c# - Ease-in and ease-out animation formula

Question

Say, if I'm doing the Ease-Out and then Ease-In animation of an object's movement from X1 coordinate to X2 coordinate over S steps at equal time intervals. Can some suggest the formula to calculate this movement's X coordinates?

question from:https://stackoverflow.com/questions/13462001/ease-in-and-ease-out-animation-formula

Answer 1

Personally, I'd rather use a function that gets a time in [0; 1] and output a value in [0; 1], so that we can apply the result to any type (2D vector, 3D vector, ...).

Solution 1

For the quadratic easing in/out, the curve is separated in two distinct functions depending on the value of t:

• when t <= 0.5: f(x) = 2 * x * x with x in [0;0.5] (graph)
• when t > 0.5: f(x) = 2 * x * (1 - x) + 0.5 with x in [0;0.5] (graph)

Here are the graphs:

Since the second function is also in [0;0.5], but t > 0.5 when we start to use it, we need to reduce t by 0.5.

This is the result, in C:

float InOutQuadBlend(float t)
{
    if(t <= 0.5f)
        return 2.0f * t * t;
    t -= 0.5f;
    return 2.0f * t * (1.0f - t) + 0.5f;
}

Solution 2 (Bézier)

Another interesting blend curve is the one given by Bézier, which have the advantage to be quite optimized (no if). Here is the curve from Wolfram:

And here is the C code:

float BezierBlend(float t)
{
    return t * t * (3.0f - 2.0f * t);
}

Solution 3 (parametric function)

Another method proposed by @DannyYaroslavski is the simple formula proposed here.

It is parametric and gets a nice in/out acceleration and deceleration.

With alpha = 2, you get this function:

Which translates in C like this:

float ParametricBlend(float t)
{
    float sqt = t * t;
    return sqt / (2.0f * (sqt - t) + 1.0f);
}

Edit 1: Add solution 3 from @DannyYaroslavski
Edit 2: Better explanation for solution 1
Edit 3: Add graphs to all solutions