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# What is the math behind AnimationCurve.Evaluate?

I understand how to use animation curves in Unity, but I'm more interested in how Unity actually evaluates an animation curve at a given point. In other words, what is the math behind the function Evaluate?

The reason I ask is that I have a lot of animation data in animation curve format: time, value, in tangent, out tangent. I want to convert this data into splines like bezier curves or other functions, but I am not sure what the formula for this conversion would be.

Anyone have ideas?

It uses Bezier Curves, like it should. If you want to evaluate those, you can do a quick google search :)

well of course is bezier curves(which can be represented as hermit also),but it's not that cubic bezier curve,which anyone knows,I mean the factors are particular for the unity implementation ... at a moment I also thought they are using only trigonometric function ... whichever way,it would be of great help the exact formula

If this is still an issue in 2017 , you could check out the Runtime Curve Editor. All the sources are available in that package, so the equation, by which that curve is plotted, is also available. Of course , the package has more to offer than just plotting a curve by the equation you are looking for .

Unity's AnimationCurve lets you modify the tangets by moving that little handle. What if I wanted to make my own class similar to animation Curve, but I wanted to set the tangents with a vector2 instead of the handle that then in turn calculated the tangents? How could I evauluate that, keeping in mind that there could be 3 intersections?

**Answer** by Varaughe
·
Aug 06, 2013 at 03:21 AM

Alright ,so empiricaly I found something that matches exactly the animation curve drawn by unity ... it can be drawn as a bezier curve ... as you know the AnimationCurve of Unity , has .time and .value for the two keyframes of the segment(lets call the two endpoints P1,P2, time is x and value is y coordinate), and a out tangent(first point) and in tangent(2nd point) , can be called tgOut and tgIn. The Bezier curve (the cubic one),its equation is well known,can be found on wiki, has aside the 2 end points(P1,P2), 2 control points ... let's call them C1 , C2 . We have to corelate C1,C2 with tgOut and tgIn ,this relation is not given by unity,and I found it empiricaly:

```
float tangLengthX = Mathf.Abs(p1.x-p2.x)*0.333333f;
float tangLengthY = tangLengthX ;
c1 = p1;
c2 = p2;
c1.x += tangLengthX ;
c1.y += tangLengthY * tgOut;
c2.x -= tangLengthX ;
c2.y -= tangLengthY * tgIn;
```

//draw now the bezier using p1,c1,c2,p2

```
remember that tgOut and tgIn need a scaling , to match the ratio of the window's width and height where they are going to be drawn ,and the ratio of gradations of which scope the curve's time and value are part ...
to be more explicit ..... let's say that **gridRect** is the rect in pixels where the curve is to be drawn and **gradRect** gives the min and max values for the gradations of the grid(e.g. the time ranges from 0 to 5, and the values from 0 to 7 ...so gradation maximum is 7 for y axis and 5 for x axis) ....
so: let's say
```

**ratio** = (gridRect.height/gridRect.width)*(gradRect.width/gradRect.height) then **tgOut** is ratio*curve[i].outTangent and **tgIn** is ratio*curve[i+1].inTangent (where curve is of type AnimationCurve)

**Answer** by Paulius-Liekis
·
Aug 06, 2013 at 09:10 AM

It should be something like this:

```
float Evaluate(float t, Keyframe keyframe0, Keyframe keyframe1)
{
float dt = keyframe1.time - keyframe0.time;
float m0 = keyframe0.outTangent * dt;
float m1 = keyframe1.inTangent * dt;
float t2 = t * t;
float t3 = t2 * t;
float a = 2 * t3 - 3 * t2 + 1;
float b = t3 - 2 * t2 + t;
float c = t3 - t2;
float d = -2 * t3 + 3 * t2;
return a * keyframe0.value + b * m0 + c * m1 + d * keyframe1.value;
}
```

have you tried this one,did you drawn twice the curve ?, me I drawn the curve twice,once using directly the evaluate method from unity and once with the formula I explained above your comment ... and the curves are matching 99,9%..of course the tangents needs scaling with the aspect ratio ... what I see strange in your method, is that you evaluate by 't' :) ... the unity evaluates by time...so perhaps from that time it's finding the 't' and then back the keyframe.value ... in my case ,what I needed was just to have bezier curve exactly matching the curve drawn by unity

t is from 0 to 1 in this case, i.e. t = Mathf.InverseLerp(keyframe0.time, keyframe1.time, time)

No, I haven't tried it myself. I copied this function from one of my projects (this is slightly modified version).

