Hi!

There is an article about car physics at http://home.planet.nl/~monstrous/tutcar.html

There are also information about drag and friction.

Greetings
McMc

quote:Original post by goltrpoat
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.

There''s a problem with your equation. It should be:

x = x'' + v*dt + 0.5*a*dt^2

You were missing the 0.5 coefficient on the last term. Also, that is not an Euler integration *unless* you say that x, v, and a on the right-hand side are all from the previous time step. If you mean for x'' to be the initial starting value of x'' then the equation doesn''t represent an Euler integration. Rather, it represents an exact solution for constant acceleration, a, and in this case the v also must be the initial velocity at the start time. And dt must be measured from the start time.

Also, a minor beef. Note that I''ve capitalized Euler. Euler is a person''s name and when a method, number, or equation is named after a person the word should be capitalized. Any technical editor, book editor, or professional scientist or engineer will correct you on this. (And I am occasionally all of these.)

quote:Original post by goltrpoat
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.

Actually, ref_area does *not* depend on the direction moving relative to the wind. Its a constant for a given shape, and it is defined based on how the drag coefficient is defined, which is independent of the orientation of the object. In fact, it is C_d that changes depending on direction of motion relative to the wind. In the case you mention, the cardboard dropped perpendicular to the ground has much less wind resistance and a much lower C_d than the cardboard dropped parallel to the ground. The ref_area is really just a constant scaling factor on drag, a convenience term that is used to convert a drag force (which has units such as pounds or Newtons) to a unitless value.

But you make a good point. If you have an object that does exhibit significantly different wind resistance (which *is* related to the dynamically changing projected area-----different than ref_area), then C_d must change for realistic drag. For some simple shapes, such as airplane wings at moderate angles to the wind, its easy to come up with an estimate (using a parabolic drag polar, possibly with an additional parabolic low-drag bucket for very low angles relative to the wind). For complex shapes (automobiles) it is not as easy.

I realize I''m going off on a tangent here, that may no longer be important to this topic. Forgive me, but I studied aerodynamics in university for 9 years and its nice to talk about sometimes, .

quote:Original post by goltrpoat
also, you seem to be confusing drag with friction.. they aren''t the same thing.

Hmmm. Of course, a component of aerodynamic drag *is* the friction force between the air and the body and so they really *are* the same thing to a degree. There is a pressure component to drag as well and that is not friction.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.