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

[LT72884]

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 . The hope in using the average values was that the “truth” was somewhere in the middle.

ok, thats what i thought. I told that to my professor and he didnt have time to think about my response during the team meeting and he looked confused haha, but i re-assured him that a $80 sensor may not be the most accurate and precise.

the "truth" being the consistent beta values that give me consistent Cd's right?

thanks

 

[erobz]

ok, thats what i thought. I told that to my professor and he didnt have time to think about my response during the team meeting and he looked confused haha, but i re-assured him that a $80 sensor may not be the most accurate and precise.

the "truth" being the consistent beta values that give me consistent Cd's right?

thanks

Correct.

So are you ready to write this last equation and find terminal velocity yet?

 

[LT72884]

Correct.

So are you ready to write this last equation and find terminal velocity yet?

i think so haha. I just got home from my 3 hour train ride from school.

Which equation should i start with? or is it a whole new equation you have in mind?

thanks

 

[cjl]

so what is happening at 200m/s with this graph? thats where the system settles too. Is this terminal velocity?

No, that's just where you're going fast enough to have a decent amount of drag-induced acceleration, giving you fairly clean data. Cd should be (largely) independent of speed in the subsonic regime for a rocket-shaped object, and that's just where your data is good enough to observe that.

(For an example of what much cleaner drag data looks like for a rocket, here is an actual measured drag profile from a friend of mine).

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 biggest problem with that is that 600m/s is well supersonic, and thus your real terminal velocity would be lower thanks to the drag rise that occurs at mach (as seen in my sample data above). I'd also note that your very high sectional density is actually making your data a lot muddier - since you're limited by accelerometer resolution, you'd get much better data with a rocket with higher drag relative to weight, but obviously that goes against the general design goals here.

Also, I'm curious what motor you used here, since your frontal area makes this seem like a 54mm MD rocket, and it's not exactly common to get to 20,000 feet on a 54mm motor.

 

[erobz]

i think so haha. I just got home from my 3 hour train ride from school.

Which equation should i start with? or is it a whole new equation you have in mind?

thanks

Newton’s 2nd Law for the rocket in free fall.

 

[LT72884]

No, that's just where you're going fast enough to have a decent amount of drag-induced acceleration, giving you fairly clean data. Cd should be (largely) independent of speed in the subsonic regime for a rocket-shaped object, and that's just where your data is good enough to observe that.

(For an example of what much cleaner drag data looks like for a rocket, here is an actual measured drag profile from a friend of mine).
The biggest problem with that is that 600m/s is well supersonic, and thus your real terminal velocity would be lower thanks to the drag rise that occurs at mach (as seen in my sample data above). I'd also note that your very high sectional density is actually making your data a lot muddier - since you're limited by accelerometer resolution, you'd get much better data with a rocket with higher drag relative to weight, but obviously that goes against the general design goals here.

Also, I'm curious what motor you used here, since your frontal area makes this seem like a 54mm MD rocket, and it's not exactly common to get to 20,000 feet on a 54mm motor.

My friend (someone from the rocket forums) used an L1030 motor.

You are correct that it is a 54mm rocket.
Ok, for the Cd and 200m/s that makes sense. I was thinking it was either that or vt. I have loved every step of this project because i have learned so much from you guys.

 

[LT72884]

Newton’s 2nd Law for the rocket in free fall.

well, no time like the present. Lets give it a go.
ok, we know that F=ma
and in free fall, a=0
so i would have 0=mg-kvt where kvt or kv is pointing up on a FBD and mg is down
but before velocity is terminal, so just kv we would have
ma=mg-kv
a=dv/dt
m(dv/dt)=mg-kv
i can seperate and solve the integration if needs be?

 

[erobz]

well, no time like the present. Lets give it a go.
ok, we know that F=ma
and in free fall, a=0
so i would have 0=mg-kvt where kvt or kv is pointing up on a FBD and mg is down
but before velocity is terminal, so just kv we would have
ma=mg-kv
a=dv/dt
m(dv/dt)=mg-kv
i can seperate and solve the integration if needs be?

where did you get $kv$? Do you not remember the whole thing was about quadratic drag?

 

[LT72884]

where did you get $kv$? Do you not remember the whole thing was about quadratic drag?

just from a basic FBD. so technically it should be kv^2 or the Bv^2.. i did forget about the quadratic drag haha

 

[erobz]

Good, so rewrite it. After you have done that set the acceleration to 0, and solve the resulting equation.

 

[LT72884]

Good, so rewrite it. After you have done that set the acceleration to 0, and solve the resulting equation.

Vt = sqrt(-mg/k)

 

[erobz]

Vt = sqrt(-mg/k)

Are you going to get a real result when you take the square root of that? What equation did you algebraically manipulate to get that result?

 

[LT72884]

Are you going to get a real result when you take the square root of that?

nope, it will be imaginary. i am trying to see which way to solve this.

 

[erobz]

nope, it will be imaginary. i am trying to see which way to solve this.

You didn't solve the equation you wrote??? Pick a direction as positive. Label all forces on rocket relative to that chosen direction. Please list that full equation in your next reply.

 

[LT72884]

You didn't solve the equation you wrote??? Pick a direction as positive. Label all forces on rocket relative to that chosen direction. Please list that full equation in your next reply.

i solved for v which then becomes imaginary due to the negative in the sqrt. so i need to solve for v a different way. ok, iw ill write soon

 

[erobz]

i solved for v which then becomes imaginary due to the negative in the sqrt. so i need to solve for v a different way. ok, iw ill write soon

you clearly solved a different equation from what you were writing ( as far as the directions of the forces go), or you made a trivial algebra mistake.

 

[LT72884]

with m(dv/dt) = mg-kv^2
and if i set dv/dt = a = 0
therefore 0=mg-kv^2
then solve for v using the correct signs
kv^2 = mg
v^2=(mg)/k
v=sqrt(mg/k)
if i am misunderstanding you, im sorry haha:)

 

[erobz]

with m(dv/dt) = mg-kv^2
and if i set dv/dt = a = 0
therefore 0=mg-kv^2
then solve for v using the correct signs
kv^2 = mg
v^2=(mg)/k
v=sqrt(mg/k)
if i am misunderstanding you, im sorry haha:)

Ok, that’s better. So what did you get for the terminal velocity?

 

[LT72884]

Ok, that’s better. So what did you get for the terminal velocity?

just making sure but K is the same as Beta right?
give me a few moments to get this calculated. might be about an hour or so. had something come up that is very important

 

[erobz]

just making sure but K is the same as Beta right?
give me a few moments to get this calculated. might be about an hour or so. had something come up that is very important

Yeah. k is β. No hurry.

 

[LT72884]

Yeah. k is β. No hurry.

average terminal velocity is 234.68. IF this was free fall... but we know its not

 

[LT72884]

average terminal velocity is 234.68. IF this was free fall... but we know its not

thanks for all the help. I really do appreciate it alot. Our actual project is to design an active drag system for our rocket. This ADS will be used to slow the rocket down to achieve as close to 10,000 feet as possible. so far our design is pretty cool.

 

[erobz]

thanks for all the help. I really do appreciate it alot. Our actual project is to design an active drag system for our rocket. This ADS will be used to slow the rocket down to achieve as close to 10,000 feet as possible. so far our design is pretty cool.

Well, best of luck out there, and have fun with the rest of it!

 

[LT72884]

Well, best of luck out there, and have fun with the rest of it!

thank you very much my friend:) your an excellent teacher

 

[erobz]

Just an amateur, but thanks for the compliment.

 

[LT72884]

well, better than me haha. i taught high school math for a few years, but physics was never my strong suite haha.