. is the Tangent 38 degr. 15 min. the Angle E that was sought.  
Otherwise, having let fall the Perpendicular, and having found CE, 14 deg. 57 min. together with the Base 18 degr. 47 min. then to find E, work as by the 12 Problem of rectangled Spherical Triangles, in this manner.  
Set the Tangent of the Base, 18 degr. 47 min. to the Tangent of that part of the side CE, 14 degr. 57 min. And then against the Radius on the first, is 38 degr. 15 min. the sine of the angle E, as before. In this manner may any of the other terms be also found out, by one or other of those Problems.   

PROBLEM. XXI. Two sides, and one angle next to the side sought being given; to find the same side.  
LEt the sides AB 30 degr. and BE 18 degr. 57 min. and the angle A 23 deg. 30 min. be given, and the side A required. Here first let fall the Perpendicular BC, and say, 

As 60 deg. the Co-sine of AB, is to 71 deg. 13 min. the Co-sine of BE, 18 degr. 47 min.  
So is the Co-sine of AC 27 degr. 54 min. viz. 62 deg. 6 min. to 75 degr. 3 min. the Co-sine of CE.   Therefore,  
Set 60 deg. the Co-sine of AB, to 71 degr. 13 min. the Co-sine of BE; and then against 62 deg. 6 min. the Co-sine of AC, is 14 degr. 3 min. the Co-sine of CE. Wherefore CE is 14 degr. 57 min. which added to AC 27 degr. 54 min. makes 42 degr. 51 min. for the side A, which was sought.   

PROBLEM XXII. Two sides, and an angle adjacent to one of them being given; to find the angle included between the same sides.  
LEt the sides given be AB, 30 degrees, and BE 18 degr. 47 min. with the angle A 23 deg. 30 min. and let the obtuse angle B included betwixt the two sides, BA, and BE, be required▪ To resolve this demand, Say, 

1 As the Radius, to the Co-sine of AB;  
So is the Tangent of the angle A, to the Tangent of aBc.   

2 As the Tangent of BE, to the Tangent of AB;  
So is the Co-sine of aBc, to the Co-sine of cBe.   Therefore,  
1 Set the Radius, to 66 degrees, the Co-sine of AB, and then against the Tangent of A 23 degr. 30 min. is the Tangent 69 degr. 22 min. the angle aBc, made by letting fall the Perpendicular, and is part of the angle aBc sought. Now to find the Remainder.  
1 Set 18 degr. 47 min. the Tangent of BE on the first, to the Tangent of AB 30 degrees; and then against the Co-sine of aBc, which is 20 deg. 38 min. is 36 deg. 46 min. whose Compliment 53 degr. 14 min. is the angle cBe, which added to 69 degr. 22 min. makes 122 deg. 36 min. the angle able, which is the thing required.   

PROBLEM XXIII. Two angles, with the side lying betwixt them being known; to find either of the other sides.  
LEt be given the angles A 23 degr. 30 min. and B 122 degr. 36 min. and the side AB 30 degrees, and let the side BE be required. In which case having let fall the Perpendicular BC, and found the angle cBe. Say, 

As the Co-sine of cBe, is to the Co-sine of aBc;  
So is the Tangent of AB, To the Tangent of BE.   Therefore,  
Set 36 degr. 46 min. the Co-sine of the angle cBe on the first, to the Co-sine of the angle aBc; and then against 30 degrees, the tangent of the side given, is the tangent 18 deg. 47 min. the side BE, which was required.   

PROBLEM. XXIV. Two angles, with the side lying betwixt them being given, to find the third angle.  
LEt the angle A 23 deg. 30 min. and the angle B 122 degr. 36 min. with the side AB 30 degrees be given, and the third angle at E, required.  
Having let fall the Perpendicular, and found the angles aBc, and cBe. Then say, 

As the Sine of aBc, is to the Sine of cBe;  
So is the Co-sine of A, the angle given, to the Co-sine of E, the angle sought.   Therefore,  
Set 69 deg. 22 min. the Sine of aBc on the first, to 53 degr. 14 min. the Sine of cBe on the second; and then against the Co-sine of A 66 deg. 30 min. on the first, is 51 degr. 45 min. the Co-sine of the angle E. Wherefore E is 38 deg. 15 min.   

