NEARLY EVERYBODY will remember having made the acquaintance of vanishing points in art classes. Normally you’d have worked with one vanishing point in the centre of your picture or with two of them being placed at its lateral borders in order to display objects from an oblique perspective. Thus positions were chosen quite arbitrarily, and hardly anyone—even hardly an art teacher—gave thought to how it would have been mathematically correct. But calculating positions of vanishing points, as is required for obtaining a perfect perspective, is quite easy. Simple knowledge of trigonometry (which one will probably also remember) shall be enough. As my projected watercolour painting “Tower Guards” is to become a downright vanishing point projection, I’d like to give a short explanation of how it works.

Trigonometric calculation of vanishing points is based on the following rationale: Imagine a duplicate of the object to be displayed behind the image layer—e. g. a cuboid rotated around several axes—being placed in front of that layer and set with its frontmost corner exactly at the viewer’s eye. (Oppress the pain in your thoughts.) Three of the cuboid’s edges now point into different directions from the eye to the image plane. The vanishing point for each edge is located at the very point where this edge’s extension meets the image plane^{1}—and be it far off the image’s borders^{2}.

One thing becomes clear thereby: Vanishing points can always be only set correctly for exactly one eye position in relation to the image plane, i. e. for a particular eye distance before a particular spot of the image. This shall be illustrated by the following consideration: Let’s imagine we’re watching a scenery commonly depicted by vanishing point projections—e. g. urban highrise architecture—through a glass plate we’re holding with outstretched arms before us. As it is a technically really awesome glass plate, we can press a button at its border, thus ‘freezing’ the view and recording it to the plate. (This can roughly be imagined as shown in this example here.) Instead of the real scenery we now merely see its image on the plate, and illusion remains perfect as long as we don’t move ourselves or the plate. As soon as we slant the plate or move it to or fro us, the image becomes more or less distorted.^{3} Perfect illusion just doesn’t exist but for a single eye position.

In order to calculate the vanishing points’ positions for a planned picture, we ﬁrst ponder on where the viewer’s eye will typically be located in relation to the picture. Usually it will be held before the middle of the horizon line.^{4} The range between eye and image should be estimated for a concentrative viewer.

Let’s now try the projection of a simple single ﬂoor hut. The position of the viewer’s eye we call A, and the eye’s distance to the image we call d. That spot of the image the eye is held in front of—where the perpendicular is dropped from the eye to the image—we refer to as L.

Now we think of the hut as a miniature located before the image plane and set with the frontmost corner of its eaves (that edge where the rainwater pipe is ﬁxed) at the viewer’s eye. The vertical edges of the hut—the house’s corners so to say—run parallel to the image plane^{5} and thus don’t need any vanishing points.^{6} The eaves instead point laterally to the horizon, and we’d like to know where their extension may meet the image plane. In order to ﬁnd out, we ﬁrst determine the angle between the eaves and the perpendicular dropped to the image and call the angle α. The tangent of α multiplied with d makes the distance between L and the wanted vanishing point F1 on the image plane. This is the vanishing point for any other edge of the hut running parallel to the eaves, e. g. the ridge or two sides of the ﬂoor and the ceiling each. However, the ﬂoor’s and the ceiling’s other two sides are aligned to another vanishing point we call F2. This point is located at the horizon line as well, but at the opposite side of L. Its position is calculated correspondingly; the angle β being relevant here is the missing rest to 90°, of course.

More difﬁcult is calculation of the slant edges of the two gable sides. Therefore we consider the following: If the frontmost edge of our imaginary miniature hut is put at the viewer’s eye, the gable side’s complete wall area points to the image plane, and its extension intersects that plane in a line running vertically through the vanishing point we’ve just calculated, F2. Any line lying plainly on the wall area now also leads to a vanishing point located somewhere at that intersection line. Hence, this applies to the gable’s roof edges as well. Calculation of their vanishing points is now performed analogously to the calculation just done, the horizon line being replaced by that intersection line, L being replaced by F2, and d by the distance between A and F2 we refer to as b. The latter is reckoned by d, divided by the cosine of β. Now we determine the pitch the hut’s roof shall have, as well as the angle between the attic’s ﬂoor and the roof area we call γ. The two wanted vanishing points’ distance to F2 now follows from b, multiplied with the tangent of γ. So these vanishing points—F3 and F4—are set at the mentionened intersection line in the just computed distance above and beneath F2.

Another, quite vivid, description of how to draw using vanishing points by Alexander Vögtli can be found here, by the way. But note that the approach adopted there is somewhat different: First, a square is depicted lying ﬂat on the ﬂoor and pointing to the perpendicular point L, and its rear edge is drawn in by eye. Then, the square’s diagonals are extended to the horizon line in order to determine the vanishing points for the angles α und β both being 45°. The distance between each of these points and the perpendicular point L equates to the eye’s distance from the image plane, as the tangent of 45° is 1. Thus, the eye distance is determined in retrospect, whereas here it is assigned in advance.

1 In programmes as for example Corel Draw objects can even be placed far off the image’s borders.

2 If some of the cuboid’s edges (i. e. four or eight) are parallel to the image plane, they only meet it in inﬁnity, of course—hence, nowhere. These edges just have no need of vanishing points, then.

3 The eye doesn’t realise much of it though—photographs, above all small ones, are as well viewed from all kinds of positions that certainly do not suit the particular perspective. But foreshortened illusion just isn’t perfect then anymore.

4 This is not imperative, though, e. g. in images depicting a view up- or downwards.

5 This would be different if we depicted a view up- or downwards and swinged the image plane with us, thus having it always orthographic to our viewing direction.

6 See above fn. 2.