Originally posted by sjs:
The topic of rotational mass/inertia, moment of inertia, etc. is discussed here from time to time but I haven't seen anyone post the precise math for calculating the equivalent HP loss. [This message has been edited by sjs (edited 05 April 2002).]
I have done the math. Rotational mass and power loss is one of the most misunderstood areas. Let me try to explain the basics and give some numbers.
Unsprung mass and rotational mass are two different concepts. Rotational mass is just like it sounds, anything that rotates: flywheel, half-shafts, rotors, wheels, tires.... Reducing unsprung mass helps the suspension conform to the road (think about trying to control a bowling ball on the end of spring versus a softball). They are confused because some of the rotational mass is also unsprung.
Rotational mass takes energy and power to spin it up to any given RPM. The location of the mass effects the mass moment of inertia. Mass further from the center takes more engergy to spin it up. Recall as a kid hand pushing the marry-go-round type equipment in the playground. If a bunch of kids were standing on the outside edge