PROBLEM. XXV. Two angles, and a side opposite to one of them being given; to find the side adjacent to both the angles.  
LEt the angles A 23 degr. 30 min and E 38 15 min. with the side BASILIUS 30 degrees, being opposite to the angle E be given, and the side A required.  
Having let fall the Perpendicular, and together with it found AC, a part of A Say then, 

As the tangent of E, is to the tangent of A.  
So is the sine of AC, to the sine of CE.   Therefore,  
Set 38 deg 15 min. the tangent of E on the first, to 23 degr. 30 min. the tangent of A on the second; and then against 27 degr. 54 min. the sine of the side of AC on the first, is 14 deg. 57 min. on the second, which is the sine of CE. This 14 degr. 57 min. added to 27 degr. 54 min. the other part AC afore found, makes 42 deg. 51 min. for the whole side A, which was sought.   

PROBLEM. XXVI. Two angles, and a side opposite to one of them being known; to find the third angle.  
LEt the angles given be A 23 deg. 30 min. and E 38 deg. 15 min. and let the side AB, being opposite to the angle E, be 30 degr. and let the third angle able be required.  
As in the former Problems, so also in this; Let fall the Perpendicular BC, and by it gain the angle aBc, which being had, to gain the Remainder of the angle able. Say, 

As the Co-sine of the angle A, is to the Co-sine of the angle E;  
So is the sine of aBc, to the sine of cBe.   Therefore,  
Set 66 deg. 30 min. the Co-sine of the angle A on the first, to 51 degr. 45 min. the Co-sine of the angle E on the second; and then against 69 degr. 22 min. the sine of aBc on the first, is 53 degr. 14 min. the sine of cBe. These two angles added together, make 122 degr. 36 min. the angle ABE, which was required.   

PROBLEM. XXVII. The three sides being given, to find any one of the angles.  
LEt the three sides be AB 30 degr. A 42 deg. 51 min. and BE 18 degr. 47 min. and let the angle B be sought.  
To resolve this Problem: First, Add all the three sides together into one sum, of which sum take the half, and then from that half take the side opposite to the angle sought, and reserve the difference. As for Example. The three sides given, are AB 30 d. 00 m. A 42 d. 51 m. BE 18 d. 47 m. The sum of the three sides  91 d. 38 m. The half of that sum  45 d. 49 m. The side A, opposite to the required angle B, substracted  42 d. 51 m. And the Differ. to be reserved is  2 d. 58 m.  
This done, to proceed to find the angle sought. Say, 

1 As the Radius, is to the Sine of one of the sides including the angle sought;  
So is the Sine of the other side, including the same angle, to a fourth Sine.   

2 As that fourth Sine, is to the Sine of the half sum;  
So is the Sine of the reserved difference, to a seventh Sine.   Therefore,  
1 Set the Radius on the first, to 30 deg. the Sine of one of the sides given, including the angle sought; and then against 18 deg. 47 min. the Sine of the other including side, is 9 degr. 16 min. for a fourth Sine.  
2 Set this fourth Sine 9 deg. 16 min. on the first, to 45 degr. 49 min. the Sine of the half sum on the second; and then against 2 degr. 58 min. the Sine of the reserved difference, is 13 deg. 20 min. for the seventh Sine.  
This being done (by the help of a pair of Compasses, or otherwise) divide the distance betwixt this point of 13 degr. 20 min. and the Radius into two equal parts, and the middle point is at 28 deg. 42 min. whose Compliment 61 deg. 18 min. doubled, makes 122 degr. 36 min. the angle a B e, which was sought.   

PROBLEM. XXVIII. The three angles being only known, to find any of the sides.  
TO resolve this Problem, the angles are to be turned into sides, and the sides into angles, (otherwise it is unresolveable) and it is thus done: In stead of the greatest angle, take its Compliment to 180 degrees, and then the angles convert themselves into sides, and the sides into angles, and then the resolution is just the same as in the last Problem.    

 diagram of an oblique-angled triangle  
 diagram of a right-angled triangle    

CHAP. XII. The Use of the Instrument in Navigation.  
THis Instrument, with these Double Scales inscribed thereon, is of most excellent use in Navigation, as I shall show in part, and leave the rest to the Ingenious to find out of themselves. And first, by the Meridian Line, and Line of equal parts joined together upon the Instrument; you may do all things that may be done by the Table of Meridional Degrees, set forth by Mr. Wright, Mr. Gunter, Mr. Wingate, and others treating of Navigation.  

