Rocket car strategy

[SomptingGuy]

My company runs a rocket car race on or around 5th November most years. Contestants build small cars to be powered by one standard rocket (firework size). Longest distance travelled wins.

My question: Why is it always the lighter cars that win? I know they have less friction, but they win by a country mile. Is there any inherent advantage to be had by getting off the line fast - i.e. does a moving vehicle somehow extract more of the available rocket power than a stationary one?

 

[ivymike]

yes, if you assume ~constant thrust (thrust not depending on velocity of vehicle):
* thrust (force) is ~ contstant
* work done is force times distance travelled
* rate of work done (power into vehicle) is force times vehicle velocity

...a vehicle held stationary converts 0% of the chemical energy into vehicle kinetic energy, despite being exposed to a constant force over a given time period.

...a vehicle moving at Ymph converts X joules of the chemical energy into kinetic energy over the same time period. (obviously if the speed is unchanging the KE is simultaneously being lost to mechanical and aero friction)

...a vehicle moving at 2Ymph converts 2X joules of the chemical energy into kinetic energy over the same time period. (same story about friction)

 

[ivymike]

...stability of such a vehicle can be problematic if the vehicle is required to remain on a track (or so I've heard)

 

[SomptingGuy]

This was the argument I was giving to some of my colleagues, but they'd have none of it and kept going on about impulse and momentum.

Stability is a problem, as also are keeping the thing in a straight line and on the ground.

 

[ivymike]

well, perhaps they could hire some consultants from the US to straighten them out. I hear that there are some pretty good rocket car designers in the Chicago area.

 

[Fabrico]

Is there any inherent advantage to be had by getting off the line fast - i.e. does a moving vehicle somehow extract more of the available rocket power than a stationary one?

It doesn't extract more power, it just moves the car farther. It simply takes more energy to accelerate the heavier car and you don't get it all back after flameout.

 

[ivymike]

Don't get it all back? In either case, the entirety of the vehicle KE will be lost to friction & drag by the time the car stops. The lighter car accelerates more rapidly, but as a result it encounters much larger drag forces (~v^2), and will decelerate more rapidly too.

How can the lighter car encounter approximately the same rolling resistance and significantly higher aero drag whilst traveling a greater distance, if the total energy input to each vehicle is the same?

 

[SomptingGuy]

This is starting to sound familiar. I have a plot of the rocket's force vs time. I may digitise it and do some proper analysis...

 

[ivymike]

I'm not sure that a simulation is really necessary - if you simply do an energy balance for the vehicle (in excel if you prefer), I think you'll find that you were right all along.

 

[SomptingGuy]

I wasn't thinking simulation, just hand calcs ... maybe throw in some aerodynamic and friction approximations too.

 

[globi5]

Friction is less relevant in this case.

You can also put it like this (similar to what ivymike said):

Power output = Thrust*velocity (Thrust is a constant)
Power input = constant_input (consumption of fuel is constant)
Efficiency = output/input = (Thrust_constant*velocity)/constant_input

So the higher your vehicle's velocity the higher its efficiency. The velocity is higher the faster it accelerates. Higher efficiency = more distance travelled.

You can use this simple equation on a wheel driven car with a CVT as well where efficiency and input power is (more or less) constant. If efficiency is constant thrust (force at the wheels) must be significantly higher at low speeds.
On a rocket power car thrust is more or less the same, thus it must waste a lot of power at low speeds.

 

[Fabrico]

Amazing, some here can push a Cadillac as far as a Beetle.

 

[ivymike]

Unless I missed something Fab, you're the closest to making that contention. Care to support your hypothesis with something other than hot air?

 

[ivymike]

here's mine, pick it apart, or just pick whatever's handy:

vehicle 1 assumptions:
mass (kg) 0.1
density (kg/m3) 1.2
frontal area (m^2) 4.00E-03
Cd 0.35
Constant Thrust (N) 1
Burn Duration (s) 3
Crr (N/kg) 0.3

at zero seconds, the state of vehicle 1 is:
Time 0
thrust (N) 1
acceleration (m/s2) 10
velocity (m/s) 0
distance (m) 0
Rocket Work (J) 0
vehicle KE (J) 0
rolling res. (N) 0
aero drag (N) 0
rolling loss (J) 0
drag loss (J) 0

at 0.01 seconds, the state of vehicle 1 is:
Time 0.01
thrust (N) 1
acceleration (m/s2) 9.699920475
velocity (m/s) 0.097299773
distance (m) 0.000487849
cumulative Rocket Work (J) 0.000487849
vehicle KE (J) 0.000473362
rolling res. (N) -0.03
aero drag (N) -7.95249E-06
cumulative rolling loss (J) -1.46355E-05
cumulative drag loss (J) -2.1977E-09

The vehicle continues to accelerate, and reaches max velocity at flame-out:
Time 3.001
thrust (N) 0
acceleration (m/s2) -4.97952904
velocity (m/s) 23.60268608
distance (m) 39.19699735
Rocket Work (J) 39.17339717
vehicle KE (J) 27.85433952
rolling res. (N) -0.03
aero drag (N) -0.467952904
rolling loss (J) -1.17590992
drag loss (J) -10.17411135

the vehicle slows to a stop, finally reaching 0 m/s much later:
Time 29.351
thrust (N) 0
acceleration (m/s2) -1.14926E-10
velocity (m/s) -0.000116969
distance (m) 206.4020208
Rocket Work (J) 39.17339717
vehicle KE (J) 6.84083E-10
rolling res. (N) 0
aero drag (N) -1.14926E-11
rolling loss (J) -6.192060622
drag loss (J) -33.00493688

forgive the fact that the cumulative losses are slightly different than the total work input - it was a quick-and-dirty spreadsheet calc and I didn't spend much time thinking about changes between the timesteps. 0.06% error is probably okay.

