GEOMETRICAL PERSPECTIVE.-XIX. SHADOWS CAST UPON INCLINED PLANES. PLANES or surfaces upon which shadows are cast may be in any position. We have in the previous lessons considered those planes only which are horizontal or vertical, and we now introduce those that are inclined. One or two important and leading principles will first engage our attention. The indefinite projection of the shadow of a given line coincides with a plane passing through the source of light (the sun) and the given line; this we call the plane of shade. Suppose in Fig. 87, s to be the sun, a b an object, say a post, casting a shadow, the ray from s through a to c will determine the length of the shadow b c (see Lesson XVI.); then the space inclosed by a b c is deprived of light by the object a b, therefore the triangle abc is the plane of shade. When the plane of shade is intersected by any surface, the form and extent of the culty presents itself, the meaning of the trace of the plane of shade, and how it represents the plane. Planes in space in projection are represented by their traces only. Thus, in Figs. 88, 89, 90, the traces h n and h c are the vertical and horizontal traces of the plane a b c d; and according to the positions of these traces, we understand the positions of the planes. In Fig. 88 the plane is at an angle with the ground, and perpendicular to the vertical plane; in Fig. 89 it is at an angle with both planes of projection; in Fig. 90 it is perpendicular to both planes of projection. In Linear Perspective the line PP, the picture plane, is the horizontal trace of an indefinite perpendicular plane; the line HL, horizontal line, is the vertical trace on the picture plane of a plane passing through the eye and parallel with the ground. To determine the plane of shade we must necessarily project its trace, by drawing a straight line through the vanishing point of the line projecting the shadow and the vanishing point of the sun's rays; because both these 40° shadow upon that surface are determined according to the inclination of the surface with the plane of shade. Thus, in Fig. 93 the trace of the plane of the shade of the pole is A B. The pole and its shadow are both lying in this plane; the zigzag form the shadow takes arises from the surfaces (the walls and roofs), which cut this plane, being irregular, or in other words, forming various angles with the plane of shade. To illustrate this change in the direction of the course of the shadow-that is, to show why the shadow of the pole is so angular-let the pupil hold a pencil in an inclined position under a lamp, and allow the shadow to fall upon a slip of cardboard, placing the board first in an horizontal position, then in a perpendicular one, then at an angle with the table, afterwards turn it, so that it shall be parallel with the pencil, he will at once see that according to the position of the cardboard, as it intersects the plane of shade, so will the inclination, position, and length of the shadow be affected, and he will also see the reason for the varied form of the shadow of the pole in Fig. 93. It will now be evident that in order to project the shadows of objects upon inclined planes we must determine the plane of shade, which is accomplished by drawing its trace; here a diffi vanishing points are in the plane of shade. Then the vanishing points for the shadow of a line, projected upon various inclined planes, will be found upon the trace of shade at the intersections of the traces of the inclined planes upon which the shadow falls. We shall refer to this again in a problem to illustrate it. PROBLEM LIII. (Fig. 91) is a square block of masonry be yond which a beam projects. The sun is in front of the picture. It will be observed that a line is drawn from the vanishing point, VP2, to which the beam, the object that causes the shadow, retires, through the vanishing point of the sun's elevation, VPSE, to a perpendicular line drawn from the vanishing point of the object upon which the shadow is cast (the block), meeting at VP3. This is the trace of the plane of shade. Consequently the shadow from the beam on the block is drawn in the direc tion of VP3, whilst the rays which determine its length are * Because the plane or surface of the