# Drag equation derivation?

1. Apr 5, 2009

### revolution200

I've been told I can derive the drag equation from momentum and force. I have also found a site that has a similar question but no solution.

http://www.physics.umd.edu/perg/abp/TPProbs/Problems/D/D30.htm

I have an idea of how to get a solution but get stuck.

2. Apr 5, 2009

### Bob S

Use the drag equation derivation given in the web site you referenced, and then recall that

a) Force (Newtons) times dx/dt (velocity, meters/sec) is joules per sec, or power.

b) Force (Newtons) times dt (sec) = mass times velocity = momentum change in time dt. (acceleration (or deceleration) times time is the velocity change in time dt). (recall that 1 Newton = 1 kilogram-meter per second^2)

You need to watch units very carefully. By the way, your reference is the drag derivation for high Reynold's number (turbulent flow). See the reference below for both types of drag.

http://en.wikipedia.org/wiki/Air_resistance

Last edited: Apr 5, 2009
3. Apr 5, 2009

### revolution200

Thank you, I've been told so many times that this doesn't exist. Thank you

4. Apr 5, 2009

### revolution200

I apologise but I've just tried using these references and I just don't understand. I'm not brilliant at these sort of problems could you please explain how these relate to the terms. Are they the description of the collisions with the wall and the particle. Do I put the force equal to the results you have given me?
m(dx/dt^2)=??
Apologies for my incompetence

5. Apr 5, 2009

### Bob S

OK. Iam not going to do your homework, but I am going to show you how to use the equations

Power
The power required to overcome the aerodynamic drag is given by:

P (power) = (1/2) rho A Cd v3

where rho = air density = 1.28 Kg per meter3

A = frontal area of car = 2.5 meters2

Cd - air drag coefficient = 0.32 (unitless, typical)

v= 100 kilometers per hour = 27.8 meters per second (about 62 mph)

Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times the work done in half the time requires eight times the power.

Let's use the numbers and units above

P = (1/2) (1.28) (2.5) (.32) (27.8)3 = 11,000 (what are units?)

units =(Kg/m3) (m2) (m3/sec3) = (Kg-m/sec2) x (m/sec) = Newton-meters/sec = Joules/sec = watts

Converting to horespower we have P = 11,000 watts x 1Hp/746 watts = 14.7 Hp

OK. Now here is a problem for you.
Gasoline conains about 44,000 joules per gram of energy (heat of combustion)
there are 2750 grams of gasoline per U. S. gallon
The internal combustion engine (plus transmission) is about 15% efficient
You are burning gasoline at the rate of 11,000 watts (62 miles per hour-see above) for air drag

If your car is getting 20 miles per gallon, what fraction of that is air drag?

6. Apr 6, 2009

### revolution200

Thank you I really do appreciate your help and time. However it isn't the terms themselves that I don't understand.
If I have a block moving through the air with a frontal area A, it is pushing A times rho (units per Volume) particles of air. The force F(drag) created from this is from the momentum from the collision of these particles. What I'm stuck with is how do describe this in terms of momentum.
Now I know momentum is P = mv

therefore

P = m(dx/dt)

Now Force is the rate that momentum is done

this is where I get stuck

do I integrate from t0 to tn?

this gives

F(drag) = A*rho*m(integral from t0 to tn of (dx/dt).dt)

It can't be this because the drag equation doesn't have m and it doesn't make much sense anyway. Hence I'm stuck.

How do I calculate the rate? You can see from the drag equation that it is some integral since we have 1/2(v^2)

Thanks again for you response

I think it is something to do with this sorry for the long post

They are related by the fact that force is the rate at which momentum changes with respect to time (F = dp/dt). Note that if p = mv and m is constant, then F = dp/dt = m*dv/dt = ma. On the other hand, you can also say that the change in momentum is equal to the force multiplied by the time in which it was applied (or the integral of force with respect to time, if the force is not constant over the time period).

7. Apr 7, 2009

### Bob S

I attach a pdf copy of my derivation of air drag for automobiles. This differs slightly from the common derivation on the web, but it is easier and gets the same result.

#### Attached Files:

• ###### Aerodynamic Drag for Autos.pdf
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Last edited: Apr 7, 2009