perspective drawing

1 Introduction

Perspective drawing refers to a particular technique of projecting a three-dimensional scene — mathemetically, some subset of ${ℝ}^3$ — onto a subset of ${ℝ}^2$. The aim is to make the scene look “natural”.

The most common model for perspective drawing consists of an ideal eye or focus at a point $𝐞\in {ℝ}^3$, located behind a planar screen that is perpendicular to the unit vector $𝐠\in {ℝ}^3$, representing the direction of gaze.

To project a point $𝐩\in {ℝ}^3$ in front of the screen onto a point $𝐪\in {ℝ}^3$ on the screen, we find the intersection, with the screen, of the light ray emitted from the point $𝐩$ towards the eye $𝐞$:

2 Basic equation for the projection

Let $𝐨\in {ℝ}^3$ be a point on the screen. Without loss of generality, assume that $𝐞-𝐨$ is anti-parallel to $𝐠$, as in the previous picture. (We can always move the point $𝐨$ on the screen.) This means that $𝐨$ will be the origin of the perspective drawing.

To calculate the projected point $𝐪$ explicitly, we can solve these equations for $t\in [0,1]$:

$𝐪$ $=(1-t)𝐩+t𝐞$ ($𝐪$ is on the ray from $𝐩$ to $𝐞$)

$0$ $=⟨𝐪-𝐨,𝐠⟩$ ($𝐪$ is on plane centered at $𝐨$ perpendicular to $𝐠$)

(The notation $⟨,⟩$ denotes the dot product in ${ℝ}^3$.) The solution is readily found to be:

$$t=\frac{⟨𝐩-𝐨,𝐠⟩}{⟨𝐩-𝐞,𝐠⟩}=\frac{\text{horiz. distance of }𝐩\text{ to the screen}}{\text{horiz. distance of }𝐩\text{ to the eye}}.$$

This $t$ satisfies , because the horizontal distance of the point $𝐩$ to the screen is less than its horizontal distance to the eye.

From the expression for $t$, we can determine $𝐪$:

$$𝐪=\frac{⟨𝐩-𝐨,𝐠⟩𝐞-⟨𝐞-𝐨,𝐠⟩𝐩}{⟨𝐩-𝐞,𝐠⟩}.$$ (1)

3 Isomorphism of the planar screen to ${ℝ}^2$

Of course, we are interested in displaying or drawing the projected point $𝐪$ on a flat screen, so we need to assign coordinates to the screen. In mathematical terms, we seek an (affine) isomorphism to ${ℝ}^2$, of the plane centered at $𝐨$ and perpendicular to $𝐠$.

The isomorphism is, obviously, not unique in general. But it is uniquely determined if we impose these reasonable conditions:

• Firstly, Euclidean distances on the screen should be preserved when it is transformed into ${ℝ}^2$.

• And secondly, if we assign an “up direction” $𝐯\in {ℝ}^3$, then this direction should correspond with the unit vector $(0,1)$ on ${ℝ}^2$.

• Similarly, if $𝐮\in {ℝ}^3$ is a “right”-pointing vector, satisfying $𝐮\times 𝐯=-𝐠$, then $𝐮$ should map to $(1,0)$ on ${ℝ}^2$. (The minus sign is there to preserve the right-hand rule: the vector $𝐮\times 𝐯$ should point towards the viewer, which is exactly the opposite of the viewer’s gaze direction.)

Let $\gamma$ be the right-handed orthonormal basis for ${ℝ}^3$ consisting of the vectors $𝐮$, $𝐯$ and $-𝐠$, where $𝐠$ is the gaze direction as before and $𝐯$ is a given “up” vector. The vector $𝐮$ can be determined by

$$𝐮=𝐯\times -𝐠=𝐠\times 𝐯.$$

Then any point $𝐪$ on the planar screen is to be mapped to the point ${𝐪}^{\prime }\in {ℝ}^2$, where

$$({𝐪}^{\prime },0)={[𝐪-𝐨]}_{\gamma },$$ (2)

and ${[𝐪-𝐨]}_{\gamma }$ is the representation of $(𝐪-𝐨)$ in the basis $\gamma$.

4 The perspective transform in eye coordinates

The perspective projection (1) takes a simple form in eye coordinates (coordinates with respect to the basis $\gamma$). We derive it now.