PROBLEM I. Two places being propounded, the one under the Equinoctial, and the other without; to find their Meridional Difference.  
LOok the Latitude of that place situate without the Equinoctial, upon the Meridian line, and right against it, on the Line of equal parts, is the Meridional Difference of those two places.  
Example. Let the entrance of the River of the Amazons, under the Equinoctial, be one place propounded, and the Lizard in the Latitude of 50 degrees be the other place, between which the Meridional Difference is desired.  
Look 50 degrees, the Latitude of the Lizard on the Meridian Line, and right against it on the Line of equal parts, is 57 degrees 90 centesimal minutes, which is 57 degrees, and 9 tenth parts of a degree, or 57 degrees, 54 sexaginary minutes. So much is the Meridional Difference of those two places.   

PROBLEM. II. Two places having both Northerly, or both Southerly Latitude; to find their Meridional Difference.  
Subtract the degrees and minutes which you find on the Line of equal parts against the lesser Latitude, from the degrees and minutes found on the same Line of equal parts against the greater Latitude, and the Remainder is the Meridional Difference of those two places.  
Example. Let St. Christopher's and the Lizard be two places, between which you would find the Meridional Difference; the Latitude of St. Christopher's is known to be 15 degrees 30 minutes North, to which answereth on the Line of equal parts 15 degr. 69 centesimal minutes: the Latitude of the Lizard is also known to be 50 degrees North. To which on the Line of equal parts answereth 57 degr. 90 centesms. Now take 15 degr. 69 cent. the lesser number of degrees found on the Line of equal parts, out of 57 deg. 90 cent. the greater number of degrees found on the Line of equal parts, and the Remainder is 42 degr. 21 cent. or 42 deg. 12 min. of the sexaginary division. So much is the Meridional Difference desired.   

PROBLEM. III. Two places being situate, the one Southerly, and the other Northerly; to find the Meridional Difference.  
TAke the degrees and parts found on the Line of equal parts, against the two Latitudes proposed, sought in the Meridian Line, and add them both together, and the sum of that addition is the Meridional Difference.  
Example. Let the two places proposed be the Cape of Good Hope, in the Latitude of 36 deg. 30 min. South, and Japan in the East Indies, whose Latitude is 30 degrees North. Now on the Meridian Line, look 36 degr. 30 min. and against it on the Line of equal parts, is 39 deg. 15 min. and against the 30 degrees is 31 degr. 28 min. These two sums 39,15, and 31,28, being added together, make 70 deg. 43 min. So much is the Meridional Difference required.   

