# Speed loss per second due to air resistance

1. Oct 15, 2014

### otester

Is there a formula covering this? (not including free fall/gravity)

I have the drag coefficient, velocity and every other variable I think I would need for the object in particular.

2. Oct 15, 2014

### billy_joule

What have you tried?
Once you draw a FBD it should become clear..

3. Oct 15, 2014

### NTW

The drag force is the product of the airspeed squared, a coefficient, and the area projected into a plane perpendicular to the wind.

4. Oct 15, 2014

### otester

So you'd multiply the drag coefficient and the velocity to get the loss per second?

5. Oct 15, 2014

### NTW

No... You get a force... Off the head, the equation is f = 0,5*v2*S*rho*Cd, where v is the airspeed, S the area, rho the air density, and Cd the coefficient of drag.

Concerning the 'loss per second', that depends on the magnitude of the force pushing the thing... If it's equal to the drag, the loss will be zero...

6. Oct 15, 2014

### otester

It's for a tank sabot shell, so once fired there wouldn't be any further force pushing it, if that changes anything?

7. Oct 15, 2014

### NTW

Not easy... The sabot will be probably be 'sucked' by the wake of the shell during the first meters of flight, and that suckimg could be a significant force. And there will be probably a tremendous supersonic turbulence there, ... A simple calculation, with equations as the one in the post above, will probably give results that are far from real...

8. Oct 15, 2014

### otester

Yes from what I've read it affects the drag coefficient and I intend to use a ballistic table for a similar shaped bullet to model it on.

I just need the equation to calculate the speed lost per second due to drag.

9. Oct 15, 2014

### billy_joule

Do you know newtons second law?

Have you drawn a FBD?

10. Oct 15, 2014

### otester

I drew it and I think I get it now.

Drag force is 408N.

Using force formula (a=f/m), force = 408N, mass = 3.6kg, the deceleration would be 113 m/s ?

Last edited: Oct 15, 2014
11. Oct 15, 2014

### billy_joule

Assuming your drag force is correct that looks right. Though, you'd lose marks for incorrect units.

Remember, your result is only valid instantaneously, in the next instant the bullet will be travelling slower; the drag force has changed so the deceleration will too.

12. Oct 16, 2014

### otester

Also by incorrect units it should be 113m/s2 ?

13. Oct 16, 2014

### billy_joule

You are welcome. Yes that's the correct acceleration unit

14. Oct 16, 2014

Once the shell leaves the area dominated by the plume of propellant gases coming from the barrel, turbukence will likely play no role.

More concerning is that the drag coefficient will be substantially different for a subsonic vs. a supersonic projectile.

15. Oct 16, 2014

### NTW

Turbulence will come not only from the plume of propellant gas, but from the wake of the shell... And the sabot will probably tumble after parting from the shell... This sabot thing would be a good candidate for a computer simulation.

16. Oct 16, 2014

In a supersonic flow, any of that nonsense occurring behind the projectile does not matter, as information about that flow field (in the form of pressure waves) cannot propagate through the air toward the shell faster than it is already traveling through the air (since it is supersonic in this case). The shell itself is rather small and unlikely to have a turbulent boundary layer, particularly if it is supersonic.

Additionally, the sabot would lose energy much more rapidly and fall away, meaning there is a sizable distance over which the projectile travels in which the only effect is its own motion. That region is the vast majority of the flight path.

Of course the bottom line here is that there is no single answer for how much velocity per second is lost during flight, as the acceleration will not be constant. With some reasonable assumptions, you could set up a pretty simple differential equation to solve to figure out the time history of the velocity over some time frame of interest assuming you know the drag coefficient already.

The drag force is equal to the mass times acceleration of the projectile.
$$F_d = \dfrac{1}{2} \rho v^2 C_d A = m_p \dfrac{dv}{dt}$$
$$\dfrac{dv}{dt} = \dfrac{\rho C_d A}{2 m_p} v^2$$

Assuming all of the constants there are truly constant, then that is easily solvable.

Last edited: Oct 16, 2014