It is often possible to shorten other perspective operations considerably, by finding the vanishing-points of the inclined lines of the object. Thus, in drawing the gabled roof in Fig. 43., if the gable A Y C be drawn in perspective, and the vanishing-point of A Y determined, it is not necessary to draw the two sides of the rectangle, A′ D′ and D′ B′, in order to determine the point Y′; but merely to draw Y Y′ to the vanishing-point of A A′ and A′ Y′ to the vanishing-point of A Y, meeting in Y′, the point required.
Again, if there be a series of gables, or other figures produced by parallel inclined lines, and retiring to the point V, as in Fig. 72.,[Footnote 34] it is not necessary to draw each separately, but merely to determine their breadths on the line A V, and draw the slopes of each to their vanishing-points, as shown in Fig. 72. Or if the gables are equal in height, and a line be drawn from Y to V, the construction resolves itself into a zigzag drawn alternately to P and Q, between the lines Y V and A V.
The student must be very cautious, in finding the vanishing-points of inclined lines, to notice their relations to the horizontals beneath them, else he may easily mistake the horizontal to which they belong.
Thus, let A B C D, Fig. 73., be a rectangular inclined plane, and let it be required to find the vanishing-point of its diagonal B D.
Find V, the vanishing-point of A D and B C.
Draw A E to the opposite vanishing-point, so that D A E may represent a right angle.
Let fall from B the vertical B E, cutting A E in E.
Join E D, and produce it to cut the sight-line in V′.
[p95]
[Geometric diagram]
Fig. 72.
[p96]
Then, since the point E is vertically under the point B, the
horizontal line E D is vertically under the inclined line B D.
[Geometric diagram]
Fig. 73.
So that if we now let fall the vertical V′ P from V′, and produce B D to cut V′ P in P, the point P will be the vanishing-point of B D, and of all lines parallel to it.[Footnote 35]
[Footnote 34: The diagram is inaccurately cut. Y V should be a right line.] Return to text
[Footnote 35: The student may perhaps understand this construction better by completing the rectangle A D F E, drawing D F to the vanishing-point of A E, and E F to V. The whole figure, B F, may then be conceived as representing half the gable roof of a house, A F the rectangle of its base, and A C the rectangle of its sloping side.
In nearly all picturesque buildings, especially on the Continent, the slopes of gables are much varied (frequently unequal on the two sides), and the vanishing-points of their inclined lines become very important, if accuracy is required in the intersections of tiling, sides of dormer windows, etc.
Obviously, also, irregular triangles and polygons in vertical planes may be more easily constructed by finding the vanishing-points of their sides, than by the construction given in the corollary to Problem IX.; and if such triangles or polygons have others concentrically inscribed within them, as often in Byzantine mosaics, etc., the use of the vanishing-points will become essential.] Return to text
Before examining the last three problems it is necessary that you should understand accurately what is meant by the position of an inclined plane.
Cut a piece of strong white pasteboard into any irregular shape, and dip it in a sloped position into water. However you hold it, the edge of the water, of course, will always draw a horizontal line across its surface. The direction of this horizontal line is the direction of the inclined plane. (In beds of rock geologists call it their “strike.”)
[Geometric diagram]
Fig. 74.
Next, draw a semicircle on the piece of pasteboard; draw its diameter, A B, Fig. 74., and a vertical line from its center, C D; and draw some other lines, C E, C F, etc., from the center to any points in the circumference.
Now dip the piece of pasteboard again into water, and, holding it at any inclination and in any direction you choose, bring the surface of the water to the line A B. Then the line C D will be the most steeply inclined of all the lines drawn to the circumference of the circle; G C and H C will be less steep; and E C and F C less steep still. The nearer the lines to C D, the steeper they will be; and the nearer to A B, the more nearly horizontal.
[p98]
When, therefore, the line A B is horizontal (or marks the
water surface), its direction is the direction of the inclined
plane, and the inclination of the line D C is the inclination
of the inclined plane. In beds of rock geologists call the
inclination of the line D C their “dip.”
To fix the position of an inclined plane, therefore, is to determine the direction of any two lines in the plane, A B and C D, of which one shall be horizontal and the other at right angles to it. Then any lines drawn in the inclined plane, parallel to A B, will be horizontal; and lines drawn parallel to C D will be as steep as C D, and are spoken of in the text as the “steepest lines” in the plane.
But farther, whatever the direction of a plane may be, if it be extended indefinitely, it will be terminated, to the eye of the observer, by a boundary line, which, in a horizontal plane, is horizontal (coinciding nearly with the visible horizon);—in a vertical plane, is vertical;—and, in an inclined plane, is inclined.
This line is properly, in each case, called the “sight-line” of such plane; but it is only properly called the “horizon” in the case of a horizontal plane: and I have preferred using always the term “sight-line,” not only because more comprehensive, but more accurate; for though the curvature of the earth’s surface is so slight that practically its visible limit always coincides with the sight-line of a horizontal plane, it does not mathematically coincide with it, and the two lines ought not to be considered as theoretically identical, though they are so in practice.
It is evident that all vanishing-points of lines in any plane must be found on its sight-line, and, therefore, that the sight-line of any plane may be found by joining any two of such vanishing-points. Hence the construction of Problem XVIII.
In Fig. 8. omit the lines C D, C′ D′, and D S; and, as here in Fig. 75., from a draw a d parallel to A B, cutting B T in d; and from d draw d e parallel to B C′.
[Geometric diagram]
Fig. 75.
