This rocket physics problem seems unrealistic

[LT72884]

TL;DR Summary

Finding max height for a rocket, then terminal velocity during free fall

Ok, so here is what i am trying to do. Find apogee of a model rocket i built using a C motor, then i need to calculate terminal velocity, but something is not adding up, i get a really really high apogee, almost a mile.

STAGE 1:
I have a c6-7 rocket motor which provides 10N-s of impulse, and 6N of thrust.
Burn time = Impulse/Thrust = 10/6 = 1.66 seconds of burn time
Rocket mass is 53g or 0.05398kg
Weight = 0.529N
a=F/m where F=Th-mg so Resultant force is 5.47N
acceleration =101.35m/ss
Burnout Velocity = v0+at = 0+(101.35)(1.66)=168.2m/s
Height of rocket after burnout velocity is y1=v0t+(at^2)(0.5) = 0+(101.35)(1.66^2)/2=139m (seems reasonable for a C motor)

STAGE 2: after burnout Acceleration is now -9.8
vi=168.241m/s
at top of apogee, v=0 so 0=v0+at t=-v0/a t=(-168.241)/-9.8 = 17.1 seconds of coasting time. this seems like ALOT OF TIME maybe because rocket is so light??
now y2 = v0t-at^2(0.5) = 168.241)(17.167)-((9.8)(17.167^2)(0.5)= 1444m
total apogee is y1+y2 = 1583m or 4749feet give or take...This seems ridiculous and un-realistic, maybe because a C motor is way to powerful for that size of rocket?

Ok, let's say these numbers do indeed work out, how do i find terminal velocity now without drag? is it sqrt(2gh)? I am trying to find Cd myself to compare with my win tunnel results by using
Cd=((2mg)/(rho(vt^2)(Area)
EDIT: i just looked in my fluid dynamics book, and i have a feeling that to find the CD or drag force without testing is a bunch of crazy integration?

thanks

 

[kuruman]

I am not sure I understand the equation

Burn time = Impulse/Thrust

If Impulse is 10 N and thrust is 6 N, it follows that 10 (N) / 6 (N) = 10/6. Where do the units of "seconds" come from? Why not microseconds or fortnights? It seems to me that burn time depends on the rate at which the fuel is consumed, but then again I am not a rocket scientist.

 

[jrmichler]

.This seems ridiculous and un-realistic

Your intuition is correct because air drag is significant at the scale of a model rocket. The $C_d$ of a model rocket is determined by experiment, or estimated from experimental results of similar shapes. I'm not a model rocket person, but Google tells me that a C motor is 18 mm diameter. Assuming that is the diameter of the rocket, then the aerodynamic drag can be estimated as follows:

Drag Force = $0.5 * C_d * A * \rho * V^2$, where
$C_d$ is estimated at 0.5 (this is just a guess to make the calculation)
$A$ is cross sectional area ($.000254 m^2$for 18 mm diameter)
$\rho$ is the density of air about (1.2 $kg / m^3$)
$V$ is the velocity in $m/sec$

Then the drag force is about 2.2 N at 168 m/sec. The quick conclusion is that the rocket will not reach 168 m/sec, and it will slow down rapidly after the motor stops firing.

There may be an analytical solution to this problem, but I solve these types of problems numerically. This can be done using a spreadsheet, although I usually use Octave, a freeware Matlab clone.

 

[jbriggs444]

If Impulse is 10 N and thrust is 6 N, it follows that 10 (N) / 6 (N) = 10/6. Where do the units of "seconds" come from?

The impulse has units of Newton-seconds (a force integrated over an interval). Divide by force (and make the assumption of constant thrust) and you get an interval.

If you prefer, impulse has dimensions of momentum or of mass times velocity. But force times time is the same thing and is more intuitive for the purposes of this calculation.

 

[kuruman]

The impulse has units of Newton-seconds (a force integrated over an interval). Divide by force (and make the assumption of constant thrust) and you get an interval.

