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<DATE>2013-07-05</DATE>
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<NAME>Sarah Wingo</NAME>
<RESP>MURP</RESP>
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<ITEM>Proofed and reviewed as part of batch B130511.pdcc.</ITEM>
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<IDG S="marc" R="UM" ID="A79434">
<STC T="W">C3735</STC>
<STC T="T">E1381_1</STC>
<STC T="C">R209392</STC>
<BIBNO T="umi">99868273</BIBNO>
<VID>169895</VID>
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<TEXT LANG="eng">
<FRONT>
<DIV1 TYPE="title page">
<PB REF="1" MS="y">
<P>THE TRIGONALL SECTOR, THE De&s;cription and u&s;e thereof: Being an In&s;trument mo&s;t aptly &s;erving for the re&s;olution of all <HI>Right lined Triangles,</HI> with great fa|cility and delight.</P>
<P><HI>By which all</HI> Planimetrical, <HI>and</HI> Altimetrical <HI>conclu&s;ions may be wrought at plea&s;ure.</HI></P>
<P>The Lines of <HI>Sines, Tangents, Secants,</HI> and <HI>Chords,</HI> pricked down on any In&s;trument: Many Arithmeticall proportions calculated, and found out in a moment. Dialls, delineated upon mo&s;t &s; <GAP DESC="illegible" RESP="MURP" EXTENT="1 letter">ts of plaines: with many other delightfull conclu&s;ions. Lately invented and now expo&s;ed to the publique view.</P>
<P>By John Chatfeilde.</P>
<P><HI>LONDON,</HI> Printed by <HI>Robert Leybourn,</HI> and are to be &s;old by <HI>James Nuthall,</HI> over again&s;t the <HI>George</HI> neer <HI>Holbourn</HI>-bridge, 1650.</P>
</DIV1>
<DIV1 TYPE="to the reader">
<PB REF="2">
<PB REF="2">
<HEAD>TO THE READER.</HEAD>
<OPENER><SALUTE>READER,</SALUTE></OPENER>
<P><HI>_THou</HI> ha&s;t here pre&s;en|ted to thy view, the de&s;cription, and u&s;e of a Geometrical In&s;trument, wher|by (if thou art yet a learner) thou mai&s;t be helped abundant|ly, in the right vnder&s;tanding of the Doctrine of Triangles,
<PB REF="3">with greater ea&s;e: but if thou ha&s;t already waded through the greate&s;t difficulties, and art a Ma&s;ter in this Art, yet doubt not, but thou &s;halt finde &s;ome|thing here, which will admi|&s;ter unto thee &s;ome delight, and perchance may inable thee to do tho&s;e things with greater &s;peed and ea&s;e, then in the way that thou walked&s;t in before, howe|ver, I intreat thee to take it in good part, from him that de&s;i|reth to be</P>
<CLOSER><SIGNED>A Servant unto all for their Advantage, J. C.</SIGNED></CLOSER>
</DIV1>
<DIV1 TYPE="illustration">
<PB REF="3">
<PB REF="4">
<P><FIGURE>
<SIGNED>GG. &s;ulps-</SIGNED>
<FIGDESC>woodcut depicting the "trigonall sector"</FIGDESC>
</FIGURE></P>
</DIV1>
</FRONT>
<BODY>
<DIV1 TYPE="text">
<PB N="1" REF="4">
<HEAD>Of the Trigonall SECTOR: The de&s;cription and u&s;e thereof.</HEAD>
<P>_THe Trigonall <HI>Sector,</HI> is an In&s;trument Geometri|call containing all varie|ty of rightlined Trian|gles, together with the proportions of every &s;ide betwixt them&s;elves and to their per|pendiculars.</P>
<P>It con&s;i&s;teth of a &s;quare plate of metal, or piece of wood, on who&s;e edges, round about, are to be fixed certain laminae, or long &s;lips that may &s;tick up a little above the plate, and two labels at the extremes of one of the &s;ides, which moving on their &s;everall centers, may be applyed un|to
<PB N="2" REF="5">each other, till cro&s;&s;ing one another they make an angle betwixt them&s;elves: which alwaies &s;hall be the complement unto 180 <HI>gr.</HI> of tho&s;e angles which both the labels make unto the Radius, which is a line drawn directly betwixt their centers.</P>
<P>The in&s;criptions on the In&s;trument are fir&s;t, three Scales of Lines divided into 100 equall parts, and an&s;wering unto one another, <HI>(viz.)</HI> on the two labels and the lower lamina that contains the Ra|dius betwixt the centers of the labels.</P>