As far as I remember Unity uses this kind of code (a bit more optimized).

What you wrote is just regular Cubic Hermite spline with tangents defined by hand. The same curve is default interpolation curve in 3d studio max (TCB controler - parameters Tension, Continuity, Bias are used to calculate tangents)

Hey guys! Thanks for this, works perfectly! If you want it to work with more than 2 keyframes or if the first and last keyframe are not exactly 0 and 1 just add this line to normalize t:

t = (t - keyframe0.time) / dt;

**Answer** by SuperPingu
·
Feb 03, 2016 at 03:39 PM

Unity define a curve with 2 keyframes, each composed of a point and a tangent. I guess the simplest curve matching that is a third degree polynomial (a cubic function). Given the 2 points and tangents, it is possible to compute the polynomial coefficients simply by solving the following equation system:

```
(1) a*p1x^3 + b*p1x^2 + c*p1x + d = p1y
(2) a*p2x^3 + b*p2x^2 + c*p2x + d = p2y
(3) 3*a*p1x^2 + 2*b*p1x + c = tp1
(4) 3*a*p2x^2 + 2*b*p2x + c = tp2
```

You can solve this manually or using a computer algebra system.

This gives you:

```
float a = (p1x * tp1 + p1x * tp2 - p2x * tp1 - p2x * tp2 - 2 * p1y + 2 * p2y) / (p1x * p1x * p1x - p2x * p2x * p2x + 3 * p1x * p2x * p2x - 3 * p1x * p1x * p2x);
float b = ((-p1x * p1x * tp1 - 2 * p1x * p1x * tp2 + 2 * p2x * p2x * tp1 + p2x * p2x * tp2 - p1x * p2x * tp1 + p1x * p2x * tp2 + 3 * p1x * p1y - 3 * p1x * p2y + 3 * p1y * p2x - 3 * p2x * p2y) / (p1x * p1x * p1x - p2x * p2x * p2x + 3 * p1x * p2x * p2x - 3 * p1x * p1x * p2x));
float c = ((p1x * p1x * p1x * tp2 - p2x * p2x * p2x * tp1 - p1x * p2x * p2x * tp1 - 2 * p1x * p2x * p2x * tp2 + 2 * p1x * p1x * p2x * tp1 + p1x * p1x * p2x * tp2 - 6 * p1x * p1y * p2x + 6 * p1x * p2x * p2y) / (p1x * p1x * p1x - p2x * p2x * p2x + 3 * p1x * p2x * p2x - 3 * p1x * p1x * p2x));
float d = ((p1x * p2x * p2x * p2x * tp1 - p1x * p1x * p2x * p2x * tp1 + p1x * p1x * p2x * p2x * tp2 - p1x * p1x * p1x * p2x * tp2 - p1y * p2x * p2x * p2x + p1x * p1x * p1x * p2y + 3 * p1x * p1y * p2x * p2x - 3 * p1x * p1x * p2x * p2y) / (p1x * p1x * p1x - p2x * p2x * p2x + 3 * p1x * p2x * p2x - 3 * p1x * p1x * p2x));
```

Then, to evaluate the value:

```
float Evaluate(float t)
{
return a*t*t*t + b*t*t + c*t + d;
}
```

