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
  • #36
erobz said:
Ok, but something is a miss here. Apogee velocity is zero. You did say this was data from burnout to apogee?
correct. i have the data from launch to apogee. I selected the data from when velocity hits its max, then begins to drop toward 0. roughly at 3 seconds after launch is when this happens.

1677524172424.png

at 20,751.7 feet, the velocity is -0.7 ft/sec meaning the rocket is falling. Thats where my data ends haha.
Looking up the aerotech L1030 motor, its burn time is 2.8 seconds,
 
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  • #37
erobz said:
. Thats the velocity 0.25 s after burnout.
Then why is it still accelerating at 200 units?
 
  • #38
Also, I probably jumped the gun on pulling terminal velocity right out of the graph. Its exponential, but its a complicated exponential. Sorry to get your hopes up. its looks like a bit of a struggle is ahead.
 
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  • #39
LT72884 said:
correct. i have the data from launch to apogee. I selected the data from when velocity hits its max, then begins to drop toward 0. roughly at 3 seconds after launch is when this happens.

View attachment 322952
at 20,751.7 feet, the velocity is -0.7 ft/sec meaning the rocket is falling. Thats where my data ends haha.
Looking up the aerotech L1030 motor, its burn time is 2.8 seconds,
We need data near to appogee ##v = 0## where is that data?
 
  • #40
Vanadium 50 said:
Then why is it still accelerating at 200 units?
I don't know exactly the numbers, but I would expect something negative greater than ##g## given the equation:

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

?
 
  • #41
erobz said:
Also, I probably jumped the gun on pulling terminal velocity right out of the graph. Its exponential, but its a complicated exponential. Sorry to get your hopes up. its looks like a bit of a struggle is ahead.
thats ok. This is a complicated project for sure. My goal is to make a "virtual wind tunnel" sort of speak. This is for my capstone project. We just built an active drag system for our rocket to compete is spaceport america cup, but our university has no wind tunnel haha. The standard value of 0.75 is a ROUGH number, so i am trying to find a more realistic number.
Using openrocket software, the number should be 0.62 for the drag coef. so i am really close. Openrocket uses a runge-kuta itteration method and since i do not know how to program, i am using excel and the solver feature to take my intial guess and then solve based on the data. you can see how close i was able to get the modeled data to match my actual
 
  • #42
erobz said:
We need data near to appogee
If the numbers were understood, we could look much later than apogee when the acceleration is close to zero. That's terminal velocity.
 
  • #43
LT72884 said:
thats ok. This is a complicated project for sure. My goal is to make a "virtual wind tunnel" sort of speak. This is for my capstone project. We just built an active drag system for our rocket to compete is spaceport america cup, but our university has no wind tunnel haha. The standard value of 0.75 is a ROUGH number, so i am trying to find a more realistic number.
Using openrocket software, the number should be 0.62 for the drag coef. so i am really close. Openrocket uses a runge-kuta itteration method and since i do not know how to program, i am using excel and the solver feature to take my intial guess and then solve based on the data. you can see how close i was able to get the modeled data to match my actual
Step 1 present the proper data. We need data after burnout in the vicinity of apogee. I suspect the terminal velocity is going to be much lower than the burnout velocity. So data around burnout is not good data for this, we need data closer to the speed that it will be falling...closer to near apogee speeds than burnout speeds. Can you present the data from when the rocket is traveling at 100 m/s onward to apogee?
 
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  • #44
Vanadium 50 said:
If the numbers were understood, we could look much later than apogee when the acceleration is close to zero. That's terminal velocity.
Yeah, if they could get fall data directly that would be best. But I'm suspecting they don't want to obliterate their rocket.
 
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  • #45
Here is from burnout to apogee. its alot of numbers, let me see if i can attach it as a file first
 

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  • #46
LT72884 said:
Here is from burnout to apogee. its alot of numbers, let me see if i can attach it as a file first
just give a filterd version. We don't need data anywhere near burnout speeds. The rocket will not have a terminal velocity anywhere near 800 m/s.
 
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  • #47
erobz said:
just give a filterd version. We don't need data anywhere near burnout speeds. The rocket will not have a terminal velocity anywhere near 800 m/s.
here is the excel file. This is from burnout to apogee. I can further filter if needs be. Starting around 500m/s?
 

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  • #48
erobz said:
Step 1 present the proper data. We need data after burnout in the vicinity of apogee. I suspect the terminal velocity is going to be much lower than the burnout velocity. So data around burnout is not good data for this, we need data closer to the speed that it will be falling...closer to near apogee speeds than burnout speeds. Can you present the data from when the rocket is traveling at 100 m/s onward to apogee?
i did not see this reply, yes i can do this real quick.. this should be much better.
 

