Finding the terminal velocity of a model rocket from a list of velocities

[LT72884]

move the formula down one cell. Then just stop it at the end. you are going to truncate the data a bit more to remove some outliers, or you could remove them point by point if you feel like it.

here is what i have for beta, using avg acel for dv/dt. B=(-ma-mg)/v^2

the numbers are odd, and when i plot it.... man it looks nasty

 

[Chestermiller]

I want you to take the average acceleration across each time step (just for the truncated data).

$a_{avg} = \frac{ v_2 - v_1 }{ t_2- t_1}$

and arithmetic average velocity over the same time step

$v_{avg} = \frac{v_1 + v_2}{2}$

You are then going to plot $\beta$ vs $v_{avg}$

This is the exact trick we used in atmospheric modeling to determine the final steady state ozone column without running the calculation out to infinite time.

 

[LT72884]

This is the exact trick we used in atmospheric modeling to determine the final steady state ozone column without running the calculation out to infinite time.

thats cool. Its been great the last couple of days with this personal challenge of mine. This entire question is not for an assignment, but for my understanding and growth. I am so used to bookwork and homework questions, that i decided to branch out and ask myself how i would do this. I will USE this data though in my senior capstone report.

 

[erobz]

here is what i have for beta, using avg acel for dv/dt. B=(-ma-mg)/v^2

the numbers are odd, and when i plot it.... man it looks nasty
View attachment 323047

$a_{avg}$ is negative. The acceleration of the rocket is opposite its direction of motion. What you have is positive. What calculation did you do for that? After you fix it, please share the plot so we can try to interpret it. Or maybe you can offer you interpretation of it... more than the "numbers are odd, and it looks nasty", think about if its average behavior aligns with any expectations about $\beta$.

 

[LT72884]

$a_{avg}$ is negative. The acceleration of the rocket is opposite its direction of motion. What you have is positive. What calculation did you do for that? After you fix it, please share the plot so we can try to interpret it. Or maybe you can offer you interpretation of it... more than the "numbers are odd, and it looks nasty", think about if its average behavior aligns with any expectations about $\beta$.

i used a=(v2-v1)/(t2-t1) but if it needs to be -, then i will switch it around

 

[erobz]

i used a=(v2-v1)/(t2-t1) but if it needs to be -, then i will switch it around
View attachment 323051

If you did use that formula you would be getting a negative value, and you aren't.

 

[LT72884]

If you did use that formula you would be getting a negative value, and you aren't.

exactly. But excel shows that i am using the correct formula. Ill just reverse the C3 and C2 values, that will produce negative numbers. I just dont know if beta is correct.
if i let B=(rho)(Cd)(Ar)(0.5) and solve for cd, i get 0.72 for the first entry. Now, Cd does change ALOT during a flight due to many factors. So a plot of Beta should look almost like a switch back type trail..

 

[LT72884]

if i did indeed solve for the Cd correctly, i can use that in the terminal velocity equation by taking the average Cd over a specific time. When velocity is 0.3m/s my drag coef is -44,000 so its gotta be the right numbers.

or did you have a different aproach to find Tv?

 

[erobz]

if i did indeed solve for the Cd correctly, i can use that in the terminal velocity equation by taking the average Cd over a specific time. When velocity is 0.3m/s my drag coef is -44,000 so its gotta be the right numbers.

or did you have a different aproach to find Tv?

Well, you need to solve for the drag coefficient $C_d$ from $\beta$. What do you calculate as $\beta$, how did you get it? Please elaborate on what you are doing to get these numbers. Can you please share the plot? I think the variation you are seeing is noise, not the variation of $\beta$ as a function of $v_{avg}$.

 

[LT72884]

Well, you need to solve for the drag coefficient $C_d$ from $\beta$. What do you calculate as $\beta$, how did you get it? Please elaborate on what you are doing to get these numbers. Can you please share the plot? I think the variation you are seeing is noise, not the variation of $\beta$ as a function of $v_{avg}$.

