How to find terminal velocity from experimental data?

AI Thread Summary
To find terminal velocity from experimental data, the equation V = Vt(1 - e^(-bt/m)) is used, where Vt is terminal velocity, m is mass, and b is a constant. The discussion suggests using a system of equations based on measured velocities at different times to solve for Vt. With 56 data points available, the user is encouraged to plot velocity against time to identify trends. A method involving squaring and dividing equations derived from the velocity data is proposed to isolate Vt. Additionally, using a data-processing program like Excel can help analyze the data further.
greco117
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Homework Statement



So I have a set of data, velocity and time, of a falling object. So the question is: how can I get the value for the terminal speed?

Homework Equations


Since we're talking about terminal velocity an ecuation that descrives a Velocity vs time graph should be of the form:
V =Vt\left(1-e^\left(\frac{-bt}{m}\right)\right)

where Vt is ther terminal velocity
m is the mass
b is a constant which depends of the caracteristics of the body
and t is the time


The Attempt at a Solution


So I need to fit my data with an ecuation of this style,my first thought was that I had to do a logistic regression but that didn't look right (the ecuation is similar but is not the same),
Then I thought doing a little of algebra
Vt-V=Vt\left(e^\left(\frac{-bt}{m}\right)\right)
And then All I needed to do was an exponential regression but again that doesn't look like the best way because I'm already giving Vt a value
 
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If you have got measurement data at time t1 and 2t1 you can determine Vt by solving the system of equations
Vt-V1=Vte-Bt1
Vt-V2=Vte-2Bt1.
(B=b/m)

Square the first equation and divide by the second. The exponentials cancels, and you get an equation for Vt in terms of V1 and V2, easy to solve.

Find as many pairs as you can, and take the average of Vt-s.

ehild
 
How many data points do you have?
Do you have enough to plot a graph of v against t?
 
ehild said:
If you have got measurement data at time t1 and 2t1 you can determine Vt by solving the system of equations
Vt-V1=Vte-Bt1
Vt-V2=Vte-2Bt1.
(B=b/m)

Square the first equation and divide by the second. The exponentials cancels, and you get an equation for Vt in terms of V1 and V2, easy to solve.

Find as many pairs as you can, and take the average of Vt-s.

ehild
So you're saying that I should do the following?
(Vt-V1)^2=Vt^2*e^-2bt
and then divided by
(VT-V2)=Vt*e^-2bt
and then I should get:
(Vt-V1)^2/(Vt-V2)=Vt
but that ecuation is not so easy to solve unless I've misunderstood the procedure
 
truesearch said:
How many data points do you have?
Do you have enough to plot a graph of v against t?
yes, I have 56 data points, the problem lies in the graph because the points don't seem to have a tendency to a certain value, I thought about taking the maximum as VT but I don't like that solution.
Any suggestion?
 
greco117 said:
So you're saying that I should do the following?
(Vt-V1)^2=Vt^2*e^-2bt
and then divided by
(VT-V2)=Vt*e^-2bt
and then I should get:
(Vt-V1)^2/(Vt-V2)=Vt
but that ecuation is not so easy to solve unless I've misunderstood the procedure

That equation is equivalent to (Vt-V1)^2=Vt(Vt-V2).
Expand both sides of the equation. : Vt^2-2VtV1+V1^2=Vt^2-VtV2.

Vt^2 cancels. -2VtV1+V1^2=-VtV2. Isolate Vt.
Vt=V1^2/(2V1-V2)


ehild
 
thank you!
seems obvious now, that would do it
 
I guess you have got some data-processing program, Excel perhaps? As you have enough data points, it is possible to get the approximate derivative of the V(t) function. V' = VtBe-Bt. The plot ln(V') vs time has to be straight line, with slope -VtB and crossing the vertical axis at VtB.

ehild
 
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