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Question by metroidsnes · Feb 14, 2018 at 04:41 PM · physicsfixedupdatetimestep

How to make code in FixedUpdate() independent from fixed timestep?

If you put a physics code into FixedUpdate(), e.g. rigidbody.AddForce(force) and your fixed timestep is set to default 0.02 (50 FPS) then your code will be executed 50 times per second. But if you decide to change fixed timestep to 0.01 (100 FPS) the same force will be applied 100 times a sec so it'll have effect at gameplay.

If this happened mid-development, it could break the game and the only way to fix it would be to figure-out all the force values (in the inspector) yet again.

Is there a way to make physics calculation in FixedUpdate() independent from fixed timestep? Similar thing can be done for Update() by using Time.deltaTime but it doesn't seem to work in FixedUpdate().

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Answer by Commoble · Feb 14, 2018 at 06:11 PM

 FORCE_PER_SECOND = 50;    // or whatever
 
 rigidbody.AddForce(FORCE_PER_SECOND * Time.deltaTime);
 // when used inside the FixedUpdate, Time.deltaTime is equivalent to Time.fixedDeltaTime
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avatar image metroidsnes · Feb 14, 2018 at 07:02 PM 0
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This is my method for applying force:

alt text

It makes a cube hover nicely above the ground. In play mode, when I change the timestep from 0.01 to 0.1, the cube starts jumping high ins$$anonymous$$d of hovering.

devenv-2018-02-14-19-57-53.png (7.4 kB)
avatar image nt314p metroidsnes · Feb 17, 2018 at 02:15 PM 0
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As @Commoble was saying, you can either call your ApplyForce() method inside FixedUpdate, or you can use Time.fixedDeltaTime.

avatar image metroidsnes nt314p · Feb 17, 2018 at 02:56 PM 0
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But how does it solve the problem? If I put my physics code in FixedUpdate() or use Time.fixedDeltaTime and then change fixed timestep, physics will get broken. The question I asked was how to make code in FixedUpdate() independent from fixed timestep.

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