PROBLEM IU. The Latitude of two places, and their Difference of Longitude being known; to find the Rumb leading from the one to the other.  
LEt the two places proposed be the Lizard and St. Christopher's; the Latitude of the Lizard 50 degrees, and of St. Christopher's 15 degr. 30 min. and their Difference of Longitude 68 deg. 30 min. And let the Rumb passing from the Lizard to St. Christopher's be required.  
First, by the second Problem, seek the Meridional Difference between the two places, which you find to be 42 deg. 12 min. by that second Problem. Then,  
Upon the Line of Numbers, Set 68 degr. 30 min. the Difference of Longitude on the first, to 42 degr. 12 min. the Meridional Difference on the second; and then upon the Lines of Tangents against the Radius on the first, is the Tangent 58 deg. 21 min on the second. So much lieth the Rumb from the Meridian, which is on the fifth Rumb from the Meridian Westwards, and two degrees 6 minutes, over and above that fifth Rumb.  
And if you be to sail from the Lizard to the Bermudas Lands, in the Latitude of 32 degr. 25 min. whose Difference of Longitude is known to be Westwards 70 degrees; and you would know on what point of the Compass you must steer your course thither.  
Here first by the second Problem, seek the Meridional Difference, and find it 23 degr. 36 min. And then,  
Set 70, the Difference of Longitude on the first, to 23 deg. 36 min. the Meridional Difference on the second; and then against the Radius on the first, is the Tangent 71 deg. 21 min. on the second; that is, on the point 71 deg. 21 min. from the Meridian, or South point, which is on the sixth Rumb, and 3 deg. 51 min. over. And because the Bermudas lie Westwards from the Meridian of the Lizard; therefore the Rumb is the West Southwest point, and 3 deg. 51 min. over.  
Here is to be noted, that every Rumb from the Meridian containeth 11 deg. 15 min. And therefore the first Rumb makes an angle with the Meridian of 11 degr. 15 min. the second Rumb an angle of 22 degr. 30 min. and so of the rest, as in the little Table following appeareth, wherein you see that the sixth Rumb makes an angle of 67 degr. 30 min. which taken from 71 deg. 21 min. in this last Example found, there remained 3 degr. 21 min. over the sixth Rumb.  
Rumbs 1 11 deg. 15 min. 2 22 deg. 30 min. 3 33 deg. 45 min. 4 45 deg. 00 min. 5 56 deg. 15 min. 6 67 deg. 30 min. 7 78 deg. 45 min. 8 90 deg. 00 min.   

PROBLEM. V The Latitudes of two places being given with the Rumb; to find the distance upon the Rumb.  
LEt the place from whence you sail be in the Latitude of 50 degrees, and the Latitude of a place to which you are come, be 52 degr. 30 min. So that the Difference of Latitude is 2 deg. 30 min. find also you and that you have sailed on the first Rumb, from the Meridian Northerly, that is, have made an angle with the Meridian of 56 degr. 15 min. The Compliment of the fifth Rumb to eight is 3, which is 33 deg. 45 min. and you would know your distance sailed upon that Rumb.  
Set 33 degr. 45 min. the Co-sine of the Rumb on the first, to the Radius on the second; and then on the Line of Numbers, against 2,50 the Difference of Latitude on the first, is 4,50 on the second, that is to say, 4 degrees 50 centesmes, or 4 deg. 30 min. So much is the distance sailed upon that course, in degrees and minutes.  
Again, Let the Lizard, lying in 50 degrees of North Latitude, be th' place from whence you are to go; and let the other place that you are to go unto be St. Christopher's, in the Latitude of 15 degr. 30 min. North. And the Rumb leading from one to the other, bearing from the Meridian of the Lizard 58 degr. 21 min. which is something more than the fifth Rumb. Now by these things known, you would know the distance upon the Rumb between those two places.  
Here you must first of all subtract 15 degr. 30 min. the lesser Latitude out of 50 degrees, the greater Latitude, and the Remainder 34 deg. 30 min. is the true Difference of Latitude; and then, Set 31 degr. 39 min. the Co-sine of the Rumb on the first, to the Radius on the second; and then on the Line of Numbers, against 34,30 representing 34 degr. 30 min. the Difference of Latitude on that first, is 65,46 on the second, that is 65 deg 46 min. So much in degrees and minutes, is the distance upon the Rumb between the Lizard and St. Christopher's.   

PROBLEM VI The Latitudes of two places, and their distance being known; to find the Rumb leading from one to the other.  
LEt the two places propounded be those mentioned in the last Problem, whose Latitudes were 50 degrees, and 52 degr. 30 min. and their distance known 4 deg. 30 min. & the Rumb you know not, but desire to know it.  
Upon the Line of Numbers, Set 4,30 representing 4 deg. 30 minutes, the distance known, on the first, to 2,30 representing 2 degrees 30 minutes, the Difference of Latitude on the second; and then upon the Lines of Sines, against the Radius on the first, is the Sine 33 deg. 45 min. on the second, whose Co-sine is 56 deg. 15 min. So much is the distance that the course lieth from the Meridian, which is upon the fifth Rumb.  
Or let the Lizard, in the Latitude of 50 deg. and St. Christopher's in the Latitude 15 degr. 30 min. be the two places proposed, whose Difference of Latitude is 34 degrees 46 min. and their distance known 65 deg. 46 min. and the Rumb that the course lieth upon, is desired.  
Upon the Line of Numbers, 65,46 the distance known on the first, to 34,46 the Difference of Latitude on the second; and then on the Lines of Sines, against the Radius on the first, is 58 degr. 21 min. on the second. So much is the distance of the Rumb from the Meridian of the Lizard, on which the Course lieth.   

