[Geometric diagram]
Fig. 28.
Let A B, Fig. 28., be the curve.
Inclose it in a rectangle, C D E F.
Fix the position of the point C or D, and draw the rectangle. (Problem VIII. Coroll. I.)[Footnote 20]
Let C D E F, Fig. 29., be the rectangle so drawn.
[Geometric diagram]
Fig. 29.
If an extremity of the curve, as A, is in a side of the rectangle, divide the side C E, Fig. 29., so that A C shall be (in perspective ratio) to A E as A C is to A E in Fig. 28. (Prob. V. Cor. II.)
Similarly determine the points of contact of the curve and rectangle e, f, g.
If an extremity of the curve, as B, is not in a side of the rectangle, let [p39] fall the perpendiculars B a, B b on the rectangle sides. Determine the correspondent points a and b in Fig. 29., as you have already determined A, B, e, and f.
From b, Fig. 29., draw b B parallel to C D,[Footnote 21] and from a draw a B to the vanishing-point of D F, cutting each other in B. Then B is the extremity of the curve.
Determine any other important point in the curve, as P, in the same way, by letting fall P q and P r on the rectangle’s sides.
Any number of points in the curve may be thus determined, and the curve drawn through the series; in most cases, three or four will be enough. Practically, complicated curves may be better drawn in perspective by an experienced eye than by rule, as the fixing of the various points in haste involves too many chances of error; but it is well to draw a good many by rule first, in order to give the eye its experience.[Footnote 22]
If the curve required be a circle, Fig. 30., the rectangle which incloses it will become a square, and the curve will have four points of contact, A B C D, in the middle of the sides of the square.
[Geometric diagram]
Fig. 30.
Draw the square, and as a square may be drawn about a circle in any position, draw it with its nearest side, E G, parallel to the sight-line.
Let E F, Fig. 31., be the square so drawn.
[p40]
Draw its diagonals E F, G H; and through the center of the
square (determined by their intersection) draw A B to the
vanishing-point of G F, and C D parallel to E G. Then the
points A B C D are the four points of the circle’s contact.
[Geometric diagram]
Fig. 31.
On E G describe a half square, E L; draw the semicircle K A L; and from its center, R, the diagonals R E, R G, cutting the circle in x, y.
From the points x y, where the circle cuts the diagonals, raise perpendiculars, P x, Q y, to E G.
From P and Q draw P P′, Q Q′, to the vanishing-point of G F, cutting the diagonals in m, n, and o, p.
Then m, n, o, p are four other points in the circle.
Through these eight points the circle may be drawn by the hand accurately enough for general purposes; but any number of points required may, of course, be determined, as in Problem XI.
The distance E P is approximately one-seventh of E G, and may be assumed to be so in quick practice, as the error involved is not greater than would be incurred in the hasty operation of drawing the circle and diagonals.
It may frequently happen that, in consequence of associated [p41] constructions, it may be inconvenient to draw E G parallel to the sight-line, the square being perhaps first constructed in some oblique direction. In such cases, Q G and E P must be determined in perspective ratio by the dividing-point, the line E G being used as a measuring-line.
[Obs. In drawing Fig. 31. the station-point has been taken much nearer the paper than is usually advisable, in order to show the character of the curve in a very distinct form.
If the student turns the book so that E G may be vertical, Fig. 31. will represent the construction for drawing a circle in a vertical plane, the sight-line being then of course parallel to G L; and the semicircles A D B, A C B, on each side of the diameter A B, will represent ordinary semicircular arches seen in perspective. In that case, if the book be held so that the line E H is the top of the square, the upper semicircle will represent a semicircular arch, above the eye, drawn in perspective. But if the book be held so that the line G F is the top of the square, the upper semicircle will represent a semicircular arch, below the eye, drawn in perspective.
If the book be turned upside down, the figure will represent a circle drawn on the ceiling, or any other horizontal plane above the eye; and the construction is, of course, accurate in every case.]
[Footnote 20: Or if the curve is in a vertical plane, Coroll. to Problem IX. As a rectangle may be drawn in any position round any given curve, its position with respect to the curve will in either case be regulated by convenience. See the Exercises on this Problem, in the Appendix, p. 85.] Return to text
[Footnote 21: Or to its vanishing-point, if C D has one.] Return to text
[Footnote 22: Of course, by dividing the original rectangle into any number of equal rectangles, and dividing the perspective rectangle similarly, the curve may be approximately drawn without any trouble; but, when accuracy is required, the points should be fixed, as in the problem.] Return to text
Let A B, Fig. 32., be the circle drawn in perspective. It is required to divide it into a given number of equal parts; in this case, 20.
