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Pixar in a Box
Unit 10: Lesson 2
Mathematics of animation curves- Start here!
- 1. Mathematics of linear interpolation
- Linear interpolation
- 2. Repeated linear interpolation
- 3. De Casteljau's algorithm
- Constructing curves using repeated linear interpolation
- 4. What degree are these curves?
- Bonus: Equations from de Casteljau's algorithm
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Bonus: Equations from de Casteljau's algorithm
Challenge question: can you work out the equations for n-degree curves generated by de Casteljau's algorithm?
Parametric equation for a line
In the first step of de Casteljau's algorithm we define a point along a line in terms of t. For example, if we have a line between two points, start color #6495ed, A, end color #6495ed and start color #6495ed, B, end color #6495ed, then we can define a point, P, left parenthesis, t, right parenthesis on that line.
The equation for the point is:
As t goes from 0 to 1, P, left parenthesis, t, right parenthesis traces out the line from start color #6495ed, A, end color #6495ed and start color #6495ed, B, end color #6495ed. The equation is linear, so the line can be considered a degree 1 curve.
Degree 2 curves
When we create a degree 2 curve (a parabola), we use three points, start color #6495ed, A, end color #6495ed, start color #6495ed, B, end color #6495ed, and start color #6495ed, C, end color #6495ed
Now we get this equation for a point on the curve:
Degree 3 curves
Degree 4 curves
Degree n curves
Now let's see if we can spot any patterns in these equations that will allow us to find a general equation that uses n, plus, 1 points, start color #6495ed, A, start subscript, 0, end subscript, end color #6495ed, comma, start color #6495ed, A, start subscript, 1, end subscript, end color #6495ed, comma, point, point, point, comma, start color #6495ed, A, start subscript, n, minus, 1, end subscript, end color #6495ed, comma, start color #6495ed, A, start subscript, n, end subscript, end color #6495ed, to define an n degree curve.
Now, the hardest part: look at the remaining terms in each of the above equations. Notice that each term includes:
- a constant
- left parenthesis, 1, minus, t, right parenthesis raised to a power
- t raised to a power
For example, for a degree 2 curve, the start color #6495ed, A, start subscript, 1, end subscript, end color #6495ed term is 2, left parenthesis, 1, minus, t, right parenthesis, t, so the constant term is 2, the exponent on left parenthesis, 1, minus, t, right parenthesis is 1, and the exponent on t is 1.
In the coefficient for the start color #6495ed, A, start subscript, i, end subscript, end color #6495ed term in an equation for an n degree curve:
Extra Super Bonus Challenge
Can you find a formula for the constant term for start color #6495ed, A, start subscript, i, end subscript, end color #6495ed? Once you have done that, can you combine all these parts into an equation for P, left parenthesis, t, right parenthesis for an n degree curve?
Want to join the conversation?
- thanks for those really interesting courses.
Maybe I missed something but how can we get the value from a spline curve given a t value from the timeline (not the t parametric value going from 0 to 1).
In the first video, it's easy with th y=mx+b equation but I can't get it with the Casteljau's equation.
Maybe I missed something ?
Thanks in advance(28 votes)- If I understand the question correctly, you want to know, for example, if you have a Bezier spline that goes from frame 1 to frame 10, how do you find the value at, say, frame 5. It's tempted to say that you put in the value of t = 0.5, but that doesn't quite work since the x-coordinate along the curve does not change linearly with t.
I don't know if you can do it directly via deCasteljau's equation. I believe it's done by finding the equation as a cubic, and solving for the cubic, which is quite difficult and involves some fairly intense algebra.(6 votes)
- (1-t)^4 A + 4(1-t)^3 t B + 4(1-t)^2 t^2 C + 4(1-t) t^3 D + t^4 E
I typed this into the 4th-degree question and it says it is wrong. Is it because of spacing maybe? From the rest of the notes, it seems it would be the correct answer.(7 votes)- You were so close!! if you look at my answer to Favour Umoru you will see the pascal's triangle at the end. In your case you have got the coeficient of the middle term wrong it should be a 6 and not a 4. This ought to be what you are looking for
(1-t)^4 A + 4(1-t)^3 t B + 6(1-t)^2 t^2 C + 4(1-t) t^3 D + t^4 E
(11 votes)
- Lessons have a Pixar part and a math part.
It would be nice to have a programming part as well; for example, drawing n-degree curves.(9 votes) - was i supposed to do this because i am only in 4 grade(7 votes)
- No, because it is only required if you were feeling incredibly bold for this gigantic challenge.(1 vote)
- this is cabrasing my brain(7 votes)
- my student seems a bit to young for this so is there any recomended age?(6 votes)
- According to the Lesson Description, it is best if your student already or is learning algebra 1, for most people, they will learn algebra in 7th, 8th, or 9th grade.(1 vote)
- i didn't quite understand these curves(3 votes)
- I'm unclear on the terms A(sub)0 or A(sub)1 or A(sub)i etc. Are these sequential terms in the equation or do they refer to specific locations in the equation? I have everything correct until what is the exponent for (1-t)? part.(4 votes)
- A0, A1 etc. refer to points to the control points in order. For a degree 2 curve, we used points A, B, C. But for a degree n curve, we use A0, A1, ... An. We use this instead of A, B, C to make it easier to refer to a point.
So, they do refer to sequential terms in the equation (but the order doesn't really matter - you could writeP = (1 - t)A + tB
orP = tB + (1 - t)A
). With the new naming scheming for the points, the equation for the degree 2 curve isP = (1 - t)A0 + tA1
.(5 votes)
- i dont know how to do this?(5 votes)
- duh you do not know this so simpa(0 votes)
- I thought this was code not math(3 votes)
- It is code, code can still have math.(1 vote)