PRO

312

PRO

i

be 50;° add 40 to 90, the sum is 130; the half of 130 is 65 ;subtract 40 from 65, and the remainder is 25° ; then thetangent of 25°, set on the other side of 40 from the centre ofthe picture, will give the distance of the vanishing pointof the line, the radial of which makes 40° with the primary-radial. Again, add 50 to 90, the sum is 140, the half is 70;subtract 50 from 70, and the remainder is 20° ; then, thetangent of 20° being set from the centre of the picture uponthe other side of the angle formed by the primary radial, theother radial, containing 50°, will give its distance.

In general, suppose the lesser angle to be called v, and

, v + 90 90— v

the greater to ; then --- v = — - -=45 — —•

£ U

and the tangent of this angle, set upon the other side, willgive its vanishing point,

, . w + 90 90— w w . ,

Again, --- w — —-— = 45 — —, gives the dis-

A £

tance of the other vanishing point, set on the contrary side

w

of the centre ; thus, if w — 60, then 45 — — = 15, therefore

set the tangent of 15° as directed ; and thus the tangent ofthe remainder of half the angle contained by the primaryradial and the radial from 45°, gives the vanishing point;or, take the tangent of half the complement of the said angle,and set it on the vanishing line from the centre of the pictureon the other side, and it will give the vanishing point.

Problem XIII. — Given the. angle made by the radials ofany two original lines with each other, and the angle whichone of them makes with the ‘primary radial; to find a divid-ing point common to the same measures or scale, so as to cutoff a portion from the indefinite representation of each line,such that the portions may be the representations of the twooriginal lines.

Note .—When one of the angles which the primary radialmakes with the one radial is given, that made by theprimary radial with the other may be found by subtractingthe one given from the whole angle contained by the radials.This being obtained, proceed as follows:—Subtract the lesserangle formed by the one radial and the primary radial fromhalf the angle contained by the radials of the original lines ;then, if the tangent of the remaining angle be set on thevanishing line, on the side of the primary radial which hasthe greatest angle, it will give the dividing point required.

A

Thus, call the whole angle A ; the half is _ ; call the

2

lesser angle made by the primary radial with one radial, b,

then — — b is the difference ; find the tangent of — — b ;2 2then the greater angle made by the primary radial with the

other radial being A — b, set the tangent of — — b from c, on

2

the vanishing line on the side of A — b, and the extremityof the distance will give the dividing point required.

Example 1.—Suppose the angle contained between the tworadials to be 100°, and the lesser angle to be 40°, then thegreater will be 60: now the half of 100 is 50 : subtract 40°,the lesser angle, from 50, there remains 10 : set the index ofthe proportional compasses to 10° on the line of tangents ;take the length of the primary radial with the legs marked r,and set off the distance contained by the opposite points fromthe centre of the picture on the vanishing line, on the side ofthe primary which has the angle of 60°, and the extremityof the distance is the dividing point.

Example 2 .—Suppose the angle contained by the radials

to be 80°, the lesser angle to be 30°, and conseQJ 1 40°greater 50°; the half of 80° is 40°: subtract 30 1and 10° remain. Set the index of the proportional 0 ^ {lieto the tangent of 10° ; take the length of the radia bj

legs marked R, and set the distance of the point coD^the opposite legs from C on the vanishing line, on ^ j. y j.

itlifo

the angle of 50°, and the extremity will gi veP oint - , „ plane,

Problem XIV .—Given the sun’s altitude on ayheight of a line, and the length of its shadow on f‘ ^ se ct l0,ithe orthographical representation of a cornice, tot ,of the same, also the seat of the sun’s rays ; to find

of the cornice. , ® n °

— ------- • - architr J ‘ u ,a. of

perpendicular to the ® r ep r ®iL h for the indefin> te poi „t*<

ise 9 >

Figure 15.—Let No. 1 be the cornice, orNo. 2 a profile, or section,edges, of the mouldings; draw b b for thesentation of the shadow of a line represented by 1 thro" 1 !No. 1, and let b b be the distance that the shadow . a

c b : 2

from the point b, so that b will be the shadow o ^ ^the external angles draw c C, e e, g G, I H. ^°. v t ^ N 0, {0is the length of the line represented by the P 01 ^ ^ jjo. 1, ^set the proportional compasses in the ratio o_^ ^ vV ;tli ^

set tne proportional compasses in uie i.*— -o b No. 2 : make c c equal to b v, : take op, f ^jgtai'shorter ends, and make e e, No. 1, equal to the ^between the points of the remote ends; take S' r > , „%;gtan cethe shorter legs, and make g G, No. 1, equal to

the

b c;

dis

draw

the lines M L, k j, parallel to b b, c C, &c., toat l and J ; then will M L

to u is, v <j, -- j oW s uP 0 !? c e :

at ij aiiti u f men win iu. u and kj be the f 1 tl>e.%other side. To find the shadows upon the pl®j^ aI1 y po' nin the representation, cm, of the arris line, ta* t jje I- eP j

the line which projects the shadow ;

esented

tot

ith

lines in the original object are also represlines. _ . c1 ilina e ' - f w°

Problem XV. — The representation of ^ ghado^ J^gsquare abacus, or cap, being given, to find e ylind e , a jipe

cap upon the cylindric surface, the axis of s f l(J dott' °Jparallel to the plane of projection ; also, j in d *

the axis of the cylinder, \ se rrd'

Parallel to the p/ tP er Pfndicttla r m ,,

Passing along t/ ie , ax>s of mefigure 16.— Tot “ axis ’ and through the lumiwm-

e ^'>der, ABCOLFGbe the plan of the s %

vat, on of the semi ' at of tI,e CI 'P1 let wXT* be** , etB be the projects and imvu that of the 0 J j,,n,

fr °m the Ooiner°of' n ,i° 1 ' re P l ' e sentation, of a rayon the P^pybnder at b • 1 ie a bucus at i, cutting the P hin ic ;,-cte

! n f> and cutting LF P^Uel tom, to touch **«%?,, 1the centre 0 oVT L ! draw Fr perpendieuf*to ,

Pcontact: j n T , t ,e semicircle ; and F will be 0 jntSi

J > M > K and draw an r number of intermediate ‘? , hg

kb, parallel to xb, cutt#*

tl» fl

C P

laf® 1 '

between the points of the remote ends ; P‘ n j;

parallel to mc, and draw e d parallel to ed; ^^jiel

iel to ex, and gf parallel to gf; draw P‘ l gjoe-and Jbcdefghi will be the shadow from ^ e t sP t iie

and draw vw parallel to b b, or c C, &c.; also, - - n t, ^sentation, ex, of the next arris line, take tmy g fic, v ^draw x y parallel to b b or c c ; take the d ,s ‘ , l( j g .

the shorter ends of the proportional com P a p S jj’ e lonS e1 'equal to the distance between the points ot ^ a „dalso, take the projection, f e, with the shorter o^. nt8 ^xy equal to the distance contained between ^ ^ u a11 JoWremote ends ; through the points w and y 0 fth e9 1parallel to m c, which will terminate the brea ^ cB

upon the face. _ _ . ginC e the f'® t o 11

. The reason of this operation is evident, en dicut® jjt,0

that a shadow will be thrown by a i, a doW 0 ,i e i to

plane, is as the length of the line; and t e n6 pitf® llejparallel to a plane will be projected on t two