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

[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

 

[LT72884]

sorry it has taken so long. I REALLY REALLY REALLY APRECIATE your mass amounts of PAITENCE with me. I am really good at book work and homework, but real world stuff gets to me and it makes me sad and nervous for my future carer.

 

[LT72884]

this reminds me of slope field diagrams in a way, im guessing where the lin flattens out is terminal velocity? not sure that is correct thinking on my end because based on a terminal V calculator online, i get 598m/s based on the same numbers i used. I am probably jumping ahead

EDIT, i added a linear trend line

 

[erobz]

sorry it has taken so long. I REALLY REALLY REALLY APRECIATE your mass amounts of PAITENCE with me. I am really good at book work and homework, but real world stuff gets to me and it makes me sad and nervous for my future carer.

You are going to make mistakes, and you'll learn. Get ready to swallow your pride. Thats probably universal for any young engineer.

 

[erobz]

At 50m/s.
View attachment 323066
At 55m/s a drastic change
View attachment 323067

This is why scale is really important. How large are these fluctuations when you step back and examine it. You are taking measurements with some sensor, that has some error. You aren't going to get perfection. How much does the average $\beta$ change from 50 -100 , vs 100- 150, vs 150-200? How much is the $C_d$ based on the average for each range?

 

[LT72884]

You are going to make mistakes, and you'll learn. Get ready to swallow your pride. Thats probably universal for any young engineer.

I agree about the pride. That is why im REALLY grateful that you showed me ALL the steps. I do not expect that from everyone or anyone, and when i have no single clue what i am doing, its so nice to see an example ALL the way through. That is how i learn. step by step and examining each section.
Now that we have this graph with a trend line, what is the next step in determining the terminal velocity?

 

[LT72884]

This is why scale is really important. How large are these fluctuations when you step back and examine it. You are taking measurements with some sensor, that has some error. You aren't going to get perfection. How much does the average $\beta$ change from 50 -100 , vs 100- 150, vs 150-200? How much is the $C_d$ based on the average for each range?

beta changes alot from 50 to 100m/s but then the system settles around 200m/s. The Drag Coef seems to be an average of 0.6 from 150 to 200m/s.

 

[cjl]

At 50m/s.
View attachment 323066
At 55m/s a drastic change
View attachment 323067

That looks like what I'd expect. In my experience (having done similar Cd estimation on some of my rockets), you get pretty noisy data but the higher speed average value ends up being pretty reasonable (and the 0.6 that you found does appear reasonable to me).

The multiple discrete lines of data points at lower speed is also expected - that's due to accelerometer resolution and the fact that at low speeds even with a fairly high Cd, acceleration is low (and thus measurement accuracy and precision are also low unless you had a much fancier and more expensive accelerometer).

 

[LT72884]

That looks like what I'd expect. In my experience (having done similar Cd estimation on some of my rockets), you get pretty noisy data but the higher speed average value ends up being pretty reasonable (and the 0.6 that you found does appear reasonable to me).

The multiple discrete lines of data points at lower speed is also expected - that's due to accelerometer resolution and the fact that at low speeds even with a fairly high Cd, acceleration is low (and thus measurement accuracy and precision are also low unless you had a much fancier and more expensive accelerometer).

so what is happening at 200m/s with this graph? thats where the system settles too. Is this terminal velocity?
I am not sure if it is or not because using the Vt calculation using my 0.62 Cd, density, etc, i get that Vt is close to 600m/s

thanks

 

[erobz]

so what is happening at 200m/s with this graph? thats where the system settles too. Is this terminal velocity?
I am not sure if it is or not because using the Vt calculation using my 0.62 Cd, density, etc, i get that Vt is close to 600m/s
View attachment 323073

thanks

The graph finds $\beta$. Now, assuming that it is a constant, to find the terminal velocity you have solve the eq in post #40.

To do that: What is the acceleration of the rocket at terminal velocity?

Note: This requires no integration. fully solving the ODE in general for velocity as a function of time is not necessary to determine $v_{term}$

 

[LT72884]

The graph finds $\beta$. Now, assuming that it is a constant, to find the terminal velocity you have solve the eq in post #40.

To do that: What is the acceleration of the rocket at terminal velocity?

Note: This requires no integration. fully solving the ODE in general for velocity as a function of time is not necessary to determine $v_{term}$

"To do that: What is the acceleration of the rocket at terminal velocity?" Well, my thinking is either 0 or 9.81 since terminal velocity is where Fd and Fg equalize. So probably 0 then

 

[erobz]

"To do that: What is the acceleration of the rocket at terminal velocity?" Well, my thinking is either 0 or 9.81 since terminal velocity is where Fd and Fg equalize. So probably 0 then

Actually, I was being a bit hasty. You need to write the DE for the falling rocket first, and then do what I said.

