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## Pixar in a Box

### Course: Pixar in a Box>Unit 3

Lesson 2: Mathematics of subdivision

# Bonus: Equations for points in subdivision

## Points on the control arm

Let's start with a shape with four control points, A, B, C and D, then use subdivision with weights left parenthesis, 1, comma, 1, right parenthesis.

### Subdivision 1

First, we add midpoints along each edge. Then each point is moved to the average of its current position and the position of the point clockwise to it.
For example, we create a midpoint, M between A and B:
M, equals, start fraction, 1, divided by, 2, end fraction, A, plus, start fraction, 1, divided by, 2, end fraction, B
Then the point at A moves to a position halfway between A and M (indicated with a pink arrow). If we call this point, P:
\begin{aligned} P &= \dfrac{1}{2}A + \dfrac{1}{2}M \\ \\ P &= \dfrac{1}{2}A + \dfrac{1}{2}\left( \dfrac{1}{2}A + \dfrac{1}{2}B\right) \\ \\ P &= \dfrac{1}{2}A + \dfrac{1}{4}A + \dfrac{1}{4}B \\ \\ P &=\dfrac{3}{4}A + \dfrac{1}{4}B \end{aligned}

### Subdivision 2

Then we apply the subdivision algorithm again, adding midpoints, then moving each point to the average of its current position and the position of the point clockwise to it.
What is the position of the point indicated by the pink arrow in terms of A and B?

### Subdivision 3

When we apply subdivision again, what is the position of the point indicated by the pink arrow in terms of A and B?

### Further steps of subdivision

If we apply subdivision n times, what is the position of the point on the control arm in terms of A, B and n?

If we apply subdivision an infinite number of times, what would be the position of the point on the control arm in terms of A and B?

## Bonus challenge

Can you find equations for other points in the shape after applying subdivision n times?

## Want to join the conversation?

• Any hints as to how to resolve the bonus challenge? • From the author:I don't want to give too much away in case other people want a chance to have a go (as far as I know, you're the first person to attempt these questions). That said, I can give you a few pointers.

What I did was to write down the coefficients for points after 1, 2, 3 and maybe 4 subdivisions. You need only look at the points in half of one corner, since each corner has the same pattern and corners are symmetrical. For example, look at the points between A and B, but nearer A to B. These will all have coefficients for the points A, B and D. If you don't want to work out the coefficients, you can mouse over the points in this interactive: https://www.khanacademy.org/partner-content/pixar/modeling-character/subdivision-averages/p/interactive-weighted-subdivision
(Make sure to use weights of (1,1)).

For each number of subdivisions (1 - 4), put the coefficients over the same denominator and write out the numerators. You can ignore the values on A: I don't know what pattern they follow, but you can figure out the value of A if you know the values for B and D, since the coefficients sum to 1. See if you can see a pattern for the numerators of the coefficients of B and D together.

That should give something to work with.
• Hi, I really liked this course that Khan Acadamey and PIxar did, and I am even considering going into this field, would you be able to give me some sites, programs, or people that can help me take off of from here? Thank You • It depends on which angle you'd like to take to get into the industry. If you're more math and science oriented, I suggest finding a public University with a good computer graphics program. Oregon State University has a pretty good one from what I hear. If you'd like an art school approach, I know quite a few people who have come through the Savannah College of Art & Design (SCAD). That can be pretty expensive, though, because most of the dedicated art colleges are private institutions (and some of them are getting a bad rap these days for predatory recruiting practices—caveat emptor).

If you're a self-starter and quite self-disciplined, you could do an independent study course like Animation Mentor. It's not an accredited program, but I've met quite a few people who swear by it. You might also take a look at the list of Computer Graphics MOOCs here: https://www.mooc-list.com/tags/computer-graphics?static=true

Early in your course of study, you'll want to decide what kind of job you'd ultimately like to have so that you can focus your attention on the skills you'll need for that job. I have a very general and somewhat incomplete overview of visual effects job titles on my blog here: http://www.bryanray.name/wordpress/?tag=jobs

The most well-employed artists I know tend to have a strong specialty skill supplemented by generalist skills in a few other areas. For instance, I am primarily a technical compositor with some knowledge of 3d layout, rendering and simulation. I'm lousy at both animation and modeling, though.
• The bonus challenge question its there to make me think about it? • hi, why this aswer to the third queston is not correct?

A*(1+(2^n-1))/2^n+B*(1-(2^n-1))/2^n • i do not get the bonus equations for points in subdivision it is confusing can some one help me please • I found 2^N+1/2^(N+1)A + 2^N-1/2^(N+1)B as the "If we apply subdivision n times, what is the position of the point on the control arm in terms of A, B and n?" answer. Why is it wrong? • hey! im currently inrested in this, and i would love a good way to continue from here, but at the same time i dont feel quite so ready to try this out in antoher website, is thier a way i could continue this her on khan? • Can you suggest an online explanation of the formulation that you alluded to in terms of linear algebra and eigensystems? • SPOILER ALERT 'n number of subdivisions' explained:

You need to look for the pattern in the equations from 1st, 2nd, and 3rd subdivision.
For second subdivision you want to represent 2 with 'n'.

(5/8)A + (3/8)B is an answer for second and
(5/8)A = (4 +1 / 2^3)A = (2^2 +1 / 2^[2+1])A
same way
(3/8)B = [2^2 -1 / 2^(2+1)]B.

In third subdivision
(9/16)A + (7/16)B is an answer for third and:
(9/16)A = [2^3 +1 / 2^(3+1)]A
and
(7/16)B = [2^3 -1 / 2^(3+1)]B

So if you represent twos in second subdivision equation with n,
(or threes in third subdivisions) you come up with

{2^n +1}/{2^{n+1}}A + {2^n -1}/{2^{n+1}}B  