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

[LT72884]

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

 

[PeroK]

I suggest you post this under Classical Physics and include some calculations.

 

[berkeman]

I suggest you post this under Classical Physics and include some calculations.

Done.

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 $v(t)$?

 

[erobz]

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?

 

[Vanadium 50]

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.

 

[cjl]

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.

 

[LT72884]

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.

 

[LT72884]

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

 

[berkeman]

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...

 

[LT72884]

Done.


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)

 

[LT72884]

Maybe you could make some extra money by contracting with the US government to knock down those pesky spy balloons...

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

 

[LT72884]

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.

 

[erobz]

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.

 

[LT72884]

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?

 

[erobz]

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.

 

[LT72884]

Maybe you could make some extra money by contracting with the US government to knock down those pesky spy balloons...

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

 

[LT72884]

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..

 

[erobz]

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?

 

[LT72884]

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.

 

[erobz]

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.

 

[LT72884]

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

 

[erobz]

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.


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

 

[erobz]

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.

 

[LT72884]

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

 

[erobz]

Select trendline by left click. then right click on the trendline, go to format trendline, and select exponential.

 

[Vanadium 50]

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.

 

[russ_watters]

Its going to descend engine first?

What is the basis for this assumption?

 

[erobz]

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.

 

[erobz]

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?

 

[russ_watters]

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.

 

[erobz]

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.

 

[erobz]

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 $500 \rm{m/s}$

 

[LT72884]

Select trendline by left click. then right click on the trendline, go to format trendline, and select exponential.

new trendline

 

[erobz]

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?

 

[russ_watters]

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.