<P>Secondly, a rever&s;ed Tangent, contain|ing 45 <HI>gr.</HI> on the lamina on the left hand of the In&s;trument, and 45 more even to 90 <HI>gr.</HI> on the lamina at the top.</P>
<P>Thirdly, the Quadrant of a Circle on the inward body of the plate concentri|call with the label on the left hand, and who&s;e limb or &s;emidiameter reacheth ex|actly unto the center of the label on the right hand: &s;o that the Radius thereof is ju&s;t the &s;ame with the Radius divided into 100 parts betwixt the centers of the
<PB N="3" REF="5">labels. This Quadrant is divided into 90. <HI>gr.</HI> reckoning from the center of the right hand label, and in the area thereof are drawn lines parallel both betwixt them&s;elves, and with the Radius, exactly corre&s;ponding to the divi&s;ions of the Scales of Lines before de&s;cribed, and marked as all the other three &s;cales are with ther proper figures 1.2.3. &amp;c. to 10.</P>
<P>Thus much of the generall de&s;cription of the In&s;trument, the lines con&s;i&s;t all of equall parts and therefore need no ta|bles for their in&s;cription, except only the rever&s;ed Tangent, which may be done, as is commonly known, by the ordinary tables. Only this mu&s;t be ob&s;erved, that the Tangent of 45 mu&s;t be &s;omwhat longer then the &s;emidiameter of the Qua|drant, or the forementioned Radius, be|cau&s;e its di&s;tance from the center of the label at the right hand is greater (by &s;o much &s;pace as the breadth of the left hand label doth contain) then the &s;emi|diameter is. The label on the right hand being to &s;upply the Secant of 45. is long|er
<PB N="4" REF="6">then the other label, and reacheth even to that angle of the In&s;trument where the Tangent of 45 is placed, and there|fore the divi&s;ions upon it are 50 more then the other label (1) 150: yet though they are more in number they corre|&s;pond unto the divi&s;ions of the other, this label being &frac12; in length more then that, that &s;o by it you may &s;ee the number of parts that a Secant of any degree under 45 &s;hal contain of a Radius containing 1000.</P>
<P>If the &s;cale upon the lower lamina or Radius be &s;o drawn as to make an angle at the center of the left hand label with the edge of the lamina, as it is in the line of lines in Mr. <HI>Gunter's Sector,</HI> and another &s;cale corre&s;pondent unto it, be in&s;cribed on the left hand label, the&s;e 2 Scales &s;hall performe all the Propo&s;itions to be per|formed by the <HI>Sector</HI> one the line of Lines.</P>
<DIV2 TYPE="part">
<HEAD>Thus much of the de&s;cription of the In|&s;trument and its parts: Now of the u&s;e thereof.</HEAD>
<P>THe u&s;e of this In&s;trument is very great and various, but it con&s;i&s;teth
<PB N="5" REF="6">e&s;pecially in the re&s;olution of all &s;orts of Rightlined Triangles, unto which all Geometricall conclu&s;ions are reducible, and in Arithmeticall proportions or the re&s;olution of the Golden Rule.</P>
<P>In the re&s;olution of Triangles, 4 things are con&s;iderable. 1 The quantity of the Angles. 2 The proportions of the &s;ides or &s;ubtendents and perpendicular betwixt them&s;elves. 3 The contents of the area in parts corre&s;pondent to the parts of the &s;ides. 4 The reduction of tho&s;e parts to a perpendicular and ba&s;is of another de|nomination: all which are performed by the In&s;trument in this manner.</P>
</DIV2>
<DIV2 TYPE="part">
<HEAD>To finde the quantity of any angle.</HEAD>
<P>IF it be in Planimetry, having three &s;ights placed on the In&s;trument, the fir&s;t at the center of the left hand label, the &s;econd at the center of right hand label, &amp; the third at the end of the right hand label, work thus. Lay the In&s;trument upon a Geo|metricall
