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

In summary: You can graphing the data, but it's easier if you have the velocity at a certain point in the flight (e.g. at apogee). You could also use a computer program to do the calculation.You experimentally measured the height as a function of time from a freefall of 20k ft?Yes, the height data goes up to apogee only. So from launch to burnout and then to apogee. After 5ish seconds of burn time, it just coasts another few thousand feet to apogee. then the recovery system deploys, so very very little free fall data.The goal is to find the drag coefficient of
  • #71
LT72884 said:
ok, now i see how you get the equation from post 40, and where did quadratic drag force come from?
I’m not sure I’m following what you are asking.
LT72884 said:
your book must be way better than my book. I just went through my section on drag which was 9.3, the rest of chp 9 was on flow. I think they picked this book because it focused more on stocks theorem and fluid flow. We spent one lecture on drag.

View attachment 322982
Different books can have different focuses sometimes. You’ll just have to Google “quadratic drag” in your spare time” and take my word for it for now.
 
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  • #72
the bv^2 term that i plug into the diff eq, thats what i meant haha. where does that bv^2 come from. once i get equation from post 40, im not to sure what values to place in for b. but i will worry about it in the morning and i will integrat first to see what i get. I have a 3 hour train ride home from school and they just had to tow my train so i am delayed an hour.... its almost midnight too and we have 4 feet of snow
 
  • #73
LT72884 said:
where does that bv^2 come from
It comes from the equation you should be already familiar with, namely, Fd = 1/2*rho*v2*Cd*A.

At first approximation, 1/2, rho, Cd, and A are all constant, so it simplifies down to Fd = some constant * v2.

Now, for this, you probably have to account for varying air density, so it would be good if you could look up an atmosphere model (or better still, if you know the temperature that day, a temperature corrected atmosphere model), and then you'll want to basically try to fit your deceleration data to a model using this formula, with 1/2 (obviously), rho (from your atmosphere model), A (from whatever your chosen reference area is), and v2 all known (you know v from your data, which incidentally you should use accelerometer based speed rather than baro based speed if possible, it's much more accurate), and thus you can solve for Cd.

You know Fd because you know the burnout mass of your rocket and you had an accelerometer onboard, though you're going to (at least in my experience) have issues getting any kind of accurate data below a few hundred mph because you start getting limited by accelerometer discretization at that point. Because of this, you can rearrange your equation to Cd = Fd/(1/2*rho*v2*A), and then just solve for your Cd vs V curve.

(Also, the Raven is a great little altimeter, so I'm glad to hear you're using it)
 
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  • #74
LT72884 said:
the bv^2 term that i plug into the diff eq, thats what i meant haha. where does that bv^2 come from. once i get equation from post 40, im not to sure what values to place in for b. but i will worry about it in the morning and i will integrat first to see what i get. I have a 3 hour train ride home from school and they just had to tow my train so i am delayed an hour.... its almost midnight too and we have 4 feet of snow
Yikes! That’s a good storm.

(As was pointed out by @cjl ) you have velocity data in sufficiently small time steps that you should be able to compute ##a_{avg}## and ##v_{avg}## for each time step. Then you will solve the eq in post #40 for ##\beta##, and make a plot using that eq and calculated values ##a_{avg}## and ##v_{avg}##. We are hoping to find that this plot for ##\beta## can be reasonably aproximated by a straight - nearly horizontal line, implying ##\beta## and hence ##C_d## are a constant. If it turns out that it isn’t straight-horizontal then we can talk about what to do.
 
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  • #75
erobz said:
Yikes! That’s a good storm.

(As was pointed out by @cjl ) you have velocity data in sufficiently small time steps that you should be able to compute ##a_{avg}## and ##v_{avg}## for each time step. Then you will solve the eq in post #40 for ##\beta##, and make a plot using that eq and calculated values ##a_{avg}## and ##v_{avg}##. We are hoping to find that this plot for ##\beta## can be reasonably aproximated by a straight - nearly horizontal line, implying ##\beta## and hence ##C_d## are a constant. If it turns out that it isn’t straight-horizontal then we can talk about what to do.
yeah, the storm was AMAZING!! we had a thundersnow. Its very rare to happen, but extreme winds and extreme lightning during a snowstorm. it was my first one i have ever seen so i freaked out a little because i thought "tornado"
Once i am awake and ready, i will begin the process of solving all this.
Solving for beta, do you want that done before or after the integration?

thanks
 
  • #76
LT72884 said:
yeah, the storm was AMAZING!! we had a thundersnow. Its very rare to happen, but extreme winds and extreme lightning during a snowstorm. it was my first one i have ever seen so i freaked out a little because i thought "tornado"
Once i am awake and ready, i will begin the process of solving all this.
Solving for beta, do you want that done before or after the integration?

thanks
You must find the functional form of ##\beta## before you do the integration to get ##v(t)##, as it may change the integrand dramatically depending on whether or not its constant or some function of ##v##.
 
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  • #77
LT72884 said:
I think they picked this book because it focused more on stocks theorem and fluid flow.
That would be Stoke's Theorem.
 
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  • #78
Mark44 said:
That would be Stoke's Theorem.
Yeah, i just spelled in wrong. One of my paitents i work with is name Stockely but its pronounced Stokely like stokes
 
  • #79
erobz said:
I don't know exactly the numbers, but I would expect something negative greater than ##g## given the equation:

$$ \frac{dv}{dt} = - \left( g + \frac{\beta}{m}v^2 \right)$$

?
solving for B i get:
1677611554978.png

is that how i needed to solve for it?
I know mass of the rocket, acceleration of the rocket, gravity, launch time and temp, rho, and velocity
where dv/dt is, do you just want me to use the acceleration values there?
 
