How to calculate the time it takes to reach terminal velocity under constant acceleration?

Question

I'm using a rigidbody in Unity3D which is under constant, not changeing acceleration. The rigidbody can have a drag value of > 0. Now I'm trying to figure out how long it will take for the object to reach a certain speed. I understand that finding the time it takes to reach terminal velocity might not be all that useful, so I'm also interested in finding the time needed to reach a certain percentage of terminal velocity.

Known values include mass, force applied to the rigidbody, terminal velocity and drag. I'm mainly looking for a solution that ignores gravity, but I'm also very interested in one that doesn't.

I hope you can help me with that, because the only thing I learned from google is that I really should take a few math classes... :/

Answer 1

With Time in seconds, Speed in m/s and Acceleration in m/s/s, with Acceleration being a constant and Speed being known

Time = Speed/(Acceleration - Drag)

If Drag is a constant as a reaction to the acceleration

If I'm not mistaken...

Answer 2

From the start of the fall (time=0) to reaching terminal velocity (time=T), drag is not constant because drag is directly proportional to the square of velocity. It is quite easy to approximate the dynamics numerically using small (say 1ms) time steps.

In the first step (t=0 to t=1ms), the object with mass M would face a force of Mg and speed up to v1 = 9.8E-3 ms-1 without air resistance or drag. During the same time interval, drag would go from 0 to D1 = k*(9.8E-3)^2, k is 0.5*CdADen (Cd is coefficient of drag - estimated/guess, A is approximately the cross sectional area in the direction of the fall, Den is air density).

So at the start of the second time step (t=1ms to t=2ms) the object faces a net acceleration force of 9.8M - k*(9.8E-3)^2 and speeds up to v2=(9.8E-3 + (9.8M - k*(9.8E-3)^2)0.001) and drag at t=2ms increases to D2 = k(v2)^2.

Eventually, D = 9.8M at some time step.

There is of course an analytical solution using integration to replace the numerical method above.

Answer 3

Sorry the above answer assumes a free fall under gravity, but the same method works for acceleration in any direction.

And k assumes that you have constant density (air or whatever fluid in which the object is moving). This may not be the case say for example if you have an object falling from a very high altitude. k breaks down to density and viscosity of the fluid which in turn changes with temperature.