The description and use of the universall quadrat.: By which is performed, with great expedition, the whole doctrine of triangles, both plain and sphericall, two severall wayes with ease and exactness. Also the resolution of such propositions as are most usefull in astronomie, navigation, and dialling. By which is also performed the proportioning of lines and superficies: the measuring of all manner of land, board, glasse; timber, stone. &c. / By Thomas Stirrup, Philomathemat.

Of the resolution of right lined Triangles

CHAP. X. Having three angles and one side to finde the other two sides.

IN all Triangles there being six parts viz. three Angles and three Sides, any three whereof being known in a sphericall Triangle, the other three may be found by the universal quadrat; but in a right lined Triangle, one of the three given termes must be a Side in the finding of the o∣ther parts.

Thus having three angles and one Side, you may finde 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 an∣gle ACB 53 degrees 8 minutes, the angle CAB 36 de∣grees 52 minutes, and the angle ABC 90 degrees, the

given side AB being 80, it is required to finde 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 threed upon the intersection of the contrary Sine of 53 degrees 8 minutes with the right parallel of 80 his opposite side; the threed 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 threed 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 threed 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 threed once placed, giveth both the sides required, and that right speedily in all Triangles; as well in oblique angled as right angled triangles.

CHAP. XI. Having two sides given, and one angle opposite to either of them; to finde 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.

[illustration] 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 finde the angle CAD which is opposite to the other side CD, for if I place the threed upon the intersection of the con∣trary 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 CAD.

Then keeping the threed in the former position, adde 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 threed, 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 finde the two other angles and the third side.

IF the angle contained betweene 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 threed 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 de∣grees 52 minutes, and such is the lesser angle CAB; the complement 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 oblique angle that is betweene the two sides given; the Triangle may be reduced into two rect∣angles and then resolved as before.

As in the Triangle ADC, where the side AC is 100,

and the side AD 35, and the angle CAD 36 degrees 52 minutes, if you let down the perpendicular DE, up∣on 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 complement: and with these angles and the side AD, you may finde both AE and DE, by the tenth Chapter.

Then taking AE out of AC, there remaines EC for the side of the rectangle DEC; and therefore with this side EC and the other DE, you may finde both the an∣gle at C, and the third side CD, by the former part of this Chapter.

Or you may finde 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 halfe sum of the opposite angles,
• To the Tangent of halfe the difference betweene those angles,

As in the former Triangle ADC, the sum of the sides AC and AD, is 135, and the difference betweene them 65; the angle contained 36 degrees 52 minutes; and therefore the sum of the two opposite angles 143 degrees 8 minutes, and the halfe sum 71 degrees 34 minutes, and because the halfe sum 71 degrees 34 minutes, is greater then the side of the quadrat 45 degrees, therefore I turne 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 halfe sum of the opposite angles 71 degrees 34 minutes.
• To the co-tangent of half the difference betweene those angles.

Therefore if you place the threed upon the intersecti∣on 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 complement of 71 de∣grees 34 minutes,) at the right Tangent of 34 degrees 42 minutes, the complement whereof is 55 degrees 18 minutes, and such is the halfe difference betweene the op∣posite angles at B and D, this halfe difference being ad∣ded 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 finde the third side CD by the tenth Chapter.

CHAP. XIII. Having the three sides of a right lined Triangle, to finde the three angles.

LEt one of the three sides given be the base, but rather the greater side, that the perpendicular may fall with∣in 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 threed upon the intersecti∣on of the contrary parallel of 100, with the right paral∣lel 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 halfe whereof is 28, and sheweth 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. Gunters upon his Crosse-staffe; whereby you may perceive the agree∣ment betwixt this quadrat and his staffe.

And now I will shew how they may be otherwise re∣solved upon this uniuersal quadrat, agreeing neerer 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 finde the other side and the base.

OPen the threed to the angle opposite to the side gi∣ven, 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 threed to the side of the quadrat, it shall there shew 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 an∣gle BAC 36 degrees 52 minutes, it is required to finde the other Side AB, and the base AC.

First, open the threed to the angle of 36 degrees 52 minutes, so shall it cut the right parallel of 60, at the con∣trary parallel of 80, and such is the side AB, which was required.

The threed lying in the same position, if you place the bead to this intersection, and then apply the threed 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 finde 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 threed; which being so placed giveth the lesser angle required. And if to the intersection you place the bead and then apply the threed to the side of the quadrat, it shall there shew 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 threed at the intersection of the contrary parallel of 80, with the right parallel of 60, so doth the threed give me the lesser angle at A, which is 36 degrees 52 minutes, which being known, the angle ACB will be found by complement, to be 53 degrees 8 minutes, thus have I both the angles required.

The threed lying in the former position, I place the bead to the intersection of the two sides; and then apply∣ing the threed to the side of the quadrat, it giveth me 100 for the base AC, which was required.

CHAP. XVI To finde a side and both the angles, by having the base and the other side given.

FIrst, lay the threed upon the side of the quadrat, and apply the bead to the length of the base; then open the threed untill the bead fall directly upon the contrary parallel of the side given, and it shall likewise lie up∣on the right parallel of the side required, and the threed shall also shew 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 threed to the side of the qua∣drat, and place the bead to 100, and then open the threed until the bead fall directly upon the contrary pa∣rallel of 80, so shall the same bead so posited, cut the right parallel of 60 the side required. And further, the threed 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 complement unto 90 degrees.

CHAP. XVII. To finde 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 threed to any of the two an∣gles, and the bead shall fall upon the right parallel of that side which is opposite to the angle, unto which the threed was opened; and also upon the contrary parallel of the o∣ther 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 an∣gle BAC 36 degrees 52 minutes, with the base AC 100, and it is required to finde the two sides AB, and BC.

