S H A

425

S H A

of vanishing point of the line to the vanishing point

gi^ etl es Perpendicular to the plane whose vanishing line istitled the vanishing line of another plane will be ob-th 6 ’"j which is the original of the seat of the point, andit)t er „ 'S. m al of the line in projection ; and, therefore, thethe S p Ctlon °f the vanishing line given of the plane on whichlin e fy the line required to be drawn, and the vanishingThe r ? n< h' s the vanishing point of the seat of the line,point We > < t , ’ aw a straight line through the seat of theanq ; n projection, to the vanishing point found,

of t(, e ® hne thus drawn will be the whole representation

Proposition is evident, since the vanishing line of>s giy^* ane perpendicular to the plane whose vanishing linePe ndj n > will pass through the vanishing point of lines per-liti e 11 ar to that plane ; and since the seat of the originalPngjj n ® the original of the plane given, is formed by a planefr through the original line perpendicular to the giventhig '^tersecting therewith ; therefore the vanishing line ofPair,t q p P en dicu 1 ar plane will pass through the vanishing

Ot ty

0f th

th fo u J anishin g tine is given, being the straight line passing. A » 1 t * lese points.

*0 ariv nera t knowledge of the shadows of lines upon planes°f lijjgPosition ought first to be acquired ; but as the relationW an d planes to the horizon is generally given, it will

U ’ryuaicuiar plane will pass through the vanishingyy? tines perpendicular to the original of the plane given ;la 6n . two points in a vanishing line are given, the whole

e »ch ot \ Ssar y to find the relation of these lines and planes toand here it will be proper to observe, that what-

«Vi

• 1 ' l, 'U IIUIO ill It UJ WO £/!. IV Kt * V-'J 111C11 VVildl-

a y tie the number of planes, the vanishing point of thewill remain unchangeable, or in the same positione rV«54' the first vanishing line, and will be common toovy n ^tferent planes ; but every different plane will have

other

Van ' s hing point for the seat of the sun’s rays in that' tl ® ^at vanishing point will be in the vanishing-

As vertical and horizontal planes occurpractice, these will require particular

inclination of a plane to thepicture, both being perpendicular to the original

•Post !, that plane.‘Ute^gquently i„

IT.— Given the

°f the

b) ■ Onrt ti — - - J r i - - -a -

ti ne nr l le seat an( ^ inclination of a straight line in the

- se a f ^ ie horizon ; to determine the vanishing point of

P 0;

' ! n<

-Clf _v» _ -* J i V

. y the line on the vertical plane, and the vanishingJ the line.

‘‘he,

^ Figure 1.—Let No. 1, represent the vanishingNo. 2, the plane of the picture. In the vanishingt ,r Plac e ° ^ et v 1 be the vanishing line, e the point of sight,'tpl . of the eye, a b the intersection of the original ver-j^tlle 4 at1 ?’ inclined to the plane of the picture in thei ‘ion ^ ’ an< l ^t AD be the seat of the line, as given int^l'tiati t0 l *‘ e horizon : make the angle d a f equal to the^ P<?ndi° n 1 °^" tire line to the plane of the horizon ; draw z> fa to jj. , ar to a d, and D b perpendicular to A b ; producePj a ke bk equal to d f, and join ak, which is theto^ firai^ I'J 6 on tl le vertical plane. Draw e l parallel to a b,v 1 e, s

r ,l >®

v'ttllel ^ an ' 8 ki n g point of the seat of the line. Draw e vf A > and v i perpendicular to v l ; make v n, in the

^ lh perpendicular to vZ; in v l make Im equal®ake the angle Imh equal to bajc ; and h will

«,e "'ir

Mun e Un 8le,

line,

equal to v e ; make the angle vni, equal to« f V A A which the original line makes with theL^'incy, 1 ? horizon; and draw e® perpendicular to v 1,ch *^ e Van’ V 1 .®' tk e plane of the picture, No. 2, let v lti|?' Ca of 1Sa ' n g line answering to v l, No. 1 : in v l make, lla] !ln y convenient point,©, for the centre of the pic-0,1 ie 0 L equal to © l, No. 1, and © v equal to © v,ph and vi perpendicular to tlj then h is5 <i

the vanishing point of the seat of the line, and i the vanish-ing point of the line itself.

The points h and i will be both on the same side of thevanishing line of the horizontal planes.

This problem is the same when the seat and altitude of aray of the sun are given, with the inclination of a verticalplane to the plane of the picture ; to find the vanishing pointof a ray of light', and the vanishing point of the seat of thesun’s rays.

When the sun is on the same side of the picture with thespectator, the vanishing point of the seat of the rays, and thevanishing point of the rays, will be below the vanishingline, v L ; but when on the other side of the picture, thevanishing point of the rays, and the vanishing point of theirseat, will be above v l.

Proposition III.— To find the shadow of a rectangidarprism upon the horizon on which it stands, and also uponanother rectangular prism; the base of the prism whichthrows the shadow being in the same horizontal plane witha side of the prism on which the shadow is thrown.

Figure 2.—Let o be the vanishing-point of the sun’srays, vs the vanishing-line of the horizontal plane on whichthe two prisms are placed, abed the base, and a! b' c d' thetop of the prism that throws the shadow. Then, becausethe edges of the prism stand parallel to the picture, they willhave no vanishing point but at an infinite distance ; thus,the line o s, drawn from o, will be parallel to the edgesa a, b b', cc', d d', of the erect prism, and will give thevanishing-point, S, of the shadows of a a', bb', c c, dd';therefore, draw as, ds, cs; and a' o, d'o, co, cuttinga s, d S, c S, at the points e,fg; then will defgc be theshadow on the ground. Let c s and d s cut the edge of theprism, opqr stu, at li and i; draw il and hk parallelto os, cutting the upper edge, r t, at k and l; from thepoints k and l draw lines to S, cutting the farther edge, q s,at m and n; then hkmnli will be the shadow upon therecumbent prism, op qs tu.

Proposition IV".— To find the shadow of a cylinder lyingwith its convex surface upon a horizontal plane.

Figure S. —Let x be one end of the cylinder, v the other.To find the shadow of the end x upon the horizon, it mustbe observed, that no line which terminates two surfaces canthrow a shadow upon a third, unless one of such two surfacesbe in shade. We have, therefore, only to find the shadowof that part of the circumference of the end x, containedbetween the point of contact on the horizontal plane, and atangent plane to the surface of the cylinder; and a sufficientnumber of points will be found by letting fall perpendicularsfrom as many points in the arc that throws the shadow, tomeet the horizontal plane ; then find the shadows of theupper extremities of these lines, which will be points in thecurve. Or, find the whole ellipses representing the shadowof the circumference of each end of the cylinder, and drawthe line cg, to touch these two ellipses ; then de eg will bethe shadow. To find the shadow of any point, c, in the edge |of the end x : draw c d perpendicular to the horizon, meet-ing the horizon in d ; also, draw d S and co, cutting eachother in c, then c is the shadow of c; in like manner, e willbe found to be the shadow of e, and so on for as many pointsas may be necessary.

Proposition V. — To find the shadow of one cylinder uponanother. The cylinder which throws the shadow, and thaton which the shadow is thrown, being placed, the formerwith its end, the latter with its convex surface, upon the samehorizontal plane.

The shadow of the cylinder, which stands upon its end onthe horizon, is obtained by finding the shadows of the tops