- #1
vettett15
- 13
- 0
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
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