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.