If you prefer, impulse has dimensions of momentum or of mass times velocity. But force times time is the same thing and is more intuitive for the purposes of this calculation.

Yes, I see it now. Thanks.

 

[LT72884]

Your intuition is correct because air drag is significant at the scale of a model rocket. The $C_d$ of a model rocket is determined by experiment, or estimated from experimental results of similar shapes. I'm not a model rocket person, but Google tells me that a C motor is 18 mm diameter. Assuming that is the diameter of the rocket, then the aerodynamic drag can be estimated as follows:

Drag Force = $0.5 * C_d * A * \rho * V^2$, where
$C_d$ is estimated at 0.5 (this is just a guess to make the calculation)
$A$ is cross sectional area ($.000254 m^2$for 18 mm diameter)
$\rho$ is the density of air about (1.2 $kg / m^3$)
$V$ is the velocity in $m/sec$

Then the drag force is about 2.2 N at 168 m/sec. The quick conclusion is that the rocket will not reach 168 m/sec, and it will slow down rapidly after the motor stops firing.

There may be an analytical solution to this problem, but I solve these types of problems numerically. This can be done using a spreadsheet, although I usually use Octave, a freeware Matlab clone.

ok, good. i thought something was fishy haha. Now, is it possible to calculate Cd without testing? IE, how did they first solve for Cd and drag without wind tunnels and testing?

or can drag and Cd only be determined via testing?

thanks

 

[gmax137]

Lots of good answers already. But no one has mentioned this:

Ok, let's say these numbers do indeed work out, how do i find terminal velocity now without drag?

Without drag, there is no terminal velocity. Maybe I am misunderstanding this question.

 

[erobz]

Summary: Finding max height for a rocket, then terminal velocity during free fall

Ok, so here is what i am trying to do. Find apogee of a model rocket i built using a C motor, then i need to calculate terminal velocity, but something is not adding up, i get a really really high apogee, almost a mile.

STAGE 1:
I have a c6-7 rocket motor which provides 10N-s of impulse, and 6N of thrust.
Burn time = Impulse/Thrust = 10/6 = 1.66 seconds of burn time
Rocket mass is 53g or 0.05398kg
Weight = 0.529N
a=F/m where F=Th-mg so Resultant force is 5.47N
acceleration =101.35m/ss
Burnout Velocity = v0+at = 0+(101.35)(1.66)=168.2m/s
Height of rocket after burnout velocity is y1=v0t+(at^2)(0.5) = 0+(101.35)(1.66^2)/2=139m (seems reasonable for a C motor)

STAGE 2: after burnout Acceleration is now -9.8
vi=168.241m/s
at top of apogee, v=0 so 0=v0+at t=-v0/a t=(-168.241)/-9.8 = 17.1 seconds of coasting time. this seems like ALOT OF TIME maybe because rocket is so light??
now y2 = v0t-at^2(0.5) = 168.241)(17.167)-((9.8)(17.167^2)(0.5)= 1444m
total apogee is y1+y2 = 1583m or 4749feet give or take...This seems ridiculous and un-realistic, maybe because a C motor is way to powerful for that size of rocket?

Ok, let's say these numbers do indeed work out, how do i find terminal velocity now without drag? is it sqrt(2gh)? I am trying to find Cd myself to compare with my win tunnel results by using
Cd=((2mg)/(rho(vt^2)(Area)
EDIT: i just looked in my fluid dynamics book, and i have a feeling that to find the CD or drag force without testing is a bunch of crazy integration?

thanks

Can you weigh a rocket motor before and after burnout and give this info? I would like to see if its change in mass is significant to the total rocket mass to determine whether or not this should be a variable mass solution.

Also, as others have already mentioned, we are going to need to consider drag in every stage from takeoff to landing.