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  • #49
An unnecessarily harsh post and response have been deleted. Please relax, guys. And if one doesn't want to help, then don't - @erobz is providing plenty as it is.
 
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  • #50
russ_watters said:
An unnecessarily harsh post and response have been deleted. Please relax, guys. And if one doesn't want to help, then don't - @erobz is providing plenty as it is.
yes, they are providing plenty of help and its much appreciated because i have never seen it approached from this way.
Im still used to bookwork and homework style questions since i am still in my senior year of engineering. This one of the first questions that are outside the scaffolding of homework. This is just the analysis portion of our project and i was trying to see if there was more than one way to accomplish it and erobz has offered to show that "other" way and im pretty excited to see it done
 
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  • #51
Ok, so this is your senior project. Thats good to know. I wasn't planning on doing the work for you, once you start you will see why.

So the path forward that I have in mind begins by writing a differential eq. that describes the motion after burnout up to apogee. I already have done that for you...I'm not going to try to hide that. That is your free bee. Do you understand where it comes from? You descibe how it is formulated back to me?
 
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  • #52
erobz said:
Ok, so this is your senior project. Thats good to know. I wasn't planning on doing the work for you, once you start you will see why.

So the path forward that I have in mind begins by writing a differential eq. that describes the motion after burnout up to apogee. I already have done that for you...I'm not going to try to hide that. That is your free bee. Do you understand where it comes from? You descibe how it is formulated back to me?
lol, my senior project is building a rocket that hits 10 to 20,000 feet, which we have done, and to build a ADS system for said rocket. This is just analysis so i have some numbers to work with that will help with design. Im not graded on the question i asked about terminal velocity.
This data however, will be used in the report so i guess it will be graded haha.
Ill try my best to describe what you have and if i understand it. I have been researching terminal velocity as a exponential and its the first time i have seen it this way. I might have seen it in calc class but it was probably not given as a terminal velocity problem.

http://www.dzre.com/alex/hp221_f03/notes/vterm/vterm.html
 
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  • #53
LT72884 said:
lol, my senior project is building a rocket that hits 10 to 20,000 feet, which we have done, and to build a ADS system for said rocket. This is just analysis so i have some numbers to work with that will help with design. Im not graded on the question i asked about terminal velocity.
Oh, you know what. It just dawned on me the way I was thinking would have taken you into wonderland, and we probably would have never got out!
 
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  • #54
erobz said:
Oh, you know what. It just dawned on me the way I was thinking would have taken you into wonderland, and we probably would have never got out!
should i start crying now?
 
  • #55
LT72884 said:
should i start crying now?
The new way will tell us with substantially less wandering in a mathematical forest!

Still, can you describe the differential equation to me. What assumptions do you think it has?
 
  • #56
LT72884 said:
and to build a ADS system for said rocket.
Sorry, what's ADS? Do you mean DAS?

Also, maybe this is unreasonable, but can you outfit and program a rocket to not deploy the parachute until a little before the ground, and gather data on velocity/altitude on the way down? That way the rocket won't be damaged, and you will have the terminal velocity data that you want.

EDIT/ADD -- BTW, I didn't see a parachute mentioned explicitly in the thread so far (I could have missed it), but the statements about falling tail down and not wanting to destroy the rocket made me assume a parachute. Otherwise, the rocket should fall nose down (fins/tail up, just like my arrows), and running the Data Acq System another minute or so should not drain the batteries...
 
  • #57
berkeman said:
Sorry, what's ADS? Do you mean DAS?

Also, maybe this is unreasonable, but can you outfit and program a rocket to not deploy the parachute until a little before the ground, and gather data on velocity/altitude on the way down? That way the rocket won't be damaged, and you will have the terminal velocity data that you want.

EDIT/ADD -- BTW, I didn't see a parachute mentioned explicitly in the thread so far (I could have missed it), but the statements about falling tail down and not wanting to destroy the rocket made me assume a parachute. Otherwise, the rocket should fall nose down (fins/tail up, just like my arrows), and running the Data Acq System another minute or so should not drain the batteries...

But the shoots are deployed out of the nose cone? If there is any susbstantial velocity I'm picturing it deploying, wrapping around the rocket, and smashing into the ground.
 
  • #58
erobz said:
But the shoots are deployed out of the nose cone?

Parashoots? LOL

So it won't deploy at 100mph nose down? Maybe boost the ejection charge for the chute?
 
  • #59
berkeman said:
Parashoots? LOL

So it won't deploy at 100mph nose down? Maybe boost the ejection charge for the chute?
:woot:

It sounds pretty risky... but I have no idea.
 
  • #60
Me neither, but it seems like a useful line to think about. For example, it may be that a small side drag feature needs to temporarily added to the nosecone to be sure it clears the rocket on the way down when it's popped. This will slow the rocket a bit during ascent, but who cares, since the data we want is at the end of the descent.

