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Vehicle Forces

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CVLDGolf

Aerospace
Jul 9, 2021
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Hi All,

Consider a typical 4-wheeled vehicle with motor/gears/axle assembly. How do I determine the force needed to start the vehicle rolling. Is rolling friction really the only force that you need to overcome to begin acceleration since aero forces are non-existent?

For example I have an empty 23,500kg railcar with .002 rolling friction coefficient and 1m diameter wheels. I come up with .002*9.81*23500=461N. So 461/r gives required torque of 922N. Even with a modest 6:1 gear reduction this would imply my motor only needs to be capable of 153N to accelerate the vehicle (Obviously very slowly). But this seems awfully low. Am I missing something? Friction in the drivetrain?
 
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There is friction in the bearings which may initially be higher when stopped due to lack of lubrication distribution while parked, but otherwise people can use box car mover to move railcars with a bit of effort once the brakes are released (probably the main thing to overcome otherwise.)
There can be friction against the sides of the rails if the wheels aren't centered up. There is also the fact that .002 is the equivalent of 2mm/meter. Slopes up to 1.5% (15mm/meter) are allowed, so it pays to be careful. Those who work around rail lines mention that when rolling slowly railcars are usually essentially silent.

So, yes, you probably can.
 
As a child of the 90's I remember watching TV shows where record setters would pull airplanes and rail cars with either their ear lobes or balls sacks (a post y2k symptom). I think this means the assumption nis essentially correct.
 
CVLDGolf said:
Consider a typical 4-wheeled vehicle with motor/gears/axle assembly. How do I determine the force needed to start the vehicle rolling.

Rather obvious isn't it?
[ol 1]
[li]Park your car on a level surface.[/li]
[li]Leave the gearshift in Neutral, parking brake off.[/li]
[li]Pull on the bumper with a scale until it moves.[/li]
[li]Take note of the number.[/li]
[/ol]

Guaranteed to be faster than a calculated result.

If you really want to do math, just take the number you measured and multiply it by the ratio of vehicle weights for any different car/truck. For a different surface, park your car on that surface and repeat the test. Keep the scale in the trunk for multiple opportunities to try the test. It will make for some interesting conversations in the mall parking lot.

Note that in my garage there is a 1 degree grade so that water drains out the door. I can easily push my car (in neutral) out the door if I don't feel like starting the engine, but pushing it back in again is really hard, so I have to drive it back in. The 2 degrees makes a big difference.
 
It's faster than an equation that has a lot of terms you have no values for or an incomplete one that gives entirely wrong results.

"Consider a typical 4-wheeled vehicle with motor/gears/axle assembly" is not describing a typical vehicle on a rail.
 
To find the force required you need to find the static friction coefficient which is usually higher than the rolling friction number you quoted. Now qqhow much more is the question and that will depend on many things such as bearings, gearboxes, surface condition of the wheels and rails/ road and whether any brakes are rubbing. These are difficult numbers to find to any degree of accuracy without doing tests.

In any strong man type event pulling things it's always getting it moving which is the hard bit...

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
A railcar is a fairly simple analysis bc you're dealing with steel wheels having relatively little friction against a steel track which are rolling against bearings. An AAR handbook may have some relevant data. OTOH, if you're doing an automobile analysis, you have rubber tires rolling against asphalt with relatively high friction (which varies with tire pressure), and spinning quite a few bearings, a differential or two, driveshafts/halfshafts, and likely some transmissions components as well. Both scenarios have force vs acceleration curves, but the later is a much more complex analysis.
 
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