thereon can only be transferred throughout the direction of that plane. Thus the height of the roof IL was transferred by the line LZ along that plane to its other extremity s; but the line rs is not the place of the ridge of the roof, which lies in the middle of the plane ghikm, proceeding from the point k; but any height on the angular line gr is easily transferred along that plane by means of its horizontal vanishing point V, by which means the height of the roof was obtained by the line rV at k. If, instead of the plane over the line AD (No. 4.) being produced for an intersection, the plane of the middle of the house in the direction of the ridge of the roof had been drawn, and the height of the roof had been set up on that line, it would at one application be transferred to its proper place.

2417. Let the line FE (No. 4.) be produced to P for an intersection, set off the distance OP at OP (No. 5.), and draw the intersecting line PR. On PR set up the height of the ridge of the roof equal XD (No. 1.), and draw the ridge line RZ, and it determines the exact ridge of the roof between the proper visual lines, and will be found to correspond exactly with the ridge obtained by the former process.

2418. The roof may, however, be found by another process, thus:-The slant lines of the roof have their vanishing points on the picture as well as any other direction of lines in the same object. The line km (No. 5.) being in the vertical plane ghikm, will have its vanishing points somewhere in the vanishing line of that plane. (Def. 15.) A vertical line drawn through the horizontal vanishing point V will be the vanishing line of the plane ghikm; therefore the vanishing point of the lines km, ki, and of all lines parallel to them, will be somewhere in the vertical GVXQ.

2419. Two lines drawn from the eye parallel to any two lines in an object, finding their vanishing points, will make the same angle at the eye as the lines in the object make with each other; for the two lines in the one instance are respectively parallel to the two lines in the other.

2420. The line SQ is drawn from the station S parallel to the line AB (No. 4.), and a line drawn from the station S, making the same angle with SQ as ED does with EC, (No. 1.), will find the vanishing point of the line ED, and this point must be evidently somewhere in a vertical line through the point Q. To obtain this point in practice, take the distance of the vanishing line it is in, that is, the length from S to Q in the compasses, and set off the same in the horizon (No. 5.) from V. to W. At the point W make an angle VWX equal to the inclination of the roof, that is, equal to the angle CED (No. 1.), and

produce the line till it intersects the vertical line through the vanishing point V in the horizon in the point X. The point X will be the vanishing point of the line of the roof ka (No. 5.), and of the line no, parallel to it. The slant lines of the roof km and no, already obtained, will, on application of a ruler, be found to tend to the point X, as above stated. 2421. In the same way the line of the roof ki (No. 5.) will also have its vanishing point, and in the same vertical line GVQ. It will be found to be as much below the horizontal vanishing point V as the point X is above it. (Def. 14.)

2422. Let the line AB (No. 6.) be the line of the horizon, and CD the vanishing line of a vertical plane, being the gable end of a house, and let the angle ABC be that of inclination, finding the vanishing point of the slant lines of a roof in one direction. Let the line BD be the line, finding the vanishing point of the slant lines in the other direction, having the same inclination to an horizontal line; then the angle ABD will be equal to the angle ABC, and the distance AD equal to the distance AC.

2423. Example III. To find the representation of a quadrangular building situated inclined to the picture, covered with a single hipped roof.

2424. Let the quadrangle GDHK (No. 7.) be the plan of the building; the line MN will represent the ridge of the roof. The former line QL may be the place of the plane of delineation, and it may be viewed from the same station S. The position and direction of the lines of this object being the same as those of the last example, the preparatory lines will also answer for this. We have then only to draw the visual rays MS, NS, CS, PS, and KS, intersecting the picture in the points m, n, g, p, and k, and to produce the line DG for an intersecting point at R.

