difference in any respect is, that the trace of the plane itself will be parallel to the ground-line instead of coincident with it.

SCHOLIUM. Moreover, there is nothing in the reasoning upon which this process is founded to confine the plane which contains the figure to be a horizontal one, or the picture to be perpendicular to the horizon. It is hence universally applicable, whatever be the inclination of the object-plane to the picture-plane.

(c) Let the plane of the figure be vertical, but not parallel to the picture.


Let the ground-plane, etc., be as before, as far as the parts are common; and let the traces of the rertical plane TU upon the ground and picture-planes be YU and UZ. Draw EV parallel to UY; and then VW parallel to UZ. This will be the vanishing line of the plane TU.

Let ABCD be any figure (a parallelogram is taken for example) in the plane TU. Find its perspective precisely as in the last case ; for which purpose the lines concerned are given, the student to fill up the description.

The part in the actual construction in the drawing, too, is so completely a repetition of the last case, that it is left for the student to execute in detail.

(d) For sloping planes, such as roofs, etc., the process is exactly the same.

SCHOLIUM. It has been no part of the plan of this section to furnish examples; or even to intimate what simplifications can be made under any special circumstances of a piece of architecture or of a system of machinery. These are often numerous and beautiful ; but they will be furnished by the drawing-master, whilst our business is confined to the subject simply as one of pure geometry.


PROPOSITION I. Assuming that the centre of the picture is in the middle of the hori

zontal line, to find the distance of the eye, there being some one regular building represented in the drawing. Let abcd be the base, and a'b'c'd' any horizontal course of the build



ing (as the eaves, the tops of windows, etc.); produce ad, ab and a'd', a'b' to meet in V and w. Then these are the vanishing points situated in the horizontal line. The horizontal line hence becomes known, and consequently its middle point 0.

On VW describe a semicircle, and draw OE perpendicular to VW meeting it in E. Then EO is the distance of the eye for which the perspective was formed.

This will be seen at once, if we suppose the semicircle and its lines to be revolved into a position at right angles to the picture.

ScHolium. If there be two independent buildings in the picture, it may be tested whether the centre be taken truly in the middle of HL.

For we have only to perform a similar operation upon the second figure : and the intersection of the semicircles will give E, and hence also 0.

It will be a good exercise for the student to determine the effects of taking a wrong point from which to view the picture. For instance:

(1.) If the eye be taken so as still to retain the same things, except the point O, then the figures abcd, a'b'c'd', etc., will still be the correct representations of parallelograms in the same planes; but they cease to be rectangular ones, and no two positions of the eye would cause them to represent the same parallelogram.

(2.) The same indeed is generally true of positions of the eye anywhere in the plane through E and HL: the exception being a certain circle in that plane, which it is left for the student to find; and even this is an exception which can be hardly considered absolute.

(3.) If the eye be taken without the plane of E and HL, the spective can in no case suggest abcd to be a parallelogram.


To restore the original figure in its proper position. Produce the sides of abcd to meet the bottom of the picture, viz,



as in the figure in X, Y, Z, U. Through X, Y draw lines parallel to EW; and through Z, U lines parallel to EV. These by their intersection give the figure ABCD, which is the original of abcd.

[This, it is obvious, is but a reversal of the process of finding the perspective abcd of the original ground plane ABCD.]

Again, if a'b'd'd' be the upper surface of the prism on ABCD (original), produce Wd to meet a perpendicular at X in R. Then XR is the height of the original prism, to the same scale as before.

Also, the height upon XR, which is represented by any part dg of the perspective dd', will be denoted on the same scale by XG. And similarly with respect to lines lying in any other direction as well as the vertical.


Two lines are given in perspective, to find whether they are in one


Join their traces, and likewise their vanishing points. If the two lines so drawn be parallel, the original lines are in one plane: but if not, they are not.


Given the perspective of a plane and likewise that of a line, to find

that of the point in which they intersect.