<PB N="6" REF="7">tripos or &s;taffe, as the Semicircle or Circumferentor is u&s;ually laid, by tho&s;e that mea&s;ure Land, horizontally, then turne it till you can behold one of the markes through the &s;ights at the center of the labels, afterward turne the right hand label till you &s;ee the other marke through the &s;ight at its center, and that which is at the end thereof, and then looking on the Tangent line, you &s;hall &s;ee the angle among the Tangents, where the inter&s;ection is of the label and the Tangent.</P>
<P>If the angle be above 90 <HI>gr.</HI> then you mu&s;t have foure &s;ights, and behold the fir&s;t marke through the &s;ights at the cen|ter, and the end of the left hand labell, and leaning the In&s;trument at that po&s;i|tion, turne the right hand label till you can al&s;o behold the &s;econd mark through its &s;ights, and then the angle which the two labels make where they cro&s;&s;e each other &s;hall be the angle &s;ought, which that you may know how great it is, adde together the angle found in the Tangent cut by the right hand label, and that on the Quadrant cut by the left hand, label and &s;ub&s;tract
<PB N="7" REF="7">them both out of 180 <HI>gr.</HI> and you have the angle required.</P>
<P>In <HI>Altimetry</HI> work thus, &s;et the In&s;tru|ment on one edge upright, &s;o as that the lower lamina, containing the di&s;tance betwixt the centers of the two labels, may &s;tand horizontall, and the right and left hand laminae Perpendicular; then lift up the right hand label, till, through the &s;ights thereof, you can behold the object, and then in the Tangent line &s;hall you &s;ee the angle. This is of u&s;e to find the altitude of Sun, Moon <GAP DESC="illegible" RESP="MURP" EXTENT="1 letter">or Stars, as al&s;o the altitude of any Tower, Steeple, Hill, &amp;c.</P>
</DIV2>
<DIV2 TYPE="part">
<HEAD>Two angles being known, to repre&s;ent any Rightlined Triangle.</HEAD>
<P>TRiangles of all &s;orts are di&s;tingui&s;hed two waies.</P>
<P>Fir&s;t, By their &s;ides, and &s;o they are &s;aid to be. 1 Equilaterall, having all &s;ides a like. 2 Equicrurall having only
<PB N="8" REF="8">two &s;ides equall; and the third unequall. 3 Scalene, <HI>viz.</HI> Triangles neither of who&s;e &s;ides are equall.</P>
<P>Secondly, By their angles, and they are, 1 Rectangular containing one Right-angle. 2 Obtu&s;eangular wherein one of the angles is obtu&s;e or greater then a Rightangle, that is above 90 <HI>gr.</HI> 3 A|cuteangular, wherein all the angles are le&s;&s;e then 90 <HI>gr.</HI></P>
<P>Note that the three angles of every Triangle are equall to two Right angles, that is 1 80 <HI>gr.</HI></P>
</DIV2>
<DIV2 TYPE="part">
<HEAD>To repre&s;ent any Rectangular Triangle.</HEAD>
<P>TUrne back the left hand label to the lamina of the Tangent, for then it maketh a right angle with the Radius, then turne the right hand label to the degree of the other known angle in the Tangent line, and &s;o the two labels and the Radius &s;hall give the Triangle, toge|ther
<PB N="9" REF="8">with the proportions of the &s;ides betwixt them&s;elves.</P>
<P>As &s;uppo&s;e, the two known angles be 90. and 45. the di&s;tance betwixt the cen|ters (1) the Radius will be 100 parts, the right hand label &s;hall cut the left hand label al&s;o in 141 parts.</P>
<P>The one of the&s;e &s;ides is the Radius, the &s;econd a Tangent of 45. equall al|ways to the Radius, the third, the Secant of 45. being a line drawn from the center through the limbe of the circle till it meet the tangent; if one of the known angles be above 45: and the other 90 <HI>gr.</HI> then place the labell at the tangent of its com|plement to 90. and then &s;hall the parts on the labell on the left hand repre&s;ent the Radius, and the Radius of the In&s;tru|ment &s;hall repre&s;ent the tangent, and the labell on the right hand &s;hall repre|&s;ent the &s;ecant.</P>