Let K A L be the semicircle used in the construction. Divide the semicircle K A L into half the number of parts required; in this case, 10.
Produce the line E G laterally, as far as may be necessary.
From O, the center of the semicircle K A L, draw radii through the points of division of the semicircle, p, q, r, etc., and produce them to cut the line E G in P, Q, R, etc.
From the points P Q R draw the lines P P′, Q Q′, R R′, etc., through the center of the circle A B, each cutting the circle in two points of its circumference.
Then these points divide the perspective circle as required.
If from each of the points p, q, r, a vertical were raised to the line E G, as in Fig. 31., and from the point where it cut E G a line were drawn to the vanishing-point, as Q Q′ in Fig. 31., this line would also determine two of the points of division.
[p43]
[Geometric diagram][Geometric diagram]
Fig. 32.
If it is required to divide a circle into any number of given unequal parts (as in the points A, B, and C, Fig. 33.), the shortest way is thus to raise vertical lines from A and B to the side of the perspective square X Y, and then draw to the vanishing-point, cutting the perspective circle in a and b, the points required. Only notice that if any point, as A, is on the nearer side of the circle A B C, its representative point, a, must be on the nearer side of the circle a b c; and if the point B is on the farther side of the circle A B C, b must be [p44] on the farther side of a b c. If any point, as C, is so much in the lateral arc of the circle as not to be easily determinable by the vertical line, draw the horizontal C P, find the correspondent p in the side of the perspective square, and draw p c parallel to X Y, cutting the perspective circle in c.
[Geometric diagram]
Fig. 33.
It is obvious that if the points P′, Q′, R, etc., by which the circle is divided in Fig. 32., be joined by right lines, the resulting figure will be a regular equilateral figure of twenty sides inscribed in the circle. And if the circle be divided into given unequal parts, and the points of division joined by right lines, the resulting figure will be an irregular polygon inscribed in the circle with sides of given length.
Thus any polygon, regular or irregular, inscribed in a circle, may be inscribed in position in a perspective circle.
[Geometric diagram]
Fig. 34.
Let A B, Fig. 34., be the sight-magnitude of the side of the smaller square, and A C that of the side of the larger square.
Draw the larger square. Let D E F G be the square so drawn.
Join E G and D F.
On either D E or D G set off, in perspective ratio, D H equal to one half of B C. Through H draw H K to the vanishing-point of D E, cutting D F in I and E G in K. Through I and K draw I M, K L, to vanishing-point of D G, cutting D F in L and E G in M. Join L M.
Then I K L M is the smaller square, inscribed as required.[Footnote 23]
[Geometric diagram]
Fig. 36.
If, instead of one square within another, it be required to draw one circle within another, the dimensions of both being given, inclose each circle in a square. Draw the squares first, and then the circles within, as in Fig. 36.
[Geometric diagram]
Fig. 35.
[Footnote 23: If either of the sides of the greater square is parallel to the plane of the picture, as D G in Fig. 35., D G of course must be equal to A C, and D H equal to B C/2, and the construction is as in Fig. 35.] Return to text
Let A B C D, Fig. 37., be the portion of the cone required.
[Geometric diagram]
Fig. 37.
As it is given in magnitude, its diameters must be given at the base and summit, A B and C D; and its vertical height, C E.[Footnote 24]
And as it is given in position, the center of its base must be given.
[Geometric diagram]
Fig. 38.
Draw in position, about this center,[Footnote 25] the square pillar [p48] a f d, Fig. 38., making its height, b g, equal to C E; and its side, a b, equal to A B.
In the square of its base, a b c d, inscribe a circle, which therefore is of the diameter of the base of the cone, A B.
In the square of its top, e f g h, inscribe concentrically a circle whose diameter shall equal C D. (Coroll. Prob. XIII.)
Join the extremities of the circles by the right lines k l, n m. Then k l n m is the portion of cone required.
If similar polygons be inscribed in similar positions in the circles k n and l m (Coroll. Prob. XII.), and the corresponding angles of the polygons joined by right lines, the resulting figure will be a portion of a polygonal pyramid. (The dotted lines in Fig. 38., connecting the extremities of two diameters and one diagonal in the respective circles, occupy the position of the three nearest angles of a regular octagonal pyramid, having its angles set on the diagonals and diameters of the square a d, inclosing its base.)
