I think I found it in link #2. Would it be these?
v = v - k * v
v = v - k * v - m * v2
With k being the drag coefficient?
Air Resistance
Yeah, try this
I removed the physics part of my engine atm, but i will try it as soon as possible too, cause i'm programming with 2d too
I removed the physics part of my engine atm, but i will try it as soon as possible too, cause i'm programming with 2d too
In the rigid body simulator I'm developpin', I use the following formula to compute the amount of force the fluid is applying on my rigidbody:
F = 0.5*C*p*A*v²
C -> coefficient of friction of the surface (I usually use something like 0.3)
p -> fluid density(air density is aproximatelly 1.29kg/m³)
A -> surface area in m²
v -> surface velocity in m/s²
This way you get the force in Newtons that the fluid resistance is applying on the surface. Since in my rigidbody simulator I use convex polyhedra objects, I compute the force that is being applied on each triangle of the object and then I just sum all them. It gives me good results. I'm satisfacted with them.
F = 0.5*C*p*A*v²
C -> coefficient of friction of the surface (I usually use something like 0.3)
p -> fluid density(air density is aproximatelly 1.29kg/m³)
A -> surface area in m²
v -> surface velocity in m/s²
This way you get the force in Newtons that the fluid resistance is applying on the surface. Since in my rigidbody simulator I use convex polyhedra objects, I compute the force that is being applied on each triangle of the object and then I just sum all them. It gives me good results. I'm satisfacted with them.
I still cant find the thread with terminal velocities discussion. It would be good to find it, cause there were suggestions for stable solver.
Author
So earths air density is 1.29 kg/m3? Does that mean I have to make the constant 1.293 or just leave it at 1.29? Since I'm working in 2D, I'm not sure if I should make it squared instead of cubed.
heheh...
'kg/m³' its just the unit of the 'density'...it means you must use 1.29...for water for example, you'd use 1000 since the water density is 1000kg/m³
'kg/m³' its just the unit of the 'density'...it means you must use 1.29...for water for example, you'd use 1000 since the water density is 1000kg/m³
Author
Ok, 1.29 it is. Since I'm not worried about friction too much in my case, I can just leave C out then? Which would be this now:
F = 0.5 * 1.29 * Surface_Area * Velocity²
F = 0.5 * 1.29 * Surface_Area * Velocity²
getting C out from the formula, you're assuming your object's surface has a coefficient of friction of 1.0. If you set C to 0.0, you'll get no friction meaning without friction, doesn't exist air resistance. You must put a value at there.
Drag is a force that is related to several things:
If you treat the drag D as just another force (using the equations from our discussion on jumping), then the (simplified) equation for drag force D is this:
D = C * A * V^2;
Where A is the cross-section area of the object
V is the velocity of the object
C is a tuning constant, which should be whatever looks best.
Implement it with your other forces so that
F = D + G + O;
Where G is gravity, and O is other stuff.
Then get your acceleration as usual.
Is this related to your jumping objects?
- The size of the object
- The speed of the object (squared)
- The thickness of whatever the object is travelling through
If you treat the drag D as just another force (using the equations from our discussion on jumping), then the (simplified) equation for drag force D is this:
D = C * A * V^2;
Where A is the cross-section area of the object
V is the velocity of the object
C is a tuning constant, which should be whatever looks best.
Implement it with your other forces so that
F = D + G + O;
Where G is gravity, and O is other stuff.
Then get your acceleration as usual.
Is this related to your jumping objects?
As for jumping, do you really need air resistance for jumping. If you make a character which jumps, then its an big effort for such a simple thing.
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