 

[LT72884]

Actually, I was being a bit hasty. You need to write the DE for the falling rocket first, and then do what I said.

ok, cool. I have noticed i do not have 0 in my data for acceleration so im guessing its when acceleration is a differnt value.

post #40 is that not the DE?

 

[erobz]

post #40 is that not the DE?

Its similar. But you have to think about the directions of the forces when its falling.

 

[erobz]

ok, cool. I have noticed i do not have 0 in my data for acceleration so im guessing its when acceleration is a differnt value.

You are not going to find terminal velocity in your data. Why do you think you would?

 

[LT72884]

You are not going to find terminal velocity in your data. Why do you think you would?

somewhere along the line we were discussing finding the terminal velocity. We discussed how its exponential and if i integrate the equation we have, i do indeed get the terminal velocity equation inside. Somehow i thought we were finding both the Cd and terminal velocity thats why haha. but maybe somewhere we changed our minds because hte math gets to crazy. You also said something about how you think Vt is around 100m/s or so...

wait, i might be taking your question to literal...

 

[erobz]

somewhere along the line we were discussing finding the terminal velocity. We discussed how its exponential and if i integrate the equation we have, i do indeed get the terminal velocity equation inside. Somehow i thought we were finding both the Cd and terminal velocity thats why haha. but maybe somewhere we changed our minds because hte math gets to crazy
View attachment 323075

Firstly, that is a different equation altogether. If you want to solve for the velocity as a function of time you will have to solve the new differential equation (that you haven't yet derived like I asked)

But yes. Minds were changed along the way. You will find that to be a very real part of real world problem solving. You dive in with some idea, and many times the problem or question itself will evolve as you begin to untangle things. Nothing in real world is "solve this using a,b,c,d, assuming d,e,f,g". It's quite literally...Here is this problem...we think, you figure out the rest! I had nothing more than an approach idea wasn't immediately sure what the exact answer was, or how the dots would connect. I started down a road that would have been a real back breaker...and then I had a moment of "clarity" along the way...a "duh" moment if you will.

If you are just after terminal velocity, then solving for $v(t)$ via an integration is not necessary. You can, if you want. Like I said. Its not that equation you've written out above. That is for linear drag models.

 

[LT72884]

Firstly, that is a different equation altogether. If you want to solve for the velocity as a function of time you will have to solve the new differential equation (that you haven't yet derived like I asked)

But yes. Minds were changed along the way. You will find that to be a very real part of real world problem solving. You dive in with some idea, and many times the problem or question itself will evolve as you begin to untangle things. Nothing in real world is "solve this using a,b,c,d, assuming d,e,f,g". It's quite literally...Here is this problem...we think, you figure out the rest! I had nothing more than an approach idea wasn't immediately sure what the exact answer was, or how the dots would connect. I started down a road that would have been a real back breaker...and then I had a moment of "clarity" along the way...a "duh" moment if you will.

If you are just after terminal velocity, then solving for $v(t)$ via an integration is not necessary. You can, if you want. Like I said. Its not that equation you've written out above. That is for linear drag models.

there are other Vt equations as well i could probably use.
now to answer your question, i thought Vt would be in my data. but maybe its not since its not completely free fall. If it were free fall, i would expect to see acceleration go to 0 at some point when the object has stopped accelerating and velocity is now constant (not increasing)

 

[LT72884]

second, i noticed some of my Beta values are negative. technically they should not be negative at all because B=(Cd)(area)(rho)(0.5)

 

[erobz]

there are other Vt equations as well i could probably use.
now to answer your question, i thought Vt would be in my data. but maybe its not since its not completely free fall. If it were free fall, i would expect to see acceleration go to 0 at some point when the object has stopped accelerating and velocity is now constant (not increasing)

Correct. It wold be in there if you dropped you rocket from 20k ft, but you launched it to 20k ft , and ejected a parachute.

 

[erobz]

second, i noticed some of my Beta values are negative. technically they should not be negative at all because B=(Cd)(area)(rho)(0.5)

Also correct, the β values fluctuate around because of the instrument you used. This is the difference between Accuracy and precision. If you want both in an instrument you usually have to pay . If you want accuracy,precision, and range .The hope in using the average values was that the “truth” was somewhere in the middle.