Now as a d is parallel to A B—
[p100]
Now because the triangles a c V, b c′ V, are similar—
and because the triangles d e T, b c′ T are similar—
d e ∶ b c′ ∷ d T ∶ b T.[eqn v]But a c is equal to d e—
∴ the two triangles a b d, b T V, are similar, and their angles are alternate;
∴ T V is parallel to a d.But a d is parallel to A B—
In Fig. 13., since a R is by construction parallel to A B in Fig. 12., and T V is by construction in Problem III. also parallel to A B—
Again, by the construction of Fig. 13., a R′ is parallel to M V—
And it has just been shown that also
But by construction, a R′ = a R—
We proceed to take up the general condition of the second problem, before left unexamined, namely, that in which the vertical distances B C′ and A C (Fig. 6. page 13), as well as the direct distances T D and T D′ are unequal.
In Fig. 6., here repeated (Fig. 76.), produce C′ B downwards, and make C′ E equal to C A.
[Geometric diagram]
Fig. 76.
Join A E.
Then, by the second Corollary of Problem II., A E is a horizontal line.
Draw T V parallel to A E, cutting the sight-line in V.
[p103]
Complete the constructions of Problem II. and its second
Corollary.
Then by Problem II. a b is the line A B drawn in perspective; and by its Corollary a e is the line A E drawn in perspective.
From V erect perpendicular V P, and produce a b to cut it in P.
Join T P, and from e draw e f parallel to A E, and cutting A T in f.
Now in triangles E B T and A E T, as e b is parallel to E B and e f to A E;—e b ∶ e f ∷ E B ∶ A E[eqn x].
But T V is also parallel to A E and P V to e b.
Therefore also in the triangles a P V and a V T,
Therefore P V ∶ V T ∷ E B ∶ A E[eqn xii].
And, by construction, angle T P V = ∠ A E B.
Therefore the triangles T V P, A E B, are similar; and T P is parallel to A B.
[p104]
Now the construction in this problem is entirely general
for any inclined line A B, and a horizontal line A E in the
same vertical plane with it.
So that if we find the vanishing-point of A E in V, and from V erect a vertical V P, and from T draw T P parallel to A B, cutting V P in P, P will be the vanishing-point of A B, and (by the same proof as that given at page 17) of all lines parallel to it.
[Geometric diagram]
Fig. 77.
Next, to find the dividing-point of the inclined line.
I remove some unnecessary lines from the last figure and repeat it here, Fig. 77., adding the measuring-line a M, that the student may observe its position with respect to the other lines before I remove any more of them.
Now if the line A B in this diagram represented the length of the line A B in reality (as A B does in Figs. 10. and 11.), we should only have to proceed to modify Corollary III. of Problem II. to this new construction. We shall see presently that A B does not represent the actual length of the inclined line A B in nature, nevertheless we shall first proceed as if it did, and modify our result afterwards.
[p105]
In Fig. 77. draw a d parallel to A B, cutting B T in d.
Therefore a d is the sight-magnitude of A B, as a R is of A B in Fig. 11.
[Geometric diagram]
Fig. 78.
Remove again from the figure all lines except P V, V T, P T, a b, a d, and the measuring-line.
Set off on the measuring-line a m equal to a d.
Draw P Q parallel to a m, and through b draw m Q, cutting P Q in Q.
Then, by the proof already given in page 20, P Q = P T.
Therefore if P is the vanishing-point of an inclined line A B, and Q P is a horizontal line drawn through it, make P Q equal to P T, and a m on the measuring-line equal to the sight-magnitude of the line A B in the diagram, and the line joining m Q will cut a P in b.
We have now, therefore, to consider what relation the length of the line A B in this diagram, Fig. 77., has to the length of the line A B in reality.
Now the line A E in Fig. 77. represents the length of A E in reality.
But the angle A E B, Fig. 77., and the corresponding angle in all the constructions of the earlier problems, is in reality a right angle, though in the diagram necessarily represented as obtuse.
[Geometric diagram]
Fig. 79.
Therefore, if from E we draw E C, as in Fig. 79., at right angles to A E, make E C = E B, and join A C, A C will be the real length of the line A B.
Now, therefore, if instead of a m in Fig. 78., we take the real length of A B, that real length will be to a m as A C to A B in Fig. 79.
And then, if the line drawn to the measuring-line P Q is still to cut a P in b, it is evident that the line P Q must be shortened in the same ratio that a m was shortened; and the true dividing-point will be Q′ in Fig. 80., fixed so that Q′ P shall be to Q P as a m′ is to a m; a m′ representing the real length of A B.
[p106]
But a m′ is therefore to a m as A C is to A B in Fig. 79.
Therefore P Q′ must be to P Q as A C is to A B.
But P Q equals P T (Fig. 78.); and P V is to V T (in Fig. 78.) as B E is to A E (Fig. 79.).
Hence we have only to substitute P V for E C, and V T for A E, in Fig. 79., and the resulting diagonal A C will be the required length of P Q′.
[Geometric diagram]
Fig. 80.
It will be seen that the construction given in the text (Fig. 46.) is the simplest means of obtaining this magnitude, for V D in Fig. 46. (or V M in Fig. 15.) = V T by construction in Problem IV. It should, however, be observed, that the distance P Q′ or P X, in Fig. 46., may be laid on the sight-line of the inclined plane itself, if the measuring-line be drawn parallel to that sight-line. And thus any form may be drawn on an inclined plane as conveniently as on a horizontal one, with the single exception of the radiation of the verticals, which have a vanishing-point, as shown in Problem XX.
THE END.
A handful of unequivocal typographical errors has been corrected.
For increased clarity, a few diagrams have been shifted from their original position in the text.
Images for sections of the text where the ∶ ratio and ∷ proportion symbols occur.