PROBLEM VII. The Latitude of the place from whence you go, the Rumb you go upon, and the distance gone, being given; to find the Difference of Latitude, and thereby the Latitude of the place you are in.  
LEt the place you go from, be in the Latitude of 50 degrees, the Rumb you have gone upon, the fifth from the Meridian, and the distance gone, 4 degr. 30 min. And by these you desire to know the Difference of Latude, with the Latitude of the place you are in.  
Set the Radius on the first, to 33 degr. 45 min. the Co-sine of the Rumb on the second; and then upon the Lines of Numbers, against 4,30 representing 4 degr. 30 min. the distance gone on the first, is 2,30 on the second, that is 2 deg. 30 min. So much is the Difference of Latitude. This 2 degr. 30 min. added to the Latitude of the place you came from, (because the Course was Northerly) makes 52 deg. 30 min. for the Latitude of the place you are in.  
Example. Suppose you were at St. Christopher's, in the Latitude of 15 degr. 30 min. and from thence had sailed Northwards near upon the fifth Rumb from the Meridian; or on the true point of 58 deg. 21 min. from the Meridian, and at length came to a place, where you found that you had sailed 65 deg. 46 min. from St. Christopher's, and now do desire to know in what Latitude you are.  
Set the Radius on the first, to 31 degr. 39 min. the Co-sine of the Rumb on the second; and then upon the Line of Numbers, against 65,46 representing 65 deg. 46 min. the distance gone on the first, is 34,30 on the second, signifying 34 deg. 30 min. So much doth the Latitude you are in, differ from that you came from, and because the Course was Northerly, the Latitude is increased. Therefore this Difference found, 34 degr. 30 min. is to be added to the known Latitude, 15 degr. 30 min. and they both make 50 degrees just: Therefore the Latitude of the place you are in, is 50 degrees of North Latitude.  
If the place you went from had been the Lizard, in 50 degrees of Latitude, and you sail from thence more Southwards upon the point 58 degr. 21 min. from the Meridian Westwards, and on this Course have sailed 65 degr. 46 min. and so found, that you have altered your Latitude 34 degr. 30 min. and are come to a place of unknown Latitude. In this case, because the Course is Southerly (the Latitude of the place you are in, is less than that you came from) you must take the Difference of Latitude found 34 deg. 30 min. from 50 degr. the Latitude known, the Remainder is 15 deg. 30. The Latitude of the place you are in.   

PROBLEM VIII. The Latitude of the place you are in, and the Latitude of the place you went from, with the Rumb being given; to find the Difference of Longitude.  
LEt the two places be, one the Lizard, in the Latitude of 50 degrees, from whence you go, and the other place St. Christopher's, in the Latitude of 15 degr. 30 min. and the Rumb 58 degr. 21 min. from the Meridian, and the thing desired is the Difference of Longitude.  
First, by the second Problem, find out the Meridional Difference, 42 deg. 12 min. which had; Set the Tangent of the Rumb, 58 degr. 21 min. on the first, to the Radius on the second; and then upon the Line of Numbers, against 42 deg. 12 min. on the first, is 68 deg. 30 min. on the second. Wherefore 68 degr. 30 min. is the Difference of Longitude betwixt your two places, the Lizard and St. Christopher's.  
And if the place you came from, be in the Latitude of 50 degrees, and of that you are in, be 52 degr. 30 min. and the Rumb you have come upon, be the fifth from the Meridian, and you desire the Difference of Longitude. Then,  
Set 56 deg. 15 min. the Tangent of the fifth Rumb on the first, to the Radius on the second, and then upon the Line of Numbers, against 2 deg. 30 min. the Meridional Difference of Latitudes on the first, is 3 deg. 45 min. on the second. So much is the Difference of Longitude desired, according to the projection of the common Sea-Chart.   

