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

[erobz]

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 ?

 

[LT72884]

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

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

 

[erobz]

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.

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?

 

[LT72884]

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

 

[erobz]

on the way up drag should be pushing down, same with g, and mg.

Correct, so what is the sign in front of the $F_d$?

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:

$F_d = \beta v^2$

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).

 

[LT72884]

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

 

[erobz]

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!

 

[LT72884]

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

 

[erobz]

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.

 

[LT72884]

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.

 

[erobz]

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.

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.

 

[LT72884]

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

 

[cjl]

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)

 

[erobz]

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.

 

[LT72884]

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

 

[erobz]

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 $v$.

 

[Mark44]

I think they picked this book because it focused more on stocks theorem and fluid flow.

That would be Stoke's Theorem.

 

[LT72884]

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

 

[LT72884]

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:

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?

 

[erobz]

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 $v_{avg}$

 

[LT72884]

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

 

[erobz]

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.

 

[LT72884]

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?

 

[erobz]

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...

 

[LT72884]

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

 

[LT72884]

ok, there are 2 ways i could do this, which one is the best? i have this way

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

 

[erobz]

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).

 

[LT72884]

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

 

[erobz]

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.

 

[LT72884]

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