<P>Thus &s;uppo&s;ing the known angles to be 90. and 63 <HI>gr.</HI> 30&prime;. becau&s;e if I put the right hand labell to 63 <HI>gr.</HI> 30&prime; it will not cro&s;&s;e the left hand labell, I therefore
<PB N="10" REF="9">take its complement (1) 26 <HI>gr.</HI> 30&prime;. and applying the right hand labell unto this degree in the Tangent, the Triangle comprehended, as in the former example, &s;hall be the Triangle required; and the parts di&s;covered in every inter&s;ection or angle &s;hall &s;hew the proportion of the Ra|dius, Tangent, and Secant, as here the Radius that is repre&s;ented by the label on the left hand, &s;hall be 500, the Tan|gent repre&s;ented by the Radius of the In|&s;trument 1000, and the Secant repre&s;ent|ed by the right hand label 1120 parts, which if reduced to a Radius of 1000, the proportion holds thus, the Radius 1000, the Tangent 2000, the Secant 2240, which agreeth to the tables of Tangents and Secants, unto a Unit, which is as neer as can be expected by In&s;trument.</P>
<P>If one of the known angles be le&s;&s;e then 45, and the other 90 <HI>gr.</HI> then you need no more but apply the right hand label to the degree of the Tangent line, and the Radius of the In&s;trument &s;hall be the Radius or ba&s;es of the Triangle, and
<PB N="12" REF="9">the proportions of the other &s;ides will be &s;hewed as above&s;aid.</P>
</DIV2>
<DIV2 TYPE="part">
<HEAD>To repre&s;ent any obtu&s;e angular Triangle.</HEAD>
<P>AS &s;uppo&s;e the two known angles be 60 &amp; 100 <HI>gr.</HI> this triangle mu&s;t needs be obtu&s;e anglar, becau&s;e 100 <HI>gr.</HI> is more then 90 <HI>gr.</HI> &amp; becau&s;e neither of the labels can make an angle more then 90 <HI>gr.</HI> therefore the obtu&s;e angle mu&s;t be found in the inter&s;ection of the two labels; there|fore to find out this triangle work thus, add the two known angles together, and they will make 160: &s;ub&s;tract this out of 180, and there remaineth 20, which is the third angle; therefore to repre&s;ent this, put one labell in 20 <HI>gr.</HI> and the other in 60 <HI>gr.</HI> and where they cro&s;&s;e each other, the inter&s;ection &s;hall be an angle of 100 <HI>gr.</HI> becau&s;e all three angles mu&s;t make ex|actly 180 <HI>gr.</HI> and the labels thus applied give you the Triangle required, and the
<PB N="12" REF="10">divi&s;ions of the labells in their inter&s;ecti|on together, with the Radius &s;hew you the proportions of the three &s;ides thereof, as in the former examples.</P>
</DIV2>
<DIV2 TYPE="part">
<HEAD>To repre&s;ent any Acuteangular Triangle.</HEAD>
<P>THis is to be done only by applying the labels unto the degrees of the angles known, both in the tangent line and quadrant.</P>
<P>But if any of the angles be above 60 <HI>gr.</HI> then it mu&s;t be &s;upplied, by putting one of the labels to the complement of both added together, unto 180 <HI>gr.</HI> as &s;uppo&s;e two known acute angles be 60 and 80: the labels being applied to tho&s;e degrees in the Tangent and Quadrant, will not meet to cro&s;&s;e each other; there|fore put the one label in 60, and the other in 40 <HI>gr.</HI> which is the complement of the &s;um of both 140 to 180: and &s;o you &s;hall have the Triangle required, the pro|portion
<PB N="13" REF="10">of who&s;e &s;ides is to be found out as afore&s;aid.</P>
</DIV2>
<DIV2 TYPE="part">
<HEAD>To find the Content of any of the&s;e Triangles.</HEAD>
<P>TO effect this, the perpendicular mu&s;t be known, which being multi|plied into half the ba&s;is, the product &s;hall give the Area.</P>
<P>I have &s;hewed before how to find the proportions of every &s;ide, and &s;o by con|&s;equence the length of the ba&s;is; with as great facility is the perpendicular al&s;o to be found: for when you have applied the labels unto each other, &s;o as that the Triangle is by them repre&s;ented, doe but ca&s;t your eye directly into that inter&s;ecti|on, and under the labels among&s;t the pa|rallel lines; you &s;hall find the di&s;tance from the ba&s;is, or the length of the per|pendicular: which multiplyed as afore|&s;aid, &s;hall give the Area.</P>
