Rocket car strategy
Rocket car strategy
(OP)
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?
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?





RE: Rocket car strategy
* 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)
RE: Rocket car strategy
RE: Rocket car strategy
Stability is a problem, as also are keeping the thing in a straight line and on the ground.
RE: Rocket car strategy
RE: Rocket car strategy
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.
RE: Rocket car strategy
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?
RE: Rocket car strategy
RE: Rocket car strategy
RE: Rocket car strategy
RE: Rocket car strategy
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.
RE: Rocket car strategy
Amazing, some here can push a Cadillac as far as a Beetle.
RE: Rocket car strategy
RE: Rocket car strategy
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.
RE: Rocket car strategy
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
RE: Rocket car strategy
RE: Rocket car strategy
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.
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RE: Rocket car strategy
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).
RE: Rocket car strategy
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.
RE: Rocket car strategy
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.
RE: Rocket car strategy
Both coast to a stop after the same amount of time (same rocket) and then the rest of the distance involves constant losses to friction, both through wheels and air.
If the light car is moving faster than the heavy car at flameout, it has that much farther to travel before losing everything to the wheel friction.
Assuming of course the wheels and axle bearings are anything like the Hot Wheels I used to play with.
If you had a magnetic bearing maybe you'd win?
RE: Rocket car strategy
Hmmm, which one is better; science based on assumptions or assumptions based on science? I'll take the latter.
RE: Rocket car strategy
RE: Rocket car strategy
ivymike, you seem to be an arguement waiting to happen. But more than anything else, you appear to be arguing with youself. Perhaps I am stupid beyond belief, but I don't think I am alone in not quite knowing what your point is.
My original reference was to the OP who does not appear, or try to appear, to be a rocket scientist.
I, and apparently a few others, need nothing more than the simple hot air already given. The lighter car accelerates easier. It rolls easier. It coasts farther. It goes farther. It wins the race. I will challange your lengthy equations when you convince me otherwise.
This analogy might be tough for you, but if I had to push a car through the quarter mile, and the choice was between a Cadillac or a Beetle, I know which one I would choose. And I would again reserve the right to not use math in my decision.
RE: Rocket car strategy
The rocket equation is based on the impulse law and it says:
v_max = v_fuel * ln((mass_empty+mass_fuel)/mass_empty)
So the lighter the vehicle the higher v_max and thus the farther the vehicle travels.
Or if you were to release all fuel at once you can use the general impulse equation and get:
v_rocket = mass_fuel/mass_empty * v_fuel
RE: Rocket car strategy
Please replace "coast to a stop" with "lose thrust" in my above post.
Forget all the air-blowing anyways, why hasn't anybody posted time-trial results of their custom rocket cars yet? Does the car have to stay on a track or can it go airborne?
I think we could have some fun with downforce and stability concerns - a Star to the first person to come up with a landfoil (what I envision as a rocket with a small trailing set of wheels/skids so it still counts as being on the ground).
RE: Rocket car strategy
RE: Rocket car strategy
http://aardvark.co.nz/pjet/gokart.htm
Cheers
Harry
RE: Rocket car strategy
RE: Rocket car strategy
While the obvious solution happens to be correct, the reason behind it is not so simple.
If it were an equal energy situation, such as if the cars were all powered by the same spring or the same gallon of gas, then I think an argument could be made that the heavier vehicle would travel more efficiently due to less drag encountered at its lower top speed. This would balance against increased rolling resistance, and you could calculate an optimal car weight. For example, which goes farther when you shoot it out of a gun: a feather or a bullet?
The difference in this situation is the nature of the rocket. The rocket produces a fixed force, so you can capture more of the power produced by the rocket the faster you are going. There would be no optimal weight, as the lightest car would always win.
I think it's an interesting problem.
-b
RE: Rocket car strategy
The rockets this year clearly had higher and longer thrust than in years past, to the point where structural integrity and directional stability were probably more important than any energy balance calcs. The winner (light and strong, with DIY solid tyres) topped 65m and was still bombing along at a high rate of knots when it left the track and stopped in the grass.