I checked with Unity with the following quick and dirty code:

```
using UnityEngine;
[ExecuteInEditMode]
public class TestAnimCurve : MonoBehaviour {
public AnimationCurve anim = AnimationCurve.EaseInOut(0, 0, 1, 1);
float a;
float b;
float c;
float d;
void Update () {
float p1x= anim.keys[0].time;
float p1y= anim.keys[0].value;
float tp1=anim.keys[0].outTangent;
float p2x=anim.keys[1].time;
float p2y= anim.keys[1].value;
float tp2= anim.keys[1].inTangent;
Debug.Log(p1x+ ", " + p1y+ ", " + tp1 + ", " + p2x + ", " + p2y + ", " + tp2);
Debug.Log("Evaluate Unity: " + anim.Evaluate(0.1f) + ", " + anim.Evaluate(0.2f) + ", " + anim.Evaluate(0.3f) + ", " + anim.Evaluate(0.4f) + ", " + anim.Evaluate(0.5f) + ", " + anim.Evaluate(0.6f) + ", " + anim.Evaluate(0.76f) + ", " + anim.Evaluate(0.88f) + ", " + anim.Evaluate(0.98f));
a = (p1x * tp1 + p1x * tp2 - p2x * tp1 - p2x * tp2 - 2 * p1y + 2 * p2y) / (p1x * p1x * p1x - p2x * p2x * p2x + 3 * p1x * p2x * p2x - 3 * p1x * p1x * p2x);
b = ((-p1x * p1x * tp1 - 2 * p1x * p1x * tp2 + 2 * p2x * p2x * tp1 + p2x * p2x * tp2 - p1x * p2x * tp1 + p1x * p2x * tp2 + 3 * p1x * p1y - 3 * p1x * p2y + 3 * p1y * p2x - 3 * p2x * p2y) / (p1x * p1x * p1x - p2x * p2x * p2x + 3 * p1x * p2x * p2x - 3 * p1x * p1x * p2x));
c = ((p1x * p1x * p1x * tp2 - p2x * p2x * p2x * tp1 - p1x * p2x * p2x * tp1 - 2 * p1x * p2x * p2x * tp2 + 2 * p1x * p1x * p2x * tp1 + p1x * p1x * p2x * tp2 - 6 * p1x * p1y * p2x + 6 * p1x * p2x * p2y) / (p1x * p1x * p1x - p2x * p2x * p2x + 3 * p1x * p2x * p2x - 3 * p1x * p1x * p2x));
d = ((p1x * p2x * p2x * p2x * tp1 - p1x * p1x * p2x * p2x * tp1 + p1x * p1x * p2x * p2x * tp2 - p1x * p1x * p1x * p2x * tp2 - p1y * p2x * p2x * p2x + p1x * p1x * p1x * p2y + 3 * p1x * p1y * p2x * p2x - 3 * p1x * p1x * p2x * p2y) / (p1x * p1x * p1x - p2x * p2x * p2x + 3 * p1x * p2x * p2x - 3 * p1x * p1x * p2x));
Debug.Log("Evaluate Cubic: " + Evaluate(0.1f) + ", " + Evaluate(0.2f) + ", " + Evaluate(0.3f) + ", " + Evaluate(0.4f) + ", " + Evaluate(0.5f) + ", " + Evaluate(0.6f) + ", " + Evaluate(0.76f) + ", " + Evaluate(0.88f) + ", " + anim.Evaluate(0.98f));
}
float Evaluate(float t)
{
return a * t * t * t + b * t * t + c * t + d;
}
}
```

After modifing tangents of the animation curve, the debug messages produced by this code are:

```
0, 0, -4.484611, 1, 1, -10.23884
Evaluate Unity: -0.2431039, -0.1423873, 0.2018093, 0.6891449, 1.219279, 1.691871, 2.077879, 1.854902, 1.193726
Evaluate Cubic: -0.2431039, -0.1423873, 0.2018092, 0.6891448, 1.219279, 1.691871, 2.077879, 1.854902, 1.193726
```

So it really seams that this approach is the math behind AnimationCurve.Evaluate ;)

Great work! Thanks to your code I was able to figure out how to get animation curves working with the job system. The only thing missing is weights (to affect the slope of the previous to current or current to next key) Do you have any pointers as to how to support weights?

**Answer** by Aru3
·
Aug 26, 2017 at 05:59 AM

AnimationCurve is a simple spline of cubic equations, where for each segment you specify the (time, value) coordinates and slope for each of two end points (the first point's "out" "tangent", second one's "in"). A cubic equation is the simplest polynomial which can meet these criteria.