<P><HI>Or to work by the In&s;trument, I &s;ay thus.</HI></P>
<PB N="14" REF="11">
<P>As 1 to the Multiplicator,</P>
<P>So the Multiplicand, to the Product.</P>
<P>Suppo&s;ing then the perpendicular to be 5, and the ba&s;is 10: who&s;e half is 5, that is to be multiplyed by the perpendi|cular.</P>
<P>I apply 10 in the left hand labell, (which al&s;o may be reckoned for 1 or 100 or 1000: as occa&s;ion &s;erveth) unto 5, a|mong&s;t the parallels, then looking for 5, upon the &s;aid labell, I finde it directly meeting with 25 among&s;t the parallels, which is the number &s;ought.</P>
<P>And thus may any number be multi|plied, only remembring to ob&s;erve that the &s;ame Figures and divi&s;ions &s;tand &s;ometimes for units, &s;ometimes for de|cads, 100. 1000, &amp;c. as occa&s;ion &s;erveth. So al&s;o working by the contrary way, may any be divided.</P>
<P>For, as the divi&s;or is to 1;</P>
<P>So the dividend, to the quotient.</P>
<P>Therefore apply the number belong|ing to the divi&s;or on the labell to 1, a|mong&s;t the parallels, and then over a|gain&s;t
<PB N="15" REF="11">the number of the dividend in the labell, &s;hall be the quotient among&s;t the parallels. As if I would divide 9 by 3; here 3. is the divi&s;or, and 9 the dividend.</P>
<P>I therefore bring down the 3: upon the labell, till it come exactly upon the pa|rallel of 1: and then over again&s;t 9: which was the dividend, I find the pa|rallel marked with the figure 3, which is the quotient</P>
<P>For 3. is contained in 9. three times.</P>
<P>But here is to be noted, that it &s;ome|times falleth out more convenient in working proportions on this In&s;trument to find the fir&s;t number among&s;t the pa|rallels, and the &s;econd upon the label, and then the third upon the parallels, will exactly an&s;wer to the 4^t^h or quotient upon the labell.</P>
</DIV2>
<DIV2 TYPE="part">
<PB N="16" REF="12">
<HEAD>To reduce the Area of a Trian|gle found of one denominati|on, to a perpendicular and ba|&s;is of another denomination.</HEAD>
<P>THe rea&s;on of the in&s;erting of this propo&s;ition is, becau&s;e that in mea&s;u|ring of Triangles, who&s;e Area we would finde either in Acres, Perches, Yards, Feet, or Inches; it &s;eldome falleth out that the perpendicular or ba&s;is mea&s;ured, giveth ju&s;t the &s;ame number as it doth upon the In&s;trument: yet notwith&s;tan|ding the Triangles and their &s;ides hold proportion unto each other; and therefore there is a nece&s;&s;ity of working by pro|portion, to reduce the one into the o|ther.</P>
<P>As fir&s;t, &s;uppo&s;ing we have a rectangle equicrurall Triangle containing 90.45, 45 <HI>gr.</HI> in its angles on the In&s;trument, the ba&s;is will be 100, and the perpendi|cular
<PB N="17" REF="12">100. which perpendicular multipl<GAP DESC="illegible" RESP="MURP" EXTENT="1 letter">y|ed into half the ba&s;is, giveth 5000 for the Area. Now &s;uppo&s;e again, that by mea&s;ure we find the ba&s;is to be but 30, in a Trian|gle con&s;i&s;ting of &s;uch angles, I &s;ay then.</P>
<Q>As 100. the ba&s;is on the In&s;trument to 30. the ba&s;is mea&s;ured.</Q>
<P>So 100 the perpendicular on the In|&s;trument, to 30 the perpendicular inquired</P>
<P>Which being multiplyed into 15. half the ba&s;is, giveth the Area 450.</P>
<P>Or 2. &s;uppo&s;ing the Triangle be &s;ca|lene, having all its &s;ides unequall, who&s;e angles &s;hall be 30 <HI>gr.</HI> 40 <HI>gr.</HI> and 110 <HI>gr.</HI> upon the In&s;trument the ba&s;is will be found to be 100, and the perpendicular about 35.</P>
