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at an angle of 120°, because we always prefer to make use of the angle formed by the nearest approach of the projection to the line of our position, or the picture plane.

4th. Again, suppose an inclined shutter, or a roof which is united horizontally with a wall, is said to be at an angle of 40° with the wall, the shutter or roof would be at an angle of 50° with the ground.

Fig. 70.

All this will be very evident if we consider that "if any number of straight lines meet in a point in another straight line on one side of it, the sum of the angles which they make with this straight line, and with each other, is equal to two right angles." (See Lessons in Geometry, V., Vol. I., page 156.) Therefore (Fig. 67), if A is 30° with the PP, and B 90° with A, then B will be 60° with the PP, the whole making two right ang les. With regard to the last supposi tion, we shall see that the lines of the wall, the roof

or shutter, and the ground, form a right-angled triangle, the three interior angles of which are together

DVP3

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wall with the ground is 90°, and the shutter or roof 40° with the wall, the shutter will be at an angle of 50° with the horizon (Fig. 68). Consequently, this angle of 50° must be constructed for the vanishing line, and the subject treated as an inclined plane. (See Pro

blems XXXI., XXXII., and XXXIII.) From all this we deduct a rule for finding vanishing points for lines or planes which are stated to be at given angles with other lines or planes not parallel with the picture plane :When the sum of

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the two angles of the given objects is greater than a right angle, it is subtracted from the sum of two right angles, and the remain der is the extent of the angle sought. This will explain the results of the first, second, and fourth suppositions above.

When two angles of the given objects are together less than a right angle, the sum will be the angle sought. This answers to the third supposition. We now propose a problem to illustrate our remarks about the wall and the shutter.

PROBLEM XLI. (Fig. 69).—A wall at an angle of 40° with our position is pierced by a window of 4 feet 3 inches high and 4 feet broad; a shutter projects from the top of the window at an angle of 40° with the wall: the window is 5 feet from the ground, and its nearest corner is 4 feet within the picture; other conditions at pleasure. Scale of feet t

n

Before proceeding to work this problem, we wish to give the student some directions about the scale. In this case we have given the representative fraction of the scale, and not the number of feet to the inch. It is a common practice with architects and engineers to name the proportion of the scale upon which the drawing is made, in the manner we have done here, leaving the scale to be constructed if necessary. The meaning of the fraction is that unity is divided into the number of equal parts expressed by the denomi nator. Thus a scale of feet signifies that one standard foot is divided into 48 equal parts, each part representing a

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inch

foot on paper, the result is to the foot. It also means that the original object, whether a building or piece of machinery, is 48 times larger than the drawing which represents it. If the scale had been written, yards, it would be the same as inch to represent a yard. The way to arrive at this is as follows:

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inches.

the foot.

of 3 inch to
the yard.

The above method
of stating the
scale ought to be
understood by
every one engaged
upon plan-draw-
ing.

To return to the problem. The principal consideration relates to the shutter. The inclination may be upwards, at an angle of 40° with the wall, or it may be downwards at the same angle. We will represent both cases. First, when inclined downwards. Draw the HL, which is 4 feet from the groundline; from PS draw

a perpendicular to
E; this will be the
radius for drawing

the semicircle meeting the HL to determine DE and DE Find the vanishing point for the wall vP1, and its distance point DVP; also find the vp2 by drawing a line from E to VP? at a right angle with the one from E to vp1, because if the shutter had projected from the wall in a horizontal position, it would have vanished at vp2; that is, if it had been perpendicular or at right angles with the wall. In short, the vanishing point for the horizontal position of a line must always be found whether the line retires to it horizontally or not, because the VP for an inclined retiring line is always over or under the VP (according to the angle of inclination) to which it would have retired if in a horizontal position. (See Prob. XXXI., Fig. 53.) Consequently, the vanishing point for an inclined retiring line is found by drawing a line from, in this case, the DVP, accord

Fig. 71.

