# Distorted Space

## Changes from 4441622622db95f42caf48f216200b13da774c88 to 3784b728a96d70eff70af10000703e72c2e3ce4d

---
title: Distorted Space
categories: session
---

**Briefing** [Distorted Lecture]()

Chapter 9 in  *OpenCV 3 Computer Vision with Python Cookbook* by
Alexey Spizhevoy (author).
Search for it in [Oria](https://oria.no/).
There is an e-book available.

# Exercises

## An Example of a Distorted space

### Step 1.  Preliminaries.

1.  Consider a triangle in the object space, formed by the three
vertices, $A=(0,0,10)$, $B=(20,0,10)$, and $(20,10,10)$.
Let
+ $\alpha$ be the angle between the vectors $\overrightarrow{AB}$
and $\overrightarrow{AC}$.
+ $\beta$ be the angle between the vectors $\overrightarrow{BA}$
and $\overrightarrow{BC}$.
**Draw** a figure to show this information.
2.  **Calculate** the angles $\alpha$ and $\beta$.
3.  Consider an ideal perspective camera, posed such that the camera
frame matches the world frame.
**Calculate** the image points corresponding to $A$, $B$, and $C$.
Assume that the image plane has $z=1$.
**Draw** an image to display this projection.
4.  **Calculate** the angles $\alpha'$ and $\beta'$
in the image, corresponding to $\alpha$ and $\beta$.
Do they match the original angles?
5.  **Calculate** the lengths of the edges of the triangles in the image.
6.  Make a new drawing of the image plane,
displaying the quantities calculated in 4 and 5.
Keep it for later reference.

Recall the formula for the angle $\theta$ between two vectors
$u$ and $v$
$$\theta = \tan^{-1}\frac{u\cdot v}{||u||\cdot||v||}.$$
$$\theta = \cos^{-1}\frac{u\cdot v}{||u||\cdot||v||}.$$
This is found in any elementary calculus book.
You did not need this above, because the triangles had an orthogonal
angle, but you will need it below.

### Step 2.  Distortion.

7.  Suppose the camera is not idea, but instead has calibration matrix
$$K = \begin{bmatrix} 2 & \frac1{\sqrt{3}} & 4 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}$$
8.  Calculate the image points corresponding to $A$, $B$, and $C$,
Suppose the camera is not ideal, but instead has calibration matrix
$$K = \begin{bmatrix} 2 & \frac1{\sqrt{3}} & 4 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}$$

7.  **Calculate** the image points corresponding to $A$, $B$, and $C$,
using the camera calibration matrix $K$.
Draw the resulting image to scale.
9.  Calculate the lengths of the edges of the image of the triangle.
(See Ma 2004:55)
8.  **Draw** the resulting image to scale.
9.  **Calculate** the lengths of the edges of the image of the triangle.
For now, we use the familiar formulæ for vector norms.
10. Let $\alpha'$ and $\beta'$ be the images of the angles $\alpha$
and $\beta$.
Calculate these to image angles and add the information to you figure.
**Calculate** these to image angles and add the information to you figure.
11. Compare your distorted image to the original image from Step 1.

### Step 3.  The Distorted Inner Product Space

Recall the textbook's definition of an inner product in distorted space
$$\langle u,v\rangle_S = u^TSv \quad\text{where } S = K^{-T}K^{-1},$$
for some `camera calibration' matrix $K$.
We are going to redo Step 2, using this distorted inner product.

12. [9] Calculate the edges (norms) of the triangle, using the
distorted inner product to define the norm.
13. [10] Calculate the angles $\alpha'$ and $\beta'$, using
the distorted inner product and the formula
$$\theta = \tan^{-1}\frac{u\cdot v}{||u||\cdot||v||}.$$
14. Make a figure illustrating the quantities calculated.
15. [11] Compare the edges and angles thus calculated with the
calculations from Step 2 and from Step 1.

### Step 4.  Reflection

Think through the following:

16. What does this exercise tell you about camera distortion and
inner product spaces?
17. Is there anything you do not quite understand?
Is there something missing in the exercise, something that should
have been tested?

## From the Textbook

+ Exercise 6.4

# Debrief

+ Step 1 **TODO** Scan notes
+ Step 2(7-10) **TODO** Scan notes
+ Step 2(11)
+ Note how the $x$-axis is stretched, doubling the length of $AB$.
+ Not also how it is skewed, bending the angle $\beta'$ from
$90°$ to $120°$.