Vehicle 2:
mass (kg) 0.3
density (kg/m3) 1.2
frontal area (m^2) 4.00E-03
Cd 0.35
Thrust (N) 1
Burn Duration (s) 3
Crr (N/kg) 0.3

@Time=0.01s
Time 0.01
thrust (N) 1
acceleration (m/s2) 3.033330706
velocity (m/s) 0.030633326
distance (m) 0.000154517
Rocket Work (J) 0.000154517
vehicle KE (J) 0.00014076
rolling res. (N) -0.09
aero drag (N) -7.88257E-07
rolling loss (J) -1.39065E-05
drag loss (J) -6.89924E-11

@max velocity:
Time 3.001
thrust (N) 0
acceleration (m/s2) -0.520703115
velocity (m/s) 8.878205952
distance (m) 13.48942788
Rocket Work (J) 13.48055108
vehicle KE (J) 11.82338114
rolling res. (N) -0.09
aero drag (N) -0.066210934
rolling loss (J) -1.214048509
drag loss (J) -0.452390269

@final stop:
Time 27.446
thrust (N) 0
acceleration (m/s2) -0.300000065
velocity (m/s) 0.004816143
distance (m) 111.9517007
Rocket Work (J) 13.48055108
vehicle KE (J) 3.47928E-06
rolling res. (N) -0.09
aero drag (N) -1.9484E-08
rolling loss (J) -10.07565306
drag loss (J) -3.413771489

Note that the rocket has delivered much less energy to the vehicle (lower efficiency), and that a larger proportion was lost to rolling resistance than to aero drag (okay, I assumed crappy bearings and crappy wheels for both cars). The heavy car has not travelled nearly as far.

 

[ivymike]

should note:
mass (kg) - total mass of vehicle

density (kg/m3) - density of air

frontal area (m^2) - (of vehicle)

Cd - drag coeff. of vehicle

Thrust (N) - thrust delivered by rocket (constant) during burn

Burn Duration (s) - duration of rocket burn

Crr (N/kg) - ratio of rolling resistance to mass (assuming crappy wheels and bearings)

Time - time in seconds since start of calc

thrust (N) - thrust delivered by rocket over preceding timestep

acceleration (m/s2) - acceleration of vehicle over preceding timestep (constant per step)

velocity (m/s) - velocity of vehicle at end of preceding timestep

distance (m) - distance travelled at end of preceding timestep

Rocket Work (J) - cumulative work done by the rocket on the car since the start of the calculation

vehicle KE (J) - approximate kinetic energy of the vehicle at the end of the preceding timestep (0.5mV2) - excluding rotational KE of the wheels+axles

rolling res. (N) - rolling resistance experienced by the vehicle at the end of the preceding timestep (Crr*m)

aero drag (N) - aero drag experienced by the vehicle at the end of the preceding timestep (0.5*density*Cd*A*V2)

rolling loss (J) - cumulative energy lost to rolling resistance since the beginning of the calc

drag loss (J) - cumulative energy lost to aero drag since the beginning of the calc

 

[ivymike]

200m would be a new record, I'll bet - try to make a 100gm rocket car!

 

[patprimmer]

For the same burn time and thrust, and all other factors being equal, the lighter car accelerates faster so covers more distance in the same time as it's average speed will be higher. It seems that simple to me.

Another factor is aero, but that is independent of weight.

Stability might be a bigger problem for lighter cars as they will reach higher speeds and any off centre thrust will be a greater percentage of car weight.

Improving stability might decrease aero.

Regards

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[SomptingGuy]

I just did a simple-minded Simulink model of this using the same equations and parameters as ivymike and (unsurprising) got the same results for the two vehicles.

0.1kg -> 206.3m
0.3kg -> 111.8m

Cut the rolling drag down to 10% of its value (i.e. 0.03) and the story is different. There is a peak at about 0.20 - 0.3 kg:

0.1kg -> 343.7m
0.2kg -> 420.8m
0.3kg -> 419.9m

Similarly, replace the thrust with the curve supplied by the manufacturer (initial short burst at 14N followed by 2 seconds at 4N) and an optimum appears between 0.2kg and 0.3kg.

These results are all likely to be an artefact of the less-than-real rolling drag model, but they do show that there is likely to be an optimum mass rather than the two opinions discussed here (i.e. mass is either bad or irrelevant).

 

[ivymike]

note that the rolling drag I used was N/kg, not N/N. If you convert to the more conventional units, you get 0.031N/N.

Those results may be due to the effect of the initial thrust burst as well - if you get the vehicle moving, then the efficiency during the 2S-4N burn will be reasonably high for a large range of masses - and the heavier vehicle encounters less aero drag during its run, for less net loss. Interesting result.

 

[SomptingGuy]

Continuing...

If you provide a constant thrust over the 3 second period you find that the optimum mass increases with that thrust. So it's not an !initial thrust burst" effect. My colleagues here are now using one of my least favourite expressions: "getting over the inertia"! You can't get over inertia, you can get over stiction, but inertia doesn't go away.

If I make a rocket car this year I'll be making it really light and then experimenting with ballast.