jon e prevo wrote:
> 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.
So is yours. (Note to team.net - I'm not pissed, I'm just being sarcastic)
<snip example from kart racing which may well be true - I wouldn't know>
> 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).
This is true, but unless you are actually spinning the rear tires, the
fronts will be accelerated at exactly the same rate. Might as well be
connected directly to the engine. Now, if you like to do burnouts at the
start, then lighter drive wheels will make a small difference.
> 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.
Well, it turns out these are exactly the same thing. An object's resistance
to rotational acceleration means that it likes to keep rolling if it is
rolling and likes to stay put if it's put. Either way, the term is
"rotational inertia."
> A heavier wheel on the front of a
> rear-wheel-drive will add to the necessary force applied but not in a
> linear fashion.
Again, unless you are spinning the rears it is quite linear.
> I don't have an equation for the relationship but one is out there.
Indeed there is:
F = A(m + kI)
where F is force, A is linear acceleration, m is the total mass of the
vehicle, I is the rotational inertia of the rotating parts, and k is a
constant that converts linear velocity to rotational velocity. Note the
absence of any non-linear terms. If you *really* want to be precise, you
need to sum kI for each rotating part, as they may require different
conversions (camshafts spin much faster than wheels). Either way, it's
linear.
> However, a heavier wheel _will_ generate more rotational
> inertia, requiring more force to stop...
True enough.
> Remember that your heavier, non-drive wheel is supported by bearings.
Now we're splitting hairs but the car is supported by the bearings. The
wheel is supported by the ground (unless you're up on two wheels).
> The additional weight adds to the weight of the car and
> therefore to the force required to move the car.
Yes, both the mass *and* the rotational inertia. That's why wheel weight is
a big deal - you pay for it twice.
> 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.
Leaving aside the "friction cheating" bearings for a moment (whatever that
means), *any* additional weight increases the torque required to turn the
drive wheels. That's because the drive wheels are connected to the rest of
the car (rather the point). Again if it's rotating weight, so much the
worse.
> 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.
You really lost me there. I never did care for the prose in my engineering
texts. I just read the formulas and the derivations.
<snip the tank stuff>
> In the end, the difference isn't going to be enough to justify buying
> only two light wheels anyway.
Well, here we agree. It would seem to me that if the weight savings from two
wheels was worth the cost, than twice the weight savings from four should be
worth twice the cost.
Eric Buckley
7 STR: 98 Integra GSR
St Louis Region
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