Evidently, the farther from the point _T_ we place the glass, making _ST_ longer, the larger will be the image; and the nearer we place it to _T_, the smaller the image, and that in a fixed ratio. Let the distance _DT_ be the direct distance from the Station-point to the foot of the object. Then, if we place the glass _AB_ at one-third of that whole distance, _P′Q′_ will be one-third of the real height of the object; if we place the glass at two-thirds of the distance, as at _EF_, _P″Q″_ (the height of the image at that point) will be two-thirds the height[5] of the object, and so on. Therefore the mathematical law is that _P′Q′_ will be to _PQ_ as _ST_ to _DT_. I put this ratio clearly by itself that you may remember it: _P′Q′_ ∶ _PQ_ ∷ _ST_ ∶ _DT_ or in words: _P_ dash _Q_ dash is to _PQ_ as _ST_ to _DT_ In which formula, recollect that _P′Q′_ is the height of the appearance of the object on the picture; _PQ_ the height of the object itself; _S_ the Sight-point; _T_ the Station-point; _D_ a point at the direct distance of the object; though the object is seldom placed actually on the line _TS_ produced, and may be far to the right or left of it, the formula is still the same. For let _S_, Fig. 3., be the Sight-point, and _AB_ the glass—here seen looking _down_ on its _upper edge_, not sideways;—then if the tower (represented now, as on a map, by the dark square), instead of being at _D_ on the line _ST_ produced, be at _E_, to the right (or left) of the spectator, still the apparent height of the tower on _AB_ will be as _S′T_ to _ET_, which is the same ratio as that of _ST_ to _DT_. [Illustration: Fig. 3.] Now in many perspective problems, the position of an object is more conveniently expressed by the two measurements _DT_ and _DE_, than by the single oblique measurement _ET_. I shall call _DT_ the “direct distance” of the object at _E_, and _DE_ its “lateral distance.” It is rather a license to call _DT_ its “direct” distance, for _ET_ is the more direct of the two; but there is no other term which would not cause confusion. Lastly, in order to complete our knowledge of the position of an object, the vertical height of some point in it, above or below the eye, must be given; that is to say, either _DP_ or _DQ_ in Fig. 2.[6]: this I shall call the “vertical distance” of the point given. In all perspective problems these three distances, and the dimensions of the object, must be stated, otherwise the problem is imperfectly given. It ought not to be required of us merely to draw _a_ room or _a_ church in perspective; but to draw _this_ room from _this_ corner, and _that_ church on _that_ spot, in perspective. For want of knowing how to base their drawings on the measurement and place of the object, I have known practiced students represent a parish church, certainly in true perspective, but with a nave about two miles and a half long.