Let ABVW be the perspective of the plane, and CZ of the line. Now every plane drawn through CZ to meet ABVW will pass through the point of intersection. Let then CDZK and CMZN be any two such planes put in perspective; then the point sought is in the inter, section DK and MN of these planes with ABVW; and hence at P their intersection.

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PROPOSITION V. To assign the relations that must subsist between the perspectives of a line and plane, that the originals may be perpendicular to each other.

All lines perpendicular to a plane are parallel to one another, and hence have the same vanishing point; that is, the vanishing point of any one of them is the vanishing point of all. In the same manner the vanishing line of any one of a set of parallel planes is the vanishing line of all. It is hence immaterial which of the lines and which of the planes, having each its own proper direction, we take for the determination of the vanishing point and line; and hence we are at liberty to take those which offer the greatest facilities for construction. These will readily appear to be the line and plane which, when produced, pass through the eye itself: the trace of each then coinciding with the vanishing point and vanishing line respectively. The relation between these (the vanishing point and vanishing line) will hence be that which expresses the condition sought.

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In the left-hand figure suppose PR the vanishing line of the plane, E the eye, and OE the perpendicular from E to the plane. Draw OA perpendicular to PR; join EA; produce A0; and in the plane EOA draw EB perpendicular to EA, meeting OA in B.

Then, since EO is perpendicular to PQ, and OA perpendicular to a line PR in it, the line AE is perpendicular to PR; and hence, the plane AOE is perpendicular to PR, and hence, again, to every plane passing through PR (Pls. 11. 13, 16), as to EPR. The line EB in the plane AOE perpendicular to AE in the plane PER is that which gives the vanishing point of the lines perpendicular to PER. Wherefore we have the following construction :

In the right-hand figure, let PR be the vanishing line of the plane, O the centre of the picture, and B the vanishing point of the

line perpendicular to the plane. Join BO and produce it to meet PR in A; draw OE perpendicular to BA, and make it equal to the distance of the picture; and join AE, BE.

Now if BĂ be perpendicular to PR and at the same time BE to AE, the given line and plane are at right angles : if not, not.

If BA be perpendicular to PR, but BEA not a right angle, still a position (the centre of the picture remaining the same) can be found for the eye, which will render PR and B the vanishing line and point of a plane and line perpendicular to each other.

But if BA be not perpendicular to PR, then the condition cannot be fulfilled without changing both the centre of the picture and place of

SCHOLIUM 1. It is unnecessary to give the details of the solution of the problem :-Given the eye, and either the vanishing point or vanishing line to find the other.

SCHOLIUM 2. It is obvious that the traces are the only things left arbitrary with respect to the line or to the plane concerned ; and that these, when once fixed upon, identify the particular line and plane under specific consideration. The trace of the line may be anywhere in the picture; and likewise that of the plane, subjected only to the condition of being parallel to the given vanishing line.

the eye.

PROPOSITION VI. Through a given point (in a given line) to draw a parallel to another given line. *

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Let A be the trace, and B tlie vanishing point of the given line in which the given point whose perspective is G is situated; it is required to draw a line through G parallel to the given line whose trace and vanishing point are Cand D respectively.

Join DG, DB, and throngh C draw CF
parallel to DB to meet DG in F. Then
F is the trace, and D the vanishing point
of the line sought.

For, since the lines CD and GD have
the same vanishing point D, they therefore represent parallel lines.

* The reader is requested to give attention to the condition included in the parenthesis. A point is not given by its perspective on the picture-plane alone : for that perspective may represent any point whatever in the line which passes through that point and the eye. Another condition is therefore necessary for fixing the point as to actual position; and the most usual, as well as the most convenient one is, that it shall be situated in a given line; that is, in a line whose trace and vanishing point are either given or can be found from the data.

The same holds good with respect to any segment of the perspective of a line. That perspective may represent innumerable different lines ; and it becomes in the same manner necessary that it shall be given in a given plane ; that is, in one whose trace and vauishing line are either given or can be found.

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