2425. Prepare the picture (No. 8.); let the line VZ be the horizon, GR the ground line, O the centre of the picture, and the points m, n, g p, and k coresponding with m, n, g, p and k. (No. 7.) Draw the visual line lines through those points and the intersecting point R, and proceed as follows:

2426. On the intersecting line RE set up the height RT, equal the height of the object HG (No. 2.), and draw the lines TV and RV, cutting the visual lines of the front of the building in the points z and o, y and p, determining the plane ypoz for the representation of the plane of the front. From the angular points z and y draw the lines zw and yr to their vanishing point Z determining the plane yzwx for the end of the building.

2427. On the intersecting line set up the height of the roof TE equal the height NK (No. 3.), and draw EV cutting the angular visual line of the building in the point e, from which point draw the line ez, cutting the visual line pa in the point a, the point of direction of the ridge of the roof. Draw the line a V, which, cutting the visual lines through the points m and n in the points t and v, determines the exact position of the ridge of the roof tr, which is the representation of OP (No. 3.), or of the ridge MN (No. 7.); draw the lines to, vz, and vw, which will complete the whole representation required. In No. 8., if the lines az and aw be drawn, they will form a gable end yzawr, of which the point a is the point of the gable, and will answer for the direction of the ridge, whether it be a gable end or a hipped roof, for in both cases it lies in the middle of the breadth of the house; wherefore the line aV answers as well the edge of a hipped roof as of a gable end.

2428. In examining the plans (Nos. 4. and 7.) of the two buildings, it will be seen that they are placed at right angles to each other, and in contact at the point D, so that the second example might have been easily accomplished from the first, without the aid of another intersection and other preparatory lines, than the additional visual rays from the angles, which the student will have surely no difficulty in carrying through, without the necessity of encumbering these pages with the detail.

2429. Example IV. In fig. 832. No. 1. is the general plan of a church similar to many country churches. ABCD is the main body of it; EFGH its tower; IKLM and MLNO subordinate parts of the building, and abcd the porch. No. 2. is its geometrical elevation; the ends and measurements, AB and BC, answering to IM and MO in No. 1., and the points of the roofs D, E, and F. (No. 2.) answering to the lines of the ridges QR, TV, and PL, No. 1. To find the perspective representation of this building on the plane of delineation YZ, the station being at S, the following is, perhaps, the readiest process.

2430. Find the vanishing points Y and Z of the horizontal lines of the building by the lines SY and SZ being drawn from the station parallel to them. O is the centre of the picture. Draw the visual rays from the visible angles of the object in direction to the station S, to intersect the plane of delineation.

2431. When a complicated object, that is, one composed of many parts, is to be drawn, it requires, of course, a great number of visual rays for the precise determination of those parts, and the whole together forms an apparently confused number of lines. The eye, however, which views them properly, does not perceive that confusion; and, if it perplex the student, different coloured inks, or of different shades of depth, may be used to particularise different parts. In the delineation of such an object as the present example, the most important consideration is the choice of a proper intersection; for though any inter

section will do, that should be chosen which unites most parts in its direction with the greatest exactness and the least trouble. In the case under consideration, none secms more eligible than the direction of the roof PLM, which produce to W.

2432. In the picture No. 3., GL is the ground line, GV the height of the horizon, the line VX being then the horizontal line. O in the horizon is then the centre of the picture, from which, place the distances of the horizontal vanishing points OV and OX equal OY and OZ, No. 1. AB (No. 3.) is the intersecting line, and all the visual lines on the plane of delineation are drawn conformably to their intersections on the ground line in the plan. On the intersecting line the height AC is made equal to the height AG of the elevation No. 2.; and the lines Cc and Aa, being drawn in direction to the vanishing point V, determine the height ac; being the height of that part of the building on the visual line answering to the ray from the point M in the plan No. 1. Through the points a and c draw the lines de and bf to their vanishing point X, determining the plane bdef, the representation of the plane AGHC, No. 2.; the visual lines bd and fe answering to the rays from the points I and O in the plan. Draw the lines dh and bg tending to their vanishing point V, to the ray from K in the plan completing the plane bghd. On the intersection make the height AD equal to the height of the roof NE of the elevation No. 2., and draw Di in direction to V. Through i draw the line kl to the vanishing point X, touching the visual lines of the. roofs in the points k and Z. Draw the lines km, mh, kd, kc, le and le, which will complete the whole of the structure over the plan IKNO, No. 1.

