THE DESCRIPTION and use of the Universal Quadrat. By which is performed, with great expedition, the whole Doctrine of Triangles, both Plain and Spherical, two several ways with ease and exactness. Also the resolution of such Propositions as are most useful in Astronomy, Navigation, and Dialling. By which is also performed the proportioning of Lines and Superficies: the measuring of all manner of Land, Board, Glass; Timber, Stone, etc.  
By Thomas Stirrup, Philomathemat.   printer's device or ornament, depicting sun in glory 
London, Printed by R. & W. Leybourn, for Tho. Pierrepont, at the Sun in Paul's Churchyard, 1655.   

 royal coat of arms HONI SOIT QVI MAL Y PENSE  
To the READER.  
Courteous Reader,  
I Have here presented to thy view, the description and use of an Instrument, which for its excellent and general use, is called the Universal Quadrat, for by it (being of a portable size) is performed with incredible expedition, plainness, facility, and pleasure; the whole Doctrine trigonometrical, as well plain as spherical; and that two several ways; from whence an innumerable sort of conclusions Mathematical may be resolved, more than I have here written, for hereby is resolved as many Propositions, as by any Quadrant, Circle, Cylinder, Dial, Horoscope, Astrolabe, Globe, Sector, Crossstaff, or any such like Instrument heretofore devised; and that generally for all Latitudes: yea, and if you fit Sights accordingly to the planisphere on the backside, (as I was intended to have showed, had not time called me so fast away) so as they move about with the Planisphere, the Horizon will show upon the limb the quantity of any angle taken in the field, as well as the Theodolite, Plain Table, Circumferentor, or Peractor, or any other such like Instrument; so that hereby it doth appear to be an excellent Companion for most sorts of men; but especially for Navigators or Travellers. For hereby they may find Altitude and Declination of Sun or Stars, and thereby the latitude of the place where they are; where also they may find the Amplitude and Azimuth, and such like conclusions; whereby they may find the variation of their Compass; so likewise may they readily find the time of Sunrising and setting, with the beginning and ending of twilight; the hour of the day by the Sun, and the hour of the Night by the fixed Stars and Planets; with the time of their Rising, Southing and Setting, and that very easily and speedily: with the Resolution of such Nautical Propositions, as are of most frequent use amongst them.  
The whole Treatise consisteth of three Books; the first whereof, showeth the construction and general use of the Instrument; and how thereby to resolve all such Propositions as may be applied to right lines only; as the proportioning of lines and superficies, the measuring of all manner of Land, Board, Glass, Timbet, Stone and such like.  
The Second, sheweth the most plentiful, easy, and speedy Resolution of the whole Doctrine Trigonometrical, both plain and Spherical, and that two several ways, with surpassing facility.  
The Third, sheweth how to resolve all such Propositions Astronomical, as are of most frequent use in the Art of Navigation and dialing, with the resolution of all such Nautical Propositions, as are of ordinary use amongst Seamen, concerning Longitude, Latitude, Rumbe, and Distance.  
All which I very well know, may more exactly be performed by the Canon of Sines, Tangents, and Logarithmes, yet nothing so speedily as on the Instrument: but for observation an Instrument must be had, wherewith (if thou likest it) thou art here fitted, both for observation and operation.  
Thus desiring thee to accept of this Instrument from me, as I here freely present it to thee, which I wish even so to further thee, as I know it sufficient for the performance of much more than I have spoken of it. Farewell.   


The Contents of the first Book.  

THe Descriptions of the fore-side of the Instrument, Page 1  
The description of the back side, Page 4  
How to place the fixed Stars in the Instrument, Page 7  
The General Use of the Instrument, Page 12  
To divide a line into any number of equal parts, Page 14  
To increase a line in a given proportion, Page 16  
To diminish a line in a given proportion, Page 17  
To find a proportion betwixt two or more right lines, Page 18  
To lay down suddenly 2, 3, or more right lines in any proportion required, Page 19  
To two lines given to find a third proportional, Page 20  
To three lines given to find a 4th. proportional, Page 22  
To find a mean proportional line between two right lines given, Page 24  
To find the square root of a number, Page 26  
To divide a line by extreme and mean proportion, Page 27  
Having either Segment of a line divided by extreme and mean proportion, to find the other segment, & so the whole line, Page 28  
To divide a line in power, Page 29  
To augment a superficies in any proportion, Page 30  
To diminish a superficies in any proporton, Page 31  
To find the perpendicular of an equilateral triangle, Page 32  
To find the perpendicular of any right angled triangle, Page 34  
To find the perpendicular of any right lined triangle, Page 36  
To find the proportion between two or more like superficies, Page 37  
To add two Geometrical Squares or Circles, together into one Square or Circle, Page 38  
To subtract one square or circle from another, and to produce the remainder in a third square or circle, Page 40  
To make a square equal to any superficies, Page 41  
The length and breadth of a Parallellogram given in perches, to to find the content in acres, Page 42  
The length and breadth of a Parallellogram given in chains, to find the content in acres, Page 43  
The perpendicular and base of a triangle given in perches, to find the content in acres, Page 44  
The perpendicular and base of a triangle given in chains, to find the content in acres, Page 45  
Of reducing of one kind of measure to another, Page 46  
To find the Scale by which any plot was plotted, Page 47  
To find the breadh of the acre at any length given, Page 48  
To find the content of any board, Page 49  
Having the breadth of a board, to find the length of a foot there of, Page 50  
To measure tapering board, Page 50  
Having the circumference of a circle to find the Diameter, Page 51  
Having the Diameter of a circle, to find the Circumference, Page 52  
To find the content of any circle, Page 54  
To find the content of any part, or segment of a circle, Page 55  
To find the side of a square equal to any circle, Page 56  
To find the content of any squared solid, Page 58  
To measure a triangular prisma, Page 60  
To measure timber whose base is a Rhombus, Page 61  
To measure timber whose base is a Rhomboides, Page 62  
Ta measure timber whose base is a Trapezia, Page 63  
To measure timber whose sides are many. Page 65  
To measure round Timber, Page 66  
To find how much in length maketh a foot solid of any round piece of timber, Page 68  
Another way to find the content of a Cylinder. Page 69  
How otherwise to find the length of a foot of any Cylinder, Page 70  
To find the solid content of any Cone, Page 73  
To find the altitude of a Cone, Page 77  
How to measure any segment of a Cone, or any round tapering piece of timber, Page 78  
How to find the solid content of any Pyramid, or segment thereof being cut off, Page 80    

The Contents of the second Book.  

To find the Chord of any arch Page 83  
To find the right sine of any arch Page 85  
To find the arch of any sine, Page 86  
The Radius being given, with a straight line resembling a sine, to find the quantity of that unknown sine, Page 89  
The use of the Parallels as they signify natural sins and tangents, Page 90  
Of the resolution of right lined triangles, Page 95  
Another way to resolve right lined triangles, Page 102  
Of the resolution of right angled Spherical triangles by the foreside of the Quadrat, Page 116  
To resolve any Spherical triangle whatsoever by the foreside, Page 131  
Of the resolution of right angled Spherical triangles by the backside Page 141  
How to resolve any Spherical triangle whatsoever by the backside of the Instrument. Page 146    

The Contents of the third Book.  

To find the Sun's altitude by the shadow of a perpendicular Gnomon, Page 151  
To find the length of the right shadow, Page 152  
To find the height of the Sun by the shadow of a Gnomon lying parallel to the Horizon, Page 153  
To find the length of the contrary shadow, Page 154  
To find the Sun's declination, Page 155  
To find the latitude of the place, Page 156  
To find the declination of the Sun, or Stars by observation, Page 158  
To find the Amplitude of the Sun or Star, Page 160  
To find the Poles elevation, Page 161  
To find the Sun's declination, Page 162  
To find the Sun's place, Page 159  
To find the Sun's height in the vertical Circle, Page 163  
To find the time when the Sun will come to the due East or west, Page 164  
To find the Sun's altitude at the hour of six, Page 165  
To find the Suns Azimuch at the hour of six, Page 165  
To find the Ascensional difference, Page 166  
To find the beginning and ending of twilight, Page 167  
To find the Sun's right Ascension, Page 170  
To find the Obliqne Ascension, Page 171  
To find the hour of the Day, Page 171  
To find the Azimuth, Page 172  
To find the hour of the Day, Page 173  
To find the Azimuth, Page 175  
To find the Sun's place and declination, Page 178  
To find the Sun's Right Ascension, Page 179  
To find the Sun's Amplitude, Page 179  
To find the amplitude of the Stars, Page 180  
To find the latitude of the place, Page 181  
To find the Sun's declination, Page 182  
To find the Sun's height in the vertical Circle, Page 182  
To find the time when the Sun will be due East or West, Page 183  
To find the Sun's height at the hour of six, Page 184  
To find the Sun's Azimuth at the hour of six, Page 185  
To find the Ascensional difference of the Sun or Stars, Page 185  
To find the beginning and ending of twilight, Page 187  
To find the hour of the Day, Page 187  
To find the Azimuth, Page 188  
To find the Right Ascension and declination of the Planets or fixed Stars, Page 190  
To find the longitude and latitude of the Stars, Page 191  
To find the Culmination of any of the fixed Stars, Moon or other Planets, Page 191  
To find the time of the rising and setting of any of the fixed Stars, Moon, or other Planets, Page 193  
To find the exact hour of the Night by the Stars, Page 194  
To find how many miles answer to one deg. of longitude in any latitude, Page 198  
To find what difference of longitude belongeth to one degree of distance in any latitude, Page 199  
To find how many leagues do answer to one degree of latitude upon any Rumb given, Page 200  
To find the difference of latitude, Page 201  
To find the distance upon the Rumb, Page 202  
To find the Rumb, Page 202  
To find the difference of longitude, Page 203  
To find the difference of longitude, Page 204  
To find the Rumb, Page 205  
To find the difference of lat. & long. Page 206  
To find the distance upon the Rumb, Page 208  
To find the distance between the Ship and the Land, Page 250  
To find the distance of any Ship from the Land, Page 207  
To find the distance of two Ships at Road, Page 029     

 depiction of the front of a quadrat or quadrant 
The fore-side of the Universal quadrat   
 depiction of the back of a quadrat or quadrant 
The backside of the Universal quadrat  Delineavit Antonius Thompson  
 depiction of a spherical measuring instrument   

The First Book.  
Showing the description and general use of the Universal Quadrat; and how thereby to resolve all such Propositions as may be applied unto right lines; as the proportioning of Lines and Superficies, the measuring of all manner of Land, Board, Glass, Timber, Stone, and such like.  

CHAP. I. The description and making of the Universal Quadrat.  
A Quadrat or Square Geometrical is a figure of four equal sides, whose angles are all right angles, within the limits of which figure, we intent to describe an Universal Instrument, for the speedy performance of most Mathematical concusions; and that generally for all Latitudes.  
First, therefore, prepare a piece of Box or Brass, and after you have made it very smooth and plain, cut it directly four square, with four equal sides, and angel's all right. Your Wood or Brass being thus prepared, draw thereupon four right lines, equally distant or parallel to the edges, all of them cutting one the other at right angles: let the parallel distance of these lines from the edges be so much as that you may place the figures to the divisions limb, like as you may perceive by the figure. This being done, divide each side of the Square into 100 equal parts, or rather 1000, if quantity will give leave, for the larger the Instrument is, the more exactly will the conclusions wrought thereby be performed, but the Instrument not so portable. Then from each division of the one side, draw parallel lines to the like division on the opposite side, the like do by the other two sides; so shall the whole face of the Instrument be filled with parallel lines both ways, and crossing each other at right angles, according to the common Cynical Quadrant, and for the better distinction of these parallels, you may prick every fifth, so may you guide your eye along upon them the better.  
This being done, make choice of one of the angles for a centre, as here we have done at A, from which you shall number the two next sides by 10, 20, 30, etc. unto 100, and place figures to them in the margin, as you see in the figure.  
Then setting one foot of your Compasses in the centre A, with the other opened to the full length of one of the sides, draw the Quadrant BC over all the parallels, this Quadrant divide into 90 equal degrees, and subdivide each degree into as many parts as quantity will give leave, and let them be numbered with 10, 20, 30, etc. unto 90, both ways, as there you see: and if upon the centre A with any convenient distance you describe another Quadrant, and divide it into six equal parts as in my other Quadrat set down, and draw slope lines to each 15th. deg. in the Quadrant BC, and number them backward, than these slope lines may be accounted as hours in the art of dialling, and will be sufficient to perform the whole Art of Dialling, as I have at large taught in my Book lately published, called the Complete Diallist.  
This done, by laying a Ruler upon the centre A, and each degree in the Quadrant. BC, you may divide the two opposite sides from the centre, each into 45 unequal degrees called Tangents, let these be numbered by 10, 20 30, etc. unto 90 degrees, from B at the left hand, to 45 in the opposite angle, and so on to 90 at C the right hand of the Quadrant, and back again as in the Quadrant, and as you see in the figure; draw also the diagonal line AD. To this Instrument, as to all other of this kind, in their use, is added Sights with a thread, bead, and plummet according to the usual manner.  
The fore-side of this Quadrat is of itself (being of a portable size, and exactly drawn) an absolute Instrument for the speedy resolution of both kinds of Trigonometria; as well spherical as plain; as shall be made manifest in the ensuing Treatise, by applying it in several kinds.  
But seeing the backside of this Universal Instrument standeth yet blank, we will make use thereof in fixing a movable Planisphere of Brass thereon; on which Planisphere, we may project the Sphere in plano with straight lines; and so by help of a fixed Horizontal line and fixed Almicanters, we shall make this side also become Universal to perform most Spherical conclusions, and that most demonstratively, and generally for all Latitudes. The description whereof is after this manner.   

CHAP. II. The description of the backside of the Universrl Quadrat.  
FIrst, upon the backside draw four strait lines near to the edges, and parallel thereto, all cutting one the other at right angles as here in the figure you may behold; upon which lines you shall insert the degrees of Altitude, as you shall be showed hereafter; within these degrees draw four other parallel lines, all cutting one the other at right angles, unto which you shall draw every tenth degree and set his number thereto as here you see.  
This being done inscribe a circle as large as you can within this Square, within which circle, inscribe two other Concentrical Circles representing the Ecliptic, the dividing whereof shall hereafter appear.  
Then upon the same Radius with this inwardmost circle, describe another Circle upon some thin smooth plate, this Circle divide into equal 24 parts or hours, and let every hour be divided into halfs and quarters, or rather minutes if quantity will give leave and let them be numbered towards the right hand as here you see.  
That done, draw another Concentrical Circle, upon the margin of the hour Circle, representing the plane of the general Meridian; which divide into four equal parts in 00, 90: 00, 90: and draw the two Diameters 00, 00: 90, 90, crossing one the other at right Angles in the Centre of the Planisphere; so shall the Diameter 00. 00, Represent the Equator; and the Diameter 90. 90. the Circle of the hour of Six; and it is also the Axis of the World; wherein 90 at the hour of 18, stands for the North pole, and 90 at the hour of 6, stands for the South pole; then let each Quarter of the Meridian be divided into 90 degrees and numbered from the Equator towards the Poles.  
The Meridian being thus divided, lay a Ruler upon each degree thereof counted from the Equator, and thereby draw parallel lines to the Equator; these parallels shall represent the Declination of the Sun Moon and Stars; and the Latitude of Towns, Cities and Countries: and shall divide the Axis according to the line of Sines.  
If again we divide the Equator and each of his Parallels, in the like sort as the Axes of the World is divided, and then carefully through each degree draw a line, so as it makes no angles; the lines so drawn shall be Elliptical, and represent the hour Circles, the Meridian Circle the hour of 12 at Noon and midnight, and that 15 degrees next unto it drawn through 75 degrees from the Centre, the hours of 11 & 1, and that which is drawn through 60 degrees from the Centre, the hours of 10 and 2; and so of the rest: let these hour Circles be pricked to know them from the other Meridian's. Then in the Meridian number 23 deg. 30 minutes, the greatest Declination of the Sun, from the Equator to ♋, Northward, and to ♑ Southward, and draw the line ♋ ♑, for the Ecliptic, then numbering 23 degrees 30 minutes from the North-pole Northward, and from the South pole , draw the Axes of the Ecliptic, which let be divided in like sort as the Axis of the World is divided; and through each degree or each fifth degree, draw-lines parallel to the Ecliptic; these lines so drawn shall represent the Latitude of the planets, and fixed stars: and for distinction sake prick every fifth or tenth as before: now if we divide the Ecliptic and his parallels, as the Equator and his parallels are divided; and then carefully draw a line through each 30th. degree, so as it makes no Angles; the lines so drawn shall represent the signs, the first 30 degrees from the Centre towards ♋, shall stand for the sign ♈; the 30 degrees next following for ♉, and so of the rest, as you may see by the figure.  
The Sphere being thus projected in plain, let us resort to the back of our Quadrat again, and draw a line through the Centre, cuiting two of the sides just in the middle, so shall this Diameter make right Angles with the sides, and represent the Horizontal line for any Region, then take the just length of the Semidiameter of your Plainisphere and set it above and below the Horizon at both ends thereof, in the side of the Quadrat as here you see; and let this line be divided in like sort as the Equator and his Axis is divided, and numbered with 10, 20, 30, etc. unto 90 from the Horizon both ways, these are the degrees of Altitude.  
All this being done, let us come to the dividing of the Ecliptic which incompasseth the Sphere, which may be done out of a Table of Right Ascensions; for placing your Sphere upon the Centre of the Quadration the back side, putting some pin through both their Centres, so that they may be kept both at a stay for slipping one besides the other, then turning the Sphere about, until the beginning of the hours come to the place where you would have the beginnng of ♈, and there make stay of it, and mark the beginning of ♈, at the beginning of the hours: then looking into your Table of Right Ascensions, & see what hour standeth against the beginning of ♉, and you shall find 1 hour 51 min. therefore number 1 hour 51 min. amongst the hours backwards towards the left hand, & there place the beginning of ♉, the like may you do with the rest of the Signs, and every degree thereof; remembering, that as you numbered the hours towards the right hand, so must you number the signs towards the left.  
Then last of all, prepare a thin narrow slip of Brass, and rivet it on with the Sphere upon the Quadrat, so that the Sphere may turn about at all times when need shall require, so likewise let it be riveted to the Quadrat at both ends, so that the upper edge may be just with the beginning of the degrees of Altitude; this being thus fastened, shall serve for a Horizon generally for all Latitudes.  
This horizontal line, must be divided as the Equator and his Axis; so shall it natural show the Amplitude in all Latitudes.   

CHAP. III. How to place the fixed Stars in the Sphere, and also in the outward Eclipticall Circle, for the speedy finding of the hour of the night in all Latitudes.  
FIrst, you must get both the Right Ascension and Declination, of such Stars as you intent to place in your Instrument; the finding whereof shall hereafter appear: these being known; upon the Stars parallel of Declination, seek his Right Ascension, amongst the Meridian's or hour Circles; and at the point of intersection, betwixt the stars Declination, and Right Ascension, there place the Star.  
As for Example, suppose I would place Arcturus in the Sphere, first, I inquire his Right Ascension and Declination, either by the propositions hereafter for that purpose, or by this ensuing Table; and so find his Right Ascension to be 13 hours 59 minutes and his Declination to be 21 degrees 10 minutes North; therefore upon the 21 parallel and 10 minutes I number 14 hours from ♈ unto ♋, and so back again unto 14, and there I place Arcturus; the like may be done with any other Star.  
Now what Stars you place in the Sphere, the same place in the outward Ecliptic; which may be done after this manner: enter your Table of Right Ascensions, with the Right Ascension of the Star, and see what degree of the Ecliptic answers thereto, and place the Star upon the same.  
As for example, the Right Ascension of Arcturus being 14 hours, I enter the Table of right Ascensions therewith, and I find 2 degrees of ♏, answer thereunto; therefore Arcturus must be placed at 2 degrees of ♏ in the outward Ecliptic; the like may be done with any other of the fixed Stars.  
For readiness sake, I have here inserted a Table of Right Ascensions in hours and minutes, whereby the outward Ecliptic may be divided, and the stars therein placed: and also a Table of the Right Ascension, and Declination, & magnitude, of most of the principal fixed Stars, Rectified for this present year, 1652  

A Table of Right Ascensions for every degree of the Ecliptic in Hours and Minutes.  Degr. ♈ ♉ ♊ ♋ ♌ ♍ ♎ ♏ ♐ ♑ ♒ ♓ H. M. H. M. H. M. H. M. H. M. H. M. H. M. H. M. H. M. H. M. H. M. H. M. 1 00 04 01 55 03 55 06 04 08 13 10 12 12 04 13 55 15 55 18 04 20 13 22 12 2 00 07 02 00 04 00 06 09 08 17 10 16 12 07 14 00 16 00 18 09 20 17 22 16 3 00 11 02 03 04 04 06 13 08 21 10 20 12 11 14 03 16 04 18 13 20 21 22 20 4 00 15 02 07 04 08 06 17 08 25 10 24 12 15 14 07 16 08 18 17 20 25 22 24 5 00 18 02 11 04 12 06 22 08 29 10 27 12 18 14 11 16 12 18 22 20 29 22 27 6 00 22 02 15 04 16 06 26 08 34 10 31 12 22 14 15 16 16 18 26 20 34 22 31 7 00 26 02 19 04 20 06 30 08 38 10 35 12 26 14 19 16 20 18 30 20 38 22 35 8 00 29 02 22 04 24 06 35 08 42 10 39 12 29 14 22 16 24 18 35 20 42 22 39 9 00 33 02 26 04 29 06 39 08 46 10 42 12 33 14 26 16 29 18 39 20 46 22 4● 10 00 37 02 30 04 33 06 43 08 50 10 46 12 37 14 30 16 33 18 43 20 50 22 46 11 00 40 02 34 04 38 06 48 08 54 10 50 12 40 14 34 16 38 18 48 20 54 22 50 12 00 44 02 38 04 42 06 52 08 58 10 54 12 44 14 38 16 42 18 52 20 58 22 54 13 00 48 02 42 04 46 06 56 09 02 10 57 12 48 14 42 16 46 18 56 21 02 22 57 14 00 52 02 46 04 50 07 01 09 06 11 01 12 52 14 46 16 50 19 01 21 06 23 01 15 00 55 02 50 04 55 07 05 09 10 11 05 12 55 14 50 16 55 19 05 21 10 23 05 16 00 59 02 54 04 59 07 09 09 14 11 08 12 59 14 54 16 59 19 09 21 14 23 08 17 01 03 02 58 05 04 07 14 09 18 11 12 13 03 14 58 17 04 19 14 21 18 23 12 18 01 06 03 02 05 08 07 18 09 22 11 16 13 06 15 02 17 08 19 18 21 22 23 16 19 01 10 03 06 05 12 07 22 09 26 11 20 13 10 15 06 17 12 19 22 21 26 23 20 20 01 14 03 10 05 16 07 27 09 30 11 23 13 14 15 10 17 16 19 27 21 30 23 23 21 01 18 03 14 05 21 07 31 09 34 11 27 13 18 15 14 17 21 19 31 21 34 23 27 22 01 21 03 18 05 25 07 35 09 37 11 31 13 21 15 18 17 25 19 35 21 37 23 31 23 01 25 03 22 05 30 07 39 09 41 11 34 13 25 15 22 17 30 19 39 21 41 23 34 24 01 29 03 26 05 34 07 44 09 45 11 38 13 29 15 26 17 34 19 44 21 45 23 38 25 01 33 03 31 05 38 07 48 09 49 11 42 13 33 15 31 17 38 19 48 21 49 23 42 26 01 36 03 35 05 42 07 52 09 53 11 49 13 36 15 35 17 42 19 52 21 53 23 45 27 01 40 03 39 05 47 07 56 09 57 11 49 13 40 15 39 17 47 19 56 21 57 23 49 28 01 44 03 43 05 51 08 00 10 01 11 53 13 44 15 43 17 51 20 00 22 01 23 53 29 01 48 03 47 05 56 08 05 10 05 11 56 13 48 15 47 17 56 20 05 22 05 23 56 30 01 52 03 51 06 00 08 09 10 08 12 00 13 52 15 51 18 00 20 09 22 08 24 00  

A Table of the Right Ascension, Declination, and Magnitude of some of the principal fixed Stars, Rectified for this present year 1652.  The Names of the Stars. Right Ascension Declination. Magnitude H. M. D. M.  The Pole Star 00 32 87 19 N 2 Andromeda's girdle 00 52 33 48 N 2 The Rams horn 01 34 17 27 N 3 The Rams head 01 48 21 44 N 3 The head of Mednsa 02 46 39 32 N 3 Aldebaran, the Bull's eye. 04 16 15 48 N 1 Hircus, the Goat 04 51 45 36 N 1 Orion's left foot 04 58 08 40 S 1 Orion's left shoulder 05 06 05 59 N 2 The first in Orion's girdle 05 14 00 38 S 2 Orion's right shoulder 05 36 07 17 N 2 Sirius, the great Dog 06 31 16 13 S 1 protion, the lesser Dog 07 21 06 07 N 2 The heart of Hydra 09 11 07 09 S 1 Cor Leonis, the Lion's heart 09 49 13 42 N 1 The Lion's neck 09 50 21 49 N 2 The Lions back 10 54 22 33 N 2 The foremost star in □ of the great Bear 10 40 58 23 N 2 The Lion's tail 11 30 16 34 N 1 The Root of the great Bear's tail, alio tail 12 38 57 51 N 2 The Virgin's spike 13 08 09 16 S 1 The middlemost star in the great Bears 13 10 56 45 N 2 The last of the said tail 13 26 51 05 N 2 Arcturus 13 59 21 10 N 1 The bright star of the Crown 15 24 27 43 N 3 Antares, the Scorpion's heart 16 08 25 34 S 1 The bright star of the Harp 18 25 38 30 N 1 The eagle's heart 19 33 07 58 N 2 The bright star of the Vulture 19 34 08 01 N 1 Fomahant 22 38 31 25 S 1 Pegasus shoulder 22 48 13 14 N 2 Pegasus leg. 22 52 26 09 N 2   

CHAP. iv To show the General Use of the universal Quadrat.  
IN the Use of this Quadrat, let us, for distinction sake, call those Parallels which make Right Angles with the line whereon the sights stand, Right Parallels; and the other which Run parallel to the same line, Contrary Parallels; the one signifying Right shadow, the other Contrary shadow: the General Use whereof consists in the solution of the Golden Rule, where three lines or numbers being given of a known denomination, a fourth proportional is to be found, and this solution is divers in regard of the signification of these parallel lines, and also of the entrance into the work.  
The signification of these Parallels is sometimes simple, and sometimes compound, sometimes they signify right lines or numbers only, such as they are simple in themselves, and sometimes they signify natural sins, which is half the Chord of the double Arch, and sometimes natural Tangents, which are perpendiculars at the end of the Diameter: sometimes the work may be begun with one kind of signification, and ended with another, it may be begun with the lines as they are in themselves, and ended as they are sins, it may begin as they are sins, and end as they are Tangents.  
The solution in regard of the entrance into the work, may be either right or contrary. I call that right entrance, or entrance on the Right Parallels, when the two lines or numbers of the first denomination are applied or found in the Right parallels, and the third line or number, and that which is sought for, on the contrary parallels; and on the contrary, I call that contrary entrance, when the two lines or numbers of the first denomination, are applied or found on the parallels of contrary shadow, and the third line or number, and that which is to be found out, in the parallels of right shadow.  
As for example, let there be given three numbers 50. 60, and 70. unto which I am to find a fourth proportional. Let the Question be this, if 50 pounds, gain 60 shillings what shall 70 pounds gain? here are numbers of two denominations, one of pounds the other of shillings, and the first with which I am to enter must be that of 50. pounds. If then I would enter my Question upon a parallel of right shadow, see where the right Parallel of my first number which is 50, cutteth the contrary parallel of the second which is 60, and upon the point of intersection there lay the thread, the thread lying in this position, look along upon the right Parallel of the third number, which is 70, and see what contrary parallel cutteth it directly under the thread, for that is the fourth proportional number required, which here is found to be 84, wherefore I conclude, that if 50 pounds' gain 60 shillings, 70 pounds will gain 84 shillings; for as 50 is to 60, so is 70 to 84: the like will appear if you enter your work upon the contrary parallels, for it mattereth not on which kind of Parallels you enter, so be it you enter your first and third number on the same kind of Parallels, so shall the third and fourth be of the other kind. Thus much for the General Use of this fore-side of the universal Quadrat, which being considered and well understood, there is nothing hard in that which followeth.   

