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. Assuming 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!!
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