PROBLEM. IX. In any Parallel of Latitude, to find how many leagues answer to one degree of Longitude in that Parallel.  
THis Problem is grounded upon this Analogy. 

As the Radius, is to the Co-sine of the Latitude;  
So is the number of leagues, in one Equinoctial degree, to the number of leagues answering to one degree in that Latitude.    
Example. In the Latitude of 18 deg. 12 min. I demand how many leagues sailing along in that Parallel, will alter the Longitude one degree?  
For Answer, to my desire, I set the Radius on the first, to 71 deg. 48 min. the Co-sine of the Latitude on the second; and then on the Line of Numbers, against 20 on the first, the number of leagues in one Equinoctial degree, I see 19 on the second: And therefore every 19 leagues sailing along in that Parallel, altereth the Longitude one degree: And on the contrary, for every degree that you altar your Longitude, you sail 19 leagues in that Parallel.  
2 In the Latitude of 51 degrees 32 minutes, it is demanded, How many leagues in that Parallel, do answer to one degree of Longitude?  
Set the Radius on the first, to 38 degr. 28 min. the Co-sine of the Latitude on the second; and then on the Line of Numbers, against 20 on the first, is 12 and 4 tenths on the second. So many leagues sailing in that Parallel of 51 degr. 30 min. altereth the Longitude one degree.  
If you would know how many miles altar one degree of Longitude in any Parallel. Then,  
Set the Radius on the first, to 38 degr. 28 min. the Co-sine of the latitude on the second; and then on the lines of Numbers, against 60 on the first, is 37 and about one third, which is 37 miles, and one third of a mile, to make one degree of longitude in that Parallel of latitude. And contrariwise, if you find you have altered your longitude one degree, you are then removed 37 miles, and about one third of a mile.   

PROBLEM X. To find how many miles answer to many degrees of Longitude in any Parallel.  
SVppose you should sail along in the Parallel of 50 degrees, until you have altered your longitude 35 degrees, and then would know how many miles you have sailed.  
First, reduce 35 Equinoctial degrees into minutes, by multiplying them by 60, and they make 2100 minutes. Then,  
Set the Radius on the first, to 40 degr. the Co-sine of the latitude on the second; and then against 2100, the Difference of longitude in minutes, upon the first on the line of Numbers, is 1350 on the second. So many miles answer to 35 degrees, in that Parallel.  
But if you would know how many leagues you have sailed, in altering your longitude 35 degrees, then reduce the 35 degr. into leagues, by multiplying them by 20, and they make 700. Now, Set the Radius to 40 degrees, as afore, and then right against 700, in the line of Numbers on the first, is 450 on the second. So many leagues you have sailed.   

PROBLEM XI. Upon any Rumb proposed, to find how many leagues do answer to one degree of latitude in the Meridian, or of any great circle.  
THis Problem is resolved by this Analogy. 

As the Co-sine of the Rumb from the Meridian,  
Is to the Radius;  
So is 20 leagues, the measure of one degree in any great circle,  
To the leagues that answer to one degree upon that Rumb.    
Example. Suppose you sail upon the fourth Rumb from the Meridian, which is the point of North-east, or Northwest; or else the point Southeast, or South-west, and desire to know upon that Rumb, how many leagues do answer to one degree of the Meridian, or alter the latitude one degree.  
Set 45 degrees, the Co-sine of the Rumb on the first, to the Radius on the second; and then on the line of Numbers, against 20, the number of leagues in a degree of a great circle on the first, is 28 and almost an half. Therefore 28 leagues and an half, make one degree, or alter the Latitude one degree, in sailing upon that Rumb.  
If you desire to know how many miles sailing on any Rumb, altar the Latitude one degree, as here in this Example of the fourth Rumb; then set 45 degrees, the Co-sine of the Rumb on the first, to the Radius on the second; and then on the lines of Numbers, against 60 on the first, is 85 and something above on the third. So many miles, on that fourth Rumb, altereth one degree of Latitude.  
I am now in the midst of the Sea, where if I should sail through all particulars that my Instrument is capable of, I should with my travel fill a great Volume, I will therefore leave off wading any further in this Subject, and leave the rest to the ingenuous Practitioners in Navigation, to whom I wish prosperity in all their honest and laudable undertake.    