  • #80
LT72884 said:
solving for B i get:
View attachment 323010
is that how i needed to solve for it?
I know mass of the rocket, acceleration of the rocket, gravity, launch time and temp, rho, and velocity
where dv/dt is, do you just want me to use the acceleration values there?
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}##
 
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  • #81
erobz said:
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}##
ok, so the v^2 term in beta, am i using the ACTUAL sensor velocities or the average velocity?
and for the average velocity what is v1 and v2 because i only have the sensor velocity as posted in the excel.

thanks
 
  • #82
LT72884 said:
ok, so the v^2 term in beta, am i using the ACTUAL sensor velocities or the average velocity?
and for the average velocity what is v1 and v2 because i only have the sensor velocity as posted in the excel.

thanks
The average values.
 
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  • #83
erobz said:
The average values.
ok, great. and i think i know what you mean now. from time 0.25 to 0.3, that would be v1 and v2, then from time 0.35 to 0.4 would be another v1 and v2? correct?
 
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  • #84
LT72884 said:
ok, great. and i think i know what you mean now. from time 0.25 to 0.3, that would be v1 and v2, the 0.35 to 0.4 would be another v1 and v2? correct?
Yep,thats it...
 
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  • #85
erobz said:
Yep,thats it...
ok, great. ill give that a go and see what happens. thanks for all the help. eventually i will see whats happening haha
 
  • #86
ok, there are 2 ways i could do this, which one is the best? i have this way
1677616250697.png

the 2 yellow velocities are averaged, the 2 blue and 2 green are as well. OR, i can do
(200.71+199.82)/2
(199.82+198.85)/2
(198.85+198.08)/2
(198.08+197.23)/2
 
  • #87
LT72884 said:
ok, there are 2 ways i could do this, which one is the best? i have this way
View attachment 323015
the 2 yellow velocities are averaged, the 2 blue and 2 green are as well. OR, i can do
(200.71+199.82)/2
(199.82+198.85)/2
(198.85+198.08)/2
(198.08+197.23)/2
Just start at the cell D3 . Write all the formulas as:

in cell D3:

Formula: = (C3 + C2)/(2)

And just drag it down. Same with the acceleration (different formula).
 
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  • #88
erobz said:
Just start at the cell D3 . Write all the formulas as:

in cell D3:

Formula: = (C3 + C2)/(2)

And just drag it down. Same with the acceleration (different formula).
thats how i did it at first, by dragging the formula down. But i noticed it used a previous value so wasnt sure that would mess up the data or not, but i will go ahead and do it the way i was going to haha
1677617287821.png
 
  • #89
LT72884 said:
thats how i did it at first, by dragging the formula down. But i noticed it used a previous value so wasnt sure that would mess up the data or not, but i will go ahead and do it the way i was going to haha
View attachment 323021
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.
 
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  • #90
cjl said:
It comes from the equation you should be already familiar with, namely, Fd = 1/2*rho*v2*Cd*A.

At first approximation, 1/2, rho, Cd, and A are all constant, so it simplifies down to Fd = some constant * v2.

Now, for this, you probably have to account for varying air density, so it would be good if you could look up an atmosphere model (or better still, if you know the temperature that day, a temperature corrected atmosphere model), and then you'll want to basically try to fit your deceleration data to a model using this formula, with 1/2 (obviously), rho (from your atmosphere model), A (from whatever your chosen reference area is), and v2 all known (you know v from your data, which incidentally you should use accelerometer based speed rather than baro based speed if possible, it's much more accurate), and thus you can solve for Cd.

You know Fd because you know the burnout mass of your rocket and you had an accelerometer onboard, though you're going to (at least in my experience) have issues getting any kind of accurate data below a few hundred mph because you start getting limited by accelerometer discretization at that point. Because of this, you can rearrange your equation to Cd = Fd/(1/2*rho*v2*A), and then just solve for your Cd vs V curve.

(Also, the Raven is a great little altimeter, so I'm glad to hear you're using it)
sorry for late reply. have had some crazy issues with my train rides. My train to school had to get towed by another train haha, then yesterday a truck fell off the overpass onto the train.
Ok, so how do i know Fd based on burnout mass?

thanks much.
I also preferer to use an accelerometer over barometric pressure readings for velocity.
for the quadratic drag, i see what they did. Beta is just a constant of all the constants
 
  • #91
erobz said:
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
1677686655169.png
 
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  • #92
erobz said:
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.
 
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  • #93
Chestermiller said:
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.
 
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  • #94
LT72884 said:
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##.
 
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  • #95
erobz said:
##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
1677693457865.png
 
  • #96
LT72884 said:
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.
 
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  • #97
erobz said:
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..

1677694033074.png
 
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  • #98
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?
 
  • #99
LT72884 said:
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}##.
 
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  • #100
erobz said:
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.

1677699326674.png
 
  • #101
LT72884 said:
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.
 
  • #102
erobz said:
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

1677700711721.png
 
  • #103
LT72884 said:
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?
 
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  • #104
erobz said:
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
1677701533230.png
 
  • #105
even plotted against velocity as the x axis or even as a data set, its still looks scattered
 

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