Pp2

wards, establishing its VP above the eye or HL.) Consequently, we must draw the vanishing line for the VP3 downwards from DVP2. The sides of the shutter, t w and m v, must be drawn in the direction of VP3, and cut off from DVP3, first by drawing a line through t to y; make y a equal to the length of the shutter; draw from a to DVP3, producing w. All the early part of the problem, relating to the wall and windows, and the remaining lines wv and t m, will be but a repetition of the shutter under the first position. We can prove the truth of this method of drawing the perspective inclination of a plane by another method. Draw the right angle cad (Fig. 68); make ab equal to the length of the shutter, and at an angle of 40° with a c or 50° with a d; draw b c parallel to a d; a c will be equal to the height of b above a. This must now be applied to Fig. 70. Draw a line from VP2 through t to e on the line

ing to the angle of inclination, to where it cuts a perpendicular line drawn through the vp; thus we find its vanishing point, whether its inclination be downwards or upwards; therefore draw a line from DVP2, at an angle of 50° with the HL, cutting the perpendicular from vp2 at VP3, the vanishing point. We have made the nearest corner of the window 2 feet to the left of the eye, represented by the distance i to b; a line from b must be ruled to PS, upon which we wish to cut off 4 feet to find a, the nearest point within; a line from c, which is 4 feet from b, must be drawn to DE', and where it cuts the line brs in a is the point required. Draw the perpendicular a hm. Draw from DVP through a to p; make pr equal to the width of the window. Draw back again from r cutting DVP in s; draw the perpendicular st; the base of the window is drawn from f, on the line of contact, 5 feet from the ground, to the VP; the height of the window, 4 feet 3 inches, is marked from ƒ to e; a line from e to vp1, cutting the perpendiculars from a and s in m and t, will give the top of the window. The opening of the window is m thn. Now we must draw the shutter; the corner nearest us is v, consequently it inclines upward towards the wall, but downwards from it; therefore, the VP for the shutter must be above the HL, which have explained.

VP3

we

m

To

measure or set off the

length of the shutter,

we have raised a line of contact for that purpose from o, found by drawing from VP2 through s to meet the DE ground-line. From t directed fromvp3 draw a line through w; this will be the further side of the shutter; its length must be determined thus:-From t directed from DVP draw a line to the line of contact, meeting it in y; make y z

DVP3

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Fig. 69.

S

equal to the length of the shutter, the same as the length of the window; draw from a back again to DVP3, cutting tw in w; draw w v, directed by vp1, and v m directed by VP3.

We will now draw the shutter at the same angle with the wall, but inclined upwards from it (Fig. 70). The important difference in working the problem under these conditions arises from the upward inclination of the shutter from the wall, but inclined downwards to meet the wall. This last view of the position of the shutter is the proper one for our purpose, because after a little consideration we shall perceive that it is a retiring plane, but downwards; therefore its VP is below the eye or HL. (In the former case the shutter was a retiring plane, but up

B

Dvp2

Pp1

of contact; make ef equal to the height of b above a, viz., c a (Fig. 68). Draw from f back to vp2; it will be found to cut the corner of the shutter in w, proving by both methods that t w is the perspective length of the further side of the shutter.

Vp1

A plan of a building may be made, having all its proportions, angles, and other measurements arranged and noted, yet nothing may be said as to its position with the pictureplane, and from this plan several perspective elevations may be raised. When such is the case, all that is

necessary will be to draw a PP across the paper in such a position with the plan, that by drawing visual rays, the picture-plane we have chosen may receive the view we wish to take of it. Suppose A (Fig. 71) is the plan of a building, and we wished to have two views of it

one taken with an end and front in sight, the other with a view of the front and the opposite side we should then place the PP at such an angle with the side or front as might be considered to be the best for our purpose. PP1 would receive the visual rays from the front and the end B; PP2 would receive those from the front and the end c. In short, any line may be drawn which represents the PP at any angle with the plan, or opposite any side we may wish to project. This will give a very useful illustration of the way to treat a subject when its proportions are given, as is frequently the case, without any reference to the view to be taken of it; in other words, the angle it forms with the picture-plane.

at an angle of 120°, because we always prefer to make use of the angle formed by the nearest approach of the projection to the line of our position, or the picture plane.

4th. Again, suppose an inclined shutter, or a roof which is united horizontally with a wall, is said to be at an angle of 40° with the wall, the shutter or roof would be at an angle of 50° with the ground.

All this will be very evident if we consider that "if any number of straight lines meet in a point in another straight line on one side of it, the sum of the angles which they make with this straight line, and with each other, is equal to two right angles." (See Lessons in Geometry, V., Vol. I., page 156.) Therefore (Fig. 67), if A is 30° with the PP, and B 90° with A, then B will be 60° with the PP, the whole making two right angles. With regard to the last supposition, we shall see that the lines of

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inch

Before proceeding to work this problem, we wish to give the student some directions about the scale. In this case we have given the representative fraction of the scale, and not the number of feet to the inch. It is a common practice with architects and engineers to name the proportion of the scale upon which the drawing is made, in the manner we have done here, leaving the scale to be constructed if necessary. The meaning of the fraction is that unity is divided into the number of equal parts expressed by the denominator. Thus a scale of feet signifies that one standard foot is divided into 48 equal parts, each part representing a foot on paper, the result is to the foot. It also means that the original object, whether a building or piece of machinery, is 48 times larger than the drawing which represents it. If the scale had been written, yards, it would be the same as inch to represent a yard. The way to arrive at this is as follows:

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the two angles of the given objects is greater than a right angle, it is subtracted from the sum of two right angles, and the remain der is the extent of the angle sought. This will explain the results of the first, second, and fourth suppositions above.