CHAP. V To divide a line given, into any number of equal parts.  
LEt the line given be AB, and let it be required to divide the same into 5 equal parts.  
Take with your compasses the given line AB, and lay it down upon any parallel, whose number may be equally divided into five parts without any remainder; as lay it upon the fifth parallel, and at the point where the line endeth, there lay the thread; the thread lying in this position, and in regard that 10 is ⅕ part of 50, therefore the segment of the tenth parallel which is cut by the thread, shall be the distance which shall divide the given line into 5 equal parts.  
As for example, take the given line AB, and lay it upon the fiftieth contrary parallel, as from A to B, & seeing the given line endeth in B, lay the thread upon the same point, as here you see in the following figure; then in regard that 10 is ⅕ of 50, therefore I take the segment cut by the thread upon the tenth parallel, viz. CD, and with it I divide the given line AB, into 5 equal parts as here you see.   diagram of the measurement of parts of a line (AB and CD)  

 depiction of the front of a quadrat or quadrant 
The fore-side of the Quadrat.    

CHAP. VI To increase a line in a given proportion.  
TAke the line given with a pair of Compasses, and lay it one the parallel of the number given, and upon the point where the given lines endeth, there lay the thread, which being kept in this position, the segment of the parallel, of the number required, intercepted by the thread, shall give the line required.   diagram of the measurement of proportional lines (A and B) 
Let A be a line given to be increased in the proportion of 3 to 5. First, I take the given line A with my compasses; and in regard that 30 and 50, is the same with 3 and 5, if from them you give o, therefore I take the given line A, and lay it upon the thirtieth parallel, as from F to G, in the last Chapter, and upon I, the point where the line endeth, there I lay the thread, which lying in this position, cutteth the parallel of 50 in the point B; so shall AB, be the length of the line B, which was required.   

CHAP. VII. To diminish a line in a given proportion.  
TO diminish a line in a given proportion, is no other than that which was showed in the last Chapter, but only whereas there you entered with the lesser line which was to be increased, upon the lesser parallel number; and so the line required, was found upon the greater parallel number; so here you must enter with the greater line which is to be diminished, upon the greater parallel number; and that which is required shall be found upon the lesser parallel number.   diagram of the measurement of proportional lines (A and B) 
Let A be a line given to be diminished in the proportion of 5 to 3, I take the given line B, and lay it upon the fiftieth parallel, and at the point where the line endeth I lay the thread, as at the point B in the figure of the fifth Chapter, the thread lying in this position, I see it cut the thirtieth parallel in the point G, so shall the distance FG, give me the line A, which was required, and is ⅗ of the given line B.  
Here note that you may use other numbers to work your proportion upon besides these: for you may multiply or divide the proportional numbers given, either by 2. 3, or 4, etc. and so work by their numbers: as for 3 and 5 we may work by 6 and 10, or else by 9 and 15, or by 12 and 20, or 15 and 25, or 18 and 30, or 24 and 40, or 30 and 50, as here we have done.   

CHAP. VIII. To find a proportion betwixt two or more Right lines.  
TAke the greater line betwixt your Compasses, and extend it upon that parallel, whose number you intent to make the denominator of the fraction; at the end of which extent place the thread; the thread resting in this position, take the lesser line betwixt your Compasses, and (keeping one foot in the side of your Quadrat,) carry it parallel unto tthe greater line, until the other foot touch the thread, which you may very well do by help of the parallels; so shall the number of that parallel whereon the Compasses shall stay, be the Numerator to the former Denominator.   diagram of the measurement of proportional lines (AB and CD) 
Let the lines given be AB, and CD, first, I take the line CD, and place it upon the fiftieth parallel, as from A to B, upon which point B, I lay the thread; which being kept in this position, I take the lesser line AB betwixt my Compasses, and keeping one foot in the side of the Quadrat EH, I carry them parallel to the greater line, until the other foot toucheth the thread at G; so shall the fixed foot rest in the parallel of 30, which is the Numerator to the former Denominator, wherefore I conclude that the line AB is 30/50 or ⅗ of the line CD, or the line AB, is in such proportion to the line CD, as 30 to 50, which was required.  
But if the line CD, be greater than these parallels can contain, take ½ thereof, and enter with the same as before you did with the whole line; so likewise take ½ the lesser line and carry it parallel as before you did the whole line, and you shall produce the same proportion as before.   

CHAP. IX. To lay down suddenly 2, 3, or more Right lines in any proportion required.  
FIrst, lay your thread by chance upon the Quadrat, so as it may cut the parallels of all the numbers given: so shall the thread cut off from each parallel, the proportional lines required: which may be taken with your Compasses, and laid down presently.  
As for Example, let it be required to lay down four straight lines, which shall bear such proportion one to the other as four given numbers; which let be these 60, 50, 30, 10. then opening the thread by chance, so as it may cut the parallels of all these given numbers: so shall the thread cut from each parallel, the proportional line required; which you may take from your Quadrat, and lay them down as here you see; viz. from the parallel of 60, the line KL, from the parallel of 50, the line AB, from the parallel of 30, the line FG; and from the parallel 10, the line CD.   diagram of the measurement of proportional lines (CD, FG, AB, and KL)  

CHAP. X. Having two lines given, to find a third proportional line to them.  
FIrst, place both the lines given on the contrary side of the Quadrat, and mark how many parallels of contrary shadow they do extend unto; then take the second line again, and place it upon the right side of the Quadrat, and note the extension amongst the parallels of right shadow; now see where the contrary parallel of the first line, cutteth the right parallel of the second line, and there lay the thread; the thread lying in this position, see what right parallel is cut by the contrary parallel of the second line, directly under the thread, for that is the third proportional line required.   diagram of the measurement of proportional lines (EO, EF, and FG) 
Let the two lines given be EO, and OF, and it is required to find a third line, which shall be in such proportion to OF, as OF is to EO: first, place the two lines upon the contrary side of the Quadrat, and you shall find the line EO, to extend unto 20, and OF unto 30, now extend OF on the right parallels also, which will be likewise unto 30; therefore where the contrary parallel of EO 20, cutteth the right parallel of OF 30, there lay the thread as at P, the thread lying still in this position I see where the contrary parallel of OF 30, cutteth the thread, which is at G, upon the right parallel of 45: therefore FG shall be the third proportional line required: and shall be 45 of such divisions as EO is 20, and OF 30.  
To perform the same more speedily with the numbers, for as 20, is to 30; so is 30, to a third number.  
Therefore, at the intersection of the twentieth contrary parallel, with the thirtieth right parallel, there lay the thread, which being kept in this position, see where the thirtieth contrary parallel cutteth the thread, which will be at G, upon the 45 parallel of right shadow, which 45. is the third proportional number as before.   

CHAP. XI. Having three lines given to find a fourth proportional to them.  
HEre the first line and the third, are to be placed on the contrary parallels; and the second on the right parallels: therefore, at the intersection of the contrary parallel of the first line, with the right parallel of the second, there lay the thread; the thread lying still in this position, whese the contrary parallel of the third line cutteth the thread, there is the right parallel of the fourth proportional line.   diagram of the measurement of proportional lines (EO, OP, EK, and KL) 
Let the three lines given be EO, OPEN, and EKE, and let it be required to find a fourth line, which shall have such proportion to EKE, as OPEN, hath to EO. First, therefore I place EO, upon the contrary parallels, and it extendeth unto 20, so likewise EKE; being extended upon the same parallels reacheth unto 60, now OPEN, being extended upon the right parallels giveth 30, therefore at the intersection of the twentieth contrary parallel, with the right parallel of 30, I lay the thread, as at P; the thread lying in this position, I see where the 60 contrary parallel cutteth it, and I find it to be at L, upon the right parallel of 90; so shall the distance KL, be the fourth proportionaull line required; and will be found to be 90 such parts, as EKE is 60, or OPEN, 30.  
If at any time your lines or numbers; out run the limits of your Quadrat; you may take ½, ⅓, ¼, or 1/10, or any other parts of your first and third lines or numbers; and keeping the second whole; you shall have the fourth number whole also, or you may keep your first and third numbers whole, and take ½, ⅓, ¼, or 1/10, or any other parts, of your second; so shall you have the like part of your fourth number, or if this will not please, you may take what part you please, of all three of your given lines or numbers; so shall you have the like part of your fourth number.  
Or if your numbers or lines be too small, that they fall to near the centre, you may increase them in the like manner.  
This is in effect, to alter the number of parallels upon the Quadrat, according as need shall require, for they being actually divided into 100 equal parts; if you shall take away (by supposition) the cipher from each tenth parallel, they will become only 10 parallels, and if you shall (by supposition) add a cipher to each tenth parallel, you shall make each parallel to signify 10, and every tenth to signify 100, and so the whole side to be 1000 thus you see you may either increase them, or diminish them, according to any proportion required which being well considered, will be a great help in this work of right lines: and for the rest which followeth we shall have no such need.   

CHAP. XII. To find a mean proportional line between two right lines given.  

A mean proportional line is that, whose square is equal to the long square, contained under his two extremes.  
First, join the two given lines together, so as they may make both one right line; the which divide into two equal parts; then take the distance between the point of meeting, and the middle of the line; and place it from the centre of your Quadrat, a long upon one of the sides; then take one half of the line betwixt your Compasses, and setting one foot thereof in the point where the former line ended, with the other touch the other side of the Quadrat; from which point to the centre, shall be the length of the mean proportional line required.   diagram of the measurement of lines A, B, CE, and D 
Let A and B be be two lines given, between which it is required to find a mean proportional line: first join the two lines togther in F, so as they may make the right line CE, which divide into two equal parts in the point G; then take with your Compasses the distance FG, and place it upon the side of the Quadrat, from the centre E unto N, as in the figure of the fifth Chapter; then take one half of the former joined line, and setting one foot in the point N, with the other, touch the other side of the Quadrat, as at m: so shall the distance betwixt the centre E and the point m, be the line D, which is the mean proportional line required, betwixt the two extremes A and B.   

To perform the some with numbers.  
Let 48 and 27, be two given numbers, between which it is required to find a mean proportional number: first, add the two given numbers together, and they will make 75, then take the half thereof which is 37½, out of which half take the lesser extreme, and there will remain 10½, now therefore take 37½ divisions betwixt your Compasses, and setting one foot in 10½ counted on the right side of the Quadrat, as at n, with the other foot touch the contrary side, as at m, which will be at 36, so shall 36 be the mean proportional number between 48 and 27, as was required.  
Or in stead of the Compasses you may make use of the bead, for if you lay the thread upon the side of the Quadrat, and set the bead to 37½ of the contrary parallels, which being fixed, open the thread until the bead falleth upon 10½ of the right parallel, and there it shall show amongst the contrary parallels, the mean proportional number to be 36, as before.    

CHAP. XIII. To find the square Root of a number given.  
FOr the performing of this, it were best to divide the given number, by some other number as you shall see cause, or like best of; but if it may be, let it be such a number as may divide your given number without leaving a fraction, so shall your divisor be your lesser extreme, and your quotient the greater; this being done, you may find a mean between these two extremes, by the last Chapter, which mean proportional number, shall be the square root of your given number.  
As for example, let the number given be 1296, and let it be required to find the square root thereof, first, divide the given number by some other number, that will divide it equally without a fraction if you can; as here we will divide the given number 1296 by 9, the quotient will be 144, for the greater extreme, and 9 shall be the lesser; now add the two extremes together and they will make 153, the half whereof is 76½, unto which number (counted on the side of the Quadrat) place the bead; then take 9 the lesser extreme, out of 76½, and there will rest 67½, therefore open the thread until the bead falleth upon the right parallel of 67½, and there you shall see it cut the contrary parallel of 36; wherefore I conclude the root of 1296 is 36, the thing which was required.   

CHAP. XIV. To divide a line by extreme and mean proportion.  
LEt the line AB be a line given to be so divided, first, take the given line AB betwixt your Compasses, and setting one foot in the Centre E, of the fifth Chapter, extend the other upon the side of the Quadrat, and see what parallel it reacheth unto, as here unto 60 at K; and mark the intersection of this parallel with his like, as at S; then take half the given line, and place it on the same side of the Quadrat, from the point where the whole line ended at K, unto R; then in the point R set one foot of your Compasses, and with the distance RS, make a mark in the side of your Quadrat, as at O; so shall your given line AB, be divided in the point O, by extreme & mean proportion; which proportion is after this manner.  
As the lesser segment thereof, is to the greater; so is the greater, to the whole line; and thus OB the lesser segment of the given line AB, is to the greater segment OA, as the greater segment OA is to the whole line AB, which was required   diagram of the division of lines (AB, C, D, and FG)  

CHAP. XV. Having either the greater or lesser segment, of a line divided by extreme and mean proportion, to find the other segment, and so consequently the whole line.  
FIrst by the last Chapter, divide any line by extreme and mean proportion, as here the line AB; then let the line C be the greater segment of a line so divided, and it is required to find the lesser segment thereof;  
Therefore having divided the line AB in the same manner, the proportion will hold. 

As the greater segment OA,  
To the lesser segment OB;  
So is the greater segment the line C,  
To his lesser segment the line D.   Which two segments C and D, being joined together in the point E, shall make up the whole line FG, whereof OF is the lesser segment, and EGLANTINE the greater, which was required; and in the like manner, having the lesser segment of a line divided by extreme and mean proportion, you may find the greater, behold the lines in the foregoing Chapter.   

CHAP. XVI. To divide a right line given in power, according to any proportion required.  
TO divide a line in power, is to find two other lines, whose squares together shall be equal to the square of the given line; but the square of the one, to the square of the other, to be in any proportion required.  
First by the seventh Chapter divide the given line according to the proportion given; and then by the twelfth Chapter find two mean proportionals between each part and the whole line; which two proportionals together shall be equal in power to the given line; and the power of the one, to the power of the other, according to the proportion given, shall be proportional.   diagram of the division of lines (AB, C, and D) 
Let the line AB be a line to be so divided, as the side of some square which is 75; and let the proportion given be as 16 to 9 First by the eleventh Chapter divide the given line AB, proportionally as 16 to 9, which will be in the point E, making A, 48, and BE 27: now by the twelfth Chapter finding a mean proportional number between 48, the segment A, and 75 the whole line, you shall produce 60 the line C: so likewise a mean proportional between 27 the segment BE, and 75 the whole line, shall be 45 the line D.  
Wherefore the square of 60, the line C; put to the square of 45, the line D; shall be equal to the square of 75, the given line AB; and yet the square of C, shall be in such proportion to the square of D, as 16 is to 9, which was required.   

CHAP. XVII. To augment a superficies in a given proportion.  
TO augment a superficies in a proportion given, is to find a line; upon which if you make a superficies like unto that which was given, they shall be in such proportion one to the other, like unto the proportion given.  
First, therefore by the sixth Chapter increase the side of the given superficies, according to the proportion given; then find a mean proportional line (by the twelfth Chapter) between the given side of the superficies, and the line so increased; upon which if you make the like superficies, it shall be in such proportion to the given superficies, as was required.   diagram of the measurement of a superficies (A, B, and C) 
Let the line A, be the side of a square to be augmented in the proportion of 16 to 25. First, by the sixth Chapter increase 48 the given side A, in the proportion of 16 to 25; so shall you find 75 for the length of the line B: now between 48 the given side A, and 75 the increased line B, find a mean proportional, which will be 60 the line C: upon which if you make a square, it shall bear such proportion to the square of A, as 25 to 16, as was required. 

For as 16, is to 25:  
So is 2384 the square of 48, the given side A,  
To the square of 60 which is 3600, the required square made upon the line B.     

CHAP. XVIII. To diminish a superficies in a given proportion.  
TO diminish a superficies in a proportion given, is to find out a line, upon which if you make a superficies like unto that which was given, it shall be less than it according to the proportion given.  
First, therefore by the seventh Chapter diminish the side of the given superficies, according to the proportion given; then by the twelfth Chapter find a mean proportional between the side of the given superficies, and the line so diminished, which shall be the like side of the superficies required.  
As for example, let the line B in the foregoing Chapter be the Diameter of a circle to be diminished in the proportion of 25 unto 16. First, by the seventh Chapter, diminish 75 the given diameter B, in the proportion of 25 unto 16, so shall you produce 48 for the length of the diminished line A: now between 75 the given diameter B, and 48 the diminished line A, find a mean proportional, which will be 60 the length of the line C: upon which diameter if you describe a circle it will be less than the Circle made upon the diameter B, in such proportion as 16  less than 25, which was required.  
For as 25 is to 16; so is 4419 9/14 the content of the given circle whose diameter is the line B, to the content of the circle required (whose Diameter is the line C,) which is 2828 4/7   

CHAP. XIX. To find the perpendicular of an equilater Triangle, the sides thereof being given.  
AN equilater Triangle is that, which hath all his three sides equal; for I need not stand to define what a Triangle is, seeing it is so well known to be a figure made of three lines, either Right or Spherical, but here I speak of Right lined Triangles. Now to find the length of the perpendicular, which is a line falling from the opposite angle to the base, making Right Angles therewith; lay the thread upon the side of the Quadrat, and rectify the bead to the length of one of the sides; then opening the thread, until the bead fall upon the right parallel of half the base; so shall it show the contrary parallel of the perpendicular.   diagram of the measurement of an equilateral triangle (ABC) 
Let the Triangle ABC, be given, wherein it is required to since the length of the perpendicular AD: First, lay the thread upon the side of the Quadrat, and rectify the bead to 8 the length of the side AB, or AC; then open the thread, until the bead falleth upon the Right parallel of 4, the one half of the base BC, which is BD, or CD; so shall the bead cut the contrary parallel of 6 9/10 which is the length of the perpendicular AD, which was required.  
Thus may you likewise find the perpendicular of an Isoscheles Triangle, which is a Triangle of two equal sides; for, first placing the bead to one of the sides, then opening the thread until the bead fall on the Right parallel of one half of the base; so shall it show the contrary parallel of the perpendicular.   

CHAP. XX. How speedily to find the perpendicular of a Right Angled Triangle, the sides being given; as also to find the segments of the base, cut by the perpendicular.  
A Right angled Triangle, is that whose sides comprehending the Right angle falleth square or perpendicularly one upon the other, thereby making the Angle contained betwixt them a Right Angle, or an Angle of 90 degrees.  
To find the perpendicular of such a Triangle, lay the thread upon the side of the Quadrat, & rectify the bead to the lesser side of the Triangle given, then open the thread to the intersection of the contrary parallel of the greater side, with the Right Parallel of the lesser side; so shall the bead show the contrary parallel of the perpendicular required.   diagram of the measurement of a right-angled triangle (ABC) 
Let the Triangle ABC, be a right angled Triangle given, whose perpendicular is required. First, laying the thread upon the side of the Quadrat, rectify the bead to 45 the lesser side BC, of the given Triangle ABC; then opening the thread to the point of intersection, betwixt the contrary parallel of 60 the greater side AB, and the Right Parallel of 45 the lesser side BC; the thread lying upon this point you shall see it cut the contrary parallel of 36, which is the length of the perpendicular BD, which was required.  
Or the bead being fixed as before, if you open the thread to the point of intersection, betwixt the contrary parallel of the lesser side BC, 45, and the right parallel of the greater side AB, 60; the bead shall fall upon the right parallel of 36, which is the length of the perpendicular BD as before: or having two beads upon your thread, if you place the one to 45 the lesser side, and the other to 60 the greater side; and opening the thread as before, unto the point of intersection, betwixt the contrary parallel of 60 the greater side, and the right parallel of 45 the lesser side; the thread lying in this position, the bead which signifieth the greater side, shall show the right parallel of 36, and the other bead shall show the contrary parallel of the same, which is the perpendicular BD, as before.  
And farther, the thread and beads lying still in this position, the bead which signifieth the greater side 60, shall cut the contrary parallel of 48, which shall be the greater segment of the base divided by the perpendicular, viz. AD; and the bead which signifieth the lesser side 45, shall cut the right parallel of 27, which is the lesser segment of the base, viz. CD.   

CHAP. XXI. To find the perpendicular of any Triangle, the three sides being given.  
FIrst, for this purpose lay the thread upon the side of the Quadrat, and rectify the bead to the first side, then take the second side with your Compasses, and setting one foot in the end of the third side, open the three until the moving foot of your Compasses, and the bead rest directly both at a point; so shall that parallel, which passeth by the point where the Compass foot, and the bead meeteth, (parallel to the side wherein the other foot of the Compasses standeth) be the perpendicular required. As for Example; let it be required to find the perpendicular of the Triangle in the last Chapter. First, lay the thread to the side of the quadrat, and rectify the bead to the greater side AB, which is 60; then take 45 betwixt your Compasses, which is the lesser side BC, and setting one foot in the side of the Quadrat in 75, which is the base AC, with the other open the thread, until the point of the Compasses and the bead meet both in one point; so shall the parallel distance, betwixt the bead and that side of the Quadrat wherein the foot of the Compasses standeth, be 36, the perpendicular required.   

CHAP. XXII. To find a proportion between two or more like superficies.  
FIrst, place the thread upon the side of the Quadrat, then rectify the beads, one to the one given side, and the other to the other given side, then at the point of intersection betwixt the greater sides contrary parallel, with the right parallel of the lesser side, there lay the thread; the thread lying in this position, the bead which signifieth the greater side, shall cut the contrary parallel of the greater terms required, and the other bead which signifieth the lesser side given, shall cut the right parallel of the lesser term required, which two terms or numbers shall bear such proportion the one to the other as the two given sides, which was required.   diagram of the measurement of proportional lines (A and B) 
Let A and B, be two sides of like superficies, as the sides of two squares, or the Diameters of two Circles; the side A being 45, and the side B 60; First therefore, lay the thread upon the side of the Quadrat, and rectify the beads one to 45, and the other to 60, then at the point where the contrary parallel of 60, intersecteth the right parallel of 45, there lay the thread, the thread lying in this position, the bead for 60, shall cut the contrary parallel of 48, and the bead for 45 shall cut the right parallel of 27, these two numbers viz. 48, and 27, shall be in such proportion one to the other, as the two given superficies, wherefore the proportion of the superficies, whose side is B, to that whose side is A, is as 48 to 27, which is in lesser terms, as 16 to 9   

CHAP. XXIII. Two Geometrical Squares being given, to add them together into one Square, and also to add two circles together into one circle.  
FIrst, from the centre of your Quadrat, lay or count the two sides of the given squares, one upon the one side, and the other upon the other side of the Quadrat; then setting one foot of your Compasses, in the extreme point of one of the given sides extend the other foot to the extreme point of the other given side; so shall you have betwixt your Compasses, a third side which shall make a square, equal to both the given squares. Note, what is said here of the sides of squares, the like may be understood of the Diameters of Circles.   diagram of the measurement of proportional lines (A, B, and C) 
Let the two lines A and B be the sides of two given squares, and let it be required to add them together into one square, or to make a third square equal to them both: seeing the side A is 45, and the side B 60, therefore I set one foot of my Compasses in 60 upon one side of my Quadrat, and extend the other unto 45 upon the other side of my Quadrat, so shall I have betwixt my Compasses the line C, which being applied to the side of the Quadrat giveth 75, upon which line C, if I make a square, it will be equal to both the squares of A and B, which was required.  
And what is said here of the sides of squares may be understood of the Diameters of circles: for if these three lines A, B and C, shall be the Diameters of three circles, the circle described upon the Diameter C, shall be equal to the two other circles, described upon the Diameters A and B.  
And thus you may add divers squares or circles together into one square or circle; if you add first two into one, and then unto that a third, and then unto the total a fourth, and so on at pleasure.   

CHAP. XXIIII. Two squares or Circles being given, to subtract the one out of the other, and to produce the remainder in a third square.  
FIrst, take betwixt your Compasses the side of the greater square given, and setting one foot in the side of the lesser given square, (counted on the side of the Quadrat) with the other foot cut the other side of the Quadrat; and where the Compasses so cutteth, there is the side of the remaining square, counted, from the centre.   diagram of the measurement of proportional lines (A, B, and C) 
Let the two lines B and C be the sides of two squares, or the Diameters of two circles; and it is required to subtract the square of B, out of the square of C, and to make a square equal to the remainder. First, I take 75 the line C (which is the side of the greater square,) betwixt my Compasses, and setting one foot in 60 the side of the lesser square, (counted on the side of the Quadrat) the other foot will reach unto 45, upon the other side of the Quadrat; which 45 is the length of the line A, the side of the remaining square upon which if you make a square, it shall be equal to the remainder of the greater given square, when the lesser is taken out of the same, as was required.   

CHAP. XXV. To make a square equal to any superficies given.  
IF the superficies given be a rectangle parallelogram, a mean proportional line betwixt the two unequal sides, shall be the side of his equal square.  
If it shall be a Triangle, a mean proportional between the perpendicular and half the base, shall be the side of his equal square.  
If it be a Rombus, or Romboydes, a mean proportional between one of his sides, and the parallel distance of the same sides, shall be the side of his equal square  
If it shall be any other rightlined figure, it may be resolved into Triangles, and so a side of a square found equal to every Triangle; and these being reduced into one equal square it shall be equal to the whole rightlined figure given.  
And if the superficies given be a circle, a mean proportional between the semidiameter and semicircumference, or betwixt the diameter and ¼ of the circumference, shall be the side of his equal square.   

CHAP. XVI. Having the length and breadth of an oblong superficies given in perches, to find the content in acres.  
IN every parallelogram right angled, the length in perches, multiplied by the breadth in perches, and the product divided by 160, the number of perches in an acre, produceth the content; therefore, 

As 160, to the breadth in perches:  
So the length in perches, to the content in acres.    
Here and in such like propositions, you may suppose one side of your Quadrat to be 1000; so shall every parallel signify 10, and the other side may rest as it is actually divided into 100; as you may perceive by the 11th, Chapter.  
This being presupposed, at the intersection of the contrary parallel of 160, with the right parallel of the breadth, there lay the thread; so shall the contrary parallel of the length, cut the thread at the right parallel of the content.   diagram of the measurement of an oblong or rectangle (ABCD) 
Let the plane given to be measured be ABCD, the length thereof being 50 perches, and the breadth 25⅖ perches: now lay the thread upon the intersection of the contrary parallel of 160, with the right parallel of 25⅖ so shall it cut the contrary parallel of 50, at the right parallel of 8, which is the number of acres contained in the long square ABCD, as was required.   

CHAP. XXVII. Having the length and breadth of an oblong superficies given in Chains, to find the content in acres,  
HEre we will suppose each Chain to contain four perches in length, and the whole Chain to be divided into 100 equal links, according to Gunter's description: and this Chain as it is most easy by Arithmetic, so is it also by this Instrument, for whereas by perches we turned one side of our Quadrat into 1000, we may here for the most part turn each side into 20, so shall every 〈◊〉 parallel signify an unite, and the proportion will be this. 

As 10, to the breadth in Chains;  
So the length in Chains,  
To the content in acres.    
And thus in the former plane ABCD, the breadth being 6 Chains 40 links, and the length 12 Chains 50 links; if I lay the thread upon the contrary parallel of 10, at the intersection with the right parallel of 6 Chains 40 links; it shall cut the contrary parallel of 12 Chains 50 links, at the right parallel of 8 the content in acres, as before.   

CHAP. XXVIII. The perpendicplar and base of a Triangle being given in perches, to find the content in acres.  
ALL Triangles of what kind soever, are equal in their superficial content, unto half that right angled parallelogram, whose length and breadth is equal to the perpendicular, and the side whereon it falleth, as the Triangle CED, in the 26 th'. Chapter is one half of the long square ABCD, whose content was found as before. Or thus without halfing. 

As 320, to the perpendicular in perches:  
So the base in perches to the content in acres.    
Thus in the former Triangle CED, the perpendicular being 25⅗ perches, and the base 50 perches, if you lay the thread upon the intersection of the contrary parallel of 320, with the right parallel of 25⅗, you shall see it cut the contrary parallel of 50 at the right parallel of 4, and so many acres is in the Triangle CED, the thing required.   

CHAP. XXIX. Having the perpendicular and base of a Triangle, given in Chains, to find the content in acres  

AS 20, to the perpendicular in Chains:  
So the base in chains, to the content in acres.   
And so in the former Triangle CED, in the 26th. Chapter the perpendicular OF being 6 Chains 40 links, and the base CD 12 Chains 50 links; if I place the thread upon the intersection of the right parallel of 6 Chains 40 links with the contrary parallel of 20, it shall cut the contrary parallel of 12 Chains 50 links, at the right parallel of 4, the content of the Triangle CED, as before.   

CHAP. XXX. Having the content of a superficies after one kind of perch, to find the content of the same superficies according to another kind of perch.  

AS the length of the second perch,  
To the content in acres:  
So the length of the first perch,  
To a fourth number;  
And that fourth number,  
To the content in acres required.   
Suppose the plane ABCD measured with a Statute perch of 16½ feet, contained 8 acres; and it were demanded how many acres it would contain if it were measured with a Chain of 18 foot to the perch: these kind of propositions are wrought by the backward Rule of three, after a duplicated proportion. But by exchanging the places of the first and third numbers, by putting each in others place, it will become the forward Rule of three, and may be resolved after this manner. Place the thread upon the intersection of the contrary parallel of 18, with the right parallel of 8, and it will cut the contrary parallel of 16⅓ at the right parallel of 7 33/100, now where this right parallel of 7 33/100, crosseth the former contrary parallel of 18, there lay the thread again, so shall it cut the contrary parallel of 16½ at the right parallel of 6 72/100, which shows the content to be 6 acres 72 parts (such as 100 makes an acre) measured by an 18 foot pole.   