<P>The Que&s;tion is what the perpendi|cular &s;hall be if the ba&s;es of &s;uch a Tri|angle being mea&s;ured &s;hall be 80 perches.</P>
<Q><P>I &s;ay then, as 100, the ba&s;es on the In|&s;trument, to 80 perches.</P>
<P>So 35 the perpendicular found, to the perpendicular inquired.</P></Q>
<P>Bring down the left hand label there|fore
<PB N="18" REF="13">till the figure of 10 come ju&s;t to the parallel of 8, and then ju&s;t over again&s;t 35 on the label &s;hall be found among&s;t the parallels, the perpendicular required, <HI>(viz.)</HI> 28; by which multiply halfe the ba&s;es, and the Area will be found 1120.</P>
<P>For, As 1. Is 28.</P>
<P>So is 40.61 halfe the ba&s;es, to 1120.</P>
<P>Bring down therefore the figure of 10 upon the label, (which here &s;tandeth but from 1) to 28 among&s;t the parallels, and over again&s;t 4. or 40. on the label &s;hall be found among&s;t the parallels, the num|ber afore&s;aid 1120; which in a large In|&s;trument may ea&s;ily be di&s;cerned.</P>
<P>And in this manner may mo&s;t Arith|meticall Proportions be found out, and Que&s;tions of the Golden Rule re&s;olved; as al&s;o Lines and Numbers may be in|crea&s;ed in continuall proportion. And Mr. <HI>Gunter's</HI> Canons for Land mea&s;ure, to find the contents of all oblong Super|ficies in Perches, Chaines, or Acres, may be made u&s;e of and re&s;olved by the In|&s;trument.</P>
</DIV2>
<DIV2 TYPE="part">
<PB N="19" REF="13">
<HEAD>Thus hath been &s;hewed its u&s;e in the re|&s;olution of Triangles, and in working proportions.</HEAD>
<P>BE&s;ides it is appliable to many other Mathematicall conclu&s;ions, as to find the length in parts of any Line of Signes, Tangents, Secants, and Cords, and &s;o by con&s;equence is of great u&s;e in the Pro|jection of Spheares, de&s;cribing of Dials, whether Sciotericall, or In&s;trumentall.</P>
</DIV2>
<DIV2 TYPE="part">
<HEAD><HI>To find the length of the Tan|gent line of any degree, to a Radius of</HI> 10000.</HEAD>
<P>TUrne backe the left hand label to the left hand lamina, &s;o as that it may make a Rightangle with the Radius, then turne the right hand label to the degree required in the Tangent line, and marking the place of its inter&s;ection with the other label, you &s;hall there &s;ee among&s;t the di|vi&s;ions of the left hand label the number of parts that &s;uch a Tangent mu&s;t be of.</P>
<PB N="20" REF="14">
<P>As &s;uppo&s;e the degree be 30, who&s;e Tangent you would find; the labels being applyed unto each other, as afore&s;aid, the parts intercepted in the left hand label, betwixt the right hand label, and the Ra|dius, &s;hall be 5773, which mu&s;t be the length of that Tangent.</P>
<P>But note, that if the Tangent be above 45 <HI>gr.</HI> then the right hand label mu&s;t be applyed to the complement thereof, and then the parts intercepted &s;hall re&s;emble the Radius, and the Radius of the In&s;tru|ment &s;hall repre&s;ent the Tangent, and the proportion that the intercepted parts beare unto the Radius of the In&s;trument 10000, the &s;ame proportion &s;hall the Radius of the In&s;trument 10000 beare unto the Tangent required.</P>
<P>As &s;uppo&s;e you would find the Tan|gent of 60, apply the right hand label un|to its complement, that is the Tangent of 30 <HI>gr.</HI> and you &s;hall find the parts inter|cepted, that repre&s;ent the Radius of 60 <HI>gr.</HI> 5773. and the Radius of the In&s;tru|ment that doth repre&s;ent the Tangent 10000. I &s;ay then.</P>
<PB N="21" REF="14">
<P>As 5773 the Radius found, to 10000, the Radius of the In&s;trument.</P>
<P>So 10000 the Radius of the In&s;tru|ment, to 17320 the Tangent of 60 <HI>gr.</HI></P>
</DIV2>
<DIV2 TYPE="part">
<HEAD><HI>And thus by proportion, may be found the length of the Tangent line for any degree even unto 90</HI> gr.</HEAD>
