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

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.

(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.

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 ```

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.

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 write `P = (1 - t)A + tB` or `P = tB + (1 - t)A`). With the new naming scheming for the points, the equation for the degree 2 curve is `P = (1 - t)A0 + tA1`.

Main content

Pixar in a Box

Unit 10: Lesson 2

Mathematics of animation curves

Bonus: Equations from de Casteljau's algorithm

Google Classroom

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:

P, left parenthesis, t, right parenthesis, equals, left parenthesis, 1, minus, t, right parenthesis, start color #6495ed, A, end color #6495ed, plus, t, start color #6495ed, B, end color #6495ed

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:

P, left parenthesis, t, right parenthesis, equals, left parenthesis, 1, minus, t, right parenthesis, squared, start color #6495ed, A, end color #6495ed, plus, 2, left parenthesis, 1, minus, t, right parenthesis, t, start color #6495ed, B, end color #6495ed, plus, t, squared, start color #6495ed, C, end color #6495ed

Degree 3 curves

If we create a degree 3 curve using four points, start color #6495ed, A, end color #6495ed, start color #6495ed, B, end color #6495ed, start color #6495ed, C, end color #6495ed, and start color #6495ed, D, end color #6495ed, is the equation for a point on the curve in terms of start color #6495ed, A, end color #6495ed, start color #6495ed, B, end color #6495ed, start color #6495ed, C, end color #6495ed, and start color #6495ed, D, end color #6495ed?

P, left parenthesis, t, right parenthesis, equals

Degree 4 curves

What about if we create a degree 4 curve using five points, start color #6495ed, A, end color #6495ed, start color #6495ed, B, end color #6495ed, start color #6495ed, C, end color #6495ed, start color #6495ed, D, end color #6495ed, and start color #6495ed, E, end color #6495ed?

P, left parenthesis, t, right parenthesis, equals

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.

Look at the first term in each of the above equations and see if you can spot a pattern.

What would be the coefficient for start color #6495ed, A, start subscript, 0, end subscript, end color #6495ed in an n degree curve?

Look at the last term in each of the above equations and see if you can spot a pattern.

What would be the coefficient for start color #6495ed, A, start subscript, n, end subscript, end color #6495ed in 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:

What is the exponent on left parenthesis, 1, minus, t, right parenthesis?

What is the exponent on t?

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?

[Hint]

[ANSWER]

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guillaume abbe
Posted 7 years ago. Direct link to guillaume abbe's post “thanks for those really i...”
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)
Answer
- Peter Collingridge
  Posted 7 years ago. Direct link to Peter Collingridge's post “If I understand the quest...”
  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)
Dana
Posted 6 years ago. Direct link to Dana's post “(1-t)^4 A + 4(1-t)^3 t B ...”
(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)
Answer
- ChrisRennick56
  Posted 6 years ago. Direct link to ChrisRennick56's post “You were so close!! if y...”
  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)
robert cook
Posted 7 years ago. Direct link to robert cook's post “Lessons have a Pixar part...”
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)
Answer
IEC.Learning.Co
Posted 6 years ago. Direct link to IEC.Learning.Co's post “was i supposed to do this...”
was i supposed to do this because i am only in 4 grade
(7 votes)
Answer
- Bettylovesroses#cool
  Posted 5 years ago. Direct link to Bettylovesroses#cool's post “No, because it is only re...”
  No, because it is only required if you were feeling incredibly bold for this gigantic challenge.
  (1 vote)
Лиам МакКарти (liam m)
Posted 5 years ago. Direct link to Лиам МакКарти (liam m)'s post “this is cabrasing my brai...”
this is cabrasing my brain
(7 votes)
Answer
Henry
Posted 4 years ago. Direct link to Henry's post “my student seems a bit to...”
my student seems a bit to young for this so is there any recomended age?
(6 votes)
Answer
- XDmysticXD
  Posted 3 years ago. Direct link to XDmysticXD's post “According to the Lesson D...”
  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)
7467522
Posted 4 years ago. Direct link to 7467522's post “i didn't quite understand...”
i didn't quite understand these curves
(3 votes)
Answer
Cailin Temple
Posted 7 years ago. Direct link to Cailin Temple's post “I'm unclear on the terms ...”
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)
Answer
- Peter Collingridge
  Posted 7 years ago. Direct link to Peter Collingridge's post “A0, A1 etc. refer to poin...”
  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 write P = (1 - t)A + tB or P = tB + (1 - t)A). With the new naming scheming for the points, the equation for the degree 2 curve is P = (1 - t)A0 + tA1.
  (5 votes)
wublablu
Posted 3 years ago. Direct link to wublablu's post “i dont know how to do thi...”
i dont know how to do this?
(5 votes)
Answer
- arunkumar.ganesan
  Posted a year ago. Direct link to arunkumar.ganesan's post “duh you do not know this ...”
  duh you do not know this so simpa
  (0 votes)
63801
Posted 4 years ago. Direct link to 63801's post “I thought this was code n...”
I thought this was code not math
(3 votes)
Answer
- ツ
  Posted a year ago. Direct link to ツ's post “It is code, code can stil...”
  It is code, code can still have math.
  (1 vote)

Pixar in a Box

Unit 10: Lesson 2

Parametric equation for a line

Degree 2222 curves

Degree 3333 curves

Degree 4444 curves