First it''s far easier to do this is you have a fixed time frame: calculations are simpler it''s especially difficult to debug a time frame depending on a frame rate that can vary greatly depending on hardware and OS. I would suggest seeing if you can put the dynamics in a seperate thread from the rendering as in the long run it will simplify things. E.g for your example instead of using expensive functions like pow you can use simpler linear expressions such as:

v = v - k * v

As for terminal velocity, this should work with a fixed time frame: with fixed frame rate drag is a constant, and as velocity increases the amount of velocity lost to drag each time step also increases. Eventually it matches the thrust/accleration, so it cannot go any faster. With variable time frame it should still work, though it may be destabilised by a variable frame rate.

One thing you can try is using quadratic drag. I.e. make the drag proportional to v2 instead of v. This is more realistic and has the benefit that it cuts off high speeds much more quickly. It has the opposite effect at low speeds though - things moving slowly are much less effected - so it''s often best to use a combination of both v and v2. Something like (in 1D)

v = v - k * v - m * v2

miles, I'm not sure why you write,

velocity = velocity/drag;

I mean, you do reduce the velocity but that really isn't a true drag or friction. That is, a drag or friction is a force that affects acceleration, and thus indirectly affects velocity. Your approach may work if you get the coefficients worked out, but is not a drag or friction force. Its great as long as it does what you want.

A true drag acts opposite velocity not acceleration.

Why not apply a viscous drag, which is like an aerodynamic drag? It may work well for a varying time step. Apply the viscous drag force along a vector direction opposite the direction of velocity over whatever time step your current frame produces. Viscous drag naturally goes to zero as velocity goes to zero, so it scales naturally.

The magnitude of viscous drag can be calculated using this formula:

  viscous_drag_magnitude = C_d * rho * vel * vel * ref_area  

Here, C_d is a drag coefficient. For blunt bodies (i.e., spheres or similar) at low speeds start with C_d = 1.0 and tweak. rho is the density of whatever fluid the object is moving in (use rho = 0.0023769 slugs/cubic foot or 1.225 kg/cubic meter for air at sea level) vel is the magnitude of velocity of the object relative to the fluid (just the magnitude of velocity in world space if the fluid itself is not moving----i.e., there is no wind). Make sure vel is consistent with your units for rho. And ref_area is a reference area, for a blunt body use the frontal projected area---there are of the object as projected onto a plane towards its front side. Make sure the units are consistent (if you use rho in slugs/cubic foot, ref_area should be square feet. If you use kg/cubic meter, ref_area should be square meters.) Drag will be in pounds if you're using slugs and feet and it will be in Newtons is you're using kg and meters.

(For units people, 1 pound is equal to 1 slug-foot/second_squared, and 1 Newton is equal to 1 kilogram-meter/second_squared----which you can figure since F = ma and the units on both sides of "=" have to be the same.)

Anyway, this approach may be more than you want to do but it might work.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

Edited by - grhodes_at_work on July 25, 2001 10:41:48 AM

ok, here's how it works. x=x'+v*dt+a*dt^2, dt being time increment. that's euler integration, which i assume is what you're using, although obviously without the time step. it's horrible for anything attempting to even remotely approximate real world physics, you should look into verlet or leapfrog integration, but i digress. dt = t - t', or current time stamp minus the last frame's time stamp. that would be your "10 milliseconds or perhaps 3 or maybe 45."
now, here's how you would apply the drag force as explained by graham. as you might remember, a=F/m. F here, for all intents and purposes, is the combined force acting on the object. in other words, you add up all the force vectors that you're dealing with and apply them to the object as follows(still using euler. definitely look into verlet, P* = 2P-P'+a*dt^2, it'll make your life a lot easier)

P += V;
V += A;

if you want the drag equation to work properly, you definitely don't want to put ref_area into a composite drag coefficient. it depends on which way the object is moving relative to its local space, in other words, a sheet of cardboard dropped parallel to the ground will encounter a lot more drag than a sheet of cardboard dropped perpendicularly to the ground. a simple approximation (an incorrect one, but it'll work) would be to project the geometry onto a plane perpendicular to the direction of travel and compute the area of the boolean OR of the resulting polygon set.

also, you seem to be confusing drag with friction.. they aren't the same thing.

-goltrpoat


PS - just realized that you're in 2d, not 3d.. so disregard what i said about the plane projections in regards to the reference area

Edited by - goltrpoat on July 25, 2001 12:30:05 AM