These are our notes to studying perspective without the crutch of cubes. They had better help our drawing skills. Please let us know any inaccuracies.

## The Perspective Projection Map

Let `$V$`

be the viewing plane and `$O$`

be the eye point. Then the perspective drawing is,

in essence, the projection to `$V$`

with respect to `$O$`

. More precisely, the image of a point `$P$`

is the point of intersection of the line `$OP$`

and the plane `$V$`

. Let `$f$`

denote this map.

## The Vanishing Point

Consider the images of parallel lines `$\ell_\alpha$`

for `$\alpha$`

running over an index set. The intersection of the images of these parallel lines is called the vanishing point. We determine its existence and its location geometrically. Let `$W_\alpha$`

be the plane that contains `$O$`

and `$\ell_\alpha$`

. Then the vanishing point lies on the `$W_\alpha \cap V$`

for each `$\alpha$`

. Thus it lies on `$(\cap_\alpha W_\alpha) \cap V$`

. The first factor `$\cap_\alpha W_\alpha$`

is obviously the line `$\ell$`

which is parallel to `$\ell_\alpha$`

and which contains `$O$`

. Thus the vanishing point is the point of intersection of `$\ell$`

and `$V$`

if `$\ell$`

is not parallel to `$V$`

.

## The Transfer of Distance

Let `$\ell_0$`

be a line parallel to `$V$`

. The perspective projection map when restricted to `$\ell_0$`

is scaling. Thus it is easy to draw equispaced points on `$f(\ell_0)$`

. We would like a way to transfer distance on `$f(\ell_0)$`

to that on images of lines not parallel to `$V$`

. Let `$P$`

and `$Q$`

be two points of `$\ell_0$`

and let `$\bar{P}$`

and `$\bar{Q}$`

be their images respectively. Let `$\ell$`

be a line not parallel to `$V$`

such that it passes through `$P$`

. Let `$Q’$`

be the point on `$\ell$`

such that `$Q’$`

is further away from the eye point and such that the lengths of `$PQ$`

and `$PQ’$`

are equal. Let `$N$`

be the vanishing point for the image of the line `$PQ’$`

, which is just `$\ell$`

, and let `$M$`

be the vanishing point of the image of the line `$QQ’$`

. Then the image of the line `$PQ’$`

is the line `$\bar{P}N$`

and that of the line `$QQ’$`

is `$\bar{Q}M$`

. Thus `$f(Q’)$`

is the intersection of the line `$\bar{P}N$`

and `$\bar{Q}M$`

. It is easy to see that the lengths of `$MN$`

and `$ON$`

are equal by observing the isosceles.