If the cone or polygonal pyramid is not truncated, its apex will be the center of the upper square, as in Fig. 26.
If equal circles, or equal and similar polygons, be inscribed in the upper and lower squares in Fig. 38., the resulting figure will be a vertical cylinder, or a vertical polygonal pillar, of given height and diameter, drawn in position. [p49]
If the circles in Fig. 38., instead of being inscribed in the squares b c and f g, be inscribed in the sides of the solid figure b e and d f, those sides being made square, and the line b d of any given length, the resulting figure will be, according to the constructions employed, a cone, polygonal pyramid, cylinder, or polygonal pillar, drawn in position about a horizontal axis parallel to b d.
Similarly, if the circles are drawn in the sides g d and e c, the resulting figures will be described about a horizontal axis parallel to a b.
[Footnote 24: Or if the length of its side, A C, is given instead, take a e, Fig. 37., equal to half the excess of A B over C D; from the point e raise the perpendicular c e. With center a, and distance A C, describe a circle cutting c e in c. Then c e is the vertical height of the portion of cone required, or C E.] Return to text
[Footnote 25: The direction of the side of the square will of course be regulated by convenience.] Return to text
We have hitherto been examining the conditions of horizontal and vertical lines only, or of curves inclosed in rectangles.
[Geometric diagram]
Fig. 39.
[Geometric diagram]
Fig. 40.
We must, in conclusion, investigate the perspective of inclined lines, beginning with a single one given in position. For the sake of completeness of system, I give in Appendix II. Article III. the development of this problem from the second. But, in practice, the position of an inclined line may be most conveniently defined by considering it as the diagonal of a rectangle, as A B in Fig. 39., and I shall therefore, though at some sacrifice of system, examine it here under that condition.
If the sides of the rectangle A C and A D are given, the slope of the line A B is determined; and then its position will depend on that of the rectangle. If, as in Fig. 39., the rectangle is parallel to the picture plane, the line A B must be so also. If, as in Fig. 40., the rectangle is inclined to the [p51] picture plane, the line A B will be so also. So that, to fix the position of A B, the line A C must be given in position and magnitude, and the height A D.
[Geometric diagram]
Fig. 41.
If these are given, and it is only required to draw the single line A B in perspective, the construction is entirely simple; thus:—
Draw the line A C by Problem I.
Let A C, Fig. 41., be the line so drawn. From a and c raise the vertical lines a d, c b. Make a d equal to the sight-magnitude of A D. From d draw d b to the vanishing-point of a c, cutting b c in b.
Join a b. Then a b is the inclined line required.
[Geometric diagram]
Fig. 42.
If the line is inclined in the opposite direction, as D C in Fig. 42., we have only to join d c instead of a b in Fig. 41., and d c will be the line required.
I shall hereafter call the line A C, when used to define the position of an inclined line A B (Fig. 40.), the “relative horizontal” of the line A B.
[Geometric diagram]
Fig. 43.
In general, inclined lines are most needed for gable roofs, in which, when the conditions are properly stated, the vertical height of the gable, X Y, Fig. 43., is given, and the base line, A C, in position. When these are given, draw A C; raise vertical A D; make A D equal to sight-magnitude of X Y; complete the perspective-rectangle A D B C; join A B and D C (as by dotted lines in figure); and through the intersection of the dotted lines draw vertical X Y, cutting D B in Y. Join A Y, C Y; and these lines are the sides of the gable. If [p52] the length of the roof A A′ is also given, draw in perspective the complete parallelopiped A′ D′ B C, and from Y draw Y Y′ to the vanishing-point of A A′, cutting D′ B′ in Y′. Join A′ Y, and you have the slope of the farther side of the roof.
[Geometric diagram]
Fig. 44.
The construction above the eye is as in Fig. 44.; the roof is reversed in direction merely to familiarize the student with the different aspects of its lines.
If, in Fig. 43. or Fig. 44., the lines A Y and A′ Y′ be produced, the student will find that they meet.
Let P, Fig. 45., be the point at which they meet.
From P let fall the vertical P V on the sight-line, cutting the sight-line in V.
Then the student will find experimentally that V is the vanishing-point of the line A C.[Footnote 26]
Complete the rectangle of the base A C′, by drawing A′ C′ to V, and C C′ to the vanishing-point of A A′.
Join Y′ C′.