 

[LT72884]

i can't at the moment, but i think i can ask someone who can. Ill try and get as close to a C6-7 motor

 

[brainpushups]

In case it's useful, you can get engine data here:

https://www.thrustcurve.org/

 

[LT72884]

Can you weigh a rocket motor before and after burnout and give this info? I would like to see if its change in mass is significant to the total rocket mass to determine whether or not this should be a variable mass solution.

Also, as others have already mentioned, we are going to need to consider drag in every stage from takeoff to landing.

before launch is 25.8 g
and 12.48g solid fuel material

 

[erobz]

The site @brainpushups linked shows 10 g ( about 20% of the total rocket mass) propellant weight. Probably significant...which means its a bit more ugly. Also, The thrust is not constant. Average thrust 6 N, max 15 N. The thrust curve vs time is given in that data set ( great find @brainpushups ). How accurate do you want to try and be? Its looking like a numerical solution either way for the powered flight stage.

For the ascent to apogee/descent under constant mass stages, we can get an analytical result.

https://www.thrustcurve.org/motors/Klima/C6/

 

[erobz]

Silly question. Why is this rocket falling without a parachute?

 

[erobz]

Anyhow, if you want my non-expert opinion, I think we should assume constant thrust for 2 reasons:

1. The thrust curve only spikes to 15 N for a very short time period and then is basically flat till very near the end.
2. Assuming constant thrust $T$ we have the following:

$$ T = \rho A_e v_e^2 \implies v_e = \sqrt{\frac{T}{\rho A_e}} = \rm{const.} $$

That means that the mass flow rate of ejecta is a constant given by:

$$ \dot m = -\rho A_e v_e = - \rho A_e \sqrt{\frac{T}{\rho A_e}} = \rm{const.}$$

The propellant mass remaining in the rocket at time $t$ is just a simple linear function then:

$$ m(t) = m_o - kt $$

If you don't go that route its going to be quite a bit more messy.

 

[LT72884]

Silly question. Why is this rocket falling without a parachute?

so i can find terminal velocity of the rocket so i could calculate drag without a wind tunnel hahaha

 

[LT72884]

Anyhow, if you want my non-expert opinion, I think we should assume constant thrust for 2 reasons:

1. The thrust curve only spikes to 15 N for a very short time period and then is basically flat till very near the end.
2. Assuming constant thrust $T$ we have the following:

$$ T = \rho A_e v_e^2 \implies v_e = \sqrt{\frac{T}{\rho A_e}} = \rm{const.} $$

That means that the mass flow rate of ejecta is a constant given by:

$$ \dot m = -\rho A_e v_e = - \rho A_e \sqrt{\frac{T}{\rho A_e}} = \rm{const.}$$

The propellant mass remaining in the rocket at time $t$ is just a simple linear function then:

$$ m(t) = m_o - kt $$

If you don't go that route its going to be quite a bit more messy.

where did you find the equation for T to be (rho)(A)(v^2)

im trying to understand where things came form haha

 

[erobz]

where did you find the equation for T to be (rho)(A)(v^2)

im trying to understand where things came form haha

That actually comes from the "Rocket Equation". You can look that up and see how its derived ( impulse/ momentum). There is also a pressure term that may not be negligible there for this chemical rocket that I left out that is from the exhaust gas being above ambient pressure. If we assume that it is basically constant also it doesn't change the outcome of that mass flow rate result.

But I think officially for a liquid fueled rocket, (and probably for a solid fueled rocket too). This is the thrust force.

$$ T = \rho A_e v_e^2 + P_e A_e $$

As is with everything, almost nothing as actually constant though.

 

[erobz]

so i can find terminal velocity of the rocket so i could calculate drag without a wind tunnel hahaha

So you are really only interested in the “theoretical” terminal velocity of the rocket? Or do you want to know whether it will get to some reasonable fraction of terminal velocity before hitting the ground? The second is the more difficult question, the first is quite straightforward and can be answered independently of all the other stuff.