Mathf.PingPong with ease in and out?

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I have the following code in Update() which counts from 0 to 100 and back to 0, but I want to make the value start slowly, build up to speed as it increases, then slow as it reaches 100, and then speed up as it starts back to 0, and again slow as it reaches 0.

I need to use the value to ease in and out from 0 to 100

Debug.Log(Mathf.PingPong(Time.time * 10, 100));

I have looked at lots of the Mathf functions but no joy... anyone able to help?

Answer 1

Answer by Eric5h5 · Jul 30, 2013 at 10:32 PM

 Mathf.SmoothStep(0, 100, Mathf.PingPong(Time.time/10, 1))

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Bunny83 · Jul 30, 2013 at 10:36 PM 0

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Just had a look and SmoothStep is just a hermit polynom:

 public static float SmoothStep(float from, float to, float t)
 {
     t = $$anonymous$$athf.Clamp01(t);
     t = -2f * t * t * t + 3f * t * t;
     return to * t + from * (1f - t);
 }

Eric5h5 · Jul 30, 2013 at 11:03 PM 0

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Yep, I used to write code which did that before they implemented SmoothStep. A $$anonymous$$or thing but it's kind of nice to have it built-in, since I tend to use hermite interpolation mostly.

Answer 2

Answer by Bunny83 · Jul 30, 2013 at 10:20 PM

Well, there are many ways to do this. One is to use Mathf.Cos along with Math.PingPong.

Something like that:

     // C#
     (Mathf.Sin(Mathf.PingPong(Time.time*Mathf.PI, Mathf.PI))+1)*0.5f*100f

Another way is a Hermite interpolation:

     float Hermite(float t)
     {
         return -t*t*t*2f + t*t*3f;
     }
     
     Hermite(Mathf.PingPong(Time.time,1.0f))*100f

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Bunny83 · Jul 30, 2013 at 10:28 PM 1

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So basically the first solution uses the cosinus (the blue graph) between 0 and PI and ping pongs this part.

The second solution uses a hermite interpolation:

Mathf.PingPong with ease in and out?

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