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Re: Lightweight wheels...drive only?

To: miket@interaccess.com
Subject: Re: Lightweight wheels...drive only?
From: jon e prevo <tcbracer@juno.com>
Date: Thu, 27 Jul 2000 16:44:35 -0500
I have waited for this point to be made.

On Thu, 27 Jul 00 15:13:34 -0500 miket <miket@interaccess.com> writes:
> I've pondered  this myself, and concluded that the drive wheels are 
> 50%. 

This is not exactly true.  Your reasoning is correct, it is your physics
which is flawed.

 The only difference is that the 
> drive 
> wheels are spun directly by the motor, and the non-drive wheels are 
> spun 
> indirectly by the motor 

That "only" difference is a big difference.

Ever watch sprint-kart racing?  On the longer straights, the drivers kind
of jump themselves forward in their seats, gaining all the momentum they
can.  No, it doesn't make the kart faster, what it does is unload the
rear wheels temporarily (a short-term version of what we are discussing,
right?) allowing the engine to revto top rpms more quickly.  This effect
only lasts for split seconds but you will be amazed at how effective it
is.

Lightenning the drive wheels will have the same sort of effect as
lightenning the flywheel or the driveshaft or the diff gears.  Energy
required to rotate a lighter wheel  _directly_  is less, allowing the
engine to more efficiently move the vehicle (see above example about
karts).  I believe you are confusing your terms.  "Rotational Inertia" is
a description of a tire's tendency to remain in motion, not it's
resistance to force applied.  A heavier wheel on the front of a
rear-wheel-drive will add to the necessary force applied but not in a
linear fashion.  I don't have an equation for the relationship but one is
out there.  However, a heavier wheel  _will_  generate more rotational
inertia, requiring more force to stop...

Remember that your heavier, non-drive wheel is supported by bearings. 
The additional weight adds to the weight of the car and therefore to the
force required to move the car.  The additional weight does not, however,
add to the torque required to turn the drive wheels, due to their
suspension on the friction-cheating bearings.  I postulate that as the
weight differential increases between drive- and non-drive wheels, the
effect of the lighter drive wheel (% of difference) will increase in a
curve similar to a typical horsepower curve, i.e. as the drive wheel gets
lighter potential difference will decrease.

dg, your hyperbole of the ground being the track on a tank is incorrect. 
On a tank, the drive wheel not only has to support the (ground) track but
also has to overcome the friction between (ground) the track, the ground
and the other wheels.  Friction is applied three times in this scenario,
always to one place, the drive wheel.  In a car, friction is applied
individually to all wheels, so the friction the engine has to deal with
is all of that acting on the drive wheels but only a fraction of that
acting on the non-drive wheels.  The wheels' bearings take up the rest of
that friction.

In the end, the difference isn't going to be enough to justify buying
only two light wheels anyway.

Jon FP 73

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