CHAP. XXXI. Having the plot of a plain with the content in acres, to find the scale by which it was plotted.  
SUppose the plane ABCD in the twenty sixth Chapter contained just 8 acres, if I should measure it with a scale of 12 in an inch; the length AB would be 60 pole, and the breadth AD 30 18/25 pole; and the content would be found by the 16 Chapter to be about 11½ acres where it should be but 8 acres. Therefore by the twelfth Chapter find a mean proportional number between 8 and 11½, which will be about 9½ and something better. Then will the proportion hold. 

As this mean proportional,  
To the scale you measured by;  
So the content of the plot,  
To the scale by which it was plotted.    
Wherefore I place the thread upon the intersection of the contrary parallel of 9½, with the right parallel of 12; so shall I see it cut the contrary parallel of 8, at the right parallel of 10; which 10 is the scale by which it was plotted.   

CHAP. XXXII. Having the length of the furlong given in perches, to find the breadth of the acre in perches.  

AS the length in perches unto 40:  
So is 4, to the breadth in perches.   
So the length of the furlong being 20 perches, I lay the thread at the intersection of the contrary parallel of 20, with the right parallel of 40, so will it cut the contrary parallel of 4; at the right parallel of 8, which is the breadth of the acre given, if the length had been 30, the breadth would have been 5⅓, and so of the rest.   

CHAP. XXXIV. Having the length of the furlong given in Chains, to find the breadth of the acre in Chains.  

AS the length in Chains unto 5;  
So is 2, to the breadth in Chain measure.   
So the length of the furlong being 4 Chains, lay the thread upon the intersection of the contrary parallel of 4, with the right parallel of 5, so shall it cut the contrary parallel of 2, at the right parallel of 2 Chains 50 links, and such is the breadth required.   

CHAP. XXXIV. Having the breadth of any Board given in inches, and the length in feet, to find the content in feet.  

AS 12 inches, to the breadth in inches:  
So the length in feet, to the content in feet.   
Thus in the figure A, the length being 10 foot, and the breadth 16 inches; If I place the thread at the intersection of the contrary parallel of 12, with the right parallel of 16, it will cut the contrary parallel of 10, at the right parallel of 13½, and so many square feet is in that board noted with the let A.  
 diagram of the measurement of a board   

CHAP. XXXV. Having the breadth of any board given in inches, to find the length of a foot superficial in inch measure.  

AS the breadth of the board given in inches,  
Is to 12 inches:  
So is 12 inches,  
To the length of a foot in inch measure.   
Thus in the figure A of the last Chapter, being a board whose breadth is 16 inches; if I place the thread at the intersection of the contrary parallel of 16, with the right parallel of 12, it will cut the contrary parallel of 12, at the right parallel of 9, and such is the length of a foot of that board, whose breadth is 16 inches.   

CHAP. XXXVI. How to measure board which is taper grown, or wider at one end then at the other.  
FIrst, add the breadth at both ends together, and take the one half thereof for a mean breadth; then say as in the 35 Chapter. 

As 12 inches, to this mean breadth in inches:  
So the length in feet, to the content in feet.    
As for example, let this figure B be a tapering board to be measured, whose length is 21 foot, and let the breadth at one end be 18 inches, and at the other 14, now adding 14 to 18, the sum will be 32, the half whereof is 16 inches for a mean breadth; then if you apply the thread to the contrary parallel of 12, with the right parallel of 16; it will cut the contrary parallel of 21, at the right parallel of 28, and so many square feet of flat measure, is contained in this tapering board.   diagram of the measurement of a board  

CHAP. XXXVII. Having the Circumference of a circle, to find the Diameter.  

AS 22 is to 7,  
So is the circumference to the Diameter.   
Thus in the following Circle A, the Circumference being 81 5/7, if I place the thread at the intersection of the contrary parallel of 22, with the right parallel of 81 5/7, it will cut the contrary parallel of 7, at the right parallel of 26, and such is the Diameter BC as was required.   

CHAP. XXXVIII. Having the Diameter of a circle, to find the circumference.  

AS 7 is to 22:  
So is the Diameter to the Circumference.   
Thus in this circle A, the Diameter being 26, if I place the thread to the intersection of the contrary parallel of 7, with the right parallel of 26, it will cut the contrary parallel of 22, at the right parallel of 81 5/7, and such is the circumference required.  
 diagram of the measurement of a circle   

CHAP. XXXIX. Having the chord of any arch, with the perpendicular falling from the middle of the said arch to the chord, to find the Diameter.  

AS the perpendicular, to half the chord line:  
So is half the said chord, to a fourth line or number.   
Which being added to the aforesaid perpendicular, giveth the Diameter required.  
Thus in the segment FEGH in the former circle, having the perpendicular or versed sine EH, 8, and the chord FG, 24, the half whereof is 12, the line GH, we may find the Diameter: for if we place the thread upon the intersection of the contrary parallel of 8, with the right parallel of 12; it will cut the contrary parallel of 12, at the right parallel of 18, unto which if you add 8, you shall have 26 for the whole Diameter DE as was required.   

CHAP. XL. Having the circumference and Diameter of a circle given in inches, to find the superficial content in feet.  
EVery circle is near equal to that right angled Triangle, of whose containing sides the one is equal to the Semidiameter, the other to the Circumference.  
Or to that right angled parallelogram, whose breadth is equal to the Semidiameter, and length to the Semicircumference.  
Or else to that right angled parallelogram, whose breadth is equal to the Diameter, and length to the one quarter of the Circumference. Or breadth equal to the one quarter of the circumference, and length to the Diameter; which is all one. Wherefore, 

As 144 inches, to the Diameter in inches:  
So one quarter of the circumference in inches,  
To the content in feet.    
Thus in the former circle BECD, having the Diameter BC 26 inches, and one quarter of the circumference BD or CD 20 3/7 inches, we may find the content in feet; for if we place the thread at the intersection of the contrary parallel of 144, with the right parallel of 26; it will cut the contrary parallel of 20 ●, at the right parallel of 3 5/8 and something more, and such is the content superficial in foot measure, as was required.   

CHAP. XLI. To find the superficial content of any part or segment of a circle:  
EVery part of a circle, whether it be semicircle, Quadrant, or any other sector of a circle, is near equal to that right angled parallelogram or long square, whose length and breadth is, the one equal to the semidiameter, the other to the one half of the arch line, or to the whole Diameter, and one quarter of the arch line. And therefore may be measured by the 26, or 27, or by this last Chapter.  
For if you would measure the semicircle ABDC in the 39 Chapter you may take the semidiameter AB, or AC for your breadth, and the one half of his arch line BDC, which is BD, or CD 20 3/7 for your length.  
Or if you would measure the Quadrant ABDELLA, you may take the semidiameter 13 for your length, and half the arch BD 10 3/14 for your breadth: the like holdeth for any sector of a circle whatsoever.  
But if you would measure the segment FDGH, you must first measure the plain of AFDGA, by the former rules, and then the Triangle AGF by the 28 or 29 Chapter or by the 41 Chapter, taking the perpendicular for the breadth, and half the base for the length of a right angled parallelogram; the content of which Triangle being added to the content of AFDGA, giveth the content of the segment FDGH.  
And if you would measure the segment FGE, you must first measure the sector EGAF, and from the content take the content of the Triangle AFG, so shall the remainder be the content of the segment FGE, as was required.   

CHAP. XLII. Having the circumference of a circle, to find the side of a square equal to the same circle.  

AS 1000 to the circumference:  
So is 282 to the side of the square.   
Thus in the circle of the 38 Chapter, having the circumference 81 5/7, we may find the side of a square equal thereunto.  
For if we apply the thread to the intersection of the contrary parallel of 1000, with the right parallel of 282, it will cut the contrary parallel of 81 5/7, at the right parallel of 23 43/1000, and such is the side of that square, which is equal to the same circle.   

CHAP. LXIII. Having the diameter of a circle, to find the side of a square equal to the same circle.  

AS 1000, to the diameter;  
So is 886, to the side of the square   
Thus in the circle of the 38 Chapter, having the Diameter 26 inches, we may find the side of a square equal thereunto.  
For if we place the thread upon the intersection of the contrary parallel of 1000, with the right parallel of 886; it will cut the contrary parallel of 26, at the right parallel of 23 43/100, and such is the side of the square required.   

CHAP. XLIV. Having the breadth and depth of a squared solid given in inches, and the length in feet, to find the content in feet.  


AS 12 inches, is to the breadth in inches:  
So is the length in feet, to the superficial content of that same side in feet.   Then again, 

As 12 inches, is to the depth in inches:  
So is this superficial content, to the solid content in feet.    
As for example, let the figure A be a squared piece of timber to be measured; whose breadth is 18 inches, and depth 14, and let the length thereof be 10 foot, then for to find the solid content.   diagram of the measurement of a block 
First place the thread to the intersection of the contrary parallel of 12, with the right parallel of 18; and it shall cut the contrary parallel of 10, at the right parallel of 15, and so many feet of flat measure is contained in the broadest side.  
Then look where this right parallel of 15, intersecteth the contrary parallel of 12, and there place the thread again, so shall it cut the contrary parallel of 14, at the right parallel of 17½ feet, and so many cubical feet is contained in this piece of squared Timber, noted with the letter A.   

CHAP. XLV. Having the breadth and depth of a squared solid given in inches, to find the length of a foot solid in inch measure.  


AS the breadth in inches,  
Is to 12 inches;  
So is 12 inches,  
To a fourth number.   Which is the length of a foot superficial at the breadth given then. 

As the depth in inches,  
Is to 12 inches:  
So is this fourth number,  
To the length of a foot solid.    
Thus in the squared piece of Timber in the last Chapter having the breadth 18 inches, and the depth 14 inches, we may find how many inches in length will make a foot solid of that piece.  
For if we place the thread to the intersection of the contrary parallel of 18, with the right parallel of 12; it will cut the contrary parallel of 12, at the right parallel of 8: and such is the length of a foot superficial, the breadth being 18 inches. Then again place the thread to the intersection of the contrary parallel of 14, with the right parallel of 12; and it will cut the contrary parallel of this 8, at the right parallel of 6 ●; and so many inches in length will make a foot solid, of that piece whose breadth is 18 inches, and depth 14.  
Having in this Chapter shown how you may by the Universal Quadrat, measure any squared solid or prisma, whose bases or ends are equal, like and parallel: I will now in some Chapters following, show how to reduce any prisma (of what fashion soever the base or end thereof be,) into a squared solid, and so by this Chapter to measure all manner of prismas, or all manner of Timber and Stone, whose bases or ends are equal, like and parallel.   

CHAP. XLVI How that prisma or piece of Timber, whose base or end is a Triangle, may be measured.  
ALL Triangles being equal in their superficial content, to half that right angled parallelogram, whose length and breadth is the one equal to the perpendicular, and the other to the side whereon it falleth, as was showed in the 28 Chapter; therefore if we take either half the perpendicular and the whole base, or the whole perpendicular and half the base whereon it falleth, the one for the breadth, and the other for the depth of a squared solid, we may measure it in all respects according to the 44 Chapter and that to any length required.   diagram of the measurement of a triangle (BCD) 
As for example, let the Triangle BCD be the base of a prisma or piece of Timber to be measured, now taking 10 the whole perpendicular AB for the breadth, and 8 the half base CD for the depth of a squared prisma or piece of Timber, we may find the solid content thereof by the 44 Chapter, and that according to any length required.   

CHAP. XLVII. How that prisma or piece of Timber, whose base or end is a Rhombus, (or Diamond form) may be measured.  
A Rhombus (or Diamond) is a figure of 4 equal sides, but no right angles, such as is the figure ABCD.  
Now seeing every Rhombus, is equal in their superficial content, to that right angled parallelogram (or long square) whose length is one of the sides, and breadth equal to the parallel distance of the same sides.  
If we take one of the sides for the breadth, and the length of the perpendicular falling thereon from the opposite angle, for the depth of a squared solid, we may find the solid content thereof by the 44 Chapter.  
As for example, let the Rhombus ABCD be the base or end of some piece of timber to be measured: now taking one of the sides which is 10 for the breadth of a squared solid, and the perpendicular which is something better than 8⅗ for the depth of the same solid, we may find the solid content thereof for any length required by the 44 Chapter.   diagram of the measurement of a rhombus (ABCD)  

CHAP. XLVIII. How that prisma or piece of Timber whose base or end is a Rhomboides (or Diamond like) may be measured.  
A Rhomboides (or Diamond like) is a figure, whose opposite sides, and opposite angles, are only equal, and it hath no right angles. Such as is the figure EFGH.  
Now seeing that the Rhomboides is also equal in his superficial content, to that right angled parallelogram whose length is one of the sides, and breadth equal to the parallel distance of the same sides.  
If we take one of the sides for the breadth of a squared solid, and the length of the perpendicular falling thereon from the opposite angle for the depth of the same solid, we may find the solid content thereof by the 45 Chapter.   diagram of the measurement of a rhomboid (EFGH) 
Let this Rhomboides EFGH be the base of a piece of Timber to be measured. Now if you take 16 the side GH or OF for the breadth of a squared solid, and 10 the length of the perpendicular EL for the depth of the same solid; you may find the solid content thereof by the 45 Chapter for any length required.   

CHAP. XLIX. How that prisma or piece of Timber, whose base or end is a trapeziam may be measured.  
A Trapeziam, is any irregular fouresided figure of what fashion soever, as this figure ABCD is a trapeziam, and may be cast into two Triangles, by drawing the diagonal line AC, and so each Triangle measured as is before showed; which being done, add the contents of them both together, & you shall have the content of the whole trapeziam. Or when you have drawn the diagonal line, and from the two opposite angles let down the perpendiculars thereon, join the two perpendiculars together, and take the sum thereof for the breadth of a squared solid, and take one half, the diagonal line, for the depth of the same solid, with which breadth and depth proceed by the 44 Chapter, according to your length given; and you shall thereby find the solid content required.   diagram of the measurement of a trapezium (ABCD) 
Let the trapeziam ABCD be the base or end of a piece of Timber: now if you take the two perpendiculars BF, and DE, and join them together they will make 10 for the breadth of a squared solid, and then take one half the diagonal AC, which will be 8 for the depth of the same solid, you may by the 44 Chapter find the solid content thereof according to any length given.   

CHAP. L. How that prisma or piece of Timber, whose base or end hath many sides, (as 5, 6, 7, 8, 9, 10, or more, so they be equal) may be measured.  

MAny sided figures are those which have more sides than four; and are generally called polygons; as this figure A, having equal sides and Angles, is called a polygon.  
Every regular polygon, being equal in the superficial content, to that long square whose length is equal to one half the perimiter, and breadth to a perpendicular drawn from the Centre to the middle of any side thereof.   diagram of the measurement of a polygon (regular pentagon) 
If we add all the sides together, and take half the sum for the breadth of a squared solid and then take the perpendicular for the depth of the same solid, we may by the 44 Chapter find the solid content thereof.  
Let the figure A be the base or end of a piece of Timber to be measured, consisting of 5 equal sides, and each side containing 12 inches, which being added together into one sum maketh 60, the half whereof is 30 for the breadth of your piece: then take the length of the perpendicular (falling from the Centre A to the midst of one of the sides,) which is 8 inches; for the depth of the same piece, with which breadth and depth, you may proceed to find the solid content by the 44 Chapter,  to any length given.   

A note to find the centres of those equiangle figures.  
If the sides be even, the Centre is found by drawing lines from one angle to his opposite angle; but if they be odd, it is found by drawing lines from the middle of a side to his opposite angle; so shall the cutting or intersection of those strait lines show the place of the centre.    

CHAP. LI. Having the circumference and Diameter of a Cylinder given in inches, and the length in feet, to find the content in feet.  


As 12 inches, to the Diameter in inches:  
So is the length in feet, to a fourth number.   Then, 

As 12 inches, to one quarter of the circumference:  
So is this fourth number, to the solid content in feet.    
As for example, let the figure A be a cylinder, or some round piece of Timber to be measured, whose circumference is 44 inches, and Diameter 14 inches; and let the length thereof be 10 foot, now for to find the solid content.  
First, place the thread to the intersection of the contrary parallel of 12, with the right parallel of 14; and it will cut the contrary parallel of 10, at the right parallel of 11⅔; now where this right parallel of 11⅔, cutteth the contrary parallel of 12, there place the thread again, so shall it cut the contrary parallel of 11, at the right parallel of 10 7/10, and so many solid feet is contained in this cylinder or round piece of Timber.   diagram of the measurement of a cylinder  

CHAP. LII. Having the circumference and Diameter of a cylinder or round piece of Timber, in inches, to find the length of a foot solid in inches.  


AS the Diameter in inches,  
Is to 12 inches:  
So is 12 inches,  
To a fourth number.   Then 

As one quarter of the Circumference,  
Is to 12 inches:  
So is this fourth number,  
To the length of a foot solid.    
Thus in the cylinder in the last Chapter, having the circumference 44 inches, and the Diameter 14 inches, we may find how many inches in length will make a foot solid of that round piece of Timber.  
For if we place the thread to the intersection of the contrary parallel of 14 inches, with the right parallel of 12 inches; it will cut the contrary parallel of 12 inches, at the right parallel of 10 2/7; then again I place the thread at the intersection of the contrary parallel of one quarter of the circumference (which is 11 inches,) with the right parallel of 12 inches; and that will cut the contrary parallel of 30 2/7 (which was the fourth number,) at the right parallel of 11 17/77, and so many inches in length will make a foot solid, of that round piece of Timber, whose circumference is 44 inches.   

CHAP. LIII. Having the circumference of a cylinder given in inches, and the length in feet, to find the content in feet.  


As 42 55/100, to the circumference in inches:  
So is the length in feet, to a fourth number.   Then 

as 42 55/100, to the circumference in inches,  
So is this fourth number, to the content in fect.    
Thus in the cylinder of the 51 Chapter, having the circumference 44 inches, and the length 10 foot, we may find the solid content.  
For if we place the thread at the intersection of the contrary parallel of 42 55/100, with the right parallel of 44; it will cut the contrary parallel of 10, at the right parallel of 10 34/100, then keeping the thread in the some position, it will cut the contrary parallel of this 10 34/100, at the right parallel of 10 ●, and so many solid feet is in that round piece of Timber.   

CHAP. LIV. Having the circumference of a cylinder given in inches, to find the length of a foot solid in inch measure.  


As the circumference in inches, is to 42 55/100;  
So is 12 inches, to a fourth number.   Then 

As the circumference in inches, is to 42 55/100:  
So is this fourth number, to the length of a foot solid  
in inches.    
Thus in the round piece of Timber of the 51 Chapter, having the circumference 44 inches, we may find the length of a foot solid in inches.  
For if we place the thread at the intersection of the contrary parallel of 44 inches; with the right parallel of 42 55/100: it will cut the contrary parallel of 12, at the right parallel of 11 6/10: now the thread lying in the same position, you shall see it cut the contrary parallel of this 11 6/10, at the right parallel of 11 22/100; and so many inches in length will make a foot solid of that round piece of Timber, whose circumference is 44 inches.   

CHAP. LV. Having the Diameter of a cylinder given in inches, and the length in feet, to find the content in feet.  


AS 13 54/100, to the Diameter in inches:  
So is the length in feet, to a fourth number.   Then, 

As 13 54/100, to the Diameter in inches:  
So is this fourth number, to the content in feet,    
Thus in the cylinder of the 52 Chapter, having the Diameter 14 inches, and the length 10 foot, we may find the content in feet.  
For if we place the thread at the intersection of the contrary parallel of 13 54/100, with the right parallel of 14 inches; it will cut the contrary parallel of 10 foot, at the right parallel of 10 34/100; then keeping the thread in the same position, it will cut the contrary parallel of this 10 34/100, at the right parallel of 10 7/10, and such is the solid content of that cylinder, whose Diameter is 14 inches.   

CHAP. LVI. Having the Diameter of a cylinder given in inches, to find the length of a foot solid.  


AS the Diameter in inches, is to 13 54/100;  
So is 12 inches, to a fourth number:   Then, 

As the Diameter in inches, is to 13 54/100:  
So is this fourth number, to the length of a foot solid in inches.    
Thus in the cylinder of the 51 Chapter, having the Diameter 14 inches, we may find how many inches in length will make a foot solid.  
For if we place the thread to the intersection of the contrary parallel of 14 inches, with the right parallel of 13 54/100; it will cut the contrary parallel of 12 inches, at the right parallel of 11 6/10; now the thread lying still in the same position, you shall see it cut the contrary parallel of this 11 6/10, at the right parallel of 11 22/100, and such is the length of a foot of that cylinder, whose Diameter is 14 inches.   

CHAP. LVII. Having the circumference and Diameter of the base of a cone given in inches, with the altitude thereof in feet, to find the solid content in feet.  


AS 12 inches,  
To the Diameter in inches:  
So is the third part of his altitude in feet,  
To a fourth number.   Then, 

As 12 inches,  
To one quarter of his circumference in inches:  
So is this fourth number,  
To the solid content in feet.    
As for example, let the figure ABC be a cone, or some round tapering piece of Timber or stone, which runneth tapering from the base thereof BC, to a very point at A like a piked Steeple.  
Let the circumference of the base BC be 132 inches, and the Diameter 42 inches, and let the altitude be 9 foot, which is the line AD, the axis or perpendicular falling from the highest point at A, to the very centre of the base at D.  
Now for to find the content, if you place the thread to the intersection of the contrary parallel of 12 inches, with the right parallel of 42 inches, it will cut the contrary parallel of 3 foot, (which is the third part of the Altitude given,) at the right parallel of 10 ●;   diagram of the measurement of a cone (ABC) 
Then again, if you place the thread to the intersection of the contrary parallel of 12 inches, with the right parallel of 33, (which is one quarter of the circumference,) it will cut the contrary parallel of 10½, at the right parallel of 28⅞, and so many foot solid is in this cone, whose circumference is 132 inches, and Diameter 42 inches, and altitude 9 foot.   

CHAP. LVIII. Having the circumference of the base of a cone given in inches, and his altitude in feet, to find the solid content in feet.  


AS 42 55/100 inches.  
To the circumference in inches:  
So is the third part of his altitude in inches,  
To a fourth number.   Then 

As 42 55/100 inches,  
To the circumference in inches;  
So is this fourth vumber,  
To the content in feet.    
Thus in the cone of the last Chapter having the circumference of the base 132 inches, and his altitude 9 foot we may find his solid content in feet.  
For if we place the the thread to the intersection of the contrary parallel of 42 55/100, with the right parallel of 132; it will cut the contrary parallel of 3, (which is the third part of his altitude) at the right parallel of 9 3/10; then keeping the thread in the same position, it will cut the contrary parallel of this 9 3/10, at the right parallel of 28⅞ and so many solid feet is in that cone, whose bases circumference is 132 inches, and altitude 9 foot.   

CHAP. LIX. Having the Diameter of the base of a cone given in inches, and his altitude in feet, to find the solid content in feet.  


AS 13 54/100 inches,  
To the Diameter in inches:  
So is the third part of his altitude in feet,  
To a fourth number.   Then, 

As 13 54/100 inches,  
To the diameter in inches:  
So is this fourth number,  
To the solid content in feet.    
Thus in the cone of the 57 Chapter, having the Diameter of his base 42 inches, and his altitude 9 foot, we may find his solid content in feet.  
For if we place the thread to the intersection of the contrary parallel of 13 54/100, with the right parallel of 42 inches; it will cut the contrary parallel of 3 foot, at the right parallel of 9 3/10; and lying in this same position it shall cut the contrary parallel of this 9 3/10, at the right parallel of 28⅞, the solid content of the cone, as before.   

CHAP. LX. Having the side and semidiameter of a cone, to find his altitude or perpendicular falling from the top thereof, perpendicularly upon his base.  
FIrst, place the bead to the side of the cone given upon the side of the Quadrat, and then open the thread till the bead fall directly upon the right parallel of the semidiameter of the base, so shall the bead cut the contrary parallel of the altitude required.  
Thus in the cone of the 57 Chapter having the semidiameter of his base 21 inches, and his side 9 foot and something better than 2 inches, that is 110 inches; if you place the bead to 110 inches, counted on the side of the quadrat, and then open the thread till the bead fall directly upon the right parallel of 21 inches, it shall cut the contrary parallel of 108 inches, which is just 9 foot for the altitude of the cone, which was required.   

CHAP. LXI. Having the side of a segment of a cone, with the semidiameter of each base or end thereof, to find the altitude of t●e whole cone, as it were before it was cut off, with the altitude of the lesser cone which was cut off.  
FIrst, take the lesser semidiameter out of the greater and note the difference, then say, 

As this difference,  
To the side of the segment of the cone:  
So is the greater semidiameter;  
To the altitude of the whole cone.  
And the less semidiameter,  
To the altitude of the lesser cone which was cut off.    
Thus in the cone of the 57 Chapter, although it should be cut off by the plane OF equidistantly to the base BC, yet having the side of the segment BEFC, of the cone ABC, which is BE or CF 6 foot and one inch, or 73 38/100 inches, with the semidiameter of the greater base 21 inches, and the lesser 7 inches, we may find the perpendicular of the whole greater cone as it were before it was cut off, and also of the lesser which was cut off.  
For if you take 7 inches out of 21, there will remain 14 inches; and than if you place the thread to the intersection of the contrary parallel of 14 inches, with the right parallel of 73 38/100 inches, it will cut the contrary parallel of 21 inches, (the semidiameter of the base of the greater cone) at the right parallel of 108 inches, which is 9 foot, the altitude of the whole greater cone, as if it had not been cut off. The thread lying still in the same position, doth likewise cut the contrary parallel of 7 inches, (which is the semidiameter of the base of the lesser cone which was cut off) at the right parallel of 36 inches, or 3 foot, the altitude of the lesser cone, which was required.   

CHAP. LXII. How the segment cut off from any cone may be measured.  
THe segment of a cone is no other than a round tapering piece of Timber cut off before it cometh to a point, and may be measured by that which before is delivered: when therefore at any time you meet with any such solid piece, (which will be more ordinarily then with any other) first get the length of his side, and then his semidiameter at each end, and so by the last Chapter, the altitude of the whole greater cone, and also of the lesser, which is supposed to be added to the segment to make it a perfect cone; this being done, by the 59, or 60 Chapters, measure the content of the whole greater cone, and also of the lesser; and taking the content of the lesser out of the greater, the remainder without out doubt must be the content of the segment given.  
As for example, let the segment BEFC of the cone in the 57 Chapter, be given to be measured, let the side thereof be 73 38/100 inches, and the semidiameter of the greater base BD 21 inches, and the semidiameter of the lesser base EKE 7 inches, and so by the last Chapter, the altitude of the whole greater cone AD 9 foot, and the altitude of the lesser added cone AK 3 foot, and then by the 59 Chapter, the content of the whole greater cone will be 28⅞ feet, and the lesser added cone 1 foot and something better, or 1 5/72 foot, the which if you take out of 28⅞ feet, the content of the whole greater cone, there will remain 27 29/36 feet, for the content of the segment BEFC which was required.   

CHAP. LXIII. To find the solid content of any pyramid.  
LEt the base of the pyramid be of what manner it will, you need do no more but take one third part of his altitude, and then suppose it to be a prisma (having equal bases, like, and parallel) whose length is equal to the third part of the pyramid, you may find the solid content by one or other of the former Chapters concerning prismas.  
As for example, let the following quadrangular pyramid be given to be measured, the side of whose base is 36 inches, and the side of the pyramid 76 36/100 inches, with the semidiameter of the circle enclosing the base, which is AB 25 45/100 inches, by which, and the side given, with the help of the 60 Chapter, we may find the altitude AD to be 72 inches or 6 foot; now having the side of the base in inches, and the altitude in feet, we may find the content in fect.  
For 

As 12 inches, to the side of the base in inches:  
So is the third part of the altitude in feet,  
To a fourth number.  
And this fourth number,  
To the solid content in feet.    
Wherefore, if you apply the thread to the intersection of the contrary parallel of 12, with the right parallel of 2 foot, the third part of the altitude; it will cut the contrary parallel of 36 inches the side of the base, at the right parallel of 6; the thread lying still in the same position, will cut the contrary parallel of this 6, at the right parallel of 18, and such is the solid content of this pyramid in feet which was required.   diagram of the measurement of a pyramid 
Now if you would measure any segment of a pyramid being cut off, observe the former rules given for the measuring of the segment of a cone, and you shall soon have your desire.  
For having the side of the segment of the pyramid, with the semidiameters of the circumscribing circles at each base, you may by the 61 Chapter, find the altitude of the whole greater pyramid, as if it had not been cut off, and also of the lesser pyramid which is supposed to be added to the segment, to make it a perfect pyramid; this being done, you may by the rules delivered, first get the content of the whole greater pyramid, and then of the lesser added pyramid, and taking the lesser out of the greater, the remainder will be the content of the given segment, as was required.   
The end of the first Book.   