CHAP. XIII. The Use of the double Scales of Sines and Tangents on the Instrument, in Dialling.  

PROBLEM. I. To find the distance of the hours from the Line of twelve a clock, in Horizontal Dial's, made for an obliqne Sphere.  
LEt an Horizontal Dial be propounded, to be drawn for the elevation of the Pole 51 deg. 30 min. For making whereof the Analogy standeth thus: 

As the Radius,  
Is to the Sine of the Elevation;  
So is the Tangent of the hours in a right Sphere from 12,  
To the Tangent of the same hours from the line of 12, in the obliqne Sphere propounded.   Therefore,  
Set the Radius on the first, to the Sine of the Elevation given, viz. to 51 deg. 30 min. on the second; and then the Instrument unremoved, against 15 deg. the Tangent of an Equinoctial hour on the first, is 11 degr. 51 min. the Tangent of one hours' distance from the line of 12 a clock in the obliqne Sphere of 51 degr. 30 min. and against 30 degrees on the first, being the distance of two Equinoctial hours, is 24 deg. 19 min. on the second, for the distance of two hours from the line of 12, in that given latitude. And likewise, against the Tangent of 45 degrees on the first, the distance of three Equinoctial hours, in a right Sphere on that first, is the Tangent of 38 degr. 3 min. on the second, for the distance of the hours of 3 and 9 a clock, from the hour-line of 12, in that given latitude.  
But now for the fourth and fifth hours, because the Tangents of those hours be above 45 degrees, you must work backwards in this manner: The Instrument not stirred, look 60 degrees, the equal distance of the fourth hour, in a right Sphere on the second; and then against that Tangent of 60 degrees on that second, is the Tangent 53 degr. 35 min. on the first, for the hours of four and eight; and also against the Tangent of 75 degrees, the fifth equal hour on that second, is the Tangent 71 degr. 6 min. on the first, for the hours of 5 in the forenoon, and 7 after noon: As for the hours of 6 and 6, they are at just 90 degr. from the hour-line of 12 a clock, and so for any other latitude: for the hours of 5 and 4 in the morning, they are equally distant from the hour of 6, as is 7 and 8; and so in the afternoon, the hours of 7 and 8 are equally distant from 6, as is 5 and 4. By the same rule shall you find half hours and quarters, as when you would have the true distance of one hour & an half from the noon-line, then work by the Tangent of 22 degr. 30 min. and in so doing, you shall have 17 degr. 38 min. to be the distance of one hour and an half from the noon-line, in the latitude of 51 degr. 30 min. and so for any other.   

PROBLEM II. A Dial being made, and the Elevation for which it was made not being known; to find for what Latitude it is made.  
FIrst of all, get the distance between the hours of 12 and 1, which had, Set the Tangent of 15 deg. on the first, to the Tangent of that distance on the second; and then against the Radius on the first, is the Sine of the Elevation sought for on the second.  
Example. There is an old Dial, which doth appear to be a good one, and I desire to know for what Latitude it was made, as also whether it be true made, or not; which to do, I first seek the distance betwixt 12 and 1, and find it 11 degr. 51 min. Therefore I set the Tangent of 15 degrees on the first, to the Tangent of that distance, 11 deg. 51 min. on the second; and then against the Radius on the first, is the Sine 51 degr. 30 min. on the second. Whereupon I conclude, that the Dial was made for the Elevation of 51 degr. 30 min. Thus may you examine any Dial, whether it be truly made, or not.   

PROBLEM III. To find the distances of the hour-lines, from the line of 12 a clock in a direct South Dial, for any Elevation propounded.  
IN the Latitude of 51 degr. 30 min. a Dial is to be made, wherein to find the distance of the hour-lines from the line of 12 a clock.  
Say as in the first Problem, 

