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David McLaurin · Jul 1, 2009, 12:18 PM

I found the answer to my terminal velocity question in a previous question asked. The problem is that I have attempted to solve this differential equation and cannot seem to come up with a reasonable answer for V(t)=...

The differential equation is -(g/VT^2)*V(t)^2 + g = dV(t)/dt PLEASE HELP!

ebaines · Jul 1, 2009, 02:11 PM

Please clarify the equation - is this it?

$ -( \frac g V T^2 ) [V(t)]^2 + g = \frac {dV(t)} {dt} $

Is T a constant, or did you mean to write it as lower case (time)? Also, what is that first V - the one in the denominator - is that V(t)?

David McLaurin · Jul 1, 2009, 02:31 PM

Originally Posted by ebaines

Please clarify the equation - is this it?

$ -( \frac g V T^2 ) [V(t)]^2 + g = \frac {dV(t)} {dt} $

Is T a constant, or did you mean to write it as lower case (time)? Also, what is that first V - the one in the denominator - is that V(t)?

VT is a velocity (terminal velocity).

David McLaurin · Jul 1, 2009, 02:32 PM

Originally Posted by ebaines

Please clarify the equation - is this it?

$ -( \frac g V T^2 ) [V(t)]^2 + g = \frac {dV(t)} {dt} $

Is T a constant, or did you mean to write it as lower case (time)? Also, what is that first V - the one in the denominator - is that V(t)?

$ -( \frac g VT^2 ) [V(t)]^2 + g = \frac {dV(t)} {dt} $

David McLaurin · Jul 1, 2009, 02:33 PM

$ -( \frac g (VT)^2 ) [V(t)]^2 + g = \frac {dV(t)} {dt} $

ebaines · Jul 1, 2009, 02:46 PM

Like this then:

$ - \frac {g*v^2(t)} {{V_T}^2} + g = \frac {dv(t)} {dt} $

ebaines · Jul 1, 2009, 03:21 PM

I think you'll find that v(t) is of the form:

$ v^2 = A^2(1- e^{-Bt}) $

Play with this a bit and you will find that:

$ a = V_T $

The value for B depends on factors such as the viscosity and densty of the air and the air friction constant for the shape of the object. You need an additional data point to determine B, such as the value of v at t = 10 seconds, say.

David McLaurin · Jul 2, 2009, 07:20 AM

Originally Posted by ebaines

I think you'll find that v(t) is of the form:

$ v^2 = A^2(1- e^{-Bt}) $

Play with this a bit and you will find that
$ a = V_T $ .

The value for B depends on factors such as the viscosity and densty of the air and the air friction constant for the shape of the object. You need an additional data point to determine B, such as the value of v at t = 10 seconds, say.

Thanks for your input. I have all the data points I need from an Excel spreadsheet I created but the spreadsheet itself works off the basic velocity equation. It moves in short timing steps of 0.01 seconds and uses outputs from the previous step to calculate the change in acceleration (as the object's velocity increases, the acceleration decreases due to drag). I'll work with this and see if i can't get this working in a more exacting manner.

The problem with this is that I'm not actually going to use this for Terminal velocity but will use it for hurricane debris impact loads. The hurricane debris velocity equation will be very similar except for the fact that the acceleration will decrease as drag decreases.

Perito · Jul 2, 2009, 08:00 AM

Originally Posted by David McLaurin

$ -( \frac g (VT)^2) [V(t)]^2 + g = \frac {dV(t)} {dt} $

David,

For the left and right parentheses, you can enter "\left(" and "\right)". That will enlarge them and make them fully fill the space. Also, when entering fractions, separate the numerator and the denominator by braces: "\frac {numerator}{denominator}". You used that on the right side of the equation. Spaces sometimes help separate things.

-\left( \frac {g}{(VT)^2} \right) \left[ V(t) \right]^2 + g = \frac {dV(t)}{dt} yields

$ -\left( \frac {g}{(VT)^2} \right) \left[ V(t) \right]^2 + g = \frac {dV(t)}{dt} $

If you click "Quote User", you can see the LaTeX that someone else wrote.

ebaines · Jul 2, 2009, 08:11 AM

For objects being pushed by wind the terminal velocity would be the wind speed, right? The differential equation for this would be similar to what you have started with, except that instead of the object falling under gravity it's being pushed by wind force:

$ \rho A C_D (V_w - v(t))^2 = m \frac {dv} {dt} $

where $\rho$ = viscosity of air, $C_D$ is the coefficient of drag (which depends on the object's shape), $A$ is the frontal area of the object that faces the wind, and $V_w$ is the wind speed. The solution is indeed of the same form as discussed earlier, but with these different constants instead of "g."

David McLaurin · Jul 7, 2009, 07:25 AM

Originally Posted by ebaines

For objects being pushed by wind the terminal velocity would be the wind speed, right? The differential equation for this would be similar to what you have started with, except that instead of the object falling under gravity it's being pushed by wind force:

$ \rho A C_D (V_w - v(t))^2 = m \frac {dv} {dt} $

where $\rho$ = viscosity of air, $C_D$ is the coefficient of drag (which depends on the object's shape), $A$ is the frontal area of the object that faces the wind, and $V_w$ is the wind speed. The solution is indeed of the same form as discussed earlier, but with these different constants instead of "g."

For some reason i can't read this equation

Unknown008 · Jul 7, 2009, 07:34 AM

Try reloading the page...

It's 'rho AC [subscript D] times (V [subscript w] - v(t))^2 = m (dv/dt)'