The Second Book.  
Showing the most plentiful, easy, and speedy use of the Universal Quadrat, in the Resolution of the whole Doctrine Trigonometrical, as well plain as spherical, and that two several ways upon the Instrument, with surpassing facility, and with the least intricacy that may be.  


CHAP. I. To find the chord of any arch, the radius being given, not exceeding the side of the quadrat.  
THe radius of a circle, is the semidiameter of the same circle, or the total sine or sine of 90 degrees; so A is the semidiameter or radius upon which the divided arch EGLANTINE was dramn; and AB is the radius upon which the arch BFC was drawn.  
And a Chord is a right line subtending an arch; so BC is the chord of the arch BFC, for the finding of which chord for any arch required, place the radius given upon the side of the quadrat from the centre, and note the point where the same endeth, and there place the bead; then opening the thread to the arch given, counted in the quadrant, the distance between the point noted in the side of the quadrat, and the head shall be the chord required.   diagram of the measurement of an arch 
As for example, let AB be the radius of a circle placed upon the side of the quadrat, and let it be required to find a chord of 40 degrees. First I place the bead at the point B where the given radius ended, and opening the thread to 40 degrees, in the quadrant, I take the distance betwixt the point B, and the bead; which is the chord of 40 degrees, viz. the chord of the arch BFC, drawn upon the given radius AB, as was required, and thus may you find the chord of any arch as well as upon the sector.   

CHAP. II. To find the right sine of any arch given, the radius being put 1000  
HAving showed the use of these parallel lines as they stand in their own proper signification; it resteth now to show the use of them, as they signify right fines and tangents: and first to find the right fine of any arch, seek the degree given upon the quadrant counted from any side of the quadrant; but here let it be from the beginning of the degrees at E, so shall the right parallel running through the degree given in the quadrant; give the number of the fine required; but if you would have the fine in a right line, take the length of the contrary parallel from the side of the quadrat A, to the degree given in the quadrat; and that shall be your desired sine; whereof the whole side is radius.  
As for example, let it be required to find the right sine of 53 degrees 8 minutes I look 53 degrees 8 minutes in the quadrant, counted from the beginning of the degree at E, unto O, through which point O runeth the right parallel of 800, this 800 is the right sine of 53 degrees 8 minutes, 100 being radius.  
But if you would have this sine in a right line, take the length of the contrary paaallel from the point O unto N, for the right sine of 53 degrees 8 minutes which was required.   

CHAP. III. To find the arch of any sine given, the radius being put 1000  
IF your sine be given in numbers, seek the right parallel of the number given, for where that parallel cutteth the quadrant, there is the arch required, counted from the beginning of the degrees at E as before.  
But if your sine be given in a right line, then from the side of your quadrat A, extend your given sine upon any of the contrary parallels; and that right parallel that shall cut through the extreme point thereof, shall cut the quadrant in the required arch.  
As for example, let 800 be a sine given, unto which it is required to find the arch belonging thereto; I seek 800 in the right parallels, and finding it, I follow it to the quadrant, and there I see it cut the arch of 53 degrees 8 minutes, counted from E, which 53 degrees 8 minutes is the arch belonging to the sine 800, the radius being 1000  
But if the sine be given in a right line, take it betwixt your compasses, and setting one foot in the side of the quadrat A, extend the other upon any of the contrary parallels, as here I set one foot in the point N, and extend the other unto O, through which point the right parallel of 800 runneth, and cutteth the quadrant in 53 degrees 8 minutes counted from E as before.   

CHAP. IU. Any radius not exceeding the side of the quadrat being given, to find the right sine of any arch or angle thereunto belonging.  
FIrst, take the given radius and place it upon the contrary parallel of 100, which is the total sine; and where it shall end there place the thread, which being so placed; see where it cutteth the contrary parallel that passeth through the given arch, counted in the quadrant from G: for the segment so cut, which lieth between the thread and the side A, is the fine required.   diagram of the measurement of an arch (lines A and B) 
Let the line A be the radius given, and let it be required to find the sine of 36 degrees 53 minutes.  
First, I take the line A, and place it upon the contrary parallel of 100, as from E to D, upon which point D, I place the thread; this done I count the given arch viz. 36 degrees 33 minutes from the end of the quadrant at G unto O, and see what contrary parallel there I find, which is 60 intersecting the thread at M; so shall the distance NM be the line B the sine of 36 degrees 53 minutes, as was required.   

CHAP. V The right sine of any arch being given, to find the radius.  
FIrst, take the given sine, and place it from the side A of your quadrat, along upon that contrary parallel which cutteth the degrees of the sine given, counted in the quadrant from G, and where the sine endeth there place the thread; so shall it show upon the contrary parallel of 100, the radius required.  
Let the line B in the foregoing Chapter, be a given sine of 36 degrees 53 minutes. First, place it from N to M, upon that parallel which runneth through the degrees of 36 and 53 minutes as at O, then upon the point M, I place the thread; so doth it cut the contrary parallel of 100, in the point D; making DE the radius, which is the line A in the foregoing Chapter, as was required.   

CHAP. VI The radius being given, with a straight line resembling a sine, to find the quantity of that unknown sine.  
FIrst, place the given radius upon the contrary parallel of 100, and thereto apply the thread; then take the right sine given, and setting one foot of your compasses in the side of the quadrat, carry it parallel to the former, till the other foot cut the thread, and there stay; for the parallel where the compasses so resteth, shall cut the quadrant in the degrees answering to that unknown sine given.  
As if the line A in the fourth Chapter, were the radius given, and B the straight line resembling a sine: First, I place the line A from E to D, upon the contrary parallel of 100, and apply the thread thereto; the thread lying in this position, I take the line B, and fixing one foot in the side of the quadrat A, I carry it parallel to the former, until the other foot touch the thread, so will one foot rest in the point N, and the other in M, upon the parallel of 60, which parallel cutteth the quadrat in the point O at 36 degrees 53 minutes, counted from G, this 36 degrees 53 minutes, is the arch, of which the given line B is the sine, the line A being radius.   

CHAP. VII. The use of these parallels, as they signifienaturall sins.  
I Told you in the fourth Chapter, of the foregoing Book, that these lines did sometimes signify themselves alone, sometimes sins, and sometimes tangents, and now for (distinction sake) when I call them parallels, I would have you to understand them as they are in their own proper signification; and when I call them Sines, I would have them understood as the Sins of those arches through which they run; and when I call them Tangents, I would have them understood as the Tangent of those degrees, against which they butt in the sides of the quadrant.  
And further for distincton, whereas in their own signification I call them right and contrary parallels, so now I call them right and contrary sins, and right and contrary tangents.  
These lines as they signify sins, hath like use in finding a fourth proportional sine, as the ordinary Canon of Natural sins: and the manner of finding it, is always such as in this example. 

As the sine of 90 degrees unto the sine of 30 degree:  
So is the sine of 23 degrees 30 minutes unto a fourth sine.    
Wherefore place the thread at the intersection of the contrary Sine of 90 degrees, with the right Sine of 30 degrees, and it will cut the contrary Sine of 23 degrees 30 minutes, at the right Sine of 11 degrees 30 minutes, and such is the fourth proportional Sine required.  
Or you may place the thread at the intersection of the contrary Sine of 90 degrees with the right Sine of 23 degrees 30 minutes, and it shall cut the contrary Sine of 30 degrees, at the right sine of 11 degrees 30 minutes, as before.  
And thus may all the rest of the sinical propositions be wrought both ways.   

CHAP. VIII. The use of these parallels, as they signify both Sins and Tangents.  
HEre the work is in a manner the same as before, as will appear by this example. 

As the Sine of 90 degrees, to the Sine of 51 degr. 30 min.  
So is the Tangent of 30 degrees, to a fourth Tangent.    
Thus placing the thread upon the intersection of the contrary Sine of 90 degrees, with the right Sine of 51 degrees 30 minutes, it will cut the contrary Tangent of 30 degrees, at the right Tangent of 24 degrees 20 minutes.  
Or if you place the thread at the intersection of the contrary Sine of 90 degrees, with the right Tangent of 30 degrees, it will cut the contrary Sine of 51 degrees 30 minutes, at the right Tangent of 24 degrees 20 minutes, which is the fourth proportional Tangent required.  
And this manner of work will hold, until the Tangents be greater than 45 degrees the side of the quadrat. But when the given tangent or tangents exceed 45 degrees the side of the quadrat, if the two first given numbers be Sines and the third a Tangent; set the second Sine in the place of the first, and the first Sine in the place of the second; and in steed of the given Tangent, take his compliment unto a quadrant; so shall the fourth proportional, be the compliment of the Tangent required. As for example, in this proportion. 

As the Sine of 90 degrees,  
To the Sine of 20 degrees;  
So the Tangent of 70 degrees,  
To a fourth Tangent.    
And because the Tangent of 70 degrees is greater than 45 degrees, the side of the quadrat, therefore I turn the proportion thus. 

As the Sine of 20 degrees,  
To the Sine of 90 degrees;  
So the Co-tangent of 70 degrees,  
To the Co-tangent of the sum required.    
Thus placing the thread upon the intersection of the contrary sine of 20 degrees, with the right Tangent of 20 degrees, (the compliment of 70 degrees,) it should cut the contrary Sine of 90 degrees, at the right Tangent of the compliment of the arch required, but the thread intersecteth not the said contrary Sine of 90 degrees upon the quadrat, but besides it; which never happeneth but when the one Tangent is greater than 45 degrees, and the other less. Wherefore you are to note, that at any such time, though the thread cutteth not the contrary Sine of 90 degrees upon the quadrat, yet is the thread (by the former position) placed at the true angle of the Tangent required; which is 46 degrees 47 minutes, the compliment whereof 43 degrees 13 minutes, is the arch required. And if the first given numbers be Tangents, and the third a Sine; take the compliments of the two Tangents and exchange their places, as in the examples following. But if one be more and the others less, as in this proportion. 

As the Tangent of 70 degrees,  
To the Tangent of 43 degrees 13 minutes;  
So the Sine of 90 degrees:  
To a fourth Sine.    
Herein regard the Tangent of 70 degrees is more than 45 degrees, and the Tangent of 43 degrees less than the same, if you open the thread to the greater Tangents angle, viz. to the angle of 70 degrees, it shall cut the right Tangent of the lesser, viz. the right Tangent of 43 degrees 13 minutes at the contrary Sine of 20 degrees, and such is the fourth proportional Sine required.  
I have been the larger in this, to prevent mistakes in the rest.   

CHAP. IX. The use of these parallels, as they are in their own signification, joined with their signification of Sines and Tangents.  
THis manner of work is the same with the former, only you must know when the lines signify themselves alone, and when Sines and Tangents; the which you may know by the proportions given, as here in this example. 

As the Sine of 53 degrees 8 minutes, unto 80:  
So is the Sine of 36 deg. 52 min. unto a fourth number.    
Here if you place the thread at the intersection of the contrary Sine of 53 degrees 8 minutes with the right parallel of 80; it shall cut the contrary Sine of 36 degrees 52 minutes at the right parallel of 60, and such is the fourth proportional number required.    

Of the resolution of right lined Triangles  

CHAP. X. Having three angles and one side to find the other two sides.  
IN all Triangles there being six parts viz. three Angles and three Sides, any three whereof being known in a spherical Triangle, the other three may be found by the universal quadrat; but in a right lined Triangle, one of the three given terms must be a Side in the finding of the other parts.  
Thus having three angles and one Side, you may find the other two Sides by this proportion. 

As the Sine of the angle opposite to the given Side,  
To the number belonging to the same Side:  
So is the Sine of the angle opposite to the Side required,  
To the number belonging to the Side required.    
Let the following Triangle ABC be given, whereof the three angles and the Side AB is known; viz. the angle ACB 53 degrees 8 minutes, the angle CAB 36 degrees 52 minutes, and the angle ABC 90 degrees, the given side AB being 80, it is required to find the other two sides viz. AC and BC.  
Knowing the angle ACB opposite to the given side AB to be 53 degrees 8 minutes, and the angle opposite to the required side BC to be 36 degrees 52 minutes, I place the thread upon the intersection of the contrary Sine of 53 degrees 8 minutes with the right parallel of 80 his opposite side; the thread lying in this position I see it cut the contrary Sine of 36 degrees 52 minutes, at the right parallel of 60 his opposite side; therefore I conclude the side BC to be 60 such parts as AB is 80.  
The thread lying still in the former position see where it cutteth the contrary Sine of the third angle, for there is the right parallel of the third side: so here the thread cutteth the contrary Sine of 90 degrees, (which is the third angle) at the right parallel of 100, which is the side AC, as was required.  
So here you see the thread once placed, giveth both the sides required, and that right speedily in all Triangles; as well in obliqne angled as right angled triangles.   

CHAP. XI. Having two sides given, and one angle opposite to either of them; to find the other two angles, and the third side.  

AS the side opposite to the angle given,  
Is to the Sine of the angle given:  
So is the other side given,  
To the sine of that angle to which it is opposite.    diagram of the measurement of a triangle (ABC) 
Thus in the Triangle ADC, having the two sides AD 35, and CD 75, with the angle ACD 16 degrees 16 minutes, being opposite to the side AD; I may find the angle GOD which is opposite to the other side CD, for if I place the thread upon the intersection of the contrary parallel of 35, with the right Sine of 16 degrees 16 minutes, it shall cut the contrary parallel of 75, at the right sine of 36 degrees 52 minutes, his opposite angle, the angle GOD.  
Then keeping the thread in the former position, add these two known angles together, and they will give you the outward angle ADC 53 degrees 8 minutes, now where the right Sine of 53 degrees 8 minutes, cutteth the thread, there is the contrary parallel of 100, which is the third side AC, as was required.   

CHAP. XII. Having two sides and the angle between them, to find the two other angles and the third side.  
IF the angle contained between the two sides be a right angle, the other two angles will be found readily by this proportion 

As the greater side given,  
Is to the lesser side;  
So is the Tangent of 45 degrees,  
To the Tangent of the lesser angle.    
So in the rectangle Triangle ABC, knowing the side AB to be 80, and the side BC to be 60. If I place the thread upon the intersection of the contrary parallel of 80, with the right parallel of 60; it shall cut the contrary Tangent of 45 degrees at the right Tangent of 36 degrees 52 minutes, and such is the lesser angle CAB; the compliment whereof unto a quadrant is the greater angle ACB 53 degrees 8 minutes, the angles being known, the third side AC may be found by the 10 Chapter.  
But if it be an obliqne angle that is between the two sides given; the Triangle may be reduced into two rectangles and then resolved as before.  
As in the Triangle ADC, where the side AC is 100, and the side AD 35, and the angle GOD 36 degrees 52 minutes, if you let down the perpendicular DE, upon the side AC, you shall have two rectangle Triangles, AED, DEC; and in the rectangle AED, the angle at A being 36 degrees 52 minutes, the other angle ADE will be 53 degrees 8 minutes, by compliment: and with these angles and the side AD, you may find both A and DE, by the tenth Chapter.  
Then taking A out of AC, there remains EC for the side of the rectangle DEC; and therefore with this side EC and the other DE, you may find both the angle at C, and the third side CD, by the former part of this Chapter.  
Or you may find the angles required, without letting down any perpendicular.  
For 

As the sum of the sides,  
Is to the difference of the sides:  
So the Tangent of the half sum of the opposite angles,  
To the Tangent of half the difference between those angles,    
As in the former Triangle ADC, the sum of the sides AC and AD, is 135, and the difference between them 65; the angle contained 36 degrees 52 minutes; and therefore the sum of the two opposite angles 143 degrees 8 minutes, and the half sum 71 degrees 34 minutes, and because the half sum 71 degrees 34 minutes, is greater than the side of the quadrat 45 degrees, therefore I turn the proportion after this manner. 

As the difference of the sides 65,  
Is to the sum of the sides 135:  
So is the co-tangent of the half sum of the opposite angles 71 degrees 34 minutes.  
To the co-tangent of half the difference between those angles.    
Therefore if you place the thread upon the intersection of the contrary parallel of 65, with the right parallel of 135, you shall see it cut the contrary Tangent of 18 degrees 26 minutes, (which is the compliment of 71 degrees 34 minutes,) at the right Tangent of 34 degrees 42 minutes, the compliment whereof is 55 degrees 18 minutes, and such is the half difference between the opposite angles at B and D, this half difference being added to the half sum, giveth 126 degrees 52 minutes, for the greater angle ADC, and being substracted, leaveth 16 degrees 16 minutes for the lesser angle ACD, the three angles being thus found, you may find the third side CD by the tenth Chapter.   

CHAP. XIII. Having the three sides of a right lined Triangle, to find the three angles.  
LEt one of the three sides given be the base, but rather the greater side, that the perpendicular may fall within the Triangle; then gather the sum and difference of the two other sides, which being done the proportion will be. 

As the base of the Triangle,  
Is to the sum of the sides;  
So the difference of the sides  
To the alternate base.    
This alternate base being taken out of the true base, if we let down the perpendicular from the opposite angle, it shall fall upon the middle of the remainder.  
As in the former Triangle ADC, where the base AC is 100, the sum of the sides AD and CD is 110, and the difference of them 40.  
Therefore if you place the thread upon the intersection of the contrary parallel of 100, with the right parallel of 110, it will cut the contrary parallel of 40, at the right parallel of 44, for the alternate base; this alternate base take out of 100 the true base, and the remainder will be 56, the half whereof is 28, and showeth the distance from A unto E, where the perpendicular shall fall, from the angle D, upon the base AC, dividing the former Triangle ACD into two right angled Triangles, viz. AED and DEC, in which the angles may be found by the 11th. Chapter.  
Thus you may see how by this universal quadrat, we have resolved the four propositions of Mr. Gunter's upon his cross-staff; whereby you may perceive the agreement betwixt this quadrat and his staff.  
And now I will show how they may be otherwise resolved upon this universal quadrat, agreeing nearer with the Sector; but more spendily performed, then by either Sector or Cross-staffe.   

CHAP. XIIII. In a rectangle Triangle having the angles and one of the sides given, to find the other side and the base.  
OPen the thread to the angle opposite to the side given, so shall it intersect the right parallel of the side given, at the contrary parallel of the side required; and if to the intersection you place the bead, and apply the thread to the side of the quadrat, it shall there show you the length of the base.  
As in the right angled Triangle ABC in the 11 Chapter, the Side BC being given, opposite to the angle BAC 36 degrees 52 minutes, it is required to find the other Side AB, and the base AC.  
First, open the thread to the angle of 36 degrees 52 minutes, so shall it cut the right parallel of 60, at the contrary parallel of 80, and such is the side AB, which was required.  
The thread lying in the same position, if you place the bead to this intersection, and then apply the thread to the side of the quadrat, it shall there give you 100, for the base AC as was required.   

CHAP. XV. Having both sides of a rectangle Triangle, to find the two angles and the base.  
AT the intersection of the contrary parallel of the greater side, with the right parallel of the lesser side, there place the thread; which being so placed giveth the lesser angle required. And if to the intersection you place the bead and then apply the thread to the side of the quadrat, it shall there show the base required.  
As in the rectangled Triangle ABC in the 11 Chapter, where the two given sides is AB 80, and BC 60: Wherefore I place the thread at the intersection of the contrary parallel of 80, with the right parallel of 60, so doth the thread give me the lesser angle at A, which is 36 degrees 52 minutes, which being known, the angle ACB will be found by compliment, to be 53 degrees 8 minutes, thus have I both the angles required.  
The thread lying in the former position, I place the bead to the intersection of the two sides; and then applying the thread to the side of the quadrat, it giveth me 100 for the base AC, which was required.   

CHAP. XVI To find a side and both the angles, by having the base and the other side given.  
FIrst, lay the thread upon the side of the quadrat, and apply the bead to the length of the base; then open the thread until the bead fall directly upon the contrary parallel of the side given, and it shall likewise lie upon the right parallel of the side required, and the thread shall also show the angle opposite to the required side.  
As in the former rectangle Triangle ABC in the 11th. Chapter, having the base AC 100, and the side AB 80; if you apply the thread to the side of the quadrat, and place the bead to 100, and then open the thread until the bead fall directly upon the contrary parallel of 80, so shall the same bead so posited, cut the right parallel of 60 the side required. And further, the thread lying in this constitution, giveth the angle opposite to the required side to be 36 degrees 52 minutes, which being known, the other is soon found out by compliment unto 90 degrees.   

CHAP. XVII. To find the sides by having the base, and the angles given.  
FIrst, place the bead to the base given on the side of the quadrat, then open the thread to any of the two angles, and the bead shall fall upon the right parallel of that side which is opposite to the angle, unto which the thread was opened; and also upon the contrary parallel of the other side.  
As for example, in the right angled Triangle ABC in the 11th. Chapter, the angles at A and C are given, viz. the angle ACB 53 degrees 8 minutes, and the angle BAC 36 degrees 52 minutes, with the base AC 100, and it is required to find the two sides AB, and BC.  
First, therefore I place the bead unto 100 upon the side of the quadrat, and opening the thread to the angle of 36 degrees 52 minutes, I find the bead to fall upon the right parallel of 60, which is the side CB subtending the said angle; through which intersection doth pass the contrary parallel of 80, which is the other side AB subtending the other angle ACB, which was required.   

CHAP. XVIII. In any right lined Triangle whatsoever, to find a side by knowing the other two sides, and the angle contained by them.  
FIrst, place the bead to the shorter side given, and open the thread to the given angle, then setting one foot of your Compasses in the other side given upon the side of the quadrat, with the other foot extended to the bead, you shall have the third side betwixt your compasses.  
Thus in the triangle ADC in the 11th. Chapter, having the sides AC 100, and BC 75, and the angle between them ACD 16 degrees 16 minutes if you place the bead to 75, and open the thread to the angle of 16 degrees 16 minutes, then setting one foot of your compasses in 100 and extending the other unto the bead, you shall have 35 betwixt your compasses for the third side AD, which was required.   

CHAP. XIX. To find an angle by knowing the three sides.  
FIrst, place the bead to the shortest of the sides containing the angle required; then take the side subtending the same angle betwixt your compasses, and setting one foot thereof in the third side counted in the side of the quadrat, with the other foot open or shut the thread, until the bead and the movable foot of the compasses meet both in one point, so shall the thread show the angle required.  
Thus having the three sides of the Triangle ADC in the 11th. Chapter. First, I place the bead to 75 the side CD, than I take 35 betwixt my compasses for the side AD, (which is opposite to the required angle ACD) and setting one foot in 100 (upon the side of the quadrat) for the side AC, I turn the other foot towards the thread, moving it to and fro until the compass point fall just upon the bead, so shall the thread cut 16 degrees 16 minutes for the angle ACD, as was required.  
And this may suffice for right lined Triangles, whereby you may perceive the agreement of this Universal quadrat, both with Sector & Crosstaff. I will set down only two or three Chapters more, for the ready reducing of hypothenusal to horizontal lines in the art of Surveying. And for the speedy searching out of all perpendicular altitudes.   

CHAP. XX. Of the ready reducing of hypothenusal to horizontal lines.  
FIrst, measure the hypothenusal line with your chain, then place your bead to the length thereof counted on the side of your quadrat; this being done, observe by your instrument the angle either ascending or descending, the thread and plummet having free liberty to play; So shall the bead fall upon the contrary parallel of the horizontal line required.  
Suppose ABCD be a hill or mountain to be protracted, and laid down in your plot amongst your other grounds.   diagram of the measurement of a triangle (ABC) 
It is apparent by the figure, that the hypothenusal lines AB, and C cannot be laid down exactly in a right line, between the other grounds which bounder on this hill at the point A and C: wherefore we are to find the true level and horizontal distance between A and C, which is a right line extending overthwart the ground whereon the hill standeth, which to do first measure the hypothenusal line BC, which in this example will be found to be 60 pole, unto which 60 place the bead; then planting your instrument at C, and observing the ascent from C to B, (by lifting up your Instrument towards B) you shall find the thread to fall upon 36 degrees 52 minutes, for the angle of the ascent BCD; and the bead to fall upon the contrary parallel of 48, which is the horizontal distance CD, which was required.  
So likewise, the thread hanging at the former angle the bead as before fixed shall fall upon the right parallel of 36, which is the length of the perpendicular BD, being let down from the top of the hill at B through the, same, unto the horizontal line AC whereon the hill standeth. And what is said here of the ascent CB, the same may be understood of the descent BASILIUS.   

CHAP. XXI. To find the height of an object accessible at one obsenvation. or the distance betwixt the top of the object and that point thereof which is level with your eye.  
FIrst, lift up the quadrat looking through both the sights to the top of the object, going nearer or farther from it, till the thread fall directly on the opposite angle of the quadrat, which is at the intersection of the right and contrary parallels of 100, or at 45 degrees, in the quadrant; then letting down the quadrat still looking through the sights, until the thread fall directly upon the side of the quadrat; then what point soever you see in the object through the sights, is level with your eye; then measure the distance betwixt your eye and the level point in the object, for that is equal to the height required, above the point levelly with your eye, this needeth no example.  
But more readily thus, take a station at any convenient distance, which distance measure and count the number thereof amongst the contrary parallels; then lifting up your quadrant looking through the sights to the top of the object, the thread shall cut the contrary parallel of the measured distance, at the right parallel of the height required.  
Let BC be the height of some Steeple or Tower whose height is required; first, I measure out any convenient distance; as here I measure from the base of the object at B, 80 paces unto A, and at AI take my station; and lifting up my quadrat until I can see the very top of the object at C, through both the sights, I find the thread to cut the contrary parallel of 80 at the right parallel of 60, which is the number of places contained in the height BC, as was required. And this of all others I hold the speediest way for attaining of all altitudes accessible: and if to this intersection you place the bead, it shall give you the length of the hypothenusal or scaling ladder AC.  
And you may also speedily search out any perpendicular heights by the rules of proportion after this manner.   diagram of the measurement of a triangle (ABC) 
Having made choice of a convenient station as at A, observe the angle at A made between the hypothenusal line AC, and the level line with your eye AB, the compliment whereof is the angle at the top of the object at C, made between the hypothenusal line AC, and the line CB, the height required. These being known, with the distance AB: Say, 

As the sine of the angle ACB,  
To the measured distance AB:  
So the sine of the angle BAC,  
To the required height BC.    
Thus having found the distance AB to be 80 yards or paces, and the angle at A to be 36 degrees 52 minutes, the compliment whereof is 53 degrees 8 minutes, for the angle at C; I place the thread at the intersection of the contrary Sine of 53 degrees 8 minutes with the right parallel of 80; so doth the thread cut the contrary Sine of 36 degrees 52 minutes, at the right parallel of 60, wherefore I conclude the height BC, to be 60 such parts as AB is 80, which is the thing required.   

CHAP. XXII. To find the height of an object inaccessible, at two observations.  
FIrst, as far off your object as conveniently you may, make choice of your first station; and observe the angle of your eye, made between the visual line to the top of the object, and the level line with your eye, this being done, measure forward in a right line towards the object, approaching as near thereunto as you may, and there make choice of your second station where observe the angle at your eye as before: now take the compliment of this second angle, out of the compliment of the first angle, and note the difference, then, say, 

As the sine of the difference of the compliments,  
To the measured distance;  
So the Sine of the first angle observed,  
To the second hypothenusa.   Then again, 

As the Sine of 90 degrees,  
To the second hypothenusa:  
So is the Sine of the second observed angle,  
To the height required    
As for example, let BC in the former Chapter, be the height of some Tower or Steeple to be measured, unto which I may not approach, by reason of some River or other obstacle betwixt the object and me; and yet it is required to find the altitude BC, and also the breadth of the river BD, with the length of the scaling ladder CD.  
First, therefore I make choice of a station with the best convenience I can, as at A, where I make observation, and find the angle at my eye to be 36 degrees 52 minutes, the compliment whereof is 53 degrees 8 minutes, then measuring in a right line towards the object, I approach as near as I may, which is at the point D, finding AD to be 35 paces; and now at D, I make a second observation, and find the angle at my eye to be 53 degrees 8 minutes, the compliment whereof is 36 degrees 52 minutes, this 36 degrees 52 minutes I take out of the former compliment 53 degrees 8 minutes, and the difference is 16 degrees 16 minutes, the angle ACD, wherefore I place the thread at the intersection of the contrary Sine of 16 degrees 16 minutes, with the right parallel of 35, and I find the thread to cut the contrary Sine of 36 degrees 52 minutes (which is the angle at A) at the right Parallel of 75, and such is the scalling ladder, or hypothenusal line CD.  
Then again, I place the thread at the intersection of the contrary Sine of 90 degrees, with the right parallel of 75, and it cutteth the contrary Sine of 53 degrees 8 minutes, (which is the angle at D) at the right parallel of 60, and such is the line BC, the height required. And the thread in the same position, cutteth the contrary Sine of the compliment of the second angle observed, viz. 36 degrees 52 minutes, at the right parallel of 45, and such is the breadth of the River or other obstacle BD.  
Or having found the length of the second hypothenusal line to be 75, if you place the bead thereto, and then open the thread to the second observed angle, viz. 53 degrees 8 minutes, the bead shall fall upon the right parallel of 60, which is the height BC; and also upon the contrary parallel of 45, which is the breadth BD, as before, and as was required.  
Or more speedily thus, at two observations and one operation.  
First, make choice of your stations as before, then make choice of some one of your right parallels, that the number thereof may signify the height required; then at your first station make observation as before, and note the parts cut by the thread upon the said right parallel; this being done, measure your stationary distance, as before; and at the second station, make observation, as you did at the first station; noting likewise the parts cut by the thread upon the said right parallel; now taking these latter parts out of the former, or the lesser out of the greater, the difference shall signify the stationary distance: wherefore, 

As the distance upon the parallel,  
To the measured distance:  
So the number of the said parallel,  
To the height required.    
Thus in the figure of the foregoing Chapter, I take my first station at A, and making choice of some one of the right parallels, (as here of the right parallel of 45) to signify the height required, I make observation towards C, and I find the thread to cut the said parallel at 60: then measuring the stationary distance AD, I find it to be 35, and at D, I make a second observation, and I find the thread to cut the foresaid parallel at about 33¾ which being taken out of 60, leaveth 26¼ for to signify the stationary distance; wherefore I place the thread at the intersection of the contrary parallel of 26¼, with the right parallel of 35, & it will cut the contrary parallel of 45, at the right parallel of 60, which is the height BC, as was required.  
Many propositions to this purpose might be framed, but these may suffice for a taste of the use of this universal quadrat.  
And if an index with sights were fitted to turn upon the centre it would serve then by the same reason, for the finding of all other distances.    