As the Radius,  
To the Co-sine of the Poles Elevation;  
So is the Tangent of any hour given,  
To the Tangent of the hour-line from the Meridian.   Therefore,  
Set the Radius on the first, to 38 degr. 30 min. the Co-sine of the Latitude on the second; and then against the Tangent of 15 deg. on the first, is the Tangent of 9 degr. 28 min. on the second, for the hours of 1 and 11; and against 30 degr. on the first, is the Tangent 19 deg. 46 min. on the second, the distance of the hours of 2 and 10, from the line of 12. And against the Tangent 45 on the first, is the Tangent 31 deg. 54 min. on the second, for the distance of the hour-lines of 9 and 3, from the line of 12. And now backwards, against the Tangent of 60 deg. on the second, is 47 degr. 10 min. on the first, the Tangent of the distances of the hour-lines of 4 and 8, from the line of 12. And lastly, against 75 degr. on the second, is 66 deg. 43 min. on the first, the Tangent of the distance of the hour-lines of 7 and 5, from the line of 12. For the hours of 6 and 6, they make right angles with the Meridian.   

PROBLEM. iv In a vertical Dial inclining, having the Elevation of the Pole above the Plane; to find the distance of the hour-lines from the hour-line of 12 a clock.  
LEt such a Plane be propounded, wh●se North part is elevated above the Horizon 16 deg. 30 min. which taken from the Elevation of the place 51 degr. 30 min. leaveth 35 degrees, for the Elevation of the Pole above the Dial's Plane: Wherefore to find the distance of the hour-lines in such a Dial from the Meridian, or hour-line of 12. Say, 

As the Radius,  
Is to the Sine of the Elevation above the Plane;  
So is the Tangent of any Equinoctial hour,  
To the distance of the same hour-line from the Meridian.   Therefore,  
Set the Radius on the first, to 35 degrees, the Sine of the Poles Elevation above the Dial's Plane; and then against the Tangent of 15 degrees, which is the Tangent of one Equinoctial hour on the first, is 8 degr. 45 min. on the second, the Tangent of the distance of the hours of 1 and 11, from the hour-line of 12: and against 30 degr. on the first, is 18 degr. 19 min. on the second, for the hours of 2 and 10. And for the rest of the hours, you shall have them, as you had those in the former Dial's.   

PROBLEM. V In any erect declining Dial, to find the distance of the Still from the Substyll, and of the Substyll from the Meridian.  
IN the Latitude of 52 degr. I would make a Dial to a wall, declining from the true South point 45 deg. wherein first to find the distance of the Still from the Substyll, the work is thus: 

As the the Radius,  
To 45 degr. the Co-sine of the declination of the Dial's Plane from the true South point.  
So is 38 degr. the Co-sine of the Latitude,  
To 25 degr. 48 min. the distance of the Still from the Substyll.   Therefore,  
Set the Radius on the first, to 45 degr. the Co-sine of the Declination on the second; and then against 38 degr. the Co-sine of the Latitude on the first, is 25 degr. 48 min. on the second. So much is the distance of the Still from the Substyll, or height of the Still.  
Or set the Radius to 38 deg. and then against 45 deg. is 25 deg. 48 min.  
And then for the distance of the Substyll from the Meridian, this is the Rule.  
Take 64 degr. 12 min. the Co-sine of that 25 degr. 48 min. the distance found, and 52 degr. the Latitude of the place, and count the greater of them two Sins for the first term in the rule of Proportion, and the lesser of them for the second term, and the Radius for the third term. And then,  
Set the Sine 64 degr. 12 min. on the first, (which is the Co-sine of that sine afore found) to the Radius on the second; and then against 52 degrees, the Latitude of the place on the first, is 61 degr. 5 min. on the second, whose Compliment is 28 deg. 55 min. So much is the distance of the Substill from the Meridian, in such a declining Dial. And so of any other.  
But I see by my Dial, that the Sun is declined from the Meridian far into the West, casting his shadow at half an hour past six a clock after noon, and Saturday night, August the first 1657, time to leave work. And for as much as my occasions call me another way for the next week, and for many weeks after, I must leave off proceeding any further in this Subject of Dialling.     

Advertisement.  
THe Reader might here expect to have found the figure of the Instrument, but it would have given him little or no satisfaction; besides, the Description thereof is so well known to Mr. Anthony Thompson, living in Hosier-Lane near West Smithfield, in London, who most accurately maketh this, and all other Mathematical Instruments, either in Brass or Wood, that I found it wholly unnecessary and useless.  
S. P.  
FINIS.