block upon which the shadow line lying upon that plane will have its vanishing point somewhere in falls vanishes in a perpendicular line through VP'. Therefore, any that perpendicular line, according to the angle of inclination; those which are horizontal, like the upper and lower edges of the face, vanish on the HL at VP; VPS is in the plane of shade. directed to VPSE. The edge a b of the shadow on the ground of the block is directed to VPSI, and its extent cut off at b by a ray from c to VPSE; b d, which is necessary to complete the outline of the visible portion of the shadow cast by the block, is directed towards vp1. PROBLEM LIV. (Fig. 92).—A cross, the face of which is inclined to the PP at an angle of 40°, casts its shadow on a plane inclined at 30° with the horizon; the horizontal trace of the plane is perpendicular to the PP. The sun is in the picture, that is, its rays are parallel with the picture, its elevation 55°; other conditions at pleasure. Draw anywhere across the HL the line a b at an angle of 55° for the directing ray of the sun's elevation. Draw the line c d at 30° with the ground line representing the inclination of the plane receiving the shadow. Draw a line through PS parallel to cd; this will be the trace of the inclined plane receiving the shadow, and upon which the vanishing points for the retiring edges of the shadow upon the inclined plane are to be found; thus, draw lines from vp and VP2 parallel to a b, producing vp3 and VP4. SIMPLE TENSES. J'ai, I have. Nous avons. Vous avez. Imperfect. Nous avions. INDICATIVE MOOD. COMPOUND TENSES. SIMPLE TENSES. COMPOUND TENSES. Past Indefinite. Nous avons eu. Pluperfect. Past Definite. Past Anterior. J'avais eu, I had J'aurai, I shall or J'aurai eu, I shall, had. will have. Nous aurous. CONDITIONAL MOOD. SIMPLE TENSE. Present. J'aurais, I should have. Tu aurais. Il aurait. will have had, Nous aurons eu. COMPOUND TENSE. Past. J'aurais eu, I should have had. Ju aurais eu. Il aurait eu. Nous aurions eu. Vous auriez eu. Ils auraient eu. The learner will naturally pause here to inquire why these lines should be drawn, and to this query we must reply as follows:-If there had been no inclined plane upon which the shadow falls, the whole of the shadow would have been horizontal, and consequently the retiring lines of the shadow would have vanished on the HL at VP and VP2 respectively; but as the plane containing the shadow becomes inclined, so will the trace of the plane be inclined also, elevating or depressing the vanishing points proportionately. Construct the perspective elevation of the cross according to previous instructions. The rays from the angles of the cross must be drawn parallel to a b; the horizontal projections of the shadow, as o m, must be drawn parallel to the PP (see Lesson XVI., page 260), as far as the horizontal trace of the inclined plane c PS. Afterwards the shadows of the perpendicular edges of the cross which fall upon the inclined plane must be drawn parallel to c d, as mn, the length of which is determined by the ray en; nf is the shadow of er, and is drawn in the direction of VP3; f g is the shadow of rh drawn parallel to ed, because rh is a perpendicular line; g k, the shadow of hi, is drawn towards VP; ks, the shadow of i p, is drawn Que j'aie, that I Que j'aie eu, towards vp3. The remaining edges of the shadow upon the inclined plane will not be difficult, if the pupil carefully considers the positions of the lines of the cross; the shadows of those which are perpendicular must be drawn parallel to c d ; those which would retire, had the shadow been on the ground, to vp must vanish at VP3; and those which would retire to VP2 must be directed to VP. After the shadow leaves the inclined plane at the horizontal trace e PS at z, it then falls upon the ground, consequently the edge I will go to VP, and the shadow of vw, which is q t, will vanish at VP2. The learner should go through this problem three or four times, taking the inclinations of ab to the HL and c d to the ground line at different angles to those which have been used in this problem. PROBLEM LV. (Fig. 93).