Of the resolution of spherical Triangles; and first of those which be right angled.  

CHAP. XXIII. To find a side by knowing the base, and the angle opposite to the required side.  
A Spherical Triangle, is a figure included by three arches of great circles; the angles whereof are measured by the arch of a great circle subtending the angle, and intercepted between the containing sides continued to quadrants.  
Now in every Triangle there being six parts, viz. three angles and three sides, any three of them being given in a spherical Triangle, the other three may be found by the Universal quadrat, and that two several ways, and as it were by two several Instruments; the former part agreeing nearest with the sector, the latter part (performed by the Planisphere on the back of the quadrat) agreeing nearest to the Mathematical Jewel of Master Blagraves; but the working hereby will appear more clear than through the thick rete of the Jewel.  
First, to find a side by knowing the base, and the angle opposite to the required side. 

As the radius, is to the sine of the base:  
So is the sine of the opposite angle,  
To the sine of the side required.     diagram of the measurement of a spherical triangle (ABC) 
Let the Triangle ABC be given, wherein let A stand for the point of East or West; AB an arch of the Horizon representing the azimuth of the Sun from the East or West points; BC an arch of an azimuth or vertical circle, making right angles with the Horizon AB, in the point B, representing the Sun's altitude; and AC an arch of the Equator representing the hour from Sun rising or setting when he hath no declination. and further, let the angle at A be the compliment of the latitude, and the angle at C the vertical angle, made between the equator and the vertical circle passing by the Sun at C, perpendicularly to the horizon AB, wherefore knowing the base AC to be 65 degrees 23 minutes, and the angle at A to be 37 degrees 30 minutes, I may find the side BC, for if I place the thread to the intersection of the contrary Sine of 90 degrees with the right Sine of 65 degrees 22 minutes it will cut the contrary Sine of 37 degrees 30 minutes, at the right Sine of 33 degrees 36 minutes, and such is the side BC, which was required.   

CHAP. XXIV. To find a side by knowing the base, and the other side.  

AS the Sine of the compliment of the side given,  
To the sine of the compliment of the base:  
So is the radius,  
To the Sine of the compliment of the side required.   
Thus in the rectangle ABC, having AC 65 degrees 23 minutes, and AB, 60 degrees, let it be required to find the side BC, place the thread at the intersection of the contrary Sine of 30 degrees, (the compliment of AB) with the right Sine of 24 degrees 37 minutes, (the compliment of AC) and it shall cut the contrary Sine of 90 degrees, at the right sine of 56 degrees 24 minutes, the compliment whereof is 33 degrees 36 minutes, the side BC, which was required.   

CHAP. XXV. To find a side, by knowing the two obliqne angles.  

AS the Sine of either angle,  
To the cousin of the other angle:  
So is the sine of 90 degrees,  
To the cousin of the side opposite to the second angle.   
So in the rectangle ABC, having ACB for the first angle 72 degrees 16 minutes, and BAC for the second 37 degrees 30 minutes, it is required to find the side BC, wherefore I place the the thread at the intersection of the contrary Sine of 72 degrees 16 minutes, with the right Sine of 52 degrees 30 minutes, (the compliment of the angle at A) and I find it cut the contrary Sine of 90 degrees, at the right Sine of 56 degrees 24 minutes, whose compliment is 33 degrees 36 minutes, the side BC, as was required.   

CHAP. XXVI To find the base, by knowing both the sides.  

AS the radius, to the cousin of the one side:  
So the cousin of the other side,  
To the cousin of the base.   
Thus in the rectangle ABC, knowing the latitude of the place, with the altitude of the Sun having no declination, we may find the hour of the day; as having BC the altitude of the Sun 33 degrees 36 minutes and the angle BAC the compliment of the latitude 37 degrees 30 minutes, the base AC which is the hour of the day counted from Sun rising or setting, will be found by the quadrat. For if you apply the thread to the intersection of the contrary Sine of 37 degrees 30 minutes with the right Sine of 33 degrees 36 minutes, it will cut the contrary Sine of 90 degrees, at the right Sine of 65 degrees 23 minutes and such is the base AC, which being reduced into time (by allowing 15 degrees to an hour, and for every 4 minutes a degree) giveth 4 hours 22 minutes for the time after Sun rise, or before Sun set: so that if it be in the forenoon it is 10 a clock and 22 minutes, the Sun then rising at 6; and if it be in the afternoon it is 38 minutes past one.   

CHAP. XXVIII. To find an angle, by knowing the other obliqne angle, and the side opposite to the inquired angle.  

A the radius,  
To the sine of the angle given:  
So the cousin of the side,  
To the cousin of the angle required.   
Thus in the Rectangle ABC, having the angle BAC 37 degrees 30 minutes, with the side AB 60 degrees, to find the angle ACB: I place the thread at the intersection of the contrary Sine of 90 degrees with the right Sine of 37 degrees 30 minutes, so will it cut the contrary sine of 30 degrees, (the compliment of the side AB) at the right Sine of 17 degrees 44 minutes, the compliment whereof is 72 degrees 16 minutes, the angle ACB, which was required.   

CHAP. XXIX. To find an angle, by knowing the other obliqne angle, and the side opposite to the angle given.  

As the cousin of the side,  
To the cousin of the angle given:  
So is the radius,  
To the Sine of the angle required.   
So in the rectangle ABC, having the angle BAC 37 degrees 30 minutes, and the side BC 33 degrees 36 minutes, to find the angle ACB: place the thread at the intersection of the contrary Sine of 56 degrees 24 minutes, (the compliment of BC) with the right sine of 52 degrees 30 minutes (the compliment of the angle BAC) and it will cut the contrary sine of 90 degrees, at the right sine of 72 degrees 16 minutes, the angle ACB, which was required.   

CHAP. XXX. To find an angle, by knowing the base, and the side opposite to the inquired angle.  

AS the sine of the base,  
To the sine of 90 degrees:  
So the sine of the side given,  
To the sine of the angle required.   
Thus in the rectangle ABC, having the base AC 65 degrees 23 minutes, and the side BC 33 degrees 36 minutes, to find the angle BAC: I place the thread to the intersection of the contrary sine of 65 degrees 23 minutes, with the right sine of 90 degrees, and I find it cut the contrary sine of 33 degrees 36 minutes, at the right sine of 37 degrees 30 minutes, the angle BAC which was required.  
Thus far by sins alone; but now we must require joint help of Tangents.  
And forasmuch as the Tangents extend but to 45 degrees, which is equal to the side of the quadrat. I shall set down the two ways for the resolution of each proposition, delivered by Master Gunter; and if one will not hold, the other will; observing the rules given to this purpose in the 8 Chapter.   

CHAP. XXXI. To find a side, by knowing the other side, and the angle opposite to the inquired side.  


AS the radius,  
To the sine of the side given:  
So is the Tangent of the angle,  
To the Tangent of the side required.   Or 

As the sine of the side given,  
Is to the radius:  
So the cotangent of the angle,  
To the cotangent of the side required.    
Thus in the rectangle ABC, having the side AB 60 degrees, and the angle BAC 37 degrees 30 minutes, to find the side BC; I place the thread at the intersection of the contrary sine of 90 degrees, with the right sine of 60 degrees, and it doth cut the contrary Tangent of 37 degrees 30 minutes, at the right Tangent of 33 degrees 36 minutes, the side BC, which was required.   

CHAP. XXXII. To find a side, by knowing the other side, and the angle next to the inquired side.  


AS the Tangent of the angle given,  
To the Tangent of the side given:  
So is the radius,  
To the sine of the side required.   Or, 

As the cotangent of the side given,  
To the cotangent of the angle given;  
So is the radius,  
To the Sine of the side required.    
Thus in the rectangle ABC having the side AB 60 degrees, and the angle ACB 72 degrees 16 minutes, it is required to find the side BC. Now here in regard the given Tangents exceed the side of the quadrat; I let pass the first proportion, and make use of the second: and opening the thread to the angle of 30 degrees, or which is all one, if I place the thread at the intersection of the contrary Sine of 90 degrees, with the right Tangent of 30 degrees, (which is the compliment of the side AB) it will cut the right Tangent of 17 degrees 44 minutes, (the compliment of the angle ACB) at the contrary 〈…〉 33 degrees 36 minutes, the side BC, or placing the thread at the intersection of the contrary Tangent of 30 degrees, (the compliment of the side AB) with the right Sine of 90 degrees, it will cut the contrary Tangent of 17 degrees 44 minutes, (the compliment of the angle ACB) at the right sine of 33 degrees 36 minutes, the side BC. Or if the thread be placed at the intersection of the contrary Tangent of 30 degrees, with the right Tangent of 17 degrees 44 minutes, it will cut the contrary Sine of 90 degrees, at the right Sine of 33 degrees 36 minutes, for the side BC, as before, which was required.   

CHAP. XXXIII. To find a side by knowing the base, and the angle next to the inquired side.  


AS the radius,  
To the cousin of the angle given:  
So is the Tangent of the base,  
To the Tangent of the side required.   Or, 

As the cousin of the angle given,  
Is to the radius:  
So is the cotangent of the base,  
To the cotangent of the side required.    
So in the rectangle ABC, having the base AC 65 degrees 23 minutes, for the hour from Sun-rise; (he having no declination) and the angle BAC 37 degrees 30 minutes, for the compliment of the latitude; we may find the Azimuth from the East, which is the side AB: for if the thread be applied to the intersection of the contrary Sine of 52 degrees 30 minutes, (which is the compliment of the angle BAC,) with the right Tangent of 24 degrees 37 minutes, (the compliment of the base AC) it shall cut the contrary Sine of 90 degrees at the right Tangent of 30 degrees, the compliment whereof is 60 degrees, the side AB, or the Azimuth from the East, which was required.   

CHAP. XXXIV. To find the base by knowing the obliqne angles.  

As the Tangent of the one angle,  
To the cotangent of the other angle:  
So is the radius,  
To the cousin of the base.   
Thus in the rectangle ACB, having the angle BAC 37. degrees 30 minutes, and the angle ACB 72 degrees 16 minutes, to find the base AC: I place the thread at the intersection of the contrary Tangent of 37 degrees 30 minutes with the right Sine of 90 degrees, and it will cut the contrary Tangent of 17 degrees 44 minutes, (the compliment of the angle ACB) at the right Sine of 24 degrees 37 minutes, the compliment whereof is 65 degrees 23 minutes, the base AC, as was required.   

CHAP. XXXV. To find the base, by knowing one of the sides, and the angle next the same side.  


AS the radius,  
To the cousin of the angle given:  
So is the cotangent of the side given,  
To the cotangent of the base.   Or 

As the cousin of the angle given,  
Is to the radius:  
So the tangent of the side given,  
To the tangent of the base required.    
Thus in the rectangle ABC, having AB 60 degrees, and the angle BAC 37 degrees 30 minutes, to find the base AC; I place the thread at the intersection of the contrary Sine of 90 degrees, with the right tangent of 30 degrees, (the compliment of the side AB) and it cutteth the contrary Sine of 52 degrees 30 minutes, (the compliment of the angle BAC) at the right tangent of 24 degrees 37 minutes, the compliment whereof 56 degrees 23 minutes, is the base AC, which was required.   

CHAP. XXXVI. To find an angle by knowing both the sides.  


As the radius,  
To the Sine of the side next the inquired angle:  
So is the cotangent of the opposite side,  
To the cotangent of the angle required.   Or, 

As the sine of the side next the inquired angle,  
Is to the radius:  
So is the tangent of the opposite side,  
To the tangent of the angle required.    
Thus in the rectangle ABC, having the side BC 33 degrees 36 minutes, and the side AB 60 degrees, to find the angle BAC; place the thread upon the intersection of the contrary Sine of 60 degrees, with the right tangent of 33 degrees 36 minutes, and it shall cut the contrary Sine of 90 degrees, at the right tangent of 37 degrees 30 minutes, and such is the angle BAC, as was required.   

CHAP. XXXVII. To find an angle, by knowing the base, and the side next to the inquired angle.  


AS the cotangent of the side given,  
To the cotangent of the base:  
So is the radius,  
To the cousin of the angle required.   Or 

As the tangent of the base,  
To the tangent of the side:  
So is the radius,  
To the cousin of the angle required.    
Thus in the rectangle ABC, having the base AC 65 degrees 23 minutes, and the side BC 33 degrees 36 minutes, it is required to find the angle ACB.  
Now in regard the tangent of the base is greater than the side of the quadrat, and the side BC is lesser than the same; I open the thread to an angle equal with the base, viz. to an angle of 65 degrees 23 minutes, and it will cut the right tangent of 33 degrees 36 minutes, at the contrary Sine of 17 degrees 44 minutes, the compliment whereof being 72 degrees 16 minutes, is the angle ACB which was required.   

CHAP. XXXVIII. To find an angle, by knowing the other obliqne angle, and the base.  


AS the radius,  
To the cousin of the base:  
So the Tangent of the angle given,  
To the cotangent of the angle required.   Or, 

As the cousin of the base,  
Is to the radius:  
So is the cotangent of the angle given,  
To the tangent of the angle required.    
Thus in the rectangle ABC, having the base AC 65 degrees 23 minutes and the angle BAC 37 degrees 30 minutes, to find the angle ACB; place the thread at the intersection of the contrary Sine of 90 degrees, with the right Sine of 24 degrees 37 minutes, (the compliment of the base) and it will cut the contrary tangent of 37 degrees 30 minutes, at the right tangent of 17 degrees 44 minutes, the compliment whereof being 72 degrees 16 minutes, is the angle ACB, which was required.  
Thus have we by the Universal quadrat, resolved all the 16 cases of a rectangle spherical triangle. Now followeth the resolution of all manner of spherical triangles whatsoever.    

In any Spherical Triangle whatsoever.  

CHAP. XLI. To find a side opposite to an angle given, by knowing one side, and two angles, the one opposite to the given side, the other to the side required.  

AS the sine of the opposite angle to the side given,  
Is to the sine of that side given:  
So is the sine of the angle opposite to the side required,  
To the sine of the side required.   
Thus in the following triangle PZS, having the side PZ 37 degrees 30 minutes, and the angle ZPS 53 degrees 40 minutes with the angle ZSP 34 degrees 47 minutes, we may find the side ZS; for if you apply the thread to the intersection of the contrary Sine of 34 degrees 47 minutes, with the right Sine of 37 degrees 30 minutes, it shall cut the contrary Sine of 53 degrees 40 minutes, at the right Sine of 59 degrees 15 minutes, and such is the side ZS, which was required.   

CHAP. XL. To find an angle opposite to a side given, by knowing one angle and two sides, the one opposite to the given angle, the other to the angle required.  

AS the sine of the side opposite to the angle given,  
Is to the sine of that angle given:  
So is the sine of the side opposite to the angle required,  
To the sine of the angle required.   
Thus in the triangle PZS, having the angle PZS 113 degrees 17 minutes, the Azimuth from the North; and the side PS 78 degrees 30 minutes, the compliment of the Sun's declination; with the side ZS 59 degrees 15 minutes, the compliment of the Sun's altitude; we may find the angle ZPS which is the hour from the meridian: for if we place the thread at the intersection of the contrary sine of 78 degrees 30 minutes, with the right sine of 66 degrees 43 minutes the outward angle at Z, (because the inward angle exceedeth a quadrant) it will cut the contrary sine of 59 degrees 15 minutes, at the right sine of 53 degrees 40 minutes, which is the hour from the meridian as was required.  
 diagram of the measurement of a spherical triangle (PZS)   

CHAP. XLI. To find an angle by knowing the three sides.  
IN respect of the three sides of a not right angled spherical triangle, let that which subtendeth the angle required be called the base; and the other two the containing sides.  
Then, first, take the sum and difference, of the lesser side and the compliment or excess of the greater side; and add the sins of the said sum and difference together, the half whereof shall be the first found sine: then out of the sine of the sum take this first found sine, and note the remainder.  
Now if every of the three sides be less than a quadrant, take the difference between the remainder noted, and the cousin of the base: but if any of the three sides be greater than a quadrant, add the remainder noted to the cousin or excess of the base; so shall the sum or difference be the second found sine. Then say, 

As the first found sine,  
To the second sound sine:  
So is the radius,  
To a fourth fine.    
Now when every side of the triangle is less than a quadrant; if the remainder noted be less than the cousin of the base, the arch of this fourth sine must be taken out of 90 degrees but if the remainder noted, be greater than the cousin of the base, then add the arch of this fourth sine to 90 degrees, so shall the sum or difference be the angle required.  
But when one side is greater than a quadrant, and that side be made the base; then always add this fourth arch to 90 degrees, to make up your angle required.  
And if the side exceeding a quadrant be one of the containing sides, than always take this fourth arch out of 90 degeees, and the remainder shall be the angle required.  
As in the triangle ZPS, having the side ZP, the compliment of the latitude 37 degrees 30 minutes, and the side PS the compliment of the declination 78 degrees 30 minutes with the base ZS the compliment of the Sun's altitude 59 degrees 15 minutes, we may find the angle ZPS, the hour of the day.  
For the side ZP, the compliment of the latitude being 37 degrees 30 minutes, and the compliment of the side PS, the declination of the Sun, being 11 degrees 30 minutes, their sum is 49 degrees 00 minutes, and their difference 26 degrees 00 minutes, and if we put the side of the quadrat 1000, the sine of the sum 49 degrees 00 minutes, is 754, and the sine of the difference 26 degrees 00 minutes, is 438, these two sins joined together is 1192, the half whereof is 596 for the first found sine, which taken out of the sine of the sum leaveth 158, which remainder I note, & because either of these sides is less than a quadrant, I take the difference between this remainder noted and the sine of the compliment of the base ZS being 30 degrees 45 minutes, the sine whereof is 511, out of which I take the remainder noted, and the remainder is 353, for the second found sine; then placing the thread at the intersection of the contrary parallel of 596, with the right parallel of 353, it will cut the contrary sine of 90 degrees, at the right sine of 36 degrees 20 minutes, and in regard the sides are all less than quadrants, and the remainder noted, less than the cousin of the base, I take this 36 degrees 20 minutes, out of 90 degrees, and there resteth 53 degrees 40 minutes, for the hour from the meridian which was required. The like may you do with any of the other angles, as to find the Azimuth, or the angle of the Sun's position in regard of Pole and Zenith.  
But you shall have a more speedy way to perform this problem, by the Planisphere on the back of the quadrat; for by knowing the three sides of any spherical Triangle, you shall find an angle as speedily as if you had the other two angles also.   

CHAP. XLII. To find a side by knowing the three angles.  
IN any spherical Triangle the sides may be changed into angles, and the angles into sides: by taking first for any one angle and his subtending side, the remainder to a semicircle: then shall the operation be the same, as in the former Chapter.  
Thus in the Triangle ZPS, having the angle ZPS 53 degrees 40 minutes, and ZSP 34 degrees 47 minutes, with S ZP 113 degrees 17 minutes, if you take the greater angle of 113 degrees 17 minutes, out of 180 degrees, there will remain 66 degrees 43 minutes, then as if you had a Triangle of three known sides, one of 53 degrees 40 minutes, another of 34 degrees 47 minutes, and a third of 66 degrees 43 minutes, you may find an angle opposite to any one of these sides by the last Chapter. So shall the angle thus found, be the side required, if it be not the angle opposite to that side which was made by substraction from 180 degrees, if it shall be that angle, then take it likewise from 180 degrees, and the remainder shall be the side required.   

CHAP. XLIII. To find a side, by knowing the other two sides, and the angle contained by them.  
HEre, first, let down a perpendicular, the which you may perform by the 23 Chapter, so shall the Triangle angle given be reduced into two right angled triangles Thus in the Triangle ZPS, having ZP the complemen of the latitude, and PS the compliment of declination with ZPS the angle of the hour from the meridian, we may find S Z the compliment of the Sun's altitude.  
For having let down the perpendicular Z X by the 23 Chapter: we have two Triangles ZXP, and ZXES, both right angled at X, then may we find the side PX, either by the 24, 32, or 33, Chapters, which taken out of PS, leaveth the side XS: now with this side XS, and the perpendicular ZX, we may find the base ZS by the 26 Chapter.  
Or if you let down the perpendicular SA, you shall have two right angled Triangles, viz. SA Z, and SAP, then having the base PS, and the perpendicular SA you may by the 24, or 32 or 33 Chapters find the side AP, from which if you take ZP, there will remain ZA: then with the sides AZ, and AS, you may by the 26 Chapter, find the base ZS.   

CHAP. XLIV. To find a side by knowing the other two sides, and one angle next the inquired side.  
THus in the Triangle ZPS, having the side ZP the compliment of the latitude, and PS the compliment of the declination, with P ZS the angle of the Azimuth, we may find the side ZS the compliment of the Sun's altitude. For having ZP, and the angle at Z, we may to S Z produced, let down the perpendicular PB by the 23 Chapter: then have we two right angled Triangles, PB Z, and PBS, then by the 24 Chapter, having the base PS of the Triangle PBS, and the perpendicular PB, we may find the side BS; and by the base P Z, of the Triangle PBZ, and the same perpendicular PB, we may find the side B Z which being taken out of BS leaveth ZS for the side required.   

CHAP. XLV. To find a side, by knowing one side, and the two angles next the inquired side.  
THus in the Triangle PZS, having PS the declination of the Sun, and PZS, the angle of the Azimuth, and ZP S the angle of the hour from the meridian, we may find the side ZP the compliment of the latitude.  
For if by the 23 Chapter, we let down the perpendicular SA, we shall have two right angled Triangles, SA Z, and SA P, then by the 24 Chapter, having the base PS, of the Triangle SAP, with the perpendicular SA, we may find the side AP; and by the angle SZA, of the triangle SA Z, and the same perpendicular SA, we may find the side A Z by the 32 Chapter, which being taken out of AP leaveth ZP the side required.   

CHAP. XLVI. To find a side by knowing two angles, and the side enclosed by them.  
THus in the triangle ZPS, having the angles at Z and P, with the side intercepted ZP, we may find the side PS, for having by the 23 Chap. let down the perpendicular PB, we have two right angles PBZ, and PBS, then may we by the 29, or 37, or 38, Chap. find the angle BPZ, which added to the angle ZPS giveth the angle BPS, then with this angle BPS, and the perpendicular BP, we may find the base PS, according to the 35 Chapter.   

CHAP. XLVII. To find an angle by knowing the other two angles, and the side enclosed by them.  
THus in the triangle ZPS, having the angles at Z and P with the side intercepted ZP, we may find the other angle ZSP, for having by the 23 Chapter, let down the perpendicular ZX, we have two right angled triangles ZXP, and ZXES; then may we find the angle PZX, by the 37 or 38 Chapter, and that taken out of the angle PZS leaveth the angle SZX: then with the angle SZX, and the perpendicular ZX, we may find the angle required ZSX, according to the 28 Chapter.   

CHAP. XLVIII. To find an angle, by knowing the other two angles, and one of the sides next the inquired angle.  
THus in the triangle ZPS, having the angles at P, and S, with the side ZP, we may find the angle PZS, for by the 23 Chapter having let down the perpendicular ZX we have two right angled triangles, PXZ, and SXZ, then may we by the 37 or 38 Chapter, find the angles PZX, and SXZ, which being both added together maketh the angle PZS which was required.   

CHAP. XLIX. To find an angle, by knowing two sides, and the angle contained by them.  
THus having the sides ZP, and P S, in the triangle ZPS, with the angle comprehended ZPS, we may find the angle PZS, for having let down the perpendicular AS, by the 23 Chapter, we have two right angled triangles SAZ, and SAP; then may we by the 14, or 32, or 33 Chapters, find the side AP, out of which if we take ZP, there will remain AZ: with which side AZ, and the perpendicular SA, we may find the angle AZS, by the 36 Chapter, this angle AZS, taken out of 180 degrees, leaveth the angle PZS, which was required.   

CHAP. L. To find an angle, by knowing the two sides next it, and one of ahe other angles.  
THus in the triangle ZPS, having the sides ZP, and PS, with the angle PZS, we may find the angle ZPS for having let down the perpendicular PB by the 23 Chapter, we have two right angled triangles PBZ, and PBS; then may we by the 37, or 38 Chapter find the angles BPZ, and BPS; and taking B PZ, out of B PS, there remaineth the angle Z PS, which was required.  
Thus you may see Master Gunters 28 cases of a spherical triangles, which he applied to his sector, all resolved by the foreside of ray Universal quadrat. And now you sh●… see them more speedily performed, by the movable planisphere on the back of the Quadrat; coming something near to that of Mr. Blagraves' jewel. But whereas he for the making up of his triangles, useth his dark and intricate reete, (which is no small let to to the eye, for beholding of the true intersections upon the jewel,) I shall only use a fine thread, which may always hang in the centre on the foreside, to serve readily for the aforesaid work, and being of a reasonable length, will be as useful for this side too.   

CHAP. LI. In a right angled triangle, having the base and one of the obliqne angles, to find the two sides, and the other angle.  
LEt the triangle ABC be given again, being the same right angled triangle as was given in the 23 Chapter, wherein knowing the base AC to be 65 deg. 23 min. and the angle BAC, to be 37 deg. 30 min. let it be required to find the sides AB, and BC, with the angle ACB.  
First move the sphere about, until the horizon cut the deg. of the angle given upon the limb, counted from the equinoctial point, viz. 37 deg. 30 min. which resting in this position, count the base AC 65 deg. 23 min. from the centre upon the horizon; and where this sum endeth, there the horizon cutteth the parallel of 33 deg. 36 min. which is the side BC subtending the given angle BAC, and farther, at this intersection of 65 deg. 23 min. on the Horizon, with the parallel of 33 deg. 36 min. doth meet the sixtieth meridian, which is the side AC next to the given angle at A, and here you may see your whole triangle upon your Instrument, as plainly as on the sphere itself, included by these three arches, viz. 65 deg. 23 min. on the Horizon, and 60 deg. in the Equinoctial, and 33 deg. 36 min. upon the said sixtieth meridian, and for the angles, that at the intersection of the meridians with the Equator must needs be right, the given angle is placed it the centre, and denominated by the Horizon upon the limb; the other resteth to be found. And thus of the six parts of a right angled triangle, five is in open view.   diagram of the measurement of a spherical triangle (ABC) 
But now to find the other angle subtended by the side AB, exchange the places of the sides, and let that which was counted upon the Equator amongst the meridians, be counted amongst the parallels; then moving the planisphere until the base 65 deg. 23 min. counted on the Horizon, meet with the parallel of 60 deg. the side AB subgnidnet rending the angle required: so shall the Horizon cut the limb at the angle required, counted from the Equator. And here is your triangle made up as before, viz. The base 65 deg. 23 min. on the Horizon, the side BC 33 deg. 36 min. on the Equator, and 60 deg. the side AB amongst the parallels.   