This is also known as a cubic Hermite spline, where the only difference is that AnimationCurve allows you to assign different slopes for each direction from a keyframe (a point), which creates discontinuities in the first derivative (sharp angles). However, this does not change the formula.

https://en.wikipedia.org/wiki/Cubic_Hermite_spline

It took me some time (much more than I'd like to admit...) but I simplified the system of cubic equations for the endpoints into this form.

```
pd = 1/(p2x-p1x)
td = t-p1x
ta = td*pd
out(t) = (((p2s+p1s)/2 - (p2y-p1y)*pd)*ta*(t*2 + p1x - p2x*3) + (p2s-p1s)/2*(t + p1x - p2x*2))*ta + p2s*td + p1y
```

Isolated t to calculate coefficients in advance, providing a fast t evaluation:

```
pd = 1/(p2x-p1x)
td = pd*(p2s-p1s)
a = pd^2*(p2s + p1s - 2*pd*(p2y-p1y))
b = (-a*3*(p2x + p1x) + td)/2
c = p2x*(a*3*p1x + -td) + p2s
d = p1x*(a/2*p1x*(-p2x*3 + p1x) + td*(p2x + -p1x/2) + -p2s) + p1y
out(t) = t*(t^2*a + t*b + c) + d
```

I had the same question, was looking for a straightforward answer. Because my search query took me here and this is the closest I could find, I assume such solutions to this exact problem are not abundant, so I thought I'd share mine. I derived it because I need to emulate the same function outside of unity. Looking at the other three answers here, they seem to be correct but not simplified (trust me this is not an easy task) (Varaughe's leaves out the evaluation of the Bezier curve he made from the AnimationCurve, as it is a well-known thing).

I tested these against AnimationCurve for correctness with a Python script which can randomize all six parameters of the cubic function:

```
# isolated t
def curve1(t,p1x,p2x,p1y,p2y,p1s,p2s):
e = 1/(p2x-p1x)
f = e*(p1s-p2s)
a = (p1s + p2s + 2*e*(p1y-p2y))*e**2
b = (a*3*(p2x+p1x) + f)/-2
c = p2x*(a*3*p1x + f) + p2s
d = p1x/2*(a*p1x*(p1x - p2x*3) + f*(p1x - p2x*2) - p2s*2) + p1y
return t*(t**2*a + t*b + c) + d
# short form
def curve2(t,p1x,p2x,p1y,p2y,p1s,p2s):
pd = 1/(p2x-p1x)
td = t-p1x
ta = td*pd
return ((t*2 + p1x - p2x*3)*((p2s+p1s)/2 - (p2y-p1y)*pd)*ta + (t + p1x - p2x*2)*(p2s-p1s)/2)*ta + p2s*td + p1y
# Paulius
def curve_Paulius(t,p1x,p2x,p1y,p2y,p1s,p2s):
xd = p2x-p1x
t = (t-p1x)/xd
return (2*t**3 - 3*t**2 + 1)*p1y + (t**3 - 2*t**2 + t)*p1s*xd + (t**3 - t**2)*p2s*xd + (-2*t**3 + 3*t**2)*p2y
def test(curve, times, curve_parameters):
out = []
for i in times:
out.append(curve(i, *curve_parameters))
return out
from random import random
rand_args = []
for i in range(4):
rand_args.append(random()*100)
for i in range(2):
rand_args.append(random()*2)
rand_times = []
e,f = rand_args[0:2]
points = 21
# interval count is points - 1
for i in range(points - 1):
td = (f - e)/(points - 1)
rand_times.append(random()*td + e + i*td)
```

you should check runtime curve editor ,there the math behind AnimationCurve is accurately replicated, I don't understand what do you mean by is not correct or outdated , did the Unity team changed the equation by which the bezier(or hermite) curves are plotted ? I don't see any need for doing that ...

Your post confused me. I looked at it again, and I understand it now, I apologize, it is correct, I will edit my post. You describe turning an AnimationCurve into a cubic Bezier curve, by calculating control point values like so:

dx3 = (p2x-p1x)/3

c1y = p1y + dx3*p1s

c2y = p2y - dx3*p2s

Then the AnimationCurve can be evaluated as the y portion of said Bezier curve, via this formula which I fetched from wikipedia:

y(t) = (1-t)^3*p1y + 3*(1-t)^2*c1y + 3*(1-t)*t^2*c2y + t^3*p2y

While my post instead calculates the coefficients of the cubic polynomial, for slightly faster evaluation of the curve at t.

Paulius's is also correct, I previously forgot to normalize t:

t = (t - p1x)/(p2x - p1x)

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