When two angles of the given objects are together less than a right angle, the sum will be the angle sought. This answers to the third supposition. We now propose a problem to illustrate our remarks about the wall and the shutter.

PROBLEM XLI. (Fig. 69).-A wall at an angle of 40° with our position is pierced by a window of 4 feet 3 inches high and 4 feet broad; a shutter projects from the top of the window at an angle of 40° with the wall: the window is 5 feet from the ground, and its nearest corner is 4 feet within the picture; other conditions at pleasure. Scale of feet t

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To return to the problem. The principal deration relates to the shutter. The inclination may be upwards, at an an gle of 40° with the wall, or it may b downwards at the same angle. W will represent bot cases. First, whe inclined down wards. Draw th HL, which is 4 fee from the ground line; from PS dra

a perpendicular t E; this will be th radius for drawing the semicircle meeting the HL to determine DE and DE Find the vanishing point for the wall vp, and its distanc point DVP; also find the vp2 by drawing a line from E to V at a right angle with the one from E to VP', because if th shutter had projected from the wall in a horizontal position, would have vanished at vp2; that is, if it had been perpen dicular or at right angles with the wall. In short, the vanishin point for the horizontal position of a line must always be foun whether the line retires to it horizontally or not, because the v for an inclined retiring line is always over or under the v (according to the angle of inclination) to which it would hav retired if in a horizontal position. (See Prob. XXXI., Fig. 53. Consequently, the vanishing point for an inclined retiring lin is found by drawing a line from, in this case, the DVP, accord

ing to the angle of inclination, to where it cuts a perpendicular line drawn through the vr2; thus we find its vanishing point, whether its inclination be downwards or upwards; therefore draw a line from DVP2, at an angle of 50° with the HL, cutting the perpendicular from VP2 at VP3, the vanishing point. We have made the nearest corner of the window 2 feet to the left of the eye, represented by the distance i to b; a line from b must be ruled to PS, upon which we wish to cut off 4 feet to find a, the nearest point within; a line from c, which is 4 feet from b, must be drawn to DE', and where it cuts the line

Ps in a is the point required. Draw the perpendicular a hm. Draw from DVP' through a to p; make pr equal to the width of the window. Draw back again from r, cutting BVP in s; draw the perpendicular st; the base of the window is drawn from f, on the line of contact, 5 feet from the ground, to the vpl; the height of the window, 4 feet 3 inches, is

marked from ƒ to e; a line from e to VP1, catting the perpendiculars from a and s in and t, will give the top of the window. The opening of the window is m thn. Now we must draw the shutter; the corper nearest us is v, consequently it inlines upward towards the wall, but downcards from it; therefore, the VP for the shutter must be above the HL, which have explained.

VP3

Fig. 71.

Pp2

m

we

To

measure or set off the length of the shutter, we have raised a line contact for that purpose from o, found by drawing from vp2 throughs to meet the DE ground-line. From t directed fromvp3 draw line through w; this will be the further ide of the shutter; its length must be determined thus-From directed from DVP3 aw a line to the ine of contact, meet

ng it in y; make y z

indow;

DVP 3

wards, establishing its VP above the eye or HL.) Consequently, we must draw the vanishing line for the vp3 downwards from DVP2. The sides of the shutter, t w and m v, must be drawn in the direction of vr3, and cut off from DVP3, first by drawing a line through t to y; make y a equal to the length of the shutter; draw from a to DVP3, producing w. All the early part of the problem, relating to the wall and windows, and the remaining lines wv and t m, will be but a repetition of the shutter under the first position. We can prove the truth of this method of drawing the perspective inclination of a plane by another method. Draw the right angle cad (Fig. 68); make a b equal to the length of the shutter, and at an angle of 40° with a c or 50° with a d; draw b c parallel to a d; ac will be equal to the height of b above a. This must now be applied to Fig. 70. Draw a line from vp2 through t to e on the line

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mal to the length of the shutter, the same as the length of the draw from a back again to DVP3, cutting tw in w; awwv, directed by VP', and v m directed by VP3. We will now draw the shutter at the same angle with the rall, but inclined upwards from it (Fig. 70). The important ifference in working the problem under these conditions arises rom the upward inclination of the shutter from the wall, but clined downwards to meet the wall. This last view of the osition of the shutter is the proper one for our purpose, because fter a little consideration we shall perceive that it is a retiring jane, but downwards; therefore its VP is below the eye or HL. In the former case the shutter was a retiring plane, but up

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B

DYP2

Pp1

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of contact; make ef equal to the height of b above a, viz., c a (Fig. 68). Draw from f back to vp2; it will be found to cut the corner of the shutter in w, proving by both methods that t w is the perspective length of the further side of the shutter.

