Air drag equation

1. Apr 2, 2008

vettett15

Hey guys, I am working on deriving some equations of motion for an object with air drag.

I am using this site to help me: http://www.grc.nasa.gov/WWW/K-12/airplane/flteqs.html

Here is my problem, at the bottom of that page they have this equation for horizontal velocity:

1. u = dx/dt = Vt^2 * Uo / (Vt^2 + g * Uo * t)

Then they integrate it to get horizontal distance

2. x = (Vt^2 / g) * ln( (Vt^2 + g * Uo * t) / Vt^2 )

When I take equation 1 and put it in matlab to integrate I get this

3. x = (Vt^2 / g) * ln( (Vt^2 + g * Uo * t) )

Now if I take either equation 3 or 2 and differentiate it, I will get equation 1.

How do they get equation 2 from integrating equation 1? Vt is terminal velocity by the way so that is just a constant. Uo is just the initial velocity. I assume in equation 1 when they took the integral on dx becomes x from x0 to x, x0 being 0 so you just get x. The right hand side you would get from t0 t, t0 being 0 so you are left with t.

Thanks,
Pete

2. Apr 2, 2008

Lojzek

Formula 3 is the correct indefinite integral. You must evaluate it at both points (t2 and t1) and subtract. If t1=0 this does not mean that the value of integral is zero. Use the correct values of indefinite integral and log(a)-log(b)=log(a/b) and you will get the same result as eq. 2.

3. Apr 2, 2008

vettett15

retarded I am

Thanks so much for the reply. I can't believe I missed that, I just assumed that everything was multiplied by t so the zero would wipe it all out. Thanks again.

4. Apr 3, 2008

vettett15

another question

I also had another question, on the same website I sent you, they are getting equations of motion in the horizontal and vertical directions. My question is, for both the horizontal and vertical directions they are using the same drag.

If you thought of the horizontal and vertical velocities as independent of each other, the object travels upwards with a velocity and a drag force acting down on it, but the Cd and Area would be totally different than that of the horizontal drag.

Originally I thought to find the drag in the direction of travel and then just get the components for the horizontal and vertical directions, but the problem is that there isn't a constant theta, the direction of travel is always changing.

If I make the assumption that the horizontal velocity is >> than the vertical velocity, can I just ignore the vertical drag?

5. Apr 3, 2008

kamerling

They have this wrong. they have horizontal drag force:

$$\frac{C_d A \rho v_x^2 } {2}$$

and vertical drag force

$$\frac{C_d A \rho v_y^2 } {2}$$

but the total drag force is

$$\frac{C_d A \rho (v_x^2+v_y^2)} {2}$$

and the horizontal and vertical components are:

$$\frac{C_d A \rho v_x (v_x^2+v_y^2)} {2 \sqrt{v_x^2+v_y^2 }}$$

$$\frac{C_d A \rho v_y (v_x^2+v_y^2) } {2 \sqrt{v_x^2+v_y^2 }}$$

6. Apr 3, 2008

vettett15

ok that makes since, going from what you have as total drag to the x and y drag components it looks like you just put a ratio of vx/v and vy/v in there. v equaling sqrt(vx^2+vy^2). That makes sense. Thanks

7. Apr 10, 2008

vettett15

question

kamerling,

I was looking at your formula again and had a question. making Cd*A*p/2 equal to C to make this more simple.

You have total drag force as:

C*v^2 which is the same as C*(vx^2+vy^2)

Then you say the horizontal component is:

C*vx*(vx^2+vy^2)/sqrt(vx^2_vy2)

Which is essentially, correct me if im wrong is saying:

C*v^2*cos(theta)

Where cos(theta) = vx/v = vx/sqrt(vx^2+vy^2)

C*(v*cos(theta))^2 which would just give what you have at the top of your post.

C*vx^2

and

C*vy^2

which if you say the x and y components of the drag force make up the total drag force:

(C*v^2)^2=(C*vx^2)^2+(C*vy^2)^2

C*v^4 = C*vx^4 + C*vy^4

8. Apr 10, 2008

kamerling

9. Apr 10, 2008

vettett15

kamerling,

Thanks for the reply, I see what you are saying if you have a force in line with the direction of motion than that force is F. The components are Fcos(theta) and Fsin(theta). I am thinking along this line of thought. The horizontal and vertical motions are independent, so in the horizontal I have Cd*A*p/2*vx^2 and the vertical I have Cd*A*p/2*vy^2. The problem I see with my thinking is that Cd and A are different between the horizontal and vertical motions.