<P>NOw then having known the length of the Tangent line in parts, it is ea|&s;ie (as is commonly known) to make Scales and prick down a Tangent line to a Cir|cle of any Semidiameter, which when it is done, the u&s;e thereof is &s;o commonly known, that I need &s;peak nothing of it, but refer to Mr. <HI>Gunter</HI> and others that have written largely of its u&s;e.</P>
</DIV2>
<DIV2 TYPE="part">
<HEAD>To find the length of a Secant for any degree.</HEAD>
<P>THis is to be done ju&s;t in the &s;ame manner as in the former propo&s;ition the Tangent was to be found, only with this difference, that whereas the parts in|tercepted betwixt the inter&s;ection of the
<PB N="22" REF="15">labels and the Radius of the In&s;trument on the left hand label were to be reckoned for the Tangent, or el&s;e for a Radius un|to the Tangent repre&s;ented by the Ra|dius of the In&s;trument: here the Secant is always to be found by ob&s;erving the number of parts intercepted betwixt the center of the right hand label, and its in|ter&s;ection with the other label. This in the fir&s;t of the two former examples, the Radius being found 10000, the Tangent 5773, the Secant &s;hall be .11547, which is a Secant of 30 <HI>gr.</HI> but if the degree be above 45: then mu&s;t we work by propor|tion as we did in the la&s;t, &amp; the proportion will hold thus. As the parts intercepted on the left hand labell, to the parts inter|cepted on the right hand label. So 10000 the Radius of the In&s;trument, to the parts of the Secant required. Thus in the forementioned example, the parts on the left hand label are 5773, on the right hand label 11547. I &s;ay then, As 5773, to 11547. So is 10000 to 20000.</P>
<P>Bring down therefore 5773 on the
<PB N="23" REF="15">left hand label, to 11547 among&s;t the pa|rallels in the quadrant; and then 10000 upon the left hand labell will meet exact|ly with 20000 the &s;ecant of 60 <HI>gr.</HI> who&s;e complement by which the worke was wrought is 30 <HI>gr.</HI></P>
<P>The u&s;e of this line in Trigonometry and Horography, I leave to be &s;earched out in their Books, that have written largely of the u&s;e thereof.</P>
</DIV2>
<DIV2 TYPE="part">
<HEAD>To find the length of a line of &s;ines for any degree.</HEAD>
<P>THere is no more required to find out this, but to look the degree in the quadrant; then ob&s;erving among&s;t the pa|rallels, ju&s;t meeeting with the degree: you &s;hall find there the parts required. As if I would know the length of the Sine of 30 <HI>gr.</HI> ca&s;ting my eye upon the arch of the circle, where it is cut with the degree of 30: I find meeting exactly in that inter&s;ection the parallel of 5: or
<PB N="24" REF="16">5000. I conclude then, if the Radius be 10000. the &s;ine of 30 <HI>gr.</HI> mu&s;t needs be 5000.</P>
<P>The u&s;e of this line I al&s;o referre to the writings of others, becau&s;e it is &s;o com|monly treated of, and known of all.</P>
</DIV2>
<DIV2 TYPE="part">
<HEAD>To find the length of a line of Chords for any degree.</HEAD>
<P>THis line &s;o often made u&s;e for the di|viding circles, or finding the quan|tity of any arch or angle, may be found with as great facility and delight as the three former; indeed it may be done with|out any more trouble, by doubling the &s;ine of halfe the arke, as if we would know the Chord of 60 <HI>gr.</HI> halfe its arke is 30 <HI>gr.</HI> the &s;ine whereof by the former propo&s;ition, as 5000. which doubled, giveth 10000, the Chord required.</P>
<P>But it may be found with great delight and plea&s;ure by the in&s;trument another way: as followeth.</P>
<P>Apply the edge of the right-hand la|bel,