CHAP. LII. By knowing the base and one of the sides, to find the other side, with both the obliqne angles.  
THus in the right angled triangle ABC, having the base AC 65 deg. 23 min. and the side BC 33 deg-36 min. we may find the other side AB, and the two obliqne angles at A and C.  
For if you count the base AC 65 deg. 23 min. upon the Horizon, and the side BC 33 deg. 36 min. among the parallels; and then move the planisphere to and fro, until the 65 deg. 23 min. do cut the parallel of 33 deg. 36 min. counted from the Equinoctial, and there shall meet you the 60 meridian to make up your triangle; which 60 degrees is the required side AB, and now at this very instant you may see the Horizon cut the limb at 37 deg. 30 min. which is one of the angles required, viz. the angle BAC. And now for to find the other angle you must shift the sides as in the last Chap. for the side subtending the angle required, must be counted among the parallels.  
Thus moving about the planisphere till the intersection of the sixtieth parallel and meridian of 33 deg. 36 min. do cut the Horizon, which will be at 65 deg. 23 min. the base AC as before, and now doth the horizon cut the limb at 72 deg. 16 min. which is the angle ACB as was required.   

CHAP. LIII. Having one side, and one of the obliqne angles of a right angled triangle, to find the base, the other side, and the other obliqne angle  
AS in the right angled triangle ABC, having the angle BAC 37 deg. 30 min. and the side AB 60 deg. we may find the base AC, and the side BC, with the angle ACB.  
For if you move about the planisphere, until the Horizon cut upon the limb, the degrees of the angle given, viz. 37 deg. 30 min. and then reckon the side given viz. 60 deg. upon the Equator among the meridians; you shall upon the same 60 meridian, find 33 deg. 36 min. intercepted between the equator and horizon; which 33 deg. 36 min. is the side BC subtending the given angle BAC.  
Now for to find the base, mark where the said sixtieth meridian cutteth the horizon, which is at 65 deg. 23 min. and such is the base AC. As was required. And for to find the other obliqne angle, you may shift the sides, as was showed in the two last Chapters.   

CHAP. LIV. Having the two sides of a right angled triangle, to find the base, and the two obliqne angles  
AS in the right angled triangle ABC, having the two sides AB 60 deg. and BC 33 deg. 36 min. we may find the base AC, and the two obliqne angles A and C.  
For if we count the two sides one among the meridians, and the other amongst the parallels; as here 60 deg. among the meridians, and 33 deg. 36 min. amongst the parallels; and then turn the planisphere about, until their intersection cut the Horizon, we shall have thereon cut 65 deg. 23 min. for the base AC. and the Horizon also cutting 37 deg 30 min. on the limb which is the angle BAC, subtended by the side BC, which was counted amongst the parallels.  
Thus have you the base, and one of the angles required, and the other is soon found by shifting the sides as in the 51 and 52 Chapters.   

CHAP. LV. Having the two obliqne angles, of a right angled Triangle, to find a side opposite to either of them.  
LEt the right angled Triangle ABC be given, wherein the two obliqne angles A and C are known, viz. the angle BAC 37 degrees 30 minutes, and the angle ACB 72 degrees 16 minutes, and let it be required to find the two sides AB and BC.  
First, place the planisphere so that the two poles thereof may both lie in the Horizon, so shall the Equator be in the zenith. This done, count the angle next the side required, amongst the parallels to the Equator, from any of the poles inward, and the angle subtended by the side required, among the almicanters, counted from the zenithward downward on both sides, upon which points place the thread, which by this means lieth parallel to the Horizon. Now where the thread crosseth the former parallel, there passeth by the meridian which giveth you the angle required, counted from the limb.  
As for example, both poles being in the Horizon, I count the angle ACB 72 deg. 16 min. among the parallels, and the angle BAC 37 deg. 30 min. among the Almicanters, being both counted from the limb, then applying the thread to the said Almicanter, I find it cut the aforesaid parallel, at the meridian of 33 deg. 36 min. which is the side BC which was required.  
Now for to find the other side, let the poles remain in the Horizon, and apply the thread to the Almicanter of 72 deg. 16 min. counted from the zenith, and you shall see it cut the parallel of 37 deg. 30 min. counted from the poles, as the sixtieth meridian counted from the limb, and such is the side AB as was required.   

CHAP. LVI. Having the three sides of any spherical Triangle, to find an angle opposite to any of them.  
LEt the Triangle ZPS be given, whereof the three sides are all known, and the three angles all unknown. First, therefore let the planisphere be so placed, that one of the sides next the angle required, be contained between the pole and zenith. Or place the sphere so that the Horizon may cut upon the limb, the degrees of one of the sides containing the angle required, which is all one thing, this being done count the other containing side from the pole among the parallels; and the side subtending the angle required, always from the zenith amongst the almicanters, and thereto place the thread now where the thread cutteth the parallel of the second containing side, there passeth by the meridian which giveth you the angle required, counted from the limb.  
Thus in the Triangle Z PS, having ZP the compliment of the latitude, and PS the compliment of the Sun's declination, with ZS the compliment of the Sun's altitude, we may find the angle of the hour from the meridian ZPS.  
For having placed the Horizon to the compliment of the latitude 37 deg. 30 min. the side ZP, and counted the compliment of declination 78 deg. 30 min. from the pole among the parallels of declination, or the declination itself from the Equator 11 deg. 30 min. and also 59 deg. 15 min. amongst the almicanters from the zenith which is the side ZS, or the compliment of the Sun's altitude from the Horizon, and, thereto applied the thread parallel to the Horizon; it shall cut the parallel of declination at the meridian or hour circle of 53 deg. 40 min. from the limb or general meridian, and such is the angle ZPS, which was required.  
In like manner may we find the angle PZS which is the Azimuth from the North, for letting the sphere rest as before, with the side ZP at the Horizon; if we count the side ZS 59 deg. 15 min. among the parallels counted from the pole, and the side PS 78 deg. 30 min. from the zenith amongst the almicanters, and thereto apply the thread parallel to the Horizon, it shall cut the said parallel of 59 deg. 15 min. at the meridian of 113 deg. 17 min. for in this manner of work, the angles are all counted from the South part of the limb. This 113 deg. 17 min. is the angle PZS, which was required.  
Then for to find the third angle, you may shift the sides, placing one of the other sides at the Horizon, and then proceed as before.  
Thus you may see this proposition which is most useful, but withal most difficult of all others, (as in Arithmetic, so by the Sector and cross-staff, and also by the fore-side of this Universal Quadrat) speedily performed by the planisphere on the back of this quadrat.   diagram of the measurement of a spherical triangle (ZPS)  

CHAP. LVII. To find a side, by knowing the other two sides, and the angle comprehended.  
FIrst, set the Horizon to one of the given sides, counted on the limb from the Equator, and count the other side from the pole among the parallels, then count the given angle amongst the meridians from the limb; and at the point of intersection between that meridian and the parallel, there place the thread parallel to the Horizon; which you may do by help of the almicanters, and note what almicanter the thread so cutteth, for that is the third side counted from the zenith.  
As in the Triangle ZPS, having the sides ZP 37 deg. 30 min. ZS 59 deg. 15 min. with the angle PZS 113 degrees 17 min. we may find the side PS, for first I place the Horizon at 37 deg. 30 min. and count the other side ZS 59 deg. 15 min. from the pole among the parallels, and the angle PZS 113 deg. 17 min. among the meridians from the south part of the limb, and where this meridian of 113 deg. 17 min. cutteth the parallel of 59 deg. 15 min. there I place the thread parallel to the Horizon, and I find it to cut the Almicanters at 78 deg. 30 min. from the zenith, and such is the side SPARKE, which was required.  
Thus having all three sides, you may find any of the other two angles by the last Chapter.   

CHAP. LVIII. To find a side, by knowing the other two sides, and one of the angles next the inquired side.  
FIrst, set the Horizon to that side given, which is next the given angle, counting on the limb from the Equator; then place the thread (parallel to the Horizon) upon the other side counted from the zenith amongst the Almicanters; now where the thread intersecteth the angle given counted among the meridians from the limb, there passeth by the parallel of the side required counted from the pole.  
As in the Triangle ZPS, having the side ZP the compliment of the latitude 37 deg. 30 min. and the side PS the compliment of the declination 78 deg. 30 min. with the angle PZS, the azimuth from the North 113 deg. 17 min. we may find the side ZS the compliment of latitude.  
For if we place the Horizon to the side ZP 37 deg. 30 min. from the Equator in the limb, and then place the thread to the almicanter of 78 deg. 30 min. counted from the zenith, it shall cut the meridian of 113 deg. 17 min. from the limb, at the parallel of 59 deg. 15 min. counted from the pole, and such is the side ZS, which was required.   

CHAP. LIX. Having two sides and one angle, to find any of the other two angles.  
FIrst, find the third side by one of the two last Chap. then having all three sides you may by the 56 Chap. find any angle required.  
Here needeth no example more than what those Chapters affords.   

CHAP. LX. Having either all three angles to find a side, or having two angles and one side, to find any of the other two sides.  
TO find a side by knowing the three angles, you must first convert the angles into sides by the 42 Chap. which being done, you may by the 56 Chap. find an angle opposite to any one of these sides; so shall the angle which is thus found, be the side which is here required. As in the said 42 Chap. so likewise having two angles & one side, you may by the said 42 Chap. convert these angles into sides, and the sides into angles; so shall you have two sides and one angle, by which and the 57 or 58 Chapters, you may find a third side, then having all three sides, you may by the 56 Chap. find an angle opposite to any of them, which angle thus found, shall be the side required    
The end of the second Book.   

The Third Book.  
Showing how by the Universal Quadrat, to resolve all such Astronomical propositions, as are of ordinary use, as well in the art of Navigation, as in the art of Dialling, with the resolution of such nautical propositions, as are of ordinary use amongst Seamen, concerning Longitude, Latitude, Rumb, and distance.  

CHAP. I. To find the Sun's altitude, by the shadow of a Gnomon set perpendicular to the Horizon.  
TO perform this Conclusion, place the thread upon the intersection of the contrary parallel of the shadow, with the right parallel of the Gnomon, so shall it show in the quadrant, the Sun's altitude required.  
As for example, let AB in the following diagram, be the Gnomon or substance yielding shadow, which we will suppose to be 60, and let AC be the length of the shadow, here found to be 45 such parts as the Gnomon is 60. Now placing the thread upon the intersection of the contrary parallel of 45, with the right parallel of 60, it will cut in the Quadrant 53 deg. 8 min. and such is the angle ACB, the Altitude of the Sun required.  
Or if the shadow had been AD which here is found to be 80, if you place the thread at the intersection of the right parallel of 60, with the contrary parallel of 80, it will cut in the Quadrant 36 deg. 52 min. and such is the angle ADB the altitude of the Sun required.   

CHAP. II. The height of the Sun being given, to find the length of the right shadow.  
BY right shadow is meant the shadow of any Staff, Post, Steeple, or any Gnomon whatsoever, that standeth at right angles with the Horizon, the one end thereof respecting the Zenith of the place, and the other the Nadir.  
Wherefore place the thread to the height of the Sun given in the Quadrant, and it shall cut the right parallel of the Gnomon, at the contrary parallel of the shadow.  
Let the Sun's height given be 53 deg. 8 min. and the length of the Gnomon 60 wherefore if you place the thread at 53 deg. 8 min. in the Quadrant, it will cut the right parallel of 60, at the contrary parallel of 45, and such is the length of the right shadow AC, which was required.  
Or if the Sun's height be but 36 deg. 52 min. if you place the thread to 36 deg. 52 min. in the Quadrant, it will cut the right parallel of 60, at the contrary parallel of 80, and such is the length of the right shadow AD which was required.   diagram of the measurement of length of the sun's shadow  

CHAP. III. To find the Altitude of the Sun by the shadow of a Gnomon, standing at right angles with any perpendicular wall; in such manner that it may lie parallel to the Horizon.  
HEre if you place the thread to the intersection of the contrary parallel of the Gnomon, with the right parallel of the shadow, it shall cut the quadrant at the Sun's altitude required.  
As for example, let AB in the next Chapter be some perpendicular wall, AC a Gnomon making right angles therewith, whose length is 60, and let AD be the length of the shadow which is here found to be 80: now if you place the thread to the intersection of the contrary parallel of 60, with the right parallel of 80, it will cut in the Quadrant 5; deg. 8 min. and such is the angle ACD the altitude of the Sun required.   

CHAP. IU. The Height of the Sun being given, to find the length of the contrary shadow.  
BY the contrary shadow is understood the length of any shadow that is made by a Staff or Gnomon, standing at right angles against any perpendicular wall, in such a manner that it may lie parallel to the Horizon; the length of the contrary shadow doth increase as the Sun riseth in height; whereas the right shadow doth decrease in length, as the Sun doth increase in height.  
Wherefore to find the length of the versed shadow, place the thread upon the height of the Sun given in the Quadrant, and it will cut the contrary parallel of the Gnomon, at the right parallel of the shadow.   diagram of the measurement of length of the sun's shadow 
Let AB be a perpendicular wall, AC a Gnomon making right angles there with, the length whereof is 60, and the Sun's Altitude 53 deg. 8 min. the angle ACD; now if you place the thread at 53 deg. 8 min. in the quadrant, it will cut the contrary parallel of 60, at the right parallel of 80, and such is the length of the contrary shadow AD which was required.   

CHAP. V Having the distance of the Sun from the next Equinoctial point, to find his declination.  

AS the Radius is in proportion,  
To the sine of the Sun's greatest declination:  
So the Sine of the Sun's distance from the next equinoctial point,  
To the sine of the declination required.   
Thus in the right angled spherical triangle ♈ X ♉ in the following Diagram, having the base ♈ ♉ 30 deg. (which is the Sun's distance from the Equinoctial point ♈, he being in the beginning of ♉;) and the angle X ♈ ♉ 23 degrees 30 minutes (the angle of the Sun's greatest declination) with the right angle at X, we may find the side ♉ X, the declination required.  
For if we place the thread at the intersection of the contrary sine of 90 degrees, with the right sine of 23 degrees 30 minutes, it will cut the contrary sine of 30 deg. at the right sineof of 11 deg. 30 minutes.  
Or if we place the thread at the intersection of the contrary sine of 90 deg. with the right sine of 30 deg. it will cut the contrary sine of 23 deg. 30 min. at the right sine of 11 deg. 30 min. & so much is the side ♉ X, the declination of the Sun in the beginning of ♉, which was required. The like declination hath the beginning of ♍, ♏, & ♓, all being of equal distance from their next equinoctial points; the like holdeth of all the other points of the Ecliptic; so that having the declination of every degree in one quarter all the rest are known.   

CHAP. VI How to find the latitude of a place, or the Poles elevation above the Horizon, the declination of the Sun being given.  
FIrst, you must get the Sun's meridian altitude, the which you may very well do by this Instrument after this manner; hold the superficies of your Instrument perpendicular to the Horizon, and the centre where the thread and plummet hangeth towards the Sun; then lifting up the quadrat towards the Sun, until his beams pass through both his sights at once, the thread will cut in the quadrant the degrees of the Sun's altitude. This observation you must begin a little before noon, still diligently observing until you perceive the Sun to begin to fall again, then marking what was his greatest altitude, will serve for this our present purpose.  
Having gotten the meridian altitude by this, and the declination by the last Chapter, you may find the elevation of the Pole above the Horizon as followeth. If the Sun have North declination, then subtract the declination out of the meridian altitude; but if the Sun hath South declination, then add the declination to the meridian altitude, so shall the sum or difference be the height of the Equinoctial; which being taken out of the quadrant or 90 degrees, leaveth the elevation of the Pole above the Horizon.   diagram of the measurement of latitude and polar elevation  

CHAP. VII. How to get the Declination of the Sun, or any Star, Planet, or Comet by observation in a known latitude.  
FIrst, take the latitude out of 90 degrees, and the remainder will be the height of the Equinoctial, this being known, take the meridian altitude of the Sun or any Star, or Planet by the last Chapter, which if it be found greater than the Equinoctials height, then to us that live on this side the Equinoctial, the Sun or Star shall have North Declination, if less, it shall be South: therefore take the lesser height out of the greater, the remainder is your desire.  
As for example, our latitude ZE, or Poles elevation BP, is 52 degrees 30 minutes, the compliment whereof is A 37 degrees 30 minutes, the height of the Equinoctial, now the meridian altitude of the Sun being found to be 49 degrees, as in the last Chapter, if you take A 37 degrees 30 minutes, out of AGNOSTUS 49 degrees, there will rest EGLANTINE 10 degrees 30 minutes, and so much is the declination of the Sun from the Equinoctial, and that Northwards, because his height was greater than the Equinoctial; and so you may do with any Star, to get his declination, if he cometh upon the Meridian on the South side of the Zenith.  
But if it be such a Star as cometh not on the South side of the Zenith, by reason of his great North declination, his Meridian altitude may be Northwards twice taken, once above the Pole, and again under it. For the declination of these Stars, take the latitude out of the greatest meridian altitude, or the lesser meridian altitude out of the latitude, and the remainder out of 90 degrees, there resteth then the declination sought for.   

CHAP. VIII. The Declination of the Sun, and the quarter of the Ecliptic which he possesseth being given, to find his true place.  

AS the Sine of the Sun's greatest Declination, to the sine of the Declination given:  
So is the Radius,  
To the sine of the Sun's distance from the next Equinoctial point.   
Thus in the Spherical triangle ♈, X, ♉, in the the Diagram of the 6th. Chapter, having the side ♉ X 11 deg. 30 minutes, the declination given, and the angle X ♈ ♉, the angle of the Sun's greatest declination 23 deg. 30 minutes with the right angle at X, we may find the base ♈, ♉, the distance of the Sun from the Equinoctial point ♈, by the 27 Chapter of the last Book.  
For if you place the thread upon the intersection of the contrary sine of 23 degrees 30 minutes, with the right sine of 11 degrees 30 minutes, it will cut the contrary sine of 90 degrees at the right sine of 30 degrees, and so much is the base ♈, ♉, the distance of the Sun from the Equinoctial point ♈, which was required.   

CHAP. IX. Having the latitude of the place, and the distance of the Sun from the next Equinoctial point, to find his Amplitude.  

AS the Cousin of the latitude,  
To the sine of the Sun's greatest declination:  
So the sine of the place of the Sun,  
To the sine of the amplitude.   
Thus in the right lined Triangle ♈ T ♉, in the diagram of the 6 Chap. having the compliment of the latitude the angle ♈ T ♉, 37 deg. 30 min. and the Sun's greatest declination the angle ♈ ♉ T, 23 deg. 30 min. with the Sun's distance from the Equinoctial point ♈ the side ♈ ♉, 30 deg. we may find the side ♈ T, the Amplitude by the 39 Chap. of the foregoing Book.  
For if we place the thread to the intersection of the contrary sine of 37 deg. 30 min. with the right sine of 23 deg. 30 min. it will cut the contrary sine of 30 deg. at the right sine of 19 deg. 7 min. and such is the side ♈ T, the Amplitude required.   

CHAP. X. Having the Latitude of the place, and the declination of the Sun or Star to find his Amplitude.  

AS the cousin of the Latitude,  
To the sine of the declination:  
So is the radius,  
To the sine of the Amplitude.   
Thus in the right Triangle ♈ QT, in the diagram of the 6 Chap. having the compliment of the Latitude, the angle TRQ, 37 deg, 30 min. and the side TQ, the Sun's declination 11 deg. 30 min. with the right angle at Q, we may find the base ♈, T, the Sun's amplitude by the 27 Chap. of the second Book.  
For if we place the thread to the intersection of the contrary sine of 37 deg. 30 min. with the right sine of 11 deg. 30 min. it will cut the contrary sine of 90 deg. at the right sine of 19 deg. 7 min. and such is the base ♈ T, the Sun's Amplitude required. The like may you do with any star whose declination you know.  
As for example, the declination of Arcturus this present year 1652 is known to be 21 deg. 10 min. North; wherefore I place the thread to the intersection of the contrary sine of 37 deg. 30 min. with the right sine of 21 deg. 10 min. and I find it to cut the contrary sine of 90 deg. at the right sine of 36 deg. 23 min. and such is the Amplitude of Arcturus, and that Northward, because his declination is to wards the North pole.   

CHAP. XI. Having the amplitude and declination of the Sun, to find the elevation of the Pole above the Horizon.  

AS the sine of the Amplitude,  
To the sine of the declination:  
So is the radius,  
To the cousin of the latitude.   
Thus in the right angled Triangle ♈ QT, in the Diagram of the 6 Chap. having the Amplitude the base ♈ T, 19 deg. 7 min. and the side TQ, the declination of the Sun 11 deg. 30 min. with the right angle at Q, we may find the compliment of the Latitude the angle T ♈ Q, by the 30 Chap. of the second Book.  
For if we place the thread at the intersection of the contrary sine of 19 deg. 7 min. with the right sine of 11 deg. 30 min. it will cut the contrary sine of 90 deg. at the right sine of 37 deg. 30 min. and such is the compliment of the Latitude the angle T ♈ Q, which being taken out of 90 deg. leaveth 52 deg. 30 min. for the angle P ♈ B, the elevation of the pole above the horizon which was required.   

CHAP. XII. The elevation of the Pole, and the amplitude of the Sun being given, to find his declination.  

AS the radius,  
To the sine of the Amplitud:  
So is the cousin of the Latitude,  
To the sine of the declination.   
Thus in the right angled Triangle ♈ QT, having the angle T ♈ Q the compliment of the Latitude 37 deg. 30 min. and the base ♈ T the Sun's Amplitude 19 deg. 7 min. with the right angle at Q, we may find the side TQ the declination by the 23 Chap. of the second Book.  
For if we place the thread upon the intersection of the contrary sine of 90 deg. with the right sine of 19 deg. 7 min. it will cut the contrary sine of 37 deg. 30 min. at the right sine of 11 deg. 30 min. and so much is the side TQ the declination required.  
The like may be done by any Star whose Amplitude is known.   

CHAP. XIII. Having the Latitude of the place, and the declination of the Sun, to find his height in the vertical circle, or when he shall come to be due East or West.  

AS the sine of the Latitude,  
To the sine of the declination:  
So is the radius,  
To the sine of the Altitude.   
Thus in the right angled Triangle CD ♈, in the diagram of the 6 Chap. having the angle C ♈ D the Latitude of the place 52 deg. 30 min. and the side CD the declination of the Sun 11 deg. 30 min. with the right angle at D, we may find the base ♈ C the Sun's height in the East or West points by the 27 Chap. of the second Book. For if we apply the thread to the intersection of the contrary sine of 52 deg. 30 min. with the right sine of 11 deg. 30 min. it will cut the contrary sine of 90 deg. at the right sine of 14 deg. 33 min. and such is the base ♈ C the Sun's height when he cometh to be due East or West as was required.   

CHAP. XIV. Having the latitude of the place, and the distance of the Sun from the Equinoctial point, to find his height in the vertical circle.  

AS the sine of the Latitude,  
To the sine of the Sun's greatest declination:  
So the sine of his distance from the Equinoctial point,  
To the sine of his Altitude.   
Thus in the Triangle ♈ C ♉, having the outward angle ♈ C ♉ the latitude of the place 52 deg. 30 min. and the angle ♈ ♉ C the angle of the Sun's greatest declination 23 deg 30 min. with the side ♈ ♉ the Sun's distance from the Equinoctial point ♈ 30 deg. we may find the Sun's height in the Vertical circle, the side ♈ C by the 39 Chap. of the second Book.  
For if we place the thread upon the intersection of the contrary sine of 52 deg. 30 min. with the right sine of 23 deg. 30 min. it will cut the contrary sine of 30 deg. at the right sine of 14 deg. 33 min. and so much is the side ♈ C the Sun's height when he cometh to be due East or West as was required.   

CHAP. XV. Having the Latitude of the place, and the declination of the Sun, to find the time when the Sun cometh to be due East or West.  

AS the Tangent of the latitude,  
To the Tangent of the declination:  
So is the radius,  
To the sine of the time from the hour of six.   
Thus in the right angled Triangle CD ♈ in the Diagram of the 6 Chap. having the angle C ♈ D the latitude of the place 52 deg. 30 min. and the side CD the declination of the Sun 11 deg. 30 min. with the right angle at D, we may find the side ♈ D, which denominateth the angle ♈ PD, (and is the distance of time betwixt the hour of six, and the Suns coming to the East or West points) by the 32 Chap. of the second Book, compared with the 8 Chap of the same, For if we place the thread to the angle of 52 deg 30 min. it shall cut the right Tangent of 11 deg. 30 min. at the contrary sine of about 9 deg. and such is the side ♈ D which was required.   

CHAP. XVI. Having the Latitude of the place, and the declination of the Sun, to find what altitude the Sun shall have at the hour of six.  

AS the sine of 90 degrees,  
To the sine of the declination:  
So is the sine of the Latitude,  
To the sine of the altitude.   
Thus in the right angled Triangle ♈ RO in the Diagram of the 6 Chap. having the base ♈ O the declination of the Sun 11 deg. 30 min. and the angle O ♈ R the latitude of the place 52 deg. 30 min. with the right angle at R, we may find the side OR the Sun's Altitude at six by the 23 Chap. of the second Book.  
For if we place the thread to the intersection of the contrary sine of 90 deg. with the right sine of 11 deg. 30 min. it will cut the contrary sine of 52 deg. 30 min. at the right sine of 9 deg. 6 min. and such is the side RO the Sun's altitude at the hour of six which was required.   

CHAP. XVII. Having the latitude of the place, and the declination of the Sun, to find what Azimuth the Sun shall have at the hour of six.  

AS the radius,  
To the cousin of the Latitude:  
So is the tangent of declination,  
To the Tangent of the Azimuth from the East or West points Northwards.   
Thus in the triangle ♈ RO in the diagram of the 6 Chap. having the angle O ♈ R the latitude of the place 52 deg. 30 min. & the base ♈ O the declination of the Snn 11 deg. 30 min. with the right angle R, we may find the side ♈ R the Azimuth by the 33 Chap. of the second Book.  
For if we place the thread upon the intersection of the contrary sine of 90 deg, with the right Tangent of 11 deg. 30 min. it will cut the contrary sine of 37 deg. 30 min. at the right Tangent of 7 deg. 4 min. and such is the side ♈ R, the Azimuth from the East or West points Northward at the hour of six which was required.   

CHAP. XVIII. Having the Latitude of the place, and the declination of the Sun, to find the Ascensional difference; and thereby the Time of Sunrising and setting, with the diurnal and nocturnal arches.  

AS the cotangent of the Latitude,  
To the Tangent of the declination:  
So is the radius,  
To the sine of the Ascensional difference.   
Thus in the right angled Triangle ♈ QT in the diagram of the 6 Chap. having the angle T ♈ Q the compliment of the latitude 37 deg. 30 min. and the side QT the declination of the Sun 11 deg. 30 min. with the right angle at Q, we may find the side ♈ Q by the 32 Chap. of the second Book.  
For if we place the thread at the intersection of the contrary tangent of 37 deg. 30 min. with the right sine of 90 deg. it will cut the contrary Tangent of 11 deg. 30 min. at the right sine of 15 deg. 22 min. and such is the side ♈ Q the difference of ascensions required: which being converted into time giveth one hour and one minute. now for to find the time of Sunrising and setting with the diurnal and nocturnal arches, do thus.  
When the Sun hath North declination add this Ascensional difference to 90 deg or 6 hours, so shall you have the semidiurnal arch or the time of Suns setting, as here the ascensional difference 1 hour 1 minute added to 6 hours maketh 7 hours 1 minute, the semidiurnal arch or the time of Sun setting; which being doubled giveth 14 hours 2 minutes for the Diurnal arch; this being taken out of 24 hours leaveth 9 hours 58 minutes, for the Nocturnal arch; the half of which 9 hours 58 minutes is 4 hours 59 minutes the time of Sun rising, or 7 hours 1 minute, taken out of 12 hours, leaveth 4 hours 59 minutes, the time of Sun rising.  
But if the declination be Southwards, then take this 1 hour 1 minute the ascensional difference, out of 6 hours and there will rest 4 hours 59 minutes, for the semidiurnal arch; which being doubled giveth 9 hours 58 minutes for the length of the day; the compliment unto 24 hours, is 14 hours 2 minutes the length of the night, the half of which is 7 hours 1 min. the time of Sunrising.   

