Re: [MV] Fastest "mil-veh" valve cover racer???

From: Brown, Herb (BRown@ida.org)
Date: Mon May 08 2000 - 09:54:23 PDT


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In general, the quick answer is the heavier valve cover, with the most
frictionless wheels/bearings, and the lightest wheels.

I believe that a 15% grade translates to only around 8.5 degrees (not sure
about this). If this is the case, the normal force (acting perpendicular to
the track) is around 0.9 times the weight of the vehicle (valve cover and
wheels); and the frictional force is essentially the coefficient of friction
times this normal force. The force acting along the track is approximately
0.15 times the weight of the vehicle. (Note, these numbers change as the
angle of the track changes.) If you have reasonable size, and very-very
light wheels, you may ignore the inertial forces to get the wheels spinning
(and besides the vehicle you are racing against will need wheels also, so
differences between vehicles may not be too great). Under these
circumstances, the greater the difference between the force heading down the
track and the friction force, the greater the vehicle speed.

For example, if the valve cover weights 1 LB and the coefficient of friction
is 0.1, then the force to propel the vehicle along the track is only about
0.06 lbs. However if the coefficient of friction is 0.01, the force to
propel the vehicle along the track is more like 0.14 lbs.

This answer could be more complex: if you begin to look at weight
distribution on wheels if the angle of the track changes, or if you begin to
consider the forces needed to spin up the mass of the wheels into this
equation (such as very small wheels must turn very fast or large - heavy
wheels that require extra forces to spin up the wheel mass), or if your
begin to consider overcoming static friction and influences of rolling
friction. -- But this level of detail should take all the fun out of any
race.



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