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

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Discussion Overview

The discussion revolves around the possibility of determining the terminal velocity of a model rocket based on data collected from a launch to 20,000 feet. Participants explore methods to analyze velocity data, calculate drag coefficients, and the implications of different phases of the rocket's flight (ascent vs. descent).

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant inquires whether it is possible to find terminal velocity from a dataset that includes times, altitudes, and velocities from a rocket launch.
  • Another participant suggests using Excel to calculate velocity as a function of height data collected during the descent.
  • Concerns are raised about the validity of the 20,000 feet launch height, with some participants expressing skepticism about the numbers presented.
  • Participants discuss the importance of distinguishing between the rocket's behavior during ascent and descent, noting that the drag coefficient may differ in these phases.
  • There is a question about whether terminal velocity can be determined during the coasting phase (burnout) or if it can only be calculated during free fall.
  • Some participants propose that calculating an upper bound for terminal velocity during free fall may be a feasible approach.
  • One participant mentions that they have collected accurate data from a Raven 4 sensor and plans to reach higher altitudes in future launches.

Areas of Agreement / Disagreement

Participants generally agree on the need to analyze the data carefully, but there is no consensus on the methods to determine terminal velocity or the implications of the rocket's orientation during descent. Multiple competing views remain regarding the feasibility of calculating terminal velocity during different flight phases.

Contextual Notes

Participants note that the rocket's drag coefficient may vary between ascent and descent, and the limited data available from the coasting phase may restrict the analysis. There is also uncertainty regarding the accuracy of the reported launch height and the implications for the calculations.

  • #151
erobz said:
Good, so rewrite it. After you have done that set the acceleration to 0, and solve the resulting equation.
Vt = sqrt(-mg/k)
 
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  • #152
LT72884 said:
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?
 
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  • #153
erobz said:
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.
 
  • #154
LT72884 said:
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.
 
  • #155
erobz said:
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
 
  • #156
LT72884 said:
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.
 
  • #157
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:)
 
  • #158
LT72884 said:
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?
 
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  • #159
erobz said:
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
 
  • #160
LT72884 said:
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.
 
  • #161
erobz said:
Yeah. k is β. No hurry.
average terminal velocity is 234.68. IF this was free fall... but we know its not
 
Last edited:
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  • #162
LT72884 said:
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 a lot. 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.
 
  • #163
LT72884 said:
thanks for all the help. I really do appreciate it a lot. 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!
 
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  • #164
erobz said:
Well, best of luck out there, and have fun with the rest of it!
thank you very much my friend:) your an excellent teacher
 
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  • #165
Just an amateur, but thanks for the compliment. :smile:
 
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  • #166
well, better than me haha. i taught high school math for a few years, but physics was never my strong suite haha.
 

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