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If I set linearDamping on a body, it's velocity will be decreased slowly, until it stops. The documentation says, that the units are 1/time, which is unclear for me.

If the velocity of a body is 10 m/s, and the linearDamping is 0.5, then how can I calculate that when will the velocity be decreased to 1 m/s ?

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2 Answers 2

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I worked out a closed formula to calculate the velocity v' after some elapsed time t with a given linearDamping D and original velocity v:

v' = v*exp(-D*t)

To calculate how much time is needed to reach a given velocity v':

t = ln(v' / v) / (-D)

Using the parameters from the question:

t = ln(1 / 10) / (-0.5) = 4.605

The object needs 4,6 seconds to decelerate from 10 m/s to 1 m/s

Note:

The original source code to calculate new velocity inside the step function:

v *= b2Clamp(1.0f - h * b->m_linearDamping, 0.0f, 1.0f);

To calculate the distance the object should move over this interval, assuming continuous physics decoupled from any timestep, we can take the integral of the speed over that interval.

$$\begin{align} v(T) &= v_0 e^{-D T}\\ \Delta p(T) &= \int_0^T v_0 e^{-D t} dt\\ \Delta p(T) &= \left[ \frac {v_0} {-D} e^{-Dt} \right] \bigg\rvert_0^T\\ \Delta p(T) &= \frac {v_0} {-D} \left(e^{-DT} - e^0\right)\\ \Delta p(T) &= \frac {v_0} {D} \left( 1 - e^{-DT}\right)\\ \end{align}$$

So we can see the object asymptotically approaches a distance of \$\frac {v_0} {D}\$ from its initial position.

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  • \$\begingroup\$ thanks for your closed formular. how about distance? I want to find a formula to calculate the distance not depending on timestep. \$\endgroup\$
    – shtse8
    Dec 7, 2022 at 18:08
  • \$\begingroup\$ @shtse8 I saw this comment late due to a proposed edit giving the wrong formula. I've edited the question to show the formula to use to calculate the distance travelled in T seconds in a timestep-independent way, in case it's still useful to you. Note that the actual Box2D simulation will deviate from this formula depending on the timestep it uses because it's only approximating the continuous physics in discrete steps. \$\endgroup\$
    – DMGregory
    Oct 23, 2023 at 15:36
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I looked at the Box2D source code, where it applies damping:

v *= b2Clamp(1.0f - h * b->m_linearDamping, 0.0f, 1.0f);

So the loss of velocity is certainly not a fixed number of units per second, despite what the manual says.

Your example of 10m/s to 1m/s would also slightly depend on timestep. Let's assume 120Hz simulation step...

#!/usr/bin/python

v = 10
step = 0
h = 1/120.0
time = 0.0

while v > 1 :
    v *= ( 1 - h * 0.5 )
    step += 1
    time += h
    print( "step %d, time %f, v %f" % ( step, time, v ) )

Then it would take 4.6 seconds to slow down by that much.

...
step 551, time 4.591667, v 1.001957
step 552, time 4.600000, v 0.997783

I checked with 1/60 timestep as well, and the result is almost the same, btw.

I am not sure if you can determine this time delay in a single expression, though, without stepping through it.

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  • \$\begingroup\$ Thank you for your help! In the meantime, I was able to work out a closed formula, which produces a result almost the same as your python script \$\endgroup\$
    – Iter Ator
    Jun 23, 2018 at 15:21

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