Now if Y C and Y′ C′ be produced downwards, the student will find that they meet.
Let them be produced, and meet in P′.
Produce P V, and it will be found to pass through the point P′.
Therefore if A Y (or C Y), Fig. 45., be any inclined line drawn in perspective by Problem XV., and A C the relative horizontal (A C in Figs. 39, 40.), also drawn in perspective.
Through V, the vanishing-point of A V, draw the vertical P P′ upwards and downwards.
Produce A Y (or C Y), cutting P P′ in P (or P′).
Then P is the vanishing-point of A Y (or P′ of C Y).
[Geometric diagram]
Fig. 45.
The student will observe that, in order to find the point P by this method, it is necessary first to draw a portion of the given inclined line by Problem XV. Practically, it is always necessary to do so, and, therefore, I give the problem in this form.
[p54]
Theoretically, as will be shown in the analysis of the problem,
the point P should be found by drawing a line from the
station-point parallel to the given inclined line: but there is
no practical means of drawing such a line; so that in whatever
terms the problem may be given, a portion of the inclined
line (A Y or C Y) must always be drawn in perspective
before P can be found.
[Footnote 26: The demonstration is in Appendix II. Article III.] Return to text
[Geometric diagram]
Fig. 46.
Let P, Fig. 46., be the vanishing-point of the inclined line, and V the vanishing-point of the relative horizontal.
Find the dividing-points of the relative horizontal, D and D′.
Through P draw the horizontal line X Y.
With center P and distance D P describe the two arcs D X and D′ Y, cutting the line X Y in X and Y.
Then X and Y are the dividing-points of the inclined line.[Footnote 27]
Obs. The dividing-points found by the above rule, used with the ordinary measuring-line, will lay off distances on the retiring inclined line, as the ordinary dividing-points lay them off on the retiring horizontal line.
Another dividing-point, peculiar in its application, is sometimes useful, and is to be found as follows:—
[p56]
[Geometric diagram]
Fig. 47.
Let A B, Fig. 47., be the given inclined line drawn in perspective, and A c the relative horizontal.
Find the vanishing-points, V and E, of A c and A B; D, the dividing-point of A c; and the sight-magnitude of A c on the measuring-line, or A C.
From D erect the perpendicular D F.
Join C B, and produce it to cut D E in F. Join E F.
Then, by similar triangles, D F is equal to E V, and E F is parallel to D V.
Hence it follows that if from D, the dividing-point of A c, we raise a perpendicular and make D F equal to E V, a line C F, drawn from any point C on the measuring-line to F, will mark the distance A B on the inclined line, A B being the portion of the given inclined line which forms the diagonal of the vertical rectangle of which A C is the base.
[Footnote 27: The demonstration is in Appendix II., p. 104.] Return to text
As in order to fix the position of a line two points in it must be given, so in order to fix the position of a plane, two lines in it must be given.
[Geometric diagram]
Fig. 48.
Let the two lines be A B and C D, Fig. 48.
[p58]
As they are given in position, the relative horizontals A E
and C F must be given.
Then by Problem XVI. the vanishing-point of A B is V, and of C D, V′.
Join V V′ and produce it to cut the sight-line in X.
Then V X is the sight-line of the inclined plane.
Like the horizontal sight-line, it is of indefinite length; and may be produced in either direction as occasion requires, crossing the horizontal line of sight, if the plane continues downward in that direction.
X is the vanishing-point of all horizontal lines in the inclined plane.
[Footnote 28: Read the Article on this problem in the Appendix, p. 97, before investigating the problem itself.] Return to text
[Geometric diagram]
Fig. 49.
Let V X, Fig. 49., be the given sight-line.
Produce it to cut the horizontal sight-line in X.
Therefore X is the vanishing-point of horizontal lines in the given inclined plane. (Problem XVIII.)
Join T X, and draw T Y at right angles to T X.
Therefore Y is the rectangular vanishing-point corresponding to X.[Footnote 29]
From Y erect the vertical Y P, cutting the sight-line of the inclined plane in P.
[p60]
Then P is the vanishing-point of steepest lines in the plane.
All lines drawn to it, as Q P, R P, N P, etc., are the steepest possible in the plane; and all lines drawn to X, as Q X, O X, etc., are horizontal, and at right angles to the lines P Q, P R, etc.
[Footnote 29: That is to say, the vanishing-point of horizontal lines drawn at right angles to the lines whose vanishing-point is X.] Return to text