S H A

426

S H A

t

I

I

l

i

of as many straight lines on the convex surface as may bethought sufficient for the purpose, which will give thepoints f, g, e. The shadow of the cylinder, which is placedwith its convex surface upon the horizon, is found in Figure 3.Let ae and gf be the lines of contact with a tangent planeto the point s, then the lines AE and gf will throw theshadow partly upon the horizon and partly upon the convexsurface, z, of the cylinder, which lies on the ground. It is,therefore, only necessary to find the sections of the cylindricsurface z, with the two planes of contact, sae and sgf; forthis purpose, produce the plane agfe, and the end, t, of therecumbent cylinder, till they meet in p Q. This will be readilydone as follows : Produce G p, and from the vanishing-point, L, of the end T, of the recumbent cylinder, draw theline Lap, through the point a, and also draw pq perpen-dicular to the vanishing-line v l ; then p q is the intersectionof the plane passing through G f and a e, with the plane t.Let a d' be a vertical diameter from the point of contact, d',of the end x ; draw L d! Q; then, in P Q take any number ofintermediate points, R, s ; draw R L, S L ; also draw linesfrom s, R, Q, to the vanishing-point of a g and e f, cuttinga e at b, c, d ; let the lines from R, s, drawn to L, cut theend t at c, b : draw b v and c v; also draw b l and c l,cutting b v, c v at b and c; then b and c are points in theshadow. In the same manner, points in the other edge willbe found, and the shadow completed.

Proposition YI.— To find the shadow of a building witha break.

Figure 5 .—Let v l be the vanishing-line of the horizon,v the vanishing-point of the horizontal lines representedby a c and b d, that form the end of the building, alsoof efig h , which represent the horizontal lines forming thesides of the break. Let the sun be supposed to be in theplane of the picture, or its rays parallel thereto, and letthe planes a b d c and e g h f be in shade ; and theplane e g Ay will throw a shadow upon the plane a b Ik, asthe plane a b d c will also upon the horizon. As the sun’srays are parallel to the picture, they will have no vanishing-point, but still the rule will hold in this case also. Throughthe vanishing-point L, draw L m perpendicular to V L ; thenl m is the vanishing-line of the plane a b l k, on which theshadow is to be thrown ; through v draw v M, parallel tothe sun’s rays, or make the angle l v m equal to the anglewhich the sun’s rays make with the plane of the horizon.Thus m is the vanishing point of the shadow of all linesvanishing in v, upon the plane a b l h : therefore, to find theshadow of the line h g, join M h, and produce it to m; anddraw g m parallel to M v ; then m will be the shadow of thepoint g, and h m of h g. Draw m n parallel to g e, and m nwill be the shadow of g e : therefore h m nf will be thewhole shadow of the plane h g efi upon the plane a b l k.To find the shadow of the end abed upon the plane of thehorizon : draw a o parallel to L v, and b o parallel to M v jthen a o is the shadow of the vertical line A b : join o v, anddraw dp parallel to M v, and op is the shadow of b d :join p l, and draw r q parallel to m v, and p q will be theshadow of the line d r, not seen : join s q, or draw it parallelto l v; then a op qs will be the shadow of the buildingupon the plane of the horizon.

Many more examples of shadows might be given; but ifthe principles here shown are understood, the artist will notbe at a loss to find the shadow of any right-lined object what-ever: for, to find the shadow of an object constituted byplanes, and consequently terminated by straight lines, is nomore than to find the shadow of those lines. If a circlebe given, the circumference may be divided by parallel linesinto parts, and the shadows of the points of division may be

-. a® 0

obtained by finding the shadows of the intercepted ^‘ ne l t e&drawing a curve round their extremities. If it wereto find the shadows upon several planes, first find the ^ e( ein the plane on which the object stands, and observe^.^,the shadow meets the next plane ; then, having theing-line of this second plane, observe where the va . n geC on<lline of the plane of shade cuts the vanishing-line of * llS j s |,jng'plane, and the point of intersection will be the vanpoint of the shadow on the second plane. y yvill

