The Theory and Practice of Perspective Part 12

This is simply a question of putting two points into perspective, instead of one, or like doing the previous problem twice over, for the two points represent the two extremities of the line. Thus we have to find the perspective of _A_ and _B_, namely _ab_. Join those points, and we have the line required.

[Ill.u.s.tration: Fig. 109.]

[Ill.u.s.tration: Fig. 110.]

If one end touches the base, as at _A_ (Fig. 110), then we have but to find one point, namely _b_. We also find the perspective of the angle _mAB_, namely the shaded triangle mAb. Note also that the perspective triangle equals the geometrical triangle.

[Ill.u.s.tration: Fig. 111.]

When the line required is parallel to the base line of the picture, then the perspective of it is also parallel to that base (see Rule 3).

LIII

TO FIND THE LENGTH OF A GIVEN PERSPECTIVE LINE

A perspective line _AB_ being given, find its actual length and the angle at which it is placed.

This is simply the reverse of the previous problem. Let _AB_ be the given line. From distance _D_ through _A_ draw _DC_, and from _S_, point of sight, through _A_ draw _SO_. Drop _OP_ at right angles to base, making it equal to _OC_. Join _PB_, and line _PB_ is the actual length of _AB_.

This problem is useful in finding the position of any given line or point on the perspective plane.

[Ill.u.s.tration: Fig. 112.]

LIV

TO FIND THESE POINTS WHEN THE DISTANCE-POINT IS INACCESSIBLE

[Ill.u.s.tration: Fig. 113.]

If the distance-point is a long way out of the picture, then the same result can be obtained by using the half distance and half base, as already shown.

From _a_, half of _mP_, draw quadrant _ab_, from _b_ (half base), draw line from _b_ to half Dist., which intersects _Sm_ at _P_, precisely the same point as would be obtained by using the whole distance.

LV

HOW TO PUT A GIVEN TRIANGLE OR OTHER RECTILINEAL FIGURE INTO PERSPECTIVE

Here we simply put three points into perspective to obtain the given triangle _A_, or five points to obtain the five-sided figure at _B_.

So can we deal with any number of figures placed at any angle.

[Ill.u.s.tration: Fig. 114.]

Both the above figures are placed in the same diagram, showing how any number can be drawn by means of the same point of sight and the same point of distance, which makes them belong to the same picture.

It is to be noted that the figures appear reversed in the perspective.

That is, in the geometrical triangle the base at _ab_ is uppermost, whereas in the perspective _ab_ is lowermost, yet both are nearest to the ground line.

LVI

HOW TO PUT A GIVEN SQUARE INTO ANGULAR PERSPECTIVE

Let _ABCD_ (Fig. 115) be the given square on the geometrical plane, where we can place it as near or as far from the base and at any angle that we wish. We then proceed to find its perspective on the picture by finding the perspective of the four points _ABCD_ as already shown. Note that the two sides of the perspective square _dc_ and _ab_ being produced, meet at point _V_ on the horizon, which is their vanishing point, but to find the point on the horizon where sides _bc_ and _ad_ meet, we should have to go a long way to the left of the figure, which by this method is not necessary.

[Ill.u.s.tration: Fig. 115.]

LVII

OF MEASURING POINTS

We now have to find certain points by which to measure those vanishing or retreating lines which are no longer at right angles to the picture plane, as in parallel perspective, and have to be measured in a different way, and here geometry comes to our a.s.sistance.

[Ill.u.s.tration: Fig. 116.]

Note that the perspective square _P_ equals the geometrical square _K_, so that side _AB_ of the one equals side _ab_ of the other. With centre _A_ and radius _AB_ describe arc _Bm_ till it cuts the base line at _m_. Now _AB_ = _Am_, and if we join _bm_ then triangle _BAm_ is an isosceles triangle. So likewise if we join _mb_ in the perspective figure will mAb be the same isosceles triangle in perspective. Continue line _mb_ till it cuts the horizon in _m_, which point will be the measuring point for the vanishing line _AbV_. For if in an isosceles triangle we draw lines across it, parallel to its base from one side to the other, we divide both sides in exactly the same quant.i.ties and proportions, so that if we measure on the base line of the picture the s.p.a.ces we require, such as 1, 2, 3, on the length _Am_, and then from these divisions draw lines to the measuring point, these lines will intersect the vanishing line _AbV_ in the lengths and proportions required. To find a measuring point for the lines that go to the other vanishing point, we proceed in the same way. Of course great accuracy is necessary.

Note that the dotted lines 1,1, 2,2, &c., are parallel in the perspective, as in the geometrical figure. In the former the lines are drawn to the same point _m_ on the horizon.

LVIII

HOW TO DIVIDE ANY GIVEN STRAIGHT LINE INTO EQUAL OR PROPORTIONATE PARTS

[Ill.u.s.tration: Fig. 117.]

Let _AB_ (Fig. 117) be the given straight line that we wish to divide into five equal parts. Draw _AC_ at any convenient angle, and measure off five equal parts with the compa.s.ses thereon, as 1, 2, 3, 4, 5. From 5C draw line to 5B. Now from each division on _AC_ draw lines 4,4, 3,3, &c., parallel to 5,5. Then _AB_ will be divided into the required number of equal parts.