2433. The height of the roofs of the low buildings is equal to the height of the upright walls of the body of the building, as shown by the line PR in the elevation No. 2. ; hence, the line mo, and the return line on, may be drawn to the visual lines corresponding with the intersections from the angles A and B of the plan From the angle g the line gs may also be drawn, which will determine the lines sr, rt, and tp of the porch. Make AE on the intersection equal to the height of the roof BF in the elevation, and draw the line EV determining the ridge of the roof between the two visual lines from the points P and L of the plan. Draw the lines of the gable end vo and vz, the point z being obtained by the line om drawn to its vanishing point X, cutting the visual line from the angle D of

2434. Make AG and AF on the intersection equal to the heights of the tower BO and BM of the elevation, and draw the lines GV and FV cutting the visual line from P in the plan, in the points a and b; through which points draw the lines ac and ef to their vanishing point X; and the lines cd and eg to their vanishing point V; the points g, e, and ƒ being in the proper visual lines from the angles of the tower F, E, and H in the plan. The tower will be completed by drawing the lines dg, de, ae, and af.

2435. This example elucidates the general practice of vanishing points, which are as well to be obtained of other positions of lines as horizontal ones. It is not always that the vanishing points of inclined lines are required, but they are often useful, and sometimes absolutely necessary. In the geometrical elevation No 2. the lines MO, PF, GD, IE are all parallel lines, as also are the lines OV, FR, EH, and DI, and though situated in different, yet they are in parallel planes, and will therefore have a common vanishing point. A line drawn perpendicularly to the horizon through the vanishing point X (fig. 3.), as LQ, will be the vanishing line of the plane of the end of the church over the line IO of the plan, also of the end of the body AD, likewise of the side of the tower EH; and a line drawn through the point V (No. 3.) perpendicularly to the horizon, as GM, will be the vanishing line of the planes over the lines (No. 1.) IK, AB, ab of the porch, and FE of the tower, and all lines in those planes, or the boundaries of those planes, will have their vanishing points somewhere in those vanishing lines.

2436. To obtain the vanishing points of the inclined lines of the roofs and tower, take the distance of the vanishing point Z from the station $ in the compasses, and apply it on the horizon from X to H. At the point H make an angle with the horizontal line equal the angle of the roofs a Pc (No. 2.); the curve KI and the distance of it from the centre H being equal to the curve ac, and distance of it from its centre P: then is the angle KHI equal to the angle of the roof a Pc (No. 2.). Produce the line HK to Q; Q will be the vanishing point of the line ea of the tower, also of the parallel lines ov, dk, and cl, which, though obtained by a different process, will all be found, by application of a ruler, to tend truly to that point, as is shown by the dotted lines in the example. Proceeding in the same way with the distance of the vanishing point Y from the station S, we obtain the vanishing point of the same inclination of lines in the other planes of the object. Take the length SY in the compasses, and set it off on the horizon from V to N. At the point N make an angle INT on the horizon equal the angle KHI, that is, equal the angle of inclination of the roof aPc (No. 2.). The line NT produced to M in the vanishing line GM will be the vanishing point of the line de of the top of the tower, also of the lines w3 and y5 of the porch (the inclination of the roof of the porch being the same as the other roofs of the body of the church), as shown by the dotted lines in the example. The walls of the porch are obtained from the height AP on the intersection, equal the height AT, No. 2., Pm being drawn to the vanishing point V, and mn to X, give the lines #5, 53, and 32. We may observe that the inclined lines af, le, kc, and vz have a common vanishing point, which, if required, may be obtained; it will be in the same vanishing line with the point Q, and as much below the horizontal vanishing point X as the point Q is above it, to which point, were it obtained, the lines already drawn will be found exactly to tend. It is seldom absolutely necessary to have both those points; in this instance one only of them, the point Q, is obtained, which answers every end required of both; for, supposing it were left to that vanishing point for finding the inclined lines, the visual lines being drawn, and the heights of the upright walls being found, the line dk being drawn in direction to the vanishing point Q determines one side of the gable end at the visual line in the middle; the other is accomplished by joining the points k and c together. So of the other gable, el being drawn, le is also had by joining together the points and e.

