The centre and distance of the picture being given, to find the representation of a point, whose seat on the pic ture, and distance from it, are given.
Let C (Fig. 6.) be the centre of the picture, and S the given scat. Join CS, and draw CO, in any direction, and equal to the distance of the picture. Through S draw SP parallel to CO ; but on the opposite side of CS, and equal to the distance of the point from its seat ; join OP : then p, the intersection of it with CS, is the representation required.
For the lines drawn from the point of sight to the cen tre of the picture, and from the proposed point to its seat, being at right angles to the picture, are parallel to each other, and in the same plane with CS ; farther, they are equal to CO, SP : therefore, by Lem. I. p is the point in which a straight line drawn from the point of sight to the point proposed cuts the picture ; p is therefore the repre sentation of the point proposed.
Con. 1. The point p may be found by means of propor tional compasses, without drawing the parallels CO, SP, by dividing CS in the ratio of the distance of the picture to the distance of the point from its seat.
Cott. 2. To find the representation of a straight line by this problem, find the representation of its extremities, and the straight line joining them will be the representation required.
The centre and distance of the picture being given ; to find the representation, vanishing point, and its distance, of a line whose seat on the picture, intersection, and the angle which it makes with its seat, are given.
Let C (Fig. 7.) be the centre of the picture, and S I the given seat. Make the angle LS I equal to that which the line proposed makes with its seat. Draw CV parallel to S I, and from C draw CO at right angles to it. Make CO equal to the distance of the picture, and from 0 draw OV parallel to SL ; join SV. V is the vanishing point, VO its distance, and VS the representation of the line propos ed. For the straight lines OV and VC being parallel to SL, S I, the angle OVC (see Geom.) is equal to the angle LS I, or to the angle which the line proposed makes with its scat. But the plane passing through the straight lines drawn from the point of sight to the centre of the picture, and to the vanishing point of the proposed line, is evi dently at right angles to the picture, and parallel to that plane which passes through the proposed line and its seat ; its intersection with the picture is therefore parallel to the given seat ; and the angles contained by parallel lines be ing equal, the angle contained by the straight line drawn from the centre to the vanishing point of the proposed line, and the straight line drawn from the point of sight to the same point, is equal to the angle which the line pro posed makes with its seat, or to the angle OVC. It is
therefore evident, that the triangle OVC is, in every re spect, equal to that contained by the lines which join the centre of the picture, the vanishing point of the proposed line, and the point of sight ; therefore, CV is equal to the distance of the vanishing point from the centre of the pic ture ; and it was already proved that the straight line which joins the centre of the picture with the vanishing point of the line proposed is parallel to the seat of that line. Hence V is the vanishing point of the line propos ed, OV its distance, and consequently, (Prop. IL) VS is the representation of the line extended ad infinitum.
Cox. 1. The same things being given, the perspective representation of any given point in the line proposed may be found. For, having made the same construction as above, make SA equal to the distance of the given point from S, the intersection of the line proposed ; join OA, meeting SV in a ; then a will be the representation re quired. This is evident from the Lemma,—the straight lines, drawn from the centre and the proposed point, to the points S, V, being parallel to each other, and equal to OV, SA.
C0R. 2. Hence also a b, the representation of AB, any portion of the proposed line, may be found.
The representation and vanishing point of any straight line being given ; to find the representation of the point which divides the original in any given ratio.
Let A B (Fig. 8.) be the given representation, and V the vanishing point. It is required to find the representation of the point which divides the original of AB in a given ratio. Draw VO anyhow through the point V, and draw a b parallel to, and at any distance from, VO. Take 0 any point in VO ; join OA, OB, and let them meet a b in a, b ; divide a b in c, so that a c may have to c b the given ratio, and join 0 c, and let it cut AB in C ; C is the repre sentation required. For through C draw AICN, parallel to a b, or VO ; and by similar triangles, viz. BCN, OBV ; ACM, AVO, we shall have (E. 4. 6.) the following pro portions,