A plan of a building may be made, having all its proportions, angles, and other measurements arranged and noted, yet nothing may be said as to its position with the pictureplane, and from this plan several perspective elevations may be raised. When such is the case, all that is Vp1 necessary will be to draw a PP across the paper in such a position with the plan, that by drawing visual rays, the picture-plane we have chosen may receive the view we wish to take of it. Suppose A (Fig. 71) is the plan of a building, and we wished to have two views of it

one taken with an end and front in sight, the other with a view of the front and the opposite side-we should then place the PP at such an angle with the side or front as might be considered to be the best for our purpose. rp would receive the visual rays from the front and the end B; PP2 would receive those from the front and the end c. In short, any line may be drawn which represents the PP at any angle with the plan, or opposite any side we may wish to project. This will give a very useful illustration of the way to treat a subject when its proportions are given, as is frequently the case, without any reference to the view to be taken of it; in other words, the angle it forms with the picture-plane.

at an angle of 120°, because we always prefer to make use of the angle formed by the nearest approach of the projection to the line of our position, or the picture plane.

4th. Again, suppose an inclined shutter, or a roof which is united horizontally with a wall, is said to be at an angle of 40° with the wall, the shutter or roof would be at an angle of 50° with the ground.

Fig. 70.

All this will be very evident if we consider that "if any number of straight lines meet in a point in another straight line on one side of it, the sum of the angles which they make with this straight line, and with each other, is equal to two right angles." (See Lessons in Geometry, V., Vol. I., page 156.) Therefore (Fig. 67), if a is 30° with the PP, and B 90° with A, then B will be 60° with the PP, the whole making two right angles. With regard to the last supposition, we shall see that the lines of the wall, the roof or shutter, and the ground, form a right-angled triangle, the three interior angles of which are together

m

DVP3

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h

Fig. 66.

PS

the two angles of the given objects is greater than a right angle, it is subtracted from the sum of two right angles, and the remainder is the extent of the angle sought. This will explain the results of the first, second, and fourth suppositions above.

When two angles of the given objects are together less than a right angle, the sum will be the angle sought. This answers to the third supposition. We now propose a problem to illustrate our remarks about the wall and the shutter.

PROBLEM XLI. (Fig. 69).—A wall at an angle of 40° with our position is pierced by a window of 4 feet 3 inches high and 4 fest broad; a shutter projects from the top of the window at an angle of 40° with the wall: the window is 5 feet from the ground, and its nearest corner is 4 feet within the picture; other conditions at pleasure. Scale of feet t

n

Before proceeding to work this problem, we wish to give the student some directions about the scale. In this case we have given the representative fraction of the scale, and not the number of feet to the inch. It is a common practice with architects and engineers to name the proportion of the scale upon which the drawing is made, in the manner we have done here, leaving the scale to be constructed if necessary. The meaning of the fraction is that unity is divided into the number of equal parts expressed by the denominator. Thus a scale of feet signifies that one standard foot is divided into 48 equal parts, each part representing a foot on paper, the result is inch to the foot. It also means that the original object, whether a building or piece of machinery, is 48 times larger than the drawing which represents it. If the scale had been written, yards, it would be the same as inch to represent a yard. The way

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to arrive at this is

as follows:

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the semicircle meeting the HL to determine DE and DE Find the vanishing point for the wall vP, and its distanc point DVP; also find the vp2 by drawing a line from E to V at a right angle with the one from E to VP1, because if the shutter had projected from the wall in a horizontal position, would have vanished at vp2; that is, if it had been perper dicular or at right angles with the wall. In short, the vanishin point for the horizontal position of a line must always be foun whether the line retires to it horizontally or not, because the v for an inclined retiring line is always over or under the v (according to the angle of inclination) to which it would hat retired if in a horizontal position. (See Prob. XXXI., Fig. 53. Consequently, the vanishing point for an inclined retiring lin is found by drawing a line from, in this case, the DVP, accon

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