2437. To complete the whole, draw the line rq on the tower from the point to the angle of the tower, in direction to the vanishing point Q; then draw the lines qh and nh to their proper visual lines and vanishing points V and Q. The putting on of the spire requires some consideration, and in it we must proceed with some thought and care. The base of it is intended to be a regular octagon. If the two external lines in the geometrical clevation of the spire be continued till they touch the sides of the tower, as is done at K and L (No. 2.), and an octagon be there constructed, extending the square of the tower, it will be the base of the spire. Set up the height of the spire BW (No. 2.) on the intersection (No. 3.) at B; also the height of the base line KL at R, and draw the lines BV and RV; the first, cutting the visual line through the centre of the tower in the point O, determines the height of the spire; the other, cutting the tower in the point u, determines its base. Through the point u draw a line round the lower, and find the points of the octagon in the middle of each face of the tower, to which let lines be drawn from the top O, and the whole will be completed, as shown in the example.

2438. Thus have we gone through the process of finding the representation of rather a complicated object with as little confusion of lines as possible; but one thing succeeding another, and each being required to remain for the student's observance, the whole unavoidably becomes intricate. Indeed, it is not now so perfectly executed but that

something remains for the student to complete, which must result from his own study or occupy more space than all we have already written on it. We allude to the intersections that take place at the lodgment of the spire on the top of the tower, to elucidate which it is drawn to a larger scale at No. 4., the mere inspection whereof will convey a full and, we hope, satisfactory idea of what we advert to. The student has been left to complete the base of the octagon, a process so simple that we cannot, if he retain what he has read, believe he will find difficulty in accomplishing, either by visual rays or otherwise. It is next to an impossibility to describe intricate matters like these so as to leave nothing for the exercise of the reader's judgment; for, however copious the instruction, there will always remain sufficient unexplained to keep his mind in action, and afford him the opportunity of exercising his own ingenuity.

2439. Example V. In fig. 833. the objects X and Y are plans of columns with bases

and capitals, whose general forms are shown at X and Y (No. 1.). YZ, as before, is the plane of the picture, S the station point. The picture, as previously, is prepared with the vanishing points VZ, and the ground line GL. 00 is the central line of the picture, and BA, BA are, it will be seen, lines of height.

2440. In the squares X and Y the dotted lines show the diagonals and boundaries of squares inscribed in the circles, by which so many more lines are gained for obtaining the curves which the circles form in the perspective representations. The visual rays are drawn as in the preceding examples, and transferred to the picture, the process being, in fact, nothing more than making squares following the profiles, which, at the different heights, guide the formation of circles within and around them, of which the upper ones only, for preventing confusion, are shown in the perspective representation. In each series, the extreme width of the appearance of the circle may be obtained by visual rays, as at b, b, b.

2441. At Z and z (Nos. 3. and 2.) are the plan and elevation of an arcade, from which it will be seen that the principle of inscribing squares and diagonals is equally applicable to the vertical representation of circles. Presuming that we have sufficiently described the diagram to enable the student to proceed in drawing the examples at large, we shall now submit an example of general application.

2442. Example VI. In fig. 834. YZ is the plane of delineation, and the plan of the building, with its projections, roof, and chimneys, is shown in No. 1. In practice, this is generally made on a separate drawing board, to enable the draughtsman to make his perspective