Car acceleration force required to size a DC motor torque

AI Thread Summary
The discussion focuses on calculating the torque required for a model car to accelerate from a stop to 1.22 m/s in one second, considering a weight of 0.85 kg and wheel diameter of 0.03175 m. The preliminary calculations indicate an acceleration of 1.22 m/s², leading to a wheel RPM of 733.4, and a power requirement of approximately 0.632 watts, which participants suggest may be underestimated. Participants emphasize the need to account for factors like friction, gear ratios, and angular momentum, as well as the importance of selecting a motor that can handle the required torque and power for higher speeds, aiming for a maximum speed of 4.5 m/s. Recommendations include using a safety factor in motor selection and considering the effects of back EMF when reversing the motor for deceleration. The conversation highlights the complexities of motor selection and the need for accurate calculations to ensure the model car performs as intended.
Mike Gaffer
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I'm working on making a model car, and am trying to calculate the torque required to accelerate the car from a stop to 1.22m/s in 1 second, so I can pick out the right DC motor...

The acceleration required is 1.22m/s/s (~ 4ft/s)
The weight of the car is 0.85 kg
The drive wheels diameters are 0.03175 m (1.25") [wheel circ 0.099746m]


I do not have to take into account wind resistance, frictional resistance, gear box ratios (yet), drive train friction, etc. (this is still a preliminary calculation)...

I WOULD like to take into account the angular momentum ... and am going to have to have some worm gear or gear box reduction (didn't get that far yet)...

So far, I did:
velocity (final) = 1.22m/s (in time = 1s)
= 1.22 m/s^2 acceleration

rpm of wheel = (velocity of car) / dist wheel travels in 1 rotation [circum.]
= (1.22m/s) / 0.099746 m
= 12.22 rps
= 733.4 rpm

ω = rpm x 2pi / 60
= 733.4 x (2pi / 60)
= 76.80093 rad/s

P = W/t = KE / t = 1/2 mv^2/t [power = work/time]
= 0.632 watts?
= E[J] = P[watts] x t[sec] = 0.632 Joules... ? <---- where does ω come into this?

ALSO, 0.632 watts seems like way too little power required... did I miss something?

THANKS IN ADVANCE!
 
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Accelerating a small car doesn't take much power. Things to consider, however:
- power requirements will increase over time, your maximal power is not the average power. For uniform acceleration, it is larger by a factor of 2.
- friction can be significant
- the power output of a motor depends on its speed / you need enough torque to accelerate the car, maximal power alone is not sufficient.
- the rotating parts in the car need some additional energy as well
 
mfb:
How would you recommend I start?
I actually wanted the MAX speed to be closer to 4.5m/s (10mph)...
I was going to take a regular gear on the rear axle (no differential) -- and use a worm gear on the motor directly coupled to the straight gear...
Just pick a motor that "looks" about the right size? I was considering modeling the drive loosely a model train engine's drive...
Any recommendations on how to start? I'm willing to do a 2nd revision if I get too small/big a motor :)
 
Use a sufficient safety factor, and check which motors are available.
The motor will also influence the mass of the car, of course.
Mike Gaffer said:
I actually wanted the MAX speed to be closer to 4.5m/s (10mph)...
How long is your acceleration track?
 
The acceleration track (including space for it to decelerate) will be about 25 feet long...
AGREED re safety factor... I was hoping to calculate exactly what I needed... then assume about 85% friction for gears and double whatever I calculate to size the motor...
 
To reach 4.5 m/s in ~8 meters you need a faster acceleration, and certainly more power.
 
good point... maybe 2m/s is more realistic...
That would give a terminal velocity needed of 4.5m/s in 5m and 3m to stop (reverse current).
 
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I'm confused on how you wanted the DC motor to stop? Like reverse the current so that it runs in the opposite direction, or a way to increase the resistance so that it slows down like that?
 
Reverse the current... However this will kick lots of back EMF that must be dealt with...
 
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