B= (Cd)(Area)(rho)(0.5)
Beta is all the constants. From here i just plug in what i have and get the Cd. Bellow is the plot of beta.

 

[erobz]

B= (Cd)(Area)(rho)(0.5)
Beta is all the constants. From here i just plug in what i have and get the Cd. Bellow is the plot of beta.

View attachment 323057

Yeah, thats noise. Just do scatter plot, no line.

 

[LT72884]

Yeah, thats noise. Just do scatter plot, no line.

that is a scatter plot. Just connected with a line is all. if i remove the line, its the same mess haha

 

[erobz]

that is a scatter plot. Just connected with a line is all. if i remove the line, its the same mess haha

View attachment 323058

How about some axis and scale?

 

[LT72884]

How about some axis and scale?

y axis is values where beta falls, x axis is the cell number. I have cells 2-39 selected

 

[LT72884]

even plotted against velocity as the x axis or even as a data set, its still looks scattered

 

[erobz]

even plotted against velocity as the x axis or even as a data set, its still looks scattered

You are supposed to be plotting $\beta$ vs $v_{avg}$? That is $\beta$ vs time; It's not really as useful. The purpose of this is to determine if $\beta$ depends on the velocity "$v_{avg}$"

 

[LT72884]

You are supposed to be plotting $\beta$ vs $v_{avg}$? That is $\beta$ vs time; It's not really useful. The purpose of this is to determine if $\beta$ depends on the velocity "$v_{avg}$"

correct, and when i select that data, the velocity is a linear function with negative slope, and B is linear with 0 slope.
So somewhere, either excel is trying to be "smart" and adjust the data or i have selected the wrong data haha, which is possible

 

[erobz]

correct, and when i select that data, the velocity is a linear function with negative slope, and B is linear with 0 slope.
So somewhere, either excel is trying to be "smart" and adjust the data or i have selected the wrong data haha, which is possible
View attachment 323060

You've plotted $v_{avg}$ vs time as a separate series. Delete the chart. Put in a blank one and right chart area - click "select data"

$v_{avg}$ is the $x$ values
$\beta$ is the $y$ values

 

[LT72884]

You've plotted $v_{avg}$ vs time as a separate series. Delete the chart. Put in a blank one and right chart area - click "select data"

$v_{avg}$ is the $x$ values
$\beta$ is the $y$ values

then what will the plotted data be? Im not sure i follow, but i will change the axis to match what you have said.

 

[LT72884]

i get the exact same thing

 

[erobz]

then what will the plotted data be? Im not sure i follow, but i will change the axis to match what you have said.

It will be $\beta$ vs $v_{avg}$

 

[LT72884]

It will be $\beta$ vs $v_{avg}$

i get the exact same thing as post 104

 

[erobz]

i get the exact same thing as post 104

I don't understand why there are so little a number of points?

This is what it should look like:

 

[LT72884]

I don't understand why there are so little a number of points?

This is what it should look like:

View attachment 323063

me niether. Let me close excel and refresh my RAM.. MAYBE some stupid glitch. I will be back shortly

 

[LT72884]

yup, something is not right. I selected all the average velocities as X and Beta as Y and this is my plot

 

[erobz]

yup, something is not right. I selected all the average velocities as X and Beta as Y and this is my plot

View attachment 323064

Thats because of those data points that are near -30,000! Just chop the data off below 20 m/s.

 

[LT72884]

Thats because of those data points that are near -30,000! Just chop the data off below 20 m/s.

hmm, its better but not the same. i made sure no crazy values were in the mix either.

 

[erobz]

hmm, its better but not the same. i made sure no crazy values were in the mix either.
View attachment 323065

Chop it off at 50 m/s

 

[LT72884]

i will keep chopping data back until it either changes or i find out of a wrong calculation

EDIT: i just saw your post stating the same thing as me.

 

[LT72884]

At 50m/s.

At 55m/s a drastic change