CHAP. XIX. Having the latitude of the place, & the declination of the Sun, to find the time of the beginning or ending of twilight.  
THe beginning of twilight is when the Sun approacheth within 18 degrees of the Horizon in the East, and the ending of the same at 18 deg. depression in the West, wherefore to find the same work thus.  
First, by the 16 Chap. get the height of the Sun at the hour of six, if the declination be North; but if it be South take his depression at six, for all is one thing, if his quantity of declination be alike, the height or depression at six being known, if the declination be North add the sine of this height to the sine of 18 deg. but if the declination be south, take the difference betwixt the sine of his depression and the sine of 18 deg. so shall the sum of difference be the second proportional number, then say, 

As the cousin of the Latitude,  
To this second proportional  
So is the radius,  
To a fourth proportional sine.   Then, 

As the cousin of the declination,  
To this fourth sine:  
So is the radius,  
To the sine of the quantity of time between the beginning or ending of twilight, and the hour of six.    
Thus in the Diagram of the 6 Chap. having the declination of the Sun ♈ O 11 deg. 30 min. North, and thereby the Sun's height at six RO 9 deg. 6 min. we have given us in the Triangle byO, first, the angle ybO the compliment of the latitude 37 deg. 30 min. and then the side O y 467 the radius being 1000, this side is made up of the sine of the Sun's height at six 158, and the sine of 18 deg. 309, or by adding of OR 158, to R y 309, we have O y 467 for the second proportional, these being known with the right angle at y, we may find the base bO by the 10 Chapter or 27 Chap. of the second Book.  
For if we place the thread upon the intersection of the contrary sine of 37 deg. 30 min. with the right parallel of 467, it will cut the contrary sine of 90 deg. at the right sine of 50 deg. 7 min. which is the base bO. But seeing the Radius of this sine bO, is but the cousin of the declination OH 78 deg. 30 min. therefore I place the thread again at the intersection of the contrary sine of 78 deg. 30 min. with the right sine of this 50 degr. 7 min. and it will cut the contrary sine of 90 deg. at the right sine of 51 deg. 33 min. and such is the distance of time between the hour of six and the beginning or ending of twilight, being 3 hours 26 min.  
Now at all times, if you take this distance out of six hours, your remainder shall be the beginning of twilight, and if you add this distance found unto 6 hours, the sum will be the ending of the same.  
Except it be in those latitudes, where the depression of 6 is greater than 18 deg. and then you must (in that Latitude and South declination, whose depression at six  more than 18 deg.) work by the contrary rule.  
And when the Sun hath North declination, if you take the ascensional difference out of the distance of time between the hour of six, and the beginning or ending of twilight, you shall have the distance of time between Snn rising or setting, and the beginning or ending of twilight.  
And when the declination is South, add the ascensional difference to the distance of time found, and you shall have the quantity of twilight as before.  
As for example, let the distance of time between the beginning or ending of twilight and the hour of six being found to be 3 hours 26 min. if I take this 3 hours 26 min. out of 6 hours, there will remain 2 hours 34 min for the beginning of twilight: and if I add this 3 hours 26 min. to six hours, it will make 9 hours 26 min. for the ending of twilight.  
And now being the Sun hath North declination, if I take the difference of ascensions 1 hour 1 min. (as it was found by the last Chapter,) out of 3 hours 26 min. there will remain 2 hours 25 min. for the quantity of twilight before or after Sun rising or setting.   

CHAP. XX. Having the distance of the Sun from the next Equinoctial point, to find his right ascension.  

AS the radius,  
To the cousin of the greatest declination:  
So is the Tangent of the distance,  
To the Tangent of the right ascension.   
Thus in the right angled Triangle ♈ X ♉ in the Diagram of the 6 Chap. having the angle X ♈ ♉ the Sun's greatest declination 23 deg. 30 min. and the base ♈ ♉ the distance of the Sun from the equinoctial point ♈ 30 deg. with the right angle at X, we may find the side ♈ X the Suns right ascension by the 33 Chap. of the second Book.  
For if we place the thread at the intersection of the contrary sine of 90 deg. with the right tangent of 30 deg. it will cut the contrary sine of 66 degrees 30 minutes, at the right tangent of 27 degrees 54 minutes, and so much is the side ♈ X the right ascension required,   

CHAP. XXI. Having the right ascension of the Sun or Star, together with the difference of their Ascensions, to find the obliqne Ascension and descension.  
IF the declination of the Sun or Star be North, subtract the ascensional difference from the right Ascension, and the residue will be the obliqne Ascension; but if you add them together, the sum will be the obliqne descension.  
But if the declination be South, add the Ascensional difference and right ascension together, so will the sum thereof be the obliqne ascension; but if you take the Ascensional difference out of the right ascension, the remainder will be the obliqne descension.  
As for Example, the right ascension of the Sun in the beginning of ♉ being found by the last Chapter, to be 27 degrees 54 minutes and the Ascensional difference of the same point by the 18 Chap. to be 15 deg. 22 min. if I take 15 deg. 22 min. out of 27 deg. 54 min. (because the declination is North) there will remain 12 deg. 32 min. for the obliqne ascension: likewise if I add 15 deg. 22 min. to 27 degrees 54 minutes, the sum will be 43 degrees 16 minutes for the obliqne descension.   

CHAP. XXII. Having the Azimuth, the Sun's Altitude, and the declination, to find the hour of the day.  

AS the cousin of the declination,  
To the sine of the Azimuth:  
So is the cousin of the Altitude,  
To the sine of the hour.   
Thus in the triangle ZP ♉ in the Diagram of the 6 Chapter, having the side P ♉ the compliment of declination 78 degrees 30 min. and the side Z ♉ the compliment of the Sun's Altitude 64 degrees 4 minutes with the outward angle ♉ ZP the Azimuth 74 degrees 22 min. we may find the angle ZP ♉ the hour from the meridian by the 40 Chapter of the second Book.  
For if we place the thread to the intersection of the contrary sine of 78 degrees 30 minutes with the right sine of 74 degrees 22 minutes it will cut the contrary sine of 64 degrees 4 minutes at the right sine of 62 degrees 6 minutes and such is the angle ZP ♉ the hour from the meridian, or in time 4 hours 8 minutes which was required   

CHAP. XXIII. Having the hour of the day, the Sun's Altitude, and the declination, to find the Azimuth.  

AS the cousin of the Sun's altitude,  
To the sine of the hour from the meridian:  
So is the cousin of the declination,  
To the sine of the Azimuth.   
Thus in the Triangle ZP ♉ of the diagram in the 6 Chapter having the side Z ♉ the compliment of the Sun's Altitude 64 degrees 4 minutes, and the side P ♉ the compliment of declination 78 degrees 30 minutes with the angle ZP ♉ the hour from the meridian 62 degrees 6 minutes we may find the angle PZ ♉ the Azimuth by the 40 Chapter of the second Book.  
For if we place the thread upon the intersection of the contrary sine of 64 degrees 4 minutes, with the right sine of 62 degrees 6 minutes it will cut the contrary sine of 78 degrees 30 minutes at the right sine of 74 degrees 22 minutes and such is the outward angle PZ ♉ the Sun's Azimuth from the South which was required.   

CHAP. XXIV. The Latitude of the place, the Altitude and declination of the Sun being given, to find the hour of the day,  
THus in the obliqne spherical Triangle ZP ♉ in the Diagram of the 6 Chapter having the three sides, viz. ZP the compliment of the Latitude 37 degrees 30 minutes and Z ♉ the compliment of the Sun's Altitude 64 degrees 4 minutes with P ♉ the compliment of declination 78 degrees 30 minutes we may find the angle ZP ♉ the hour from the meridian by the 41 Chapter of the second Book, but by this Instrument more readily thus,  
First, by the 16 Chapter, get the Sun's height or depression at the hour of 6, and if the declination be North take the distance betwixt this sine and the sine of the Altitude given, either by substraction, or with your compasses, this distance either count or place upon the contrary sine of the compliment of the latitude, and thereto place the thread; the thread being thus placed, will cut the contrary sine of 90 degrees, at a right sine or parallel, which follow to the contrary sine of the Compliment of declination, and thereto place the thread, so shall it cut the contrary sine of 90 degrees at the right sine of the hour from six. For, 

As the cousin of the Latitude,  
To the distance betwixt the sine of the height at six, and the sine of the Altitude given:  
So is the radius,  
To a fourth sine.   Then, 

As the cousin of declination,  
To this fourth sine:  
So is the radius,  
To the sine of the hour from six.    
Thus in the Diagram of the 6 Chapter, having the Sun's declination EGLANTINE or ♈ O 11 degrees 30 minutes and the angle P ♈ B the latitude of the place 52 degrees 30 minutes, we have by the 16 Chapter the Sun's height at 6 RO 9 deg. 6 min. the sine whereof is 158, the radius being 1000; this sine being compared with the sine of 25 deg. 56 min. 437 the sine of the Sun's altitude R a; and taking RO 158, out of R a 437, there will remain O a 279; by which and the angle a ♉ oh the compliment of the Latitude, with the right angle at a, we may find ♉ O the hour from six.  
For if we apply the thread to the intersection of the contrary sine of 37 deg. 30 min. with the right parallel of 279, it will cut the contrary sine of 90 degrees at the right sine of about 27¼ degrees, but seeing this 27¼ degrees which is O ♉, is but upon the parallel of declination GO, therefore I place the thread again at the intersection of this 27¼ with the contrary sine of 78 deg. 30 min. the compliment of declination GO, and it will cut the contrary sine of 90 deg. at the right sine of 27 deg. 54 min. and so much is ♈ X or the angle ♈ PX, the hour from six, which was required.  
And here note that if the Altitude given be greater than the Altitude at six, then is the time found to the southward of six; but if it be lesser, then is it to the northward.  
But if the declination be Southward, then add the sine of his depression at six, to the sine of the Altitude given, and with the same proceed in all respects as before, and you shall have your desire.  
All these kind of propositions upon this Instrument, when you are well acquainted therewith, are sooner wrought then spoken.   

CHAP. XXV. The Latitude of the place, the Declination and Altitude of the Sun being given, to find the Azimuth.  
THus in the obliquespherical Triang. ZP ♉ in the diagram of the 6 Chapter, having the three sides, viz. ZP, the compliment of Latitude 37 degrees 30 minutes and Z ♉ the compliment of the Sun's Altitude 64 degrees 4 minutes with the side P ♉ the compliment of the declination 78 degrees 30 minutes, we may find the angle PZ ♉ by the 41 Chapter of the second Book. But we will proceed as followeth little differing from the former Chapter.  
First, therefore if the Sun have North declination, get his height in the vertical circle by the 13 or 14 Chapters, and comparing the sine thereof with the sine of the Altitude given, take the lesser out of the greater and note the remainder, Then say, 

As the cousin of the Latitude,  
To this remainder:  
So is the sine of the Latitude,  
To a fourth sine.   Then, 

As the cousin of the Sun's Altitude,  
To this fourth sine:  
So is the radius,  
To the sine of the Azimuth from the East or West points, either Northward or Southward.    
For whereas when the Altitude of the Sun given, and his height in the vertical circle is equal, he is directly in the East or West points; so when his Altitude given is greatest, then is the Azimth towards the South; and when his altitude given is the least, than is the azimuth towards the North.  
Thus in the Diagram of the 6 Chapter, having by the 13 or 14 Chapter found the height of the Sun in the vertical circle ♈ C 14 deg. 33 min. the sine whereof 251, I take out of 437 the sine of 25 deg. 56 min. ♈ W, the height of the Sun given, and there remaineth CW 186; by which, and the angel's WC ♉ the latitude 52 deg. 30 min. and W ♉ C the compliment of the latitude 37 deg. 30 min. we may find the side W ♉ by the 10th. or 25th. Chapter of the second Book.  
For if we place the thread to the intersection of the contrary sine of 37 deg. 30 min. with the right parallel or sine 186, it will cut the contrary sine of 52 deg. 30 min. at the right sine of 14 deg. which is the side W ♉, but seeing the Radius of this 14 deg. is but the cousin of the altitude IW, therefore I place the thread again at the intersection of the contrary sine of 64 deg. 4 min. (the compliment of the altitude) with this right sine of 14 deg. and it cutteth the contrary sine of 90 deg. at the right sine of 15 deg. 38 min. and such is ♈ Y, or the angle ♈ ZY, the Azimuth from the East or West, and that Southward, because the given altitude of the Sun were more than his altitude in the vertical Circle?  
But if the declination be Southward, first, get the amplitude by the ninth or tenth Chapter, then say. 

As the cousin of the Latitude,  
To the sine of the Sun's Altitude:  
So is the sine of the Latitude,  
To a fourth sine.    
Unto which fourth Sine add the Sine of the Amplitude and note the sum thereof, then say again, 

As the cousin of the Altitude,  
To this sum last found:  
So is the radius,  
To the sine of the Azimuth from the East or West southward.    
As for example, in the Latitude of 52 deg. 30 min. the Sun's place given being in the beginning of m, we may by the 9 Chapter find his Amplitude to be 19 degrees 7 minutes the sine whereof is 327, and let the Sun's Altitude be 23 degrees 20 minutes; now placing the thread to the intersection of the contrary Sine of 37 deg. 30 min. with the right sine of 13 deg. 20 min. it will cut the contrary sine of 52 deg. 30 minutes, at the right parallel of 301; unto which 301, put 327 the sine of the Amplitude, and the sum will be 628.  
Wherefore place the thread again at the intersection of the contrary sine of 76 degrees 40 minutes, (the compliment of the Sun's Altitude) with the right parallel of this sum 628, and it will cut the contrary Sine of 90 deg. at the right Sine of 40 deg. 11 min. which is the Azimuth from the East or West points Southward.  
And thus have I by the fore-side of this Universal Quadrat, shown the working of some of the most useful propositions, both for Navigation and Dialling; yet very many more might be resolved by the same side, as the ingenuous may perceive by considering well of that which is written in the second Book, which of itself, is able to resolve all propositions of this nature. And now we will proceed to show a more speedy resolution, by the movable Planisphere on the backside of the Quadrat.   

CHAP. XXVI. The place of the Sun being given, to find his declination.  
HEre you have nothing to do but only to seek the Sun's place given in the Ecliptic line, for the parallel passing thereby showeth both at the limb and also at the Axletree-line the declination required.  
As for example, let the Sun's place given be the beginning of ♉, now if you seek the beginning of ♉ on the Ecliptic line you shall find the parallel of 11 degrees 30 minutes, to pass thereby, and such is the declination required.  
Thus may you speedily find the Sun's declination for any day in the year his true place being given, or the declination of any point in the Ecliptic by the inspection of the instrument.   

CHAP. XXVII. The declination of the Sun being given, to find his place in the Ecliptic.  
SEek his declination upon the Axletree-line or hour of six, and follow the parallel thereof to the Ecliptic; so shall it there show you the place of the Sun required.  
As for example, let the declination of the Sun be 11 deg. 30 min. which I seek upon the Axletree-line, and finding it, I follow the parallel thereof to the Ecliptic, and there I find it cut the beginning of ♉, which is the Sun's place required, in regard he possesseth that quarter of the Ecliptic, betwixt the head of ♈ and ♋. But if the Sun had been in that quarter, betwixt ♋ and ♎, then had his place been in the beginning of ♍, or if it had been the quarter betwixt ♎ and ♑, then had his place been in the beginning of ♍, or had he been between ♑ and ♈, then his place had been in the beginning of ♓.   

CHAP. XXVIII. The place of the Sun being given, to find his Right Ascension.  
Seek the Sun's place given in the Ecliptic, and note what Meridian Circle passeth thereby, for that Meridian followed to the Equator, showeth the Right Ascension required.  
As for example: let the Sun's place given be in the beginning of ♉, the which if you seek upon the Ecliptic, you shall findepass thereby a Meridian, which followed to the Equator, giveth 27 degrees 54 min. for the Right Ascension which was required, counted from the beginning of ♈.   

CHAP. XXIX. Having the latitude of the place, with the place of the Sun in the Ecliptic, to find his Amplitude.  
HEre you must first place the planisphere to the latitude of the place proposed, the which you may do, by turning the Sphere about until the North part of the Horizon, cut the limb at the quantity of the latitude counted from the North Pole; or rather (for readiness sake of the numbers set to the limb) till the South end of the Horizon cut the limb at the quantity of the Equinoctials height, counted from the said Equinoctial, which is always to the compliment of the latitude. This being done, your Instrument is ready fixed to perform a number of Conclusions in that latitude. But now for the Amplitude, seek the Sun's place in the Ecliptic line, and look what parallel meeteth you there, the same follow to the Horizon, whereon it cutteth the Amplitude counted from the Centre; showing you by the way the Declination also.  
As for example, let the place of the Sun given be in the beginning of ♉, which point I seek in the Ecliptic, and following the parallel that passeth thereby to the Horizon, I find it there to cut 19 deg. 7 min. counted from the Centre, and such is the Amplitude required, the Sphere being set to the latitude of 52 deg. 30 min.   

CHAP. XXX. Having the latitude of the place, with the declination of any Star, to find his Amplitude.  
FIrst, the Sphere to the latitude being set, if the star be therein placed, follow the parallel of Declination whereon he standeth to the Horizon, and if he be not in the Instrument, get his Declination by the seventh Chap. or otherwise by some Table for that purpose, which found, seek it in the Axletree-line among the parallels of declination, & follow the parallel thereof to the Horizon, and if it cutteth the Horizon, it there showeth the Amplitude; but if it cutteth it not, then doth not that star set or go down under our Horizon.  
As for example, let it be required to find the Amplitude of Arcturus, I look in my Instrument for Arcturus, and I find him placed upon the parallel of 21 d. 10 min. of North Declination, as it were just in the Horizon at 36 deg. 23 min. from the Centre, and such is his Amplitude.  
And if you would have the Amplitude of the Lion's Heart, look in your Instrument and you shall find him placed upon the parallel of 13 deg. 42 min. of North Declination, which if you follow to the Horizon, you shall find it there cut 22 deg. 54 min. and such is the Amplitude of the Lion's Heart, in the latitude of 52 deg. 30 min.  
But if you would seek the Amplitude of the Bright Star, called the Harp; you shall find him placed upon the parallel of 38 deg. 30 min. North Declination, but this parallel cutteth not the Horizon, therefore he goeth not under our Horizon at all, in the aforesaid latitude of 52 deg. 30 min. but when he is in his lowest, directly in the North, he is just one degree of height.   

CHAP. XXXII. Having the Declination and Amplitude of the Sun, to find the latitude of the place.  
COunt the Declination upon the Axletree-line among the parallels of Declination, and the Amplitude upon the Horizon from the Centre, the same way with the declination; then moving the Sphere to and fro, until the Amplitude, and the parallel of Declination, meet both in one point; the South end of the Horizon shall cut the Meridian, at the compliment of the latitude required.  
As for example, let the Sun's Declination be 11 deg. 30 min. and his amplitude 19 deg. 7 min. if you move the Sphere until the 19th. deg, 7 min. one the Horizon, fall directly on the parallel of 11 deg. 30 min. the South end of the Horizon will cut the limb at 37 deg. 30. min. and such is the height of the Equinoctial, which being taken out of 90 deg. leaveth 52 deg. 30 min. for the latitude of the place which was required.   

CHAP. XXXII. The latitude of the place, and the amplitude of the Sun being given, to find his Declination.  
FIrst, set the Sphere to the latitude proposed, then count the amplitude upon the Horizon, and where it shall end, shall meet you a parallel; which being followed to the Axletree-line, will there show you the Declination required.  
As in the latitude of 52 deg. 30 min. the Amplitude being 19 deg. 7 min. if you first place the Sphere to the latitude, and then count 19 deg. 7 min. upon the Horizon, there shall meet you a parallel, which being followed to the Axletree line, shall there give you 11 deg. 30 min. the Declination required.   

CHAP. XXXIII. Having the latitude of the place, and the Declination of the Sun, to find his height in the vertical Circle.  
FIrst, set the Sphere to the latitude given, then count the Declination given among the Almicanters from both ends of the Horizon towards the Zenith, and thereto place the thread parallel to the Horizon, and where the thread so placed cutteth the Axletree-line, or six a clock hour-line, there is the altitude required.  
As in the latitude of 52 deg. 30 min. the Declination of the Sun being 11 deg. 30 min. if you count this 11. deg. 30 min. from both ends of the Horizon among the Almicanters, and place the thread thereto, it shall lie parallel to the Horizon, and cut the Axle-tree-line at 14 deg. 33 min. and such is the Sun's height, when he is directly in the East or West points which was required.  
Or if you place the Horizon to 52 deg. 30 min. in the limb counted from the Equator, and count the Declination upon the Axletree-line, viz. 11 deg. 30 min. the parallel thereof shall cut the Horizon at 14 deg. 33 min. for the height of the Sun in the vertical Circle, the same as before.   

CHAP. XXXIV. Having the latitude of the place, and the Declination of the Sun, to find the time of his coming to the East or West Points.  
FIrst, set the Planisphere, to the latitude given, then place the thread upon the Zenith and the Nadir, so shall it cut the Horizon at right angles in the Centre, and the parallel of Declination at the time from six, counted among the Meridian's.  
As for example, let the latitude given be 52 deg. 30 min. and the Sun's Declination 11 deg. 30 min. First, place the Horizon to 37 deg. 30 min. the compliment of the latitude below the South end of the Equator, then place the thread to the Zenith and Nadir, that is to 52 deg. 30 min. above the South end of the Equator, and as much below the North end thereof; the thread being fised, shall cut the Horizon at right angles in the Centre, and also the parallel of 11 deg. 30 min at the Meridian of 9 deg. or in time, at 36 minutes from the hour of six, and such is the distance of time betwixt the hour of six, and the Suns coming to the East or West points, which was required.   

CHAP. XXXV. Having the latitude of the place, and the Declination of the Sun, to find his height at the hour of six.  
FIrst, set the Horizon to the compliment of the latititude below the South end of the Equator, or the Equator to the compliment of the latitude above the South end of the Horizon which is all one; then seek the Declination upon the Axletree-line, and place the thread thereunto parallel to the Horizon, by help of the Almicanters at both ends thereof; so shall the thread show among those Almicanters, the height of the Sun required.  
Thus in the latitude of 52 deg. 30 min. the Sun's declination being 11 deg. 30 min. I first set the Equator 37 deg. 30 min. above the Horizon, and then place the thread to 11 deg. 30 min. counted in the Axletree-line, and moving the ends thereof to and fro until the thread lie parallel to the Horizon, which will be when both ends of the thread cut like degrees from the Horizon upon the Almicanters; the thread being thus placed upon the 11 deg. 30 min. in the Axletree-line, and parallel to the Horizon, will cut at each end thereof 9 deg. 6 min. among the Almicanters, and such is the height of the Sun at the hour of six.  
Or if you place the Equator 52 deg. 30 min. above the South end of the Horizon, and then count 11 deg. 30 min. upon the Horizon, where you shall find meet you a parallel, which being followed to the Axletree-line, shall there give you 9 deg. 6 min. for the altitude of the Sun at six as before.   

CHAP. XXXVI. Having the latitude of the place, and the Declination of the Sun, to find his Azimuth at the hour of six.  
FIrst, set the Sphere according to the latitude given, and count the Declination upon the horizon, and look what parallel there meeteth you, the same follow to the Axtree-line, where it will show you the Azimuth desired.  
As in the latitude of 52 deg. 30 min. the declination of the Sun being 11 deg. 30 min. First, I place the Equator at 37 deg. 30 min. above the South end of the horizon, and count the Declination 11 deg. 30 min upon the horizon, where I find to meet me the parallel of 7 deg. 4 min. and such is the Azimuth from the East or West points Northwards.   

CHAP. XXXVII. Having the latitude of the place, and the Declination of the Sun, to find the Ascensional difference, and consequently the time of Sunrising and setting, with the Diurnal and Nocturnal arches.  
FIrst, set the Planisphere to the latitude given, and reckoning the declination upon the Axletree-line, follow low the parallel thereof to the Horizon, where you shall find meet you the meridian of the ascensional difference, counted from the hour of six: and this same meridian giveth you the time of Sun setting counted from the South part of the limb or general meridian, and likewise the time of his rising counted from the North part of the limb: and now if you still note this intersection of the Horizon and Sun's parallel, and count the hours thereupon between the South part of the limb and the Horizon, you shall have the Sun's semidiurnal arch, and from the Horizon to the North part of the limb his seminocturnal arch, both which being doubled, giveth you both the diurnal and nocturnal arches.  
As for example, let the latitude given be 52 degrees 30 minutes, and the declination 11 degrees 30 minutes first, I set the Equator 37 degrees 30 minutes, above the Horizon, then counting 11 degrees 30 minutes upon the Axtree-line, I follow the parallel thereof to the Horizon, where I find meet me the meridian of 15 deg. 22 min. and such is the difference of Ascensionals required.  
Now upon the same parallel of declination, I seek how many hours I can find between the South part of the limb and the Horizon, which is 7 hours 1 minute and so much is the semidiurnal arch, and the time of Sun setting; which being doubled maketh 14 hours 2 minutes, the diurnal arch; so likewise counting the hours between the North part of the limb and the horizon, I find them 4 hours 59 minutes and such is the seminocturnal arch, and the time of sunrising; which being doubled giveth 9 hours 58 min. for the Nocturnal arch, the semidiurual and seminocturnal arches, of any Star, whose declination is known, may be found in the self same manner.   

CHAP. XXXVIII. Having the latitude of the place, and the declination of the Sun, to find the time of the beginning and ending of twilight.  
FIrst, set the sphere to the latitude given, then count the declination upon the axtree-line either Northward or Southward according as it is, and then place the thread (parallel to the Horizon) to the 18 deg. of depression below the same, and note where it cutteth the parallel of declination, for those hours to the Southward of the thread giveth the time of twilight ending; and the other to the Northward the time of its beginning.  
As for example, let the latitude given be 52 degrees 30 minutes, and the declination of the Sun 11 deg. 30 min. first, I set the Equator 37 deg. 30 min. above the horizon, and counting 11 deg. 30 min. upon the axletree-line Northward, (because his declination is towards that pole;) I place the thread parallel to the horizon 18 deg. below the same, and I find it cut the said parallel of 11 deg. 30 min. at 2 hours 34 minutes, from the North part of the limb, and such is the time of the beginning of twilight; so doth it likewise cut the same parallel at 9 hours 26 minutes from the South part of the meridian, and such is the time of the ending of twilight, which was required.   

CHAP. XXXIX. The latitude of the place, the altitude and declination of the Sun being given, to find the hour of the day.  
FIrst, place the planisphere to the latitude given, and seek his declination on the axletree-line among the parallels of Declination, then place the thread (parallel to the horizon) to the degree of the Sun's altitude counted from the horizon among the Almicanters, and where the thread so placed shall cut the parallel of Declination, there passeth by the Meridian which giveth you the hour of the day.  
As for example, in the latitude of 52 deg. 30 min. the Declination of the Sun being 11 deg. 30 min. and his altitude 25 deg. 56 min. I set the Equator 37 deg. 30 min. above the horizon, and count 11 deg. 30 min. upon the Axletree-line; then placing the thread to 25 deg. 56 min. among the Almicanters, I find it cut the parallel of 11 deg. 30 min. at four hours 8 minutes from the Meridian, and so much was the hour of the day if the observation was made in the afternoon, but if the observation was made before noon, than was it 7 a clock, and 52 min. in the forenoon.  
Again, let the latitude and declination be the same, as before, and the altitude given 42 deg. 26 min. now if I place the thread to 42 deg. 26 min. among the Almicanters, it will cut the parallel of 11 deg. 30 min. at just two hours from the Meridian, so shall it be either ten a clock in the morning, or two afternoon, according to the observation.   

CHAP. XL. The latitude of the place, the Sun's altitude and Declination being given, to find his Azimuth.  
FIrst, set the Sphere to the latitude given, and then count your altitude among the parallels of Declination, and your Declination among the Almicanters, so shall your Meridian's become Azimuths, and the work will be the same as in the former Chapter. Only here by the way, I would have you note, that as your parallels and Almicanters changed offices, so must your North and South part of the general Meridian change names, so that when your thread cutteth any parallel at any Azimuth, if you count that Azimuth from the South part of the limb, it shall be the Azimuth from the North; and if you count it from the North part of the limb, it shall be the Azimuth from the South.  
And here also note, that your altitude given must be always counted among the parallels of North Declination; and when the Declination is South, the thread must lie below the horizon, and when the declination is North, it must lie above the same.  
As for example, in the latitude of 52 deg. 30 min. let the Declination of the Sun be 11 deg. 30 min. and his altitude 25 deg. 56 min. First, I set the Equator 37 deg. 30 min. above the South end of the horizon, and count 25 deg. 56 min. among the parallels of Declination, than I place the thread to 11 deg. 30 min. above the horizon, because the Declination is North, and I find it cut the parallel of 25 deg. 56 min. at the Azimuth of 74 deg. 22 minutes, counted from the North part of the limb, and such is the Azimuth from the South, and if you count the Azimuths from the South part of the limb, you shall find the thread cut the Azimuth of 105 deg. 38 min. which is the Azimuth from the North.  
Again, let the latitude and the declination be the same as before, and the altitude of the Sun 30 deg. 45 min. now the thread being placed to 11 deg. 30 min. counted among the Almicanters above the Horizon, will cut the parallel of 30 deg. 45 min. at the Azimuth of 66 deg. 30 min. counted from the North part of the limb, and such is the Azimuth from the South.  
A third example, let the latitude be the same, and the Sun's declination 11 deg. 30 min. South, and his Altitude 13 deg. 20 min. now if you place the thread 11 deg. 30 min. below the Horizon, it will cut the parallel of 13 deg. 20 min. at the Azimuth of 49 deg. 49 min. counted from the North part of the limb, and such is the Azimuth from the south, as was required.   