Let ${𝐩}_o=𝐩-𝐨$, and ${𝐞}_o=𝐞-𝐨$. Then equation (1) can be rewritten:

$𝐪$ $={\displaystyle \frac{-⟨{𝐩}_o,-𝐠⟩𝐞+⟨{𝐞}_o,-𝐠⟩𝐩}{⟨{𝐞}_o-{𝐩}_o,-𝐠⟩}},$

$𝐪-𝐨$ $={\displaystyle \frac{-⟨{𝐩}_o,-𝐠⟩{𝐞}_o+⟨{𝐞}_o,-𝐠⟩{𝐩}_o}{⟨{𝐞}_o-{𝐩}_o,-𝐠⟩}}.$

We can write out the last vector equation in $\gamma$ coordinates. Let ${[{𝐩}_o]}_{\gamma }=(x,y,z)$ and ${[{𝐞}_o]}_{\gamma }=(0,0,a)$ for some $a>0$ (the distance of the eye from the screen). Then

${[𝐪-𝐨]}_{\gamma }$ $={\displaystyle \frac{-z(0,0,a)+a(x,y,z)}{a-z}}={\displaystyle \frac{(ax,ay,0)}{a-z}}$

$=({\displaystyle \frac{x}{1-z/a}},{\displaystyle \frac{y}{1-z/a}},\mathrm{\hspace{0.17em}0}).$

Thus, according to equation (2), the perspective transform of $𝐩$ onto ${ℝ}^2$ is given by

$${𝐪}^{\prime }=(\frac{x}{1-z/a},\frac{y}{1-z/a}).$$ (3)

5 Computing the perspective transform in arbitrary coordinates

Finally, we want explicit expressions for the perspective transform in terms of coordinates of an arbitrarily given orthonormal basis $\beta$ for ${ℝ}^3$. (For example, $\beta$ might be the standard basis.)

The matrix that changes from $\gamma$ coordinates to $\beta$ coordinates this matrix composed of three column vectors:

The matrix ${[I]}_{\beta }^{\gamma }$ that changes from $\beta$ coordinates to $\gamma$ coordinates is the inverse to ${[I]}_{\gamma }^{\beta }$; but both matrices are orthogonal, so the inverse simplifies to the transpose:

$${[I]}_{\beta }^{\gamma }={\left({[I]}_{\gamma }^{\beta }\right)}^{-1}={\left({[I]}_{\gamma }^{\beta }\right)}^{\mathrm{t}}=\left(\begin{array}{c}\hfill {[𝐮]}_{\beta }^{\mathrm{t}}\hfill \\ \hfill {[𝐯]}_{\beta }^{\mathrm{t}}\hfill \\ \hfill {[-𝐠]}_{\beta }^{\mathrm{t}}\hfill \end{array}\right).$$

Therefore, to obtain the perspective transform of a given point $𝐩$ given in $\beta$ coordinates, we are to compute the matrix product

$$\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \\ \hfill z\hfill \end{array}\right)={[I]}_{\beta }^{\gamma }\cdot {[𝐩-𝐨]}_{\beta },$$ (4)

and apply formula (3) right afterwards.

6 Example

This figure was created using the MetaPost programming language. MetaPost by itself has no facilities to produce three-dimensional graphics, so the source code for this drawingcube.mp implements formulae (3) and (4) directly.

7 The perspective transform in homogeneous coordinates

The operation described by formula (3) is not linear in $𝐩$ or $𝐩-𝐨$, so it cannot be represented by a $3\times 3$ matrix. But using homogeneous coordinates, the perspective transform can be represented by the following $4\times 4$ matrix (with respect to eye coordinates):

8 Properties of the perspective transform

(To be written. Talk about the fact that lines are mapped to lines by the perspective transform, and vanishing points.)

9 Other models for three-dimensional drawing

(To be written. Talk about orthographic projection (and relate it to the special case where $a\to \mathrm{\infty }$). Could also mention the pin-hole camera model, or even more complicated models with non-infinitesimal lens.)

Title	perspective drawing
Canonical name	PerspectiveDrawing
Date of creation	2013-03-22 15:41:10
Last modified on	2013-03-22 15:41:10
Owner	stevecheng (10074)
Last modified by	stevecheng (10074)
Numerical id	12
Author	stevecheng (10074)
Entry type	Topic
Classification	msc 51N20
Classification	msc 15A90