# Opposite of air resistance - Calculating the force

For a project on school I am making a program that can visualize physics in a two dimension space. Not the relativistic and quantum physics, but just the standard high school stuff. One of the things I want to add is air resistance. I know the formula to calculate the air resistance force on an object is: $$F=\frac{1}{2}*\rho*v^2*C*A$$ When the air is going in the opposite direction of the object, it is all fine, but when the object is going in the same direction as the air, problems arise. When the object is going faster then the air, relative you could say the object is going against the air, and I can use the formula. But what if the air is equally as fast as the object, and what if the air is going faster (in the same direction). I would personally think that with a relative velocity of 0, the force would be 0?

My question is: Is my thinking about the force being 0 when the relative speed equals 0 right and what formula can I use to calculate the force of the air molecules pushing on the object when the air is going faster in the same direction, or could i just use the same formula but change the sign of velocity?

• You are correct and the force is the same if the wind air direction is opposite. The $v$ in the formula is namely the relative velocity. Jun 29 '17 at 17:17
• I'll add that if there's air flowing in the direction of travel, you may generate additional thrust instead of drag. I have my doubts that just changing the sign of the force would work though.
– JMac
Jun 29 '17 at 17:24
• @JMac The drag coefficient $C$ naturally has to be changed to fit the geometry of the new incident face, but apart from that I don't see why the situation would change just because of the object's own velocity. Jun 29 '17 at 17:38

You are correct and the force is the same if the wind air direction is opposite. The $v$ in the formula is namely the relative velocity.
The only thing to be aware of, when the air shifts from coming from the front (causing drag) to from behind (causing thrust) is that the incident face suddenly changes as well. The drag coefficient $C$ must fit the incident face.