Yesterday, I submitted an analysis of the horsepower required to maintain a
given speed. I intimated that the "real" requirement was getting to speed
and the power needed for that action. I think I was wrong! The excess
horsepower above what is needed to maintain a given speed is what is
available for acceleration. When the horsepower requirements are maximum
for maintaining a top speed, there is no power left for acceleration,
q.e.d. You simply use the excess to accelerate to whatever speed you want
to drive at and back off the pedal enough to maintain the speed.
Now, I also indicated that bearing friction was a small requirement. Here
is why.
Horsepower loss due to rolling friction (HP) = Mt * N / 63000
where Mt = F * f * (D/2)
F = bearing radial load (lbs)
f = friction factor (.0018)
D = bearing bore (inches)
N = RPM
For our case, the Alpine/Tiger weighs about 2400 lbs, give or take. And the
weight distribution is about 50/50 putting about 1200 pounds n the front
and rear bearing sets. Take the rear axle and each side has about 600
pounds riding on each axle bearing. So F = 600. I havent measured the
bearing, but, I suspect it is about 1.125 inches in diameter. So D = 1.125
inches. Lets assume our tires are about 24 inches in diameter, so the
distance rolled in one revolution is 6.28 feet (pi x Diameter). If we use
120 mph, then we are travelling 2 miles per minute. The revolutions per
minute, then for the rear axle/bearing is 1681.5 rpm at 120 mph. Putting it
all together then...
HP loss = 600 * (.0018) * (1.125/2) * 1681.5 / 63000
= 0.016.
Since there are two rear axle bearings double this. Oh, and there are four
front bearings, but at half load each, so double again. Wow, we still have
less than 1 horsepower lost to wheel bearing friction. So wheel bearing
loss is not significant.
What about toe-in/out problems. At 1/8 inch toe in and a contact patch
about 6 inches long, the slip angle is about 1.2 degrees. At 600 pounds
normal load, then the off angle load would be about 12.6 pounds per front
wheel or about 26 pounds total drag. Assumin a coefficient of friction of,
say, 0.5, the the total drag load would be about 13 pounds. Horsepower is
drag times velocity divided by 550. Assume 130 mph or 190.71 fps, then the
loss is 4.5 horse power. So if we add this to the power needed at 130 mph
to overcome aerodrag the it is about 138 horsepower to maintain.
If we can again asuume some efficiencies of the tranny and differential of
say 95% each then the power at the flywheel is 138/(.95)*(.95) or about 150
dyno hp.
This is FUN!!
L.E. Mayfield
124 Maximillion Drive
Madison, Al. 35758-8171
205-837-1051
DrMayf(at)AOL.com
mayfield(at)traveller.com <<<<preferred
"Thought is the software of the mind;
rational thought is bug free!"
This archive was generated by hypermail 2b30 : Tue Sep 05 2000 - 10:07:24 CDT