It's up to the OP/"Rocketman" to let us know if this might work. :smile:
 
  • #61
berkeman said:
Parashoots? LOL

So it won't deploy at 100mph nose down? Maybe boost the ejection charge for the chute?
Anyhow, I honestly thing we can figure it out from the data they have. But maybe they could just do an uncontrolled decent into earth on graduation day to verify calulations o0)?
 
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  • #62
the chute/shoot doesnt deploy for 24.6 seconds and apogee hits at 27.1 seconds. So the chute pops right before apogee. The chute is below the nose cone.

This image was a test launch to 1 mile
15 foot shock cord, 5 foot chute
HSfhvJiy412IS3IddecxIHPw2Q=w352-h625-no?authuser=0.jpg

i can get permission to allow a rocket to go ballistic but it has to be a smaller rocket haha.
as for the equation... Im not even sure where to start of what it would look like.
F=m(dv/dt)
F=drag -gravity
m(dv/dt)=Fd-mg

then not sure where to go from there haha
 
  • #63
LT72884 said:
the chute/shoot doesnt deploy for 24.6 seconds and apogee hits at 27.1 seconds. So the chute pops right before apogee. The chute is below the nose cone.

This image was a test launch to 1 mile
15 foot shock cord, 5 foot chute
View attachment 322976
i can get permission to allow a rocket to go ballistic but it has to be a smaller rocket haha.
:-p
LT72884 said:
as for the equation... Im not even sure where to start of what it would look like.
F=m(dv/dt)
F=drag -gravity
m(dv/dt)=Fd-mg

then not sure where to go from there haha
You're on the right track. On the way up which way is the drag force acting?
 
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  • #64
erobz said:
:-p

You're on the right track. On the way up which way is the drag force acting?
on the way up drag should be pushing down, same with g, and mg.
some equations have Fr=bv, what is b?

if mass cancels out, then i can separate and then integrate. but from here, not sure
dv/(g-bv)=dt

then i get a crazy equation with e in it. if i take the limit as time gets large, vt is left
 
  • #65
LT72884 said:
on the way up drag should be pushing down, same with g, and mg.
Correct, so what is the sign in front of the ##F_d##?
LT72884 said:
some equations have Fr=bv, what is b?
Thats linear drag. Your rocket is only going to be in that regime for a short period of time.

For rockets (and most thing that fall through the air ) we typically use quadratic drag. They are both there, but quadratic drag dominates for higher speeds:

##F_d = \beta v^2##
LT72884 said:
if mass cancels out, then i can separate and then integrate. but from here, not sure
dv/(g-bv)=dt
Don't jump too far ahead. We won't actually have to solve the ODE to get what you are after (unless you want to afterward).
 
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  • #66
erobz said:
Correct, so what is the sign in front of the ##F_d##?

Thats linear drag. Your rocket is only going to be in that regime for a short period of time.

For rockets (and most thing that fall through the air ) we typically use quadratic drag. They are both there, but quadratic drag dominates for higher speeds:

##F_d = \beta v^2##

Don't jump to far ahead. We won't actually have to solve the ODE to get what you are after (unless you want to afterward).
i have never seen the Fd=bv^2 equation... or i was not paying attention. I checked my fluids book and i do not see it....

ok, so lets go back. Fd should be negitive. so
m(dv/dt) = -Fd - mg
 
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  • #67
LT72884 said:
i have never seen the Fd=bv^2 equation... or i was not paying attention. I checked my fluids book and i do not see it....

ok, so lets go back. Fd should be negitive. so
m(dv/dt) = -Fd - mg
It's in your fluid book somewhere. You should have a chapter about lift and drag. In mine its Chapter 11!
 
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  • #68
erobz said:
It's in your fluid book somewhere. You should have a chapter about lift and drag.
ill try and find it. For now, i have the diff eq as shown above. Whats the next step?
thanks for all the help. Seeing this in a new way really solidifies the knowledge and understanding of what is really going on
 
  • #69
LT72884 said:
ill try and find it. For now, i have the diff eq as shown above. Whats the next step?
thanks for all the help. Seeing this in a new way really solidifies the knowledge and understanding of what is really going on
So you plug in the quadratic drag force to equation and divide through by ##m## and you will arrive at what I have written in post #40.

After that take a moment to study it and identify all the data that you have, or could calculate from the data you captured.
 
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  • #70
erobz said:
So you plug in the quadratic drag force to equation and divide through by ##m## and you will arrive at what I have written in post #40.

After that take a moment to study it and identify all the data that you have, or could calculate from the data you captured.
ok, now i see how you get the equation from post 40, and where did quadratic drag force come from?
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.

1677558597669.png
 

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