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## DeltaTime and acceleration

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### #1panicq  Members

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Posted 21 August 2014 - 04:29 AM

Hello,

I've a little problem of logic in a game i'm currently working on.

I've vX += acceleration * dT which represent the velocity in X

and x += vX * dT;

I've done so because of the reading of book. But after thinking about it and tying with values It seams not really accurate:

imagine acceleration = 1 and two computers A with dT = 1s and B with dt = 2s.

With A:

at the beginning vX = x = 0;

vX += 1 * 1; // = 1

x += 1 * 1; // = 1

vX += 1 * 1; // = 2

x += 2 * 1; // = 3

vX += 1 * 1; // = 3

x += 6 * 1;  // = 6

so at 1s x = 1 px

at 2s x = 3 px

at 3s x = 6 px

at 4s x = 10 px

With B:

vX += 1 * 2; // = 2

x += 2 * 2; // = 4

vX += 1 * 2; // = 4

x += 4 * 2; // = 12

so at 2s x = 4 px

at 4s x = 12 px

So here we clearly see that at 4s both x values are not at the same position, wich is clearly not ideal for my gameplay.

Any Idea on what should be really done ?

Thx

### #2Lactose!  GDNet+

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Posted 21 August 2014 - 04:37 AM

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Any Idea on what should be really done ?

The solution is to fix your timestep.

This will keep deltaTime consistent across multiple machines, meaning all calculations using it end up the same.

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### #3panicq  Members

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Posted 21 August 2014 - 04:41 AM

Hum ok, i'll see. And if I don't wan to fix the timestep, wich is the case in many games ?

Edited by panicq, 21 August 2014 - 04:42 AM.

### #4Hodgman  Moderators

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Posted 21 August 2014 - 06:30 AM

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What you're doing here is called "numerical integration" -- integrating a function by summing up small steps, instead of solving an equation analytically (e.g. o=u*t+1/2*a*t^2)

The particular technique you're using is "Eulers method". It's one of the most common in games, but yes, does give different answers depending on your step sizes.

You can actually notice in many games that the distance you can jump depends on your frame rate!!!
As above, a foolproof solution to make it deterministic is to choose a fixed simulation time-step.

You can also look into alternative integration schemes, such as RK4.

Edited by Hodgman, 23 August 2014 - 07:23 PM.

### #5DiegoSLTS  Members

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Posted 21 August 2014 - 02:55 PM

You can keep track of the elapsed time and instead of doing:

v += a*dt;
x += v*dt;


you can do:

elapsedTime += dt;
v = v0 + a*elapsedTime;
x = x0 + v*elapsedTime;


where v0 and x0 are the initial values of velocity and speed when the object starts moving.

If the object can change it's direction (meaning a different acceleration), you set a to the new value, v0 to the velocity at that moment and x0 to the position at that moment.

Edited by DiegoSLTS, 21 August 2014 - 02:58 PM.

### #6Zaoshi Kaba  Members

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Posted 22 August 2014 - 01:23 AM

Had you used correct formula you'd have identical result.

// dT = 1

x += (vX + vX + acceleration * dT) * dT * 1/2; // == 0.5
vX += acceleration * dT; // == 1

x += (vX + vX + acceleration * dT) * dT * 1/2; // == 2
vX += acceleration * dT; // == 2

x += (vX + vX + acceleration * dT) * dT * 1/2; // == 4.5
vX += acceleration * dT; // == 3

x += (vX + vX + acceleration * dT) * dT * 1/2; // == 8
vX += acceleration * dT; // == 4
// dT = 2

x += (vX + vX + acceleration * dT) * dT * 1/2; // == 2
vX += acceleration * dT; // == 2

x += (vX + vX + acceleration * dT) * dT * 1/2; // == 8
vX += acceleration * dT; // == 4

However as other's mentioned this isn't the only place that gets undetermined without fixed timestep.

Edited by Zaoshi Kaba, 22 August 2014 - 03:37 AM.

### #7Lactose!  GDNet+

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Posted 22 August 2014 - 03:35 AM

Had you used correct formula you'd have identical result.

You've got a typo, 5.5 should be 4.5.

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Apart from that, one of the reasons you (panicq) should use a fixed timestep is related to collision checking/handling. You can actually see it from Zaoshi Kaba's example; what if there's something you'd collide with at x = 4.5 (after 3 seconds), but which you would NOT collide with at x = 8 (after 4 seconds)?

How would you solve this issue? Now imagine having to deal with/special case all other issues like this. OR you could fix your timestep (a 1 time thing), and have the results deterministic regardless of what framerate the game is running at.

Note that even though the logical simulation (handing movement, collision, etc.) should have a fixed timestep, the render part does not need to have a fixed timestep. If you wanted (for some reason), you could force rendering to occur every 3 seconds, without impacting any of the logical/gameplay stuff.

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Can you give an example of why a game would not want to have a fixed timestep? You claim this is the case for many games, but I can't think of a single game for which this is true.

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### #8Aardvajk  Members

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Posted 22 August 2014 - 05:27 AM

Hum ok, i'll see. And if I don't wan to fix the timestep, wich is the case in many games ?

This was the case in at least one mainstream game I heard of (I think it was Quake 2) and as a result, there became a known exploit that players on faster computers could jump higher and further, which became an issue in multiplayer.

I'm not sure why you think a fixed time step is not the case in "many games"? Where are you getting that from? This isn't something you can tell from playing a game (other than deducing its absence as the cause of bugs).

Fix your physics timestep. Seriously. There are no good reasons not to and many good reasons to do so.

Edited by Aardvajk, 22 August 2014 - 05:29 AM.

### #9SeanMiddleditch  Members

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Posted 22 August 2014 - 12:06 PM

I'm not sure why you think a fixed time step is not the case in "many games"? Where are you getting that from? This isn't something you can tell from playing a game (other than deducing its absence as the cause of bugs).

Indeed, the common physics middleware most games use all but mandate a fixed timestep.

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### #10panicq  Members

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Posted 27 August 2014 - 09:37 AM

Ok, thank for advises, I'll fix my frame rate for physics and let a free frame rate for my logic.

Have a nice day.

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