CHAP. XLI. The longitude and latitude of any planet or fixed star being given, to finde his Declination and right Ascension.  
FIrst, seek the stars latitude among the parallels of latitude, and upon the parallel of latitude count his longitude, and where this count endeth, there is the true place of the Star, both in respect of longitude and latitude, this point being found, look what parallel of declination passeth thereby, for that is the declination of the Star from the Equinoctial; and also look what meridian meeteth you there, for that is the right ascension of the star counted from the equinoctial point Aries.  
As for example, let the declination and right ascension of the bright Star Arcturus be required, whose latitude from the ecliptic is 31 degrees North, and longitude in the 19 degrees 22 minutes, ♎ now looking the 19 degrees 22 minutes ♎ upon the parallel of 31 degrees, North latitude, I find there to meet me the parallel of 21 deg. 10 min. North declination, and also the meridian of just 14 hours of time from the equinoctial point Aries, and such is the declination and right ascension of Arcturus, which was required.   

CHAP. XLII. Having the declination and right ascension of any fixed Star, to find his longitude and latitude.  
FIrst, seek the Stars declination among the parallels of declination, upon which count his right ascension, and where this count endeth, shall meet you both the circle of his longitude, and the parallel of his latitude.  
As for example, let the longitude and latitude of Acturus be required, whose declination is 21 degrees 10 North, and right ascension 14 hours now upon the parallel of 21 degrees 10 minutes North declination, I count 14 hours from the equinoctial point Aries, and there I find to meet me the parallel of 31 degrees North latitude, and also the circle of longitude for 19 degrees 22 minutes ♎, and such is the longitude and latitude of Arcturus, which was required.   

CHAP. XLIII. To find the culmination, or southing, of any of the fixed Stars, as also of the Moon and other Planets. First, for the fixed Stars.  
IF it be a Star which is placed in your Instrument, his culmination is speedily to be found after this manner, at the estimate time of the Stars culmination, seck the Sun's place in some Ephemeris; then seek the Star whose culmination you desire in the outward Ecliptic which encompasseth the Planisphere, and bring the South point of the hour Circle (which is the point where the hours both begin and end, at 24, for I count the hours from the noon of the one day, to 24 at the noon of the day following) directly to the Star; so shall the place of the Sun in the same Ecliptic, point out the time of culmination upon the hour circle.  
As for example, upon the 20th. of September 1652, it is required to find the time of the 7 Stars coming to the South, which I guess will be about two a clock in the morning, at which time the Sun is about 7 deg. 20 min. ♎, now finding the 7 Stars in the instrument, I move about the hour circle, until the South point thereof, or 24 hours stand directly against the 7 Stars, and then looking for 7 deg. 20 min. ♎, I find it stand directly against 15 hours in the hour circle, wherefore I conclude, that the 7 Stars cometh to the South that morning at three of the clock.  
Again, upon the 20th. of December, I would know the time of the 7 Stars culmination, which I suppose may be about 8 at night; at which time I find the Sun in 10 deg. ♑, now the South point of the hour-circle being set against the 7 Stars as before, I find 10 deg. of ♑ to stand against 8 hours 45 min. in the hour-circle, whereby it doth appear, that the 7 Stars cometh to the South that afternoon at 8 of the clock, and 45 minutes.  
But if you would have the culmination, of any of the Planets, or of any other Star not placed in your instrument, you must first get their right Ascension by the 41 Chap. which being found, follow the circle thereof to the ecliptic line, and it will there show you the sine and degree whereunto you must set the South point of the hour circle, which being done; the place of the Sun will point out the hour required.  
As for example, let it be required to find the time of the Moons coming to the South upon the 30 of December 1652, which I suppose may be about 8 of the clock at night; at which time I find the Moon in 1 deg. 47 min. ♊, with latitude 4 deg. 36 min. North, and so by the 41 Chapter her right Ascension is 58 deg 40 min. which circle of 58 deg. 40 min. right ascension, I follow to the ecliptic line, and there I find it cut 00 deg. 50 min. ♊, therefore I set the South point of the hour circle; to 00 deg. 50 min. ♊, and then looking for the Sun's place, (which at the estimate time was in 20 deg. 12 min. ♑) I find it to stand against 8 hours 30 min. wherefore I conclude that the Moon cometh to the South that night at 8 of the clock and 30 min.   

CHAP. XLIV. To find the time of the rising or setting, of any of the fixed Stars, and also of the moon or other planets.  
WHen you would find the time of any Stars rising, first get his seminocturnal arch according to the 37 Chap. which put unto 12 hours, and if the Star be placed in the Instrument, bring that number of hours to the Star; but if the Star be not in the Instrument, find his place by the last Chap. and bring this Sum of 12 hours and the Stars seminocturnal arch thereto, so shall the place of the Sun point out the the time of his rising.  
As for example, the 30 of December 1652, let the rising of Arcturus be required, which I suppose may be about 10 of the clock in the night, at which time the Sun is in 20 deg. 22 min. of ♑, first, by the 37 Chap. I find his seminocturnal arch to be 4 hours 3 min. which put unto 12 hours maketh 16 hours 3 min. and in regard Arcturus is placed in the Instrument, I bring 16 hours 3 minunto him, so shall the Sun in 20 deg. 22 min. ♑, point-out 10½ hours for the time of Arcturus rising the said 30 of December in the night.  
And if you would have the time of his setting, bring unto him his semidiurnal arch, which you may find to be 7 hours 59 min. so shall the Sun in 19 deg. 56 min. ♑, (which is his place of the estimate time of the Stars setting) point to 2 hours 32 min. in the hour circle, for the time of Arcturus setting afternoon, in the day and year aforesaid.  
And thus may you find the time of any other Star, or Planets rising or setting, if you first get his declination by the 41 Chap. and then his semidiurnal and seminocturnal arches by the 37 Chap. and then by the last Chap. find the place whereunto you must bring his semidiurnal or seminocturnal arch, as you there did the South point for his culmination, so shall the Sun's place at the estimate time, point out the time required.   

CHAP. XLV. To find the exact hour of the night speedily by the Stars.  

FIrst, with your instrument observe the Stars horary distance from the meridian, that is, first take his height, and then find his hour from the meridian, as you did for the Sun in the 39 Chap. in all respects. but here I would have you note, that in taking the height of a Star by the common sights, the holes are so small that you cannot perceive a Star through them; and therefore for this purpose you may have two great sight holes for the Stars, and two lesser ones for the Sun, or rather as I use to make them for myself and my friends in all my night dials, which is thus: in the sight next my eye I make a great hole, and for the sight by the centre, I set up a flat pin of brass perpendicular to the plate, the breadth of this pin would be such, that the shadow thereof may not fully fill the hole of the other sight, so that when the Sun shineth, and the shadow of the pin fall upon the hole of the other sight, there may some small light of the Sun pass through the hole, on both sides of the shadow: by which means you may take the height of the Sun as exactly as by fine holes; and you may take the height of a Star better with this pin, then with great sight holes. But now to the matter in hand, the horary distance of the Star from the meridian being found, seek it in the hour circle, if the Star be to the Westward of the Meridian; but if the Star be to the Eastward, take the compliment of his horary distance to 24 hours and seek that in the hour circle; which being found, bring it to the Star or to his place found as in 43 Chap. so shall the place of the Sun point out the hour required.  
As for example, upon the 16 of November 1652, it being serene air, I desired to see what time of the morning it were; for which purpose I made use of Arcturus, and by observation I sound his height to be 27 deg. 12 min. towards the East; now his declination being 21 deg. 10 min. North, I find by the 39 Chap. that he was 4 hours 50 min. from the South, and in regard he is on the East side of the meridian I take the compliment of 4 hours 50 min. to 24 hours, which is 19 hours 10 min. this 19 hours 10 min. I seek in the hour circle and bring it to Arcturus; now the Sun at the estimate timebeing in 4 deg. 40 min. ♐, pointeth to 17 hours in the hour circle, which is directly 5 of the clock in the morning, the like may be done by any of the planets, as I will make appear by these two examples following, which I made in the day time by the glittering planet Venus.   

The 16 of November 1652, I made these two following observations by the planet Venus in the day time, as to the finding of the hour by the Stars, she being about her greatest distance from the Sun.  
The said 16 day being serene, and I being abroad, and in a convenient place where I might see a clear Horizon, and seeing Venus shining so bright directly at the Suns rising, I could do no less then make observation by her, to see how her longitude, latitude, right ascension, declination and likewise the hour found by her would agree with the time of Sun rising wherefore first, I observed her altitude by my Instrument which being but small I could not make so exact observation as otherwise Imight have done; but as it were I found her altitude just 30 deg. Now the estimate time being Sun rising, (which was that morning at 8 a clock and 1 min. in our latitude of 52 deg. 30 min.) I looked in my Almanac for the longitude and latitude of ♀, which I found according to Mr. Vincent Wings Ephemeris, to be in 18 deg. ♎, with 2 deg. North latitude; then by the 41 Chap. I find her declination to be 5 deg. 14 min. South, and so by the 39 Chap. her hour from the meridian to be no more but 57 min. so likewise by the 41 Chap. I find her right Ascension to be 13 hours 9 min. which circle I follow to the Ecliptic line, where it giveth me 18 deg. 45 min. ♎, for the place of ♀ in the outward ecliptic; unto which 18 deg. 45 min. ♎, I bring the compliment of 0 hours 57 min. which is 23 hours 3 min. so doth the Sun in 4 deg. 47 min. ♐ being his place at the time of his rising, point out 20 hours 1 min. for the time required, agreeing exactly with the time of Sun rising: being also a good proof, that ♀ was than in the aforesaid place, both in respect of longitude and latitude.  
A second example, the same day a while after noon I observed the altitude of ♀ again, which I found to be 16 deg. 56 min. the Sun being then in 5 deg ♐, and ♀ in 18 deg. 10 min. ♎, which 2 deg. North latitude, wherefore by the 41 Chap. I found her declination to be 5 deg. 18 min. South, and so by the 39 Chap. her distance from the meridian 3 hours 32 min. so likewise by the said 41 Chap. her right ascension was 13 hours 10 min. which circle of right ascension I follow to the ecliptic line, where it giveth me the 19th deg. of ♎ for the place of ♀ in the outward ecliptic; unto which 19 deg. ♎, I bring 3 hours 32 min. of the hour circle, so did the Sun in 5 deg. ♐, point to 0 hours 30 min. in the hour circle; whereby it doth appear, that it was then half an hour past 12 of the clock in the day time; and in deed so it was by my fixed Sun Dial, being an exact Dial of a yard square. perhaps some will think it strange, that the hour of the day should be found by this Star, or that this Star should appear so bright at the noon time of the day, that the hour should thereby be found, there being no Eclipse of the Sun at that time; but indeed it is not so strange as its true, as some of my friends can testify.  
Thus having showed on both sides of the Universal Quadrat, how to resolve such Propositions Astronomical, as are most useful for Seamen and Diallers and such like Artists; I will now show likewise how to resolve such Nautical propositions, as are of ordinary use, concerning longitude. latitude rumb, and distance.    

CHAP. XLVI. To find how many minutes or miles, answer to one degree of longitude, in any latitude required.  
IN sailing by the compass, the course holds sometime upon a great circle, sometime upon a parallel to the Equator; but most commonly upon crooked lines winding towards one of the Poles, which lines are well known by the name of Rombs.  
If the course hold upon a great circle, it is either North or South, under some one of the meridians, or else East or West under the Equator: and in these cases, every degree requires an allowance of 20 leagues or 60 miles, every 20 leagues making a degree difference in sailing upon those circles; so that here needs no further precept than this, divide your leagues sailed by 20, and you shall have the degrees of distance; and contrarily, multiply the degrees of distance by 20, so shall you have the number of leagues sailed.  
But if the course hold East or West, upon any of the parallels to the Equator, to find how many miles do answer to one degree of the Equinoctial take this proportion. 

As the radius,  
To the cousin of the latitude:  
So is 60 miles the measure of one degree at the equator,  
To the number of miles answering to one degree of longitude in the given latitude.    
Thus in the latitude of 50 deg. if I place the thread upon the intersection of the contrary sine of 90 deg. with the right sine of 40 deg. (which is the compliment to 50 deg.) it will cut the contrary parallel of 60, at the right parallel of 38 34/60, and so many miles do answer to one degree of longitude, in the latitude of 50 deg.  
Again, in the latitude of 52 deg. 30 min. I place the thread to the intersection of the contrary sine of 90 deg. with the right sine of 37 deg. 30 min. and it cutteth the contrary parallel of 60, at the right parallel of 36 31/60, and so many miles do answer to one degree of longitude, in the latitude of 52 deg 30 min. as was required.  
Or if you place the bead to 60 on the side of the quadrat, and then open the thread to the angle of the latitude, as here to 50 deg. the bead will cut the contrary parallel of 38 34/60, as in the first example.  
And if you open the thread to the angle of 52 deg 30 min. the bead will cut the contrary parallel of 36 31/60, as in the latter example.   

CHAP. XLVII. To find what difference of longitude be longeth to one degree or 20 leagues of distance, upon any parallel of latitude.  

AS the cousin of the latitude,  
To the sine of 90 degrees.  
So is the distance of 20 leagues on the parallel,  
To the difference of longitude required.   
Thus in the latitude of 50 deg. if you place the thread to the intersection of the contrary sine of 40 deg. with the right sine of 90 deg. it will cut the contrary parallel of 20 leagues, at the right parallel of 31 leagues and something better, or 1 deg. 33 min. and such is the difference of longitude at the Equator, answering to 20 leagues of distance on the parallel of 50 deg. of latitude.  
Again in the latitude of 52 deg. 30 min. if you place the thread to the intersection of the contrary sine of the compliment of the latitude 37 deg. 30 min. with the right sine of 90 deg. it will cut the contrary parallel of 20 leagues, at the right parallel of almost 33 leagues, or 1 deg. 39 min. and such is the difference of longitude, answering to 20 leagues of distance in the parallel of 52 degrees 30 min.   

CHAP. XLVIII. To find how many leagues do answer to one degree of latitude, upon any Romb required.  

AS the Cousin of the Romb from the Meridian,  
To 20 leagues, or one degree upon the Meridian:  
So is the Radius,  
To the leagues answering to one degree upon the Romb.   
Sailing from the North parallel of 40 degrees upon a course Northeast and by East, which is the fifth Romb from the Meridian, or 56 deg. 15 min. the compliment whereof is 33 deg. 45 min. it is required how many leagues the ship should run, before it should raise the Pole one degree.  
Wherefore I place the thread at the intersection of the contrary sine of 33 deg: 45 min: with the right sine of 90 deg: and it will cut the contrary parallel of 20 leagues, at at the right parallel of 36 leagues; and so much way must the Ship make upon the fift Rumb from the Meridian, before she can raise or lay the Pole one degree.   

CHAP. XLIX. To find how much any ship hath raised or depressed the Pole, knowing the course she hath made, and the leagues she hath sailed.  

AS the sine of 90 degrees,  
To the cousin of the Rumb from the Meridian.  
So is the distance upon the Rumb,  
To the difference between both latitudes.   
Thus knowing the course to be on the fift Rumb or 56 degr. 15 min. from the Meridian, the compliment whereof is 33 deg. 45 min. and the leagues sailed 120, we may find the difference of latitude.  
For if we place the thread upon the intersection of the contrary sine of 90 deg. with the right sine of 33 deg. 45 min: it will cut the contrary parallel of 120, at the right parallel of 66⅔ leagues, or 3 deg. 20 min. and so much have you altered your latitude.  
Or if you lay the thread upon the side of the quadrat and apply the bead to the leagues sailed which is 120, and then open the thread to the angle of the Rumb, which here is the fift, or 56 deg 15 min. from the Meridian, so will the bead cut the contrary parallel of 66 leagues and 2 miles, or 3 deg. 20 min. as before, which is the difference of latitude here required.   

CHAP. L. By the Rumb and both latitudes to find the distance upon the Rumb.  

AS the cousin of the Rumb from the Meridian,  
To the sine of 90 deg.  
So the difference between both latitudes,  
To the distance upon the rumb.   
Let the Latitude of the first place be 50 deg. and the latitude of the second 53 deg 20 min. and the Rumb the fifth from the Meridian, the compliment whereof is 33 deg. 45 min. then taking the lesser latitude out of the greater, there will remain 3 deg. 20 min. or 66 leagues 2 miles; now for to find the distance upon the Rumb or the leagues the Ship hath sailed, place the thread to the intersection of the contrary sine of 33 deg. 45 min. with the right sine of 90 deg. and it will cut the contrary parallel of 66 leagues 2 miles, at the right parallel of 120 leagues.  
Or if you place the thread to the angle of the fifth Rumb from the Meridian 56 deg. 15 min. and to the intersection of the thread with the contrary parallel of 66⅔ leagues place the bead, then moving the thread to the side of the quadrat, the bead will there give you 120 for the leagues sailed as before.   

CHAP. LI. By the distance and both latitudes, to find the rumb.  

AS the distance on the rumb,  
To the difference between both latitudes:  
So is the sine of 90 degrees,  
To the cousin of the Rumb from the Meridian.   
Thus the one place being in 50 deg. of latitude, and the other in the latitude of 53 deg. 20 min. the difference of latitude will be 3 deg. 20 min. or 66⅔ leagues, and the distance between them being 120 leagues, we may find the Rumb.  
For if we place the thread to the intersection of the contrary parallel of 120, with the right parallel of 66⅔; it will cut the contrary sine of 90 deg. at the right sine of 33 deg. 45 min. the compliment whereof being 56 deg. 15 min. is the inclination of the Rumb to the Meridian as was required. And if you allow 11 deg. 15 min. to each Rumb, you shall find it to be the fifth from the Meridian.  
Or if you place the bead to 120, and then open the thread until the bead cut the contrary porallel of 66⅔ so shall the thread cut in the quadrant 56 deg. 15 min. for the inclination of the Rumb to the Meridian as before.   

CHAP. LII. By the Rumb and difference of latitudes, to find the difference of longitude.  


AS the cousin of the Rumb from the Meridian,  
To the difference of latitudes:  
So is the sine of the Rumb from the Meridian,  
To the difference of longitude in the commor Sea-chart.   Then, 

As the cousin of the middle latitude,  
Is to the radius:  
So is the difference of longitude in the common Sea-chard.  
To the difference of longitude required.    
Thus sailing upon the third rumb or 33 deg. 45 min. from the Meridian, from the North parallel of 50 deg. until the Pole was raised 4 deg. 10 min. the difference of longitude is desired.  
First, therefore place the thread to the angle of 33 deg. 45 min. and it will cut the contrary parallel of 4 deg. 10 min. which is 83⅓ leagues, at the right parallel of 55⅔ leagues, which maketh 2 degrees 47 minutes, and such is the difference of longitude according to the common Sea-card, the which the foreside of this quadrat doth lively represent. But this difference of longitude so found, is always lesser than it should be, and therefore to be enlarged after this manner; put one half of the difference of latitude, to the lesser latitude, or take it out of the greater, so shall the sum or difference be the middle latitude.  
As here the difference of latitude being 4 degrees 10 min: the half thereof 2 deg: 55 min: put to 50 deg: maketh 52 deg: 5 min: for the middle latitude, the compliment whereof is 37 deg. 55 min:  
Wherefore if you place the thread to the intersection of the contrary sine of 37 deg: 55 min: with the right sine of 90 deg: it will cut the contrary parallel of 55⅔, at the right parallel of 90⅓ leagues, or 4 deg: 31 min: and such is the difference of longitude required:   

CHAP. LIII. By the distance of latitude and leagues sailed, to find the difference of longitude.  
SAiling 100 leagues between the North and the East, from the North parallel of 50 deg: until the Pole be raised 4 deg: 10 min: the difference of longitude is required:  
First place the bead to 100 leagues on the side of the quadrant, then open the thread till the bead cutteth the contrary parallel of the difference of latitudes 4 deg: 10 min: or 83⅓ leagues, so shall the thread cut in the quadrant, the degrees of the Rumb from the Meridian, which is 33 deg: 45 min: and the bead shall cut the right parallel of 55⅔ leagues, which is the difference of meridians according to the common Sea-card: now this difference being always too little, I place the thread again to the intersection of the contrary sine of the compliment of the middle latitude 52 deg: 5 min: that is to the contrary sine of 37 deg: 55 min: with the right sine of 90 deg: and it will cut the contrary parallel of 55⅔ leagues, at the right parallel of 90⅓ leagues, or 4 deg: 31 min: and such is the difference of longitude required:   

CHAP. LIV. By the difference of latitude, and difference of longitude, to find the Rumb leading from the one to the other.  


AS the sine of 90 deg,  
To the difference of Meridian's:  
So the cousin of the middle latitude,  
To the difference of longitude in the common Sea-card.   Then, 

As the difference of latitudes,  
To the difference of longitude in the common Sea-card  
So is the Tangent of 45 degrees,  
To the Tangent of the Rumb from the meridian.    
Sailing from the North parallel of 50 deg. betwixt the North and the East, until the Pole be raised 4 deg. 10 min. and altered the Meridian 4 deg. 31 min.  
Now the course the ship hath run from the first place to the second, is demanded.  
First, turn the degrees of difference into leagues, so shall you have 83⅓ leagues for the difference of latitude, and 90⅓ leagues for the difference of longitude.  
Now placing the thread to the intersection of the contrary sine of 90 deg. with the right parallel of 90⅓ leagues it will cut the contrary sine of 37 deg. 55 min. the compliment of the middle latitude, at the right parallel of 55⅔ leagues, and such is the difference of longitude in the common Sea-card.  
Then again I place the thread to the intersection of the contrary parallel of 83⅓ leagues, with the right parallel of 55⅔ leagues; and it will cut the contrary tangent of 45 deg. at the right tangent of 33 deg. 45 min. and such is the inclination of the rumb to the Meridian which was required.  
Or if you place the bead to 90⅓ leagues on the side of the quadrat, and then open the thread to the angle of 37 deg. 55 min. the bead will cut the right parallel of 55⅔ leagues; now where this parallel cutteth the contrary parallel of 83⅓ leagues, there place the thread; so shall it give in the quadrant 33 deg. 45 min: for the inclination of the rumbe to the Meridian, as before, whereby it doth appear that the Ships course was North-east and by North the thing required.   

CHAP. LV. By one latitude, Rumb, and distance, to find the difference of latitude, and also of longitude.  


AS the sine of 90 degrees,  
To the cousin of the Rumb from the Meridian:  
So is the distance on the Rumb,  
To the difference of latitudes.   Then, 

As the cousin of the Rumb from the Meridian,  
To the difference of latitudes:  
So is the sine of the Rumb from the Meridian,  
To the difference of longitude in the common Sea-card.   Then again, 

As the cousin of the middle latitude,  
Is to the radius:  
So is the difference of longitude in the common Sea-card,  
To the difference of longitude required.    
Sailing from the North parallel of 50 deg. 100 leagues, upon the course Northeast and by North, the difference of latitude, and also of longitude is required.  
First, by the first proportion according to the 49 Chap. you shall find the difference of latitudes to be 83⅓ leagues, or 4 deg: 10 min: and then by the two latter proportions according to the 52 Chapter, you shall find the difference of longitude to be 90⅓ leagues, or 4 deg: 31 min. as was required.  
Or more speedily thus, place the bead to 100 leagues on the side of the quadrat, and then open the thread to 33 deg: 45 min: the Rumb from the Merician, so shall the bead cut the contrary parallel of 83⅓ leagues, or 4 deg: 10 min: and such is the difference of latitude required  
The thread lying still in this position, the bead shall likewise cut the right parallel of 55⅔ leagues, for the difference of longitude according to the common Sea-card: now where this right parallel of 55⅔ leagues, intersecteth the contrary sine of 37 deg: 55 min: (which is the compliment of the middle) there place the thread again, so shall it cut the contrary sine of 90 deg. at the right parallel of 90⅓ leagues, or 4 deg. 31 min. as before.   

CHAP. LVI. By the longitude and latitude of two places, to find their distance upon the Rumb.  
FIrst, place the thread to the intersection of the contrary sine of 90 deg. with the right sine of the compliment of the middle latitude; and it shall cut the contrary parallel of the difference of longitude, at the right parallel of of the difference of longitude in the common Sea-card: then at the intersection of this last right parallel, with the contrary parallel of the difference of latitudes, there place the Bead; which being applied to the side of the Quadrat shall give you the distance upon the Rumb as was required.  
Let two Cities be given, one in the latitude of 50 deg. and the other in the latitude of 54 deg. 10 min. and let the difference of longitude be 4 deg. 31 min. now to find their distance, first turn their difference of longitude, and also of latitude into leagues, so will their difference of longitude be 90⅓ leagues, and their difference of latitude 83⅓ leagues; now if you place the thread to the intersection of the contrary sine of 90 deg. with the right sine of 37 deg. 55 min. (which is the compliment of the middle latitude,) it will cut the contrary parallel of 90⅓ leagues, at the right parallel of 55⅔ leagues, for the difference of longitude in the common Sea-card: now where this right parallel of 55⅔ leagues, cutteth the contrary parallel of 83⅓ leagues there place the bead, which being applied to the side of the Quadrat, giveth 100 leagues for the distance of the two Cities which was required.  
And if you mind this last position of the theed, you shall see it cut in the Quadrant, 33 deg. 45 min. for the Rumb from the Meridian, which leadeth from the one to the other, which is more than was required.   

CHAP. LVII. By the way of the Ship, and two angles of position, to find the distance between the ship and the land.  
SUppose being at A, I had sight of some land at B, the Ship going from A towards C; and observing the first angle of the Ships position BAC, I find it to be 18 deg. 26 min. and when the Ship had made 3⅓ leagues or 10 miles of way from A unto D, I observed again, and I found the second angle of the ships position BDC to be 36 deg. 52 min. now may I find the third angle ABDELLA to be 18 deg. 26 min. either by substraction, or by compliment unto 180 degrees,  diagram of the measurement of distance  
In this and the like cases, I have a right tlined triangle, in which there is one side with the three angles known, and it is required to find the other two sides, the which you may do by this proportion. 

As the sine of the angle opposite to the known side,  
Is to that known side:  
So is the sine of the angle opposite to the side required,  
To the side required.   Wherefore I place the thread to the intersection of the contrary sine of 18 deg. 26 min. with the right parallel of 10; and it will cut the contrary sine of 36 deg: 52 min. at the right parallel of very near 19; and such is the distance AB, which is the distance between the place of your first observation at A, and the land at B. The thread lying in the same position, doth likewise cut the contrary sine of 18 deg: 26 min: the angle BAC, at the right parallel of 10; and such is the distance from the place of the second observation at D, unto the land at B, which is equal to the Ships way AD, because the angles at A and B happened to be equal.   

CHAP. LVIII. To find the distance of any Ship from you, yourself standing upon some high cleft or platform by the Sea coast, the height of the said platform above the water being known.  
TO perform this conclusion and such like, turn the centre of your quadrat towards your eye, and then looking through the sights thereof, till you can see the very lower part of the hull of the Ship next the Water, note where the thread cutteth the right parallel of the height of the platform above the Water, for there passeth by the contrary parallel of the distance of the ship from the base of the platform, and if you place the bead to that intersection, and then apply it to the side of the Quadrat, it shall there give you the length of the visual line running from your eye to the ship. As for example,  
Let B be the top of some high platform by the Sea coast, whose height is known to be 60 paces or any other measure; and let it be required from B, to find the distance of a ship lying at Road at A: first, therefore I turn the Centre towards my eye, and looking through the sights towards the ship at A, moving it up and down till I can through the sights espy the very lowest part of the hull of the ship next the Water, I find the thread to cut the right parallel of 60, at the contrary parallel of 180; and so much is AC the distance of the ship at A, from the base of the cleft or platform, lying perpendicularly under the Centre of your instrument at C: and if to this intersection of the parallel of 60, with the parallel of 180, you place the bead, and then apply it to the side of the Quadrat, it shall there give you very near 190, and such is the length of the visual line, passing from your eye at B, to the ship at A as was required.  
This conclusion serveth most commodiously for all such as shall have committed to their charge any platform with ordnance, for hereby you may exactly at the first view, tell the distance of any ship or other mark from you; so that no Vessel can pass by your platform, but you may with your Ordnance at the first bough her, and never bestow vain shot, this Instrument serveth also (and is no less commodious) for all such as shall keep amount within any Town of War, whence you desire with shot to beat enemy aloof of, before he shall approach nigh, with many more such like conclusions, which the ingenious will soon conceive.   

CHAP. LIX. How by this Instrument, from a high Platform, to measure the distance of any two ships on the sea, or other marks on the land, howsoever they be situated, and that right speedily.  
FOr this purpose and the like, let there be an Index with sights fitted to turn upon the centre, and let it be divided into such equal divisions as the side of the Quadrat is. This index being in a readiness, get the distance of each ship from your station, by the last Chap. and if they be both in a right line from you, take the lesser distance out of the greater, and the remainder will be their distance.  
But if they be otherwise situate, find the angle between them by help of your Index, which being found, count the lesser distance upon the side of the Quadrat, and the greater on the Index, so shall the distance taken with a pair of Compasses, give you the distance of the two ships required.   
FINIS.