<PB N="25" REF="16">exactly to the degree in the limbe of the quadrant; who&s;e Chord you de&s;ire, and the parts intercepted betwixt that degree and the center of the label, are the Chord of that degree: thus in the former exam|ple, the labell being applied to the 60 <HI>gr.</HI> the parts intercepted &s;hal be exactly 10000, that is the radius, if applyed to the 30 <HI>gr.</HI> the parts intercepted will be 5176: which is the Chord of 30 <HI>gr.</HI> directly dou|ble to the &s;ine of 15 <HI>gr.</HI> 2588. which is half the arke.</P>
</DIV2>
<DIV2 TYPE="part">
<HEAD>To find a line or number in con|tinuall proportion, having two numbers given</HEAD>
<P>SUppo&s;e the numbers be 4: and 8: I &s;ay then.</P>
<P>As 4. to 8. So 8. to 16. So 16. to 32. &amp;c.</P>
<P>Bring down 8. on the left hand labell till it meet with 4 among&s;t the parallels, then again&s;t 8 among&s;t the parallels, &s;hal be found 16, on the label, &amp; again&s;t 16 a|mong&s;t the parallels &s;hal be found 32 on the labell, and &s;o forward.</P>
</DIV2>
<DIV2 TYPE="part">
<PB N="26" REF="17">
<HEAD>Having the Square and its Root to find the Cube there|of.</HEAD>
<P>FOr as the Root is to the Square,</P>
<P>So is the Square unto the Cube.</P>
<P>As &s;uppo&s;e, the Root be 3: and the Square 9: the Cube will be found to be 27.</P>
<P> work this.</P>
<P>Apply 9, on the left hand labell to 3: among&s;t the parallels, and then 9, a|mong&s;t the parallels &s;hall meet exactly with 27: the Cube required.</P>
</DIV2>
<DIV2 TYPE="part">
<HEAD>Having the Cube and its Square whereof it was made to find the Root of both.</HEAD>
<P>THis is to be wrought by working di|rectly contrary to the former pro|po&s;ition.</P>
<P>For.</P>
<P>As the Cube is to the &s;quare, &s;o is the
<PB N="27" REF="17">&s;quare unto the root, apply therefore the number, &s;tanding for the cube and &s;quare together, finding the one on the label the other among&s;t the parallels, and then ju&s;t over again&s;t the number for the &s;quare &s;hall &s;tand the root. Always remembring that if the fir&s;t number be found upon the label, the third number mu&s;t then be &s;earched there al&s;o: but if it be among&s;t the parallels then the third mu&s;t be a|mong&s;t them to, for the third number in the&s;e proportions is always to be found on that line or Scale where the fir&s;t was; and the fourth on that line where the &s;e|cond was: whether it be on the label or among&s;t the parallels.</P>
<P>Many other things might be wrought by this In&s;trument; I have only given hints to tho&s;e that are more curious and have greater lea&s;ure to follow the&s;e &s;tu|dies, who I doubt not, but by their in|du&s;try will be able from what I have &s;aid to draw new conclu&s;ions, and from many plea&s;ing propo&s;itions conducing much to the &s;atisfaction of tho&s;e that are ingeni|ous.</P>
</DIV2>
<TRAILER>FINIS.</TRAILER>
</DIV1>
</BODY>
<BACK>
<DIV1 TYPE="postscript">
<PB REF="18">
<HEAD>Po&s;t&s;cript.</HEAD>
<P><HI>_REader, thou ha&s;te had the In&s;tru|ment with its u&s;e, the Author now takes leave of thee for a time; if a favourable a&s;pect &s;hine on this, thou mai&s;t eare long expect more, he hath another In&s;trument called an Horographi|call Spheare, whereby the whole Art of Di|alling, will mo&s;t plainly be &s;et forth to the view of all that under&s;tand the u&s;e of it; and Dialls of all &s;orts may be delineated in all kind of Plaines with great &s;peed and plea|&s;ure, though they are never &s;o full of Gibo&s;i|ties or Concavities, which would puzzle the be&s;t Mathematician to reduce to form: he is now in hand with it, thou &s;halt heare more from him as he perceiveth this to find ex|ceptance.</HI></P>
</DIV1>
<DIV1 TYPE="advert for mathematical instruments">
<P>_THis In&s;trument called the Tri|gonall Sector, as al&s;o all other Mathematicall In&s;truments what&s;oever, are exactly made by Mr, <HI>Anthony Thomp&s;on</HI> in Ho&s;ier-lane neer Smithfield, <HI>London.</HI></P>
<TRAILER>FINIS.</TRAILER>
<PB REF="18">
</DIV1>
</BACK>
</TEXT>
</EEBO>
</ETS>