The principles exhibited in the article Project ’ t ;ye,apply equally to the representation of objects in P® 1S P inter'particularly where the planes which throw the shade.sect the plane on which the shadow is to be thrown, inter-continuing the line that throws the shadow, an® J ere th esection of the planes, to meet each other, the point . noint beshadow terminates is found; and therefore, d ‘ l P jj[ b egiven in the shadow, the direction of the shadow ^ a °known. Thus, in the last example, suppose ^ ie j ia j 0 w °f

a n°

obtained: since the point o is the beginning of the 1the line bd, produce a c and b d, to meet in V; jo" 1 ^draw the ray of the sun, dp ; then op is the shadow ra yproduce d r and c s to meet in l, and join pi; ^ ra 'p n<r q from r, and p q will be the shadow of d r, not se ^ ^ 0 j eProposition VII.— To find the vanislung-l ineupon several planes.

Plate II. Figure l.

-Let abcdefghik

b,.t«

Nr o:

of a building, with a lean-to, or penthouse, vEthe vanishing - point of all horizontal lines, m

ible»

a b l i k, of the main house, and also of the S a ' J !f’ ( ^ e bo r >'of the penthouse; l‘ is the vanishing-point of a ^ n( j a szontal lines in the parallel fronts, bpge and d E N* ^ . g t beall vertical planes have vertical vanishing-lines, c pil 0 ’vanishing-line of the parallel gables, a b l i k an ^ g s ji;l u the vanishing-line of the fronts, b f G e aIU j n bu*^"i l g h is the representation of the roof of the wing, and omnp that of the penthouse. _ . s: dr# ,v

Produce l i to meet v r, its vanishing-poi nt ’ ’ , plan®

s l 1 , which will be the vanishing-line of the 1I)C ^ jj n es j aL G h I, for s and l‘ are the vanishing-points °\. tV Y. a w 1 ’

that plane: produce mo to meet v R in T, and ^itf 0.1 t - . - - ■ name *

then T l

ic . jjiuuuuo my tv * a* -z J e .

is the vanishing-line of the inclined p ‘ , . v ani s)1

of the roof of the penthouse, because t and l areing-points of two lines in that plane. , j^nse,

Let wx be a pole resting upon the end of ^ ^ Q [, e t>„

in

and let <2

\Y,

to

or

the same plane with the gable, ablik; «*■- , x

vanishing-point of the sun’s rays : produce the/°| ie ’ pole,,meet v R in r ; then r is the vanishing-point ot ( jra' v ‘ D jof the line that throws the shadow: there* 1 ^ a( j e> «n^

a nd

Q r, Q r will be the vanishing-line of the plane oi y .let it cut v l 1 , the vanishing-line of the ^ F fit

iei cm v r. , me vauiaiiiug-ime m — - B E w " he

l 1 u, the vanishing-line of the vertical P.' lin . eS ^jjne of 1demn, of the walls, in u; VR, the vanishing ^ jn

" -’e roam . ,i

XJ CJ 41A AT, \JL CUV *AA W J f a*, *- . yQ() 1 1 1,

gables, in R; s l‘, the vanishing-line of the niai a j lw b'0and t li, the vanishing-line of the penthouse, > n ’ | 6) vVdoes but prepare for drawing the shadow of 1 e pr°^upon the horizontal plane and upon the building- o j )lt vvh e ^duce a b to meet wx in x, then x will k e , , e J,. dra"'^ it ,’

tiTl S o D ini n 7. dni ' V ' Va ’ cuttin S ™ »> . / utS

GL°at l • an yJ aW CZ< ’ cutt, ng PO in d; draw v d, o ,

x a bcdl y -nT\ LZ > cutti 4 the ridge, ift «*/' ‘For since t h t le w)loIe sll “dow of the pole. 0 \e,

or line in r ‘ e , shado ' v first begin# at the foot of ‘flec-tion of’the ,^ ne °*' l * le horizon, and since the u jg

to be thrown a wUh n fhe n - e f 0f * pk " e whi ° b ^ °!the nLne r l the ,nf ersection of the vanish

Plane of shade, gives the vanishing-line of the sh“

the pole rests upon the ground, or horizonta^ P*^ ravV