—Again, to show how to determine the vanishing points of shadows which fall upon inclined planes, we have borrowed a subject from "Malton's Perspective." In that work the subject is a ladder inclined against a house; we have chosen a pole, a b, instead, to make the explanation more simple. c VP is the trace of the inclined plane of the lower roof; D VP is the trace of the inclined plane of the upper roof. Je n'ai pas, I These are found by drawing a line from the vanishing point of the horizontal edge of the roof to the vanishing point of the inclination (see Lesson X., Problem XXXII., Vol. III., p. 333). The trace of the plane of the shadow is from A to B, found by drawing a line from the vanishing point of the object, the pole, casting the shadow, through the vanishing point of the sun's elevation, VPSE; this contains the vanishing points for the shadow of the pole, projected upon the inclined roofs, and are found where the traces of the inclined planes intersect the trace of the plane of shade. To begin with the shadow on the ground:Because this portion, a c, is horizontal, therefore its vanishing point is on the HL at VP3; c d vanishes at B, because the plane of the wall containing c d vanishes through VP'; de vanishes at A, the vanishing point of the pole, because the plane of the wall containing de is parallel with the pole; ef at VP, where the trace of the plane of the roof intersects the plane of shade; similarly, g h to vr, and g ƒ similarly to c d at B. Present.-Avoir, to have. IT is obvious that powers may be added, like other quantities, by writing them one after another, with their signs. EXAMPLES.-The sum of a3 and b2 is a3 + b2; and the sum of a2bn and h5 - d' is a2 - ba + h3 — d1. The same powers of the same letters are like quantities, hence their co-efficients may be added or subtracted. EXAMPLE. Thus the sum of 2a2 and 3a2 is 5a2. But powers of different letters, and different powers of the same letter, are unlike quantities; hence they can be added only by writing them down with their signs. EXAMPLE. The sum of a2 and a3 is a2+ a3.· SUBTRACTION OF POWERS. RULE.-Subtraction of powers is performed in the same manner as addition, except that the signs of the subtrahend must be changed as in simple subtraction. EXAMPLE. From 2a take - 6a'. Ans. 8a'. EXERCISE 43. 1. From -36" take 4b". 2. From 3hab take 4h3b". 3. From a3b" take a3b”. 4. From 5(a-h) take 2(a-h)". 5. From 6a(a+b) take a(a + b)*. 6. From 174x3 + 5xy take 12a3x3 — 4xy3. 7. From 3a3(ba — 8)3 take a3(b3 — 8)3. 8. From 5(x + y1)3 — 3(aa — b3)3 take − 3(a2 — b3)! + 4(x3 + y1)3, 9. From ab3+x3y* take a3b® — xay3. 10. From 2x(a - b)3 + 3(a — b)3 ̧take x(a - b)3 + 3(a − b)3. 11. From (x + y)2 + }(a + b)3 take }(x + y)2 + Z(a + b)3. MULTIPLICATION OF POWERS. Powers may be multiplied, like other quantities, by writing the factors one after another, either with or without the sign of multiplication between them. EXAMPLES. The product of a3 into b2, is a3b2; and r3 into am, is ama3. If the quantities to be multiplied are powers of the same root, instead of writing the factors one after another, as in the last article, we may add their exponents, and the sum placed at the right hand of the root will be the product required. The reason of this operation may be illustrated thus :— a2 X a3 is a3a3; but a = aa, and a3= aaa; and aa x aaa = aaaaa = a3. The sum of the exponents 2 + 3 is also 5; so dm x du dm+n ̧ N.B.-The same principles hold true in all other powers of the same root. If a + b be multiplied into a that isThe product of the sum and difference of two quantities is equal to the difference of their squares. This is an instance of the facility with which general truths are demonstrated in algebra. If the sum and difference of the squares be multiplied, the that is, (a2 + b2) × (a2 − b2) = (a1 — b1). It is evident that the square of a, and the cube of a, are product will be equal to the difference of the fourth powers; EXERCISE 44. 1. Multiply hab" into a1. 6. Multiply 3r into 2r3. 8. Add 5abc3, 3a2bc3, a2bc3, and 2a2bc3. 10. 9. Add ab + x°y⋆ + a3b3 and — x3y* + a$b®. 10. Add 3a3+ bcs + 5a3 + 2bc and a3+5bc2 to 6a3+2bc". 11. Add (ry- cm), 3(xy - cm)®, — }(xy — cm)®, and ?(xy' — cm)®. Multiply a3+x3y+xy3+ y3 into x-y. 11. Multiply 4x2y + 3xy 2x2-x. 12. Multiply 23+2-5 into 22 + x + 1. 13. Multiply y" into y-" into y1. 14. Multiply as into a3 into a-". |