First, therefore I place the bead unto 100 upon the side of the quadrat, and opening the threed to the angle of 36 degrees 52 minutes, I finde the bead to fall upon the right parallel of 60, which is the side CB subtending the said angle; through which intersection doth passe the contrary parallel of 80, which is the other side AB sub∣tending the other angle ACB, which was required.

CHAP. XVIII. In any right lined Triangle whatsoever, to finde a side by knowing the other two sides, and the angle contained by them.

FIrst, place the bead to the shorter side given, and o∣pen the threed to the given angle, then setting one foot of your Compasses in the other side given up∣on 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 be∣tweene them ACD 16 degrees 16 minutes if you place the bead to 75, and open the threed to the angle of 16 degrees 16 minutes, then setting one foot of your com∣passes 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 finde an angle by knowing the three sides.

FIrst, place the bead to the shortest of the sides contain∣ing 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 threed, un∣til the bead and the movable foot of the compasses meet both in one point, so shall the threed shew 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, then 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 qua∣drat) for the side AC, I turne the other foot towards the threed, moving it to and fro until the compass point fall just upon the bead, so shall the threed cut 16 degrees 16 minutes for the angle ACD, as was required.

And this may suffice for right lined Triangles, where∣by you may perceive the agreement of this Universal quadrat, both with Sector & Cros-staff. I will set down only two or three Chapters more, for the ready redu∣cing of hypothenusal to horizontal lines in the art of Sur∣veying. And for the speedy searching out of all perpen∣dicular 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 threed and plummet having free liberty to play; So shall the bead fall upon the contrary parallel of the hori∣zontal line required.

Suppose ABCD be a hill or mountain to be pro∣tracted, and laid down in your plot amongst your other grounds.

[illustration] 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 finde the true level and horizontal distance betweene 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 finde the threed to fall upon 36 degrees 52 mi∣nutes, for the angle of the ascent BCD; and the bead to fall upon the contrary parallel of 48, which is the hori∣zontal distance CD, which was required.

So likewise, the threed 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, be∣ing 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 BA.

CHAP. XXI. To finde the height of an object accessible at one obsen∣vation. 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 neerer or farther from it, till the threed 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 threed 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 requi∣red, above the point level 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 obiect, the threed 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 conve∣nient distance; as here I measure from the base of the ob∣ject 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 obiect at C, through both the sights, I finde the threed to cut the contrary parallel of 80 at the right pa∣rallel 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 ac∣cessible: 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 perpendi∣cular heights by the rules of proportion after this man∣ner.

[illustration] diagram of the measurement of a triangle (ABC)

Having made choice of a convenient station as at A, observe the angle at A made betweene the hypothenusal line AC, and the level line with your eye AB, the com∣plement 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 complement whereof is 53 degrees 8 minutes, for the angle at C; I place the threed at the intersection of the contrary Sine of 53 degrees 8 minutes with the right pa∣rallel of 80; so doth the threed 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 finde 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 an∣gle 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 neere thereunto as you may, and there make choice of your second station where observe the an∣gle at your eye as before: now take the complement of this second angle, out of the complement of the first an∣gle, and note the difference, then, say,

• As the sine of the difference of the complements,
• To the measured distance;
• So the Sine of the first angle observed,
• To the second hypothenusa.

Then againe,

• 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 o∣ther obstacle betwixt the object and me; and yet it is re∣quired to finde 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 finde the angle at my eye to be 36 degrees 52 mi∣nutes, the complement whereof is 53 degrees 8 minutes, then measuring in a right line towards the object, I ap∣proach as neere as I may, which is at the point D, finding AD to be 35 paces; and now at D, I make a second ob∣servation, and finde the angle at my eye to be 53 degrees 8 minutes, the complement whereof is 36 degrees 52 mi∣nutes, this 36 degrees 52 minutes I take out of the for∣mer complement 53 degrees 8 minutes, and the differ∣ence is 16 degrees 16 minutes, the angle ACD, where∣fore I place the threed at the intersection of the contrary Sine of 16 degrees 16 minutes, with the right parallel of 35, and I finde the threed 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 hypo∣thenusal line CD.

Then again, I place the threed 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 threed in the same position, cutteth the contrary Sine of the complement of the second angle observed, viz. 36 de∣grees 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 hypothenu∣sal line to be 75, if you place the bead thereto, and then open the threed to the second observed angle, viz. 53 degrees 8 minutes, the bead shall fall upon the right pa∣rallel of 60, which is the height BC; and also upon the contrary parallel of 45, which is the breadth BD, as be∣fore, and as was required.

Or more speedily thus, at two observations and one opperation.

First, make choice of your stations as before, then make choice of some one of your right parallels, that the num∣ber thereof may signifie the height required; then at your first station make observation as before, and note the parts cut by the threed 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 threed upon the said right parallel; now taking these latter parts out of the former, or the lesser out of the greater, the dif∣ference shall signifie 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 signifie the height required, I make observation to∣wards C, and I finde the threed to cut the said parallel at 60: then measuring the stationary distance AD, I finde it to be 35, and at D, I make a second observation, and I finde the threed to cut the foresaid parallel at about 33¾ which being taken out of 60, leaveth 26¼ for to signifie the stationary distance; wherefore I place the threed at the intersection of the contrary paral∣lel 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 tast of the use of this universal quadrat.

And if an index with sights were fitted to turne upon the center it would serve then by the same reason, for the findeing of all other distances.