   Terminal Velocities
  The terminal velocity of a sky diver is maybe 120 mph when the limbs are spread in the classical sky diving pose.  But, how about other more aerodynamically shaped objects like a bomb, a jet fighter, or a metallic ball?  What speeds might these objects reach when falling un-powered through the atmosphere?Does anyone know, from experience or calculations, what these values might be?  My physics engine is crude, but has basic thrust, drag, and gravity factors.  If I can tweak it against a few real-world values, it will be more than adequate for game play.  Thanks.   
  Terminal velocity depends on the mass and cross-section area of the object.There's an equation at http://en.wikipedia.org/wiki/Terminal_velocity. The page doesn't state what C_d or rho is, but you can easily calculate them by putting in the values for the skydiver.edit: Instead of explicitly setting a terminal velocity for every object, you may want to implement drag instead (ie calculate the air resistance based on speed and cross-section area). See http://en.wikipedia.org/wiki/Drag_equation.   ;Quote:Original post by eightyTerminal velocity depends on the mass and cross-section area of the object.There's an equation at http://en.wikipedia.org/wiki/Terminal_velocity. The page doesn't state what C_d or rho is, but you can easily calculate them by putting in the values for the skydiver.edit: Instead of explicitly setting a terminal velocity for every object, you may want to implement drag instead (ie calculate the air resistance based on speed and cross-section area). See http://en.wikipedia.org/wiki/Drag_equation.According to that page Cd is the drag (coefficient, this links to the second link you posted), and Rho is the classical representation (of which I learned it) it's the density of the "fluid".   ;Quote:Original post by mldaalderAccording to that page Cd is the drag (coefficient, this links to the second link you posted), and Rho is the classical representation (of which I learned it) it's the density of the "fluid".What I meant was that their their typical numeric values weren't stated.   ;Quote:Original post by eightyTerminal velocity depends on the mass and cross-section area of the object.Thanks for the link.  The terminal velocity equation also requires a drag coefficient, which I don't have.  What is the drag coefficient of a sphere, a bullet, or an F-18 jet?When I drop my jet dead-stick from 50,000 feet, it reaches about 820 mph, then begins to slow down at around 20,000 feet due to the thicker atmosphere.  Is this close to real-world?  Any pilots out there?  Thanks.;  I don't know how you can come up with the drag coefficient for your objects. I'm sure this is done using experimental methods, and furthermore it depends on the nature of the surface of the object along its direction of motion, so you probably wouldn't make much of it even if you knew it already. Unless your object moves on a straight line, doesn't rotate and the wind doesn't change direction.I have never tried to implement what I'm going to suggest, but it seems quite reasonable to me, and I believe it should work.My suggestion is:Approximate the shape of your object with a bunch of planar quads/triangles. The approximation doesn't have to be fine -as in shadowing or collision check proxy meshes. Something coarse will do, it doesn't even have to be convex...E.g. if you want to test that F-18, use a couple of triangles for its front wings, and a bunch of quads for the hull.  Find which of these shapes are facing towards the direction of motion at the given time. You can do this by taking the dot product of the shape's normal and the normalized vector of speed. Keep only those primitives that yield positive dot product.  Multiply each dot product with the area of the respective primitive. This gives an idea of how much of that primitive's area is exposed to atmospheric air as your object moves.If your object has moved by a distance 'dx' since previous frame, along its direction of motion, then each of the above primitives had dislocated a volume of air, equal to: (primitive's area)*(abs(dot product with speed vector))*dxNow, atmpospheric air in sea level weighs around 1.29 grams/litre. This is 20% molecular weight of oxygen (32) and 80% molecular weight of nitrogen (28) divided by 22.4, which is the number of litres in one mole.(I think this implies temperature of 15 degrees Celcius, but it's not that important.)Remember the fact that grams and pounds are merely measurements of force, and -therefore- you have calculated how much force is applied to your object, from atmospheric resistance, at the given time, and as an extra bonus, you also know (roughly) *where* on your object it is applied. (allows to calculate torques and stuff)You can also find the air pressure on your objects surface. This is pretty cool, you can even simulate an object "breaking" under extreme pressure!I don't know if this is used in physics simulations but I'd deginitely go for it if I wanted accurate air drag modelling.edit: the force of weight F on a mass 'm', is m*g, where F is calculated in newtons, mass in kilograms, and gravity in meters/second^2.You'll need that to convert the air mass to actual force.You may also want to model air density decreasing with height.The page linked by 'eighty' provides a formula for calculating it.[Edited by - someusername on December 10, 2005 6:56:53 PM];  oops! Ignore the edit notes at the end of my previous reply. I'm obviously wrong about calculating the force that way! You can't use the air's weight. It doesn't even depend on the air speed/direction!You'll probably have to apply the principle of preservation of momentum to solve a different collision for each primitive in your proxy mesh.E.g. the air volume with mass m1 and velocity v1 collides with the F-18 mass m2 and velocity v2.Then m1*v1 +m1*v2== m1*v1'+ m2*v2'where v1',v2' are the velocities after collision and (v1'/v2')=(m2/m1)Of course don't expect any air particles to bounce off at the speed of light after collision! I doubt this is can be considered as elastic collision!Anyway, the whole point of this was that, you can use the values you'll come up with, and plug them in equations to solve for the coefficient of drag.   