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
  • #1
LT72884
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Ok, i have some rocket data from a 20,000 foot launch. I have the times, altitudes, and velocity at said time. Is it possible to find the terminal velocity from this long list of numbers?
I tried graphing the velocities to see where the curve flattens out, since usually that is where terminal velocity happens, but i couldn't get it to work.
i feel like i am missing a few things. My MAIN goal is to calculate the drag constant from raw data. Wind tunnel testing i have done states that a drag constant average for a rocket is 0.75. I want to use this raw data to see if i get the same number.

I have excel, and thats it.

thanks
 
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  • #2
I suggest you post this under Classical Physics and include some calculations.
 
  • #3
PeroK said:
I suggest you post this under Classical Physics and include some calculations.
Done. :smile:

LT72884 said:
Ok, i have some rocket data from a 20,000 foot launch. I have the times, altitudes, and velocity at said time. Is it possible to find the terminal velocity from this long list of numbers?
So you have ##h(t)## height data for the whole descent too? What you want is in those values, right? Just use Excel to calculate the velocity as a function of those height numbers, and make a graph out of the velocity results. Do you know how to use adjacent columns in Excel for ##t## and ##h(t)## to make a 3rd column for ##v(t)##?
 
  • #4
LT72884 said:
Ok, i have some rocket data from a 20,000 foot launch. I have the times, altitudes, and velocity at said time. Is it possible to find the terminal velocity from this long list of numbers?
I tried graphing the velocities to see where the curve flattens out, since usually that is where terminal velocity happens, but i couldn't get it to work.
i feel like i am missing a few things. My MAIN goal is to calculate the drag constant from raw data. Wind tunnel testing i have done states that a drag constant average for a rocket is 0.75. I want to use this raw data to see if i get the same number.

I have excel, and thats it.

thanks
So you experimentally measured the height as a function of time from a freefall of 20k ft?
 
  • #5
Your last thread had crazy numbers in it. 20,000 feet is well above what hobbiests routinely do, so it looks like a crazy number too.

We can't help you when the numbers are all crazy.
 
  • #6
Vanadium 50 said:
Your last thread had crazy numbers in it. 20,000 feet is well above what hobbiests routinely do, so it looks like a crazy number too.
It's well within what's doable at the higher end of the hobby. I have yet to reach it myself, but my personal best was 18,200 feet so I'm not far off.

I would be curious for more detail about this data though.
 
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  • #7
SO SORRY GUYS!! we had a HUGE winter storm come in so i lost this thread in the process of trying to get work, school and other things done and back to normal.

as for the numbers, yes, 20,000 feet is normal. My buddy just hit 293,000 feet.. 60 miles with his rocket.
 
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  • #8
cjl said:
It's well within what's doable at the higher end of the hobby. I have yet to reach it myself, but my personal best was 18,200 feet so I'm not far off.

I would be curious for more detail about this data though.
the data came from a raven 4 sensor. Its very accurate. This week, i will be hitting close to 12,000 feet. Then in the summer we are going for the 20-30k range for our level 3 federal cert
 
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  • #9
LT72884 said:
My buddy just hit 293,000 feet.. 60 miles with his rocket.
Maybe you could make some extra money by contracting with the US government to knock down those pesky spy balloons... :smile:
 
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  • #10
berkeman said:
Done. :smile:So you have ##h(t)## height data for the whole descent too? What you want is in those values, right? Just use Excel to calculate the velocity as a function of those height numbers, and make a graph out of the velocity results. Do you know how to use adjacent columns in Excel for ##t## and ##h(t)## to make a 3rd column for ##v(t)##?
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.

I do have velocity at time and altitude in the sensor data IE:
5.75 seconds | 1625 m/s | 9531.64 feet
my goal is to find the drag coefficient of the rocket and i THINK i was able to do so using some back substitution. BUT i would like to find terminal velocity if possible as well, but not sure i can from coasting speed (mach should be 2.8, not 4.8)

1677518023676.png
 
  • #11
berkeman said:
Maybe you could make some extra money by contracting with the US government to knock down those pesky spy balloons... :smile:
LOL, we had a very similar discussion with the club a couple weeks ago about this hahaha. Our only limitation is we cant use ANY type of guidance systems... so maybe we will use pigeons like they did in ww2 to guide our rockets
 
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  • #12
erobz said:
So you experimentally measured the height as a function of time from a freefall of 20k ft?
no, i got the data from our sensor of the rocket.
 
  • #13
LT72884 said:
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.

I do have velocity at time and altitude in the sensor data IE:
5.75 seconds | 1625 m/s | 9531.64 feet
my goal is to find the drag coefficient of the rocket and i THINK i was able to do so using some back substitution.

View attachment 322945
Ok, so this is on the way up. My opinion is that you should focus on burnout to apogee. The problem is that the coefficient of drag is different on the way up, than it is on the way down. Its not falling in the same orientation as it is climbing.
 
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  • #14
erobz said:
Ok, so this is on the way up. My opinion is that you should focus on burnout to apogee.
thats exactly what i did :) thanks for validating that for me. im glad my thinking was correct. That graph i have pictured is from burnout to apogee.
So here is my next question. Is it possible to find terminal velocity during burnout (coasting speed)? or is it only possible to find TV on freefall?
 
  • #15
LT72884 said:
thats exactly what i did :) thanks for validating that for me. im glad my thinking was correct. That graph i have pictured is from burnout to apogee.
So here is my next question. Is it possible to find terminal velocity during burnout (coasting speed)? or is it only possible to find TV on freefall?
The problem is that it doesn't fall nose first. Anything you calculate will be w.r.t. it climbing ( nose first ). Its going to descend engine first? The drag will be larger on the way down.
 
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  • #16
berkeman said:
Maybe you could make some extra money by contracting with the US government to knock down those pesky spy balloons... :smile:
ok, i want to try out your method. I have the burnout (coasting) velocity, times, and altitudes from the sensor data at this stage. What are you meaning by "us velocity as a function of height" ? do you mean v(h)=h/t

if its a different equation, i do not remember all of them from my physics class :)

thanks
 
  • #17
erobz said:
The problem is that it doesn't fall nose first. Anything you calculate will be w.r.t. it climbing ( nose first ). Its going to descend engine first? The drag will be larger on the way down.
thats my reasoning as well. I might have to do this a different way..
 
  • #18
LT72884 said:
thats my reasoning as well. I might have to do this a different way..
Are you ok with just finding an reasonable upper bound for the free fall terminal velocity?
 
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  • #19
erobz said:
Are you ok with just finding an reasonable upper bound for the free fall terminal velocity?
at this stage, yes:) i have learned lots the last few days with this project haha.
 
  • #20
LT72884 said:
at this stage, yes:) i have learned lots the last few days with this project haha.
Ok. What we can calculate on the way up, we can use on the way down under the assumption it will result in a higher terminal velocity than the rocket would actually fall.
 
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  • #21
erobz said:
Ok. What we can calculate on the way up, we can use on the way down under the assumption it will result in a higher terminal velocity than the rocket would actually fall.
ok, great. now, how do we account for the Cd (drag coef) in the equation? that is one of the things i am wanting to solve for is that. I know that rockets have a Cd of 0.75, BUT thats not always true. My current rocket is 0.35 for its Cd. So my goal is to find TV and then use that to calculate the Cd
 
  • #22
erobz said:
Ok. What we can calculate on the way up, we can use on the way down under the assumption it will result in a higher terminal velocity than the rocket would actually fall.

First thing you should do is lose the polynomial, and find an exponential trendline. This is because it is an exponential equation which solves the relationship below describing flight from burnout to apogee. $$ m \frac{dv}{dt} = - ( mg + \beta v^2)$$
 
  • #23
LT72884 said:
ok, great. now, how do we account for the Cd (drag coef) in the equation? that is one of the things i am wanting to solve for is that. I know that rockets have a Cd of 0.75, BUT thats not always true. My current rocket is 0.35 for its Cd. So my goal is to find TV and then use that to calculate the Cd
Well be able to go straight to terminal velocity from the data.
 
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  • #24
erobz said:
Well be able to go straight to terminal velocity from the data.
perfect. how do we do this? do you need a pic of the data? Also, give me about 20 minutes to get to lab, just got off the train.

Thanks a mill for the help
 
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  • #25
Select trendline by left click. then right click on the trendline, go to format trendline, and select exponential.
 
  • #26
The numbers in your table don't look right. To get to the velocity you claim in a quarter second requires an acceleration of 350 g's. That's a lot. That's a ton of force on your seven pound rocket.

Then for the rest of the powered phase, the acceleration is of order 200 in some units - but the rocket isn't getting any faster.
 
  • #27
erobz said:
Its going to descend engine first?
What is the basis for this assumption?
 
  • #28
Vanadium 50 said:
The numbers in your table don't look right. To get to the velocity you claim in a quarter second requires an acceleration of 350 g's. That's a lot. That's a ton of force on your seven pound rocket.

Then for the rest of the powered phase, the acceleration is of order 200 in some units - but the rocket isn't getting any faster.
The data they have presented is from burnout to apogee. Thats the velocity 0.25 s after burnout.
 
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  • #29
russ_watters said:
What is the basis for this assumption?
Which assumption: That it falls nose up, or that if it falls nose up the ##C_D## drag will be larger than for it falling nose down?
 
  • #30
erobz said:
Which assumption: That it falls nose up, or that if it falls nose up the ##C_D## drag will be larger than for it falling nose down?
That it falls nose up.
 
  • #31
russ_watters said:
That it falls nose up.
I just dropped a pencil. That could be wrong. Rockets play with center of pressure and center of mass with fins for stability. I could see how it could flip too. Either way, that's not an issue as far as I can tell. If it does fall nose down, we will be calculating the drag coefficient for that state regardless.
 
  • #32
Vanadium 50 said:
The numbers in your table don't look right. To get to the velocity you claim in a quarter second requires an acceleration of 350 g's. That's a lot. That's a ton of force on your seven pound rocket.

Then for the rest of the powered phase, the acceleration is of order 200 in some units - but the rocket isn't getting any faster.
What I now notice about the data, is that if this is from burnout to apogee, then we have a problem. Where is ##v=0##? The graph stops at ##500 \rm{m/s}## :bugeye:
 
  • #33
erobz said:
Select trendline by left click. then right click on the trendline, go to format trendline, and select exponential.
new trendline

1677523799168.png
 
  • #34
LT72884 said:
new trendline

View attachment 322951
Ok, but something is a miss here. Apogee velocity is zero. You did say this was data from burnout to apogee?
 
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  • #35
erobz said:
I just dropped a pencil. That could be wrong. Rockets play with center of pressure and center of mass with fins for stability. I could see how it could flip too. Either way, that's not an issue as far as I can tell. If it does fall nose down, we will be calculating the drag coefficient for that state regardless.
Yeah, I was assuming would have fins and therefore would fall nose down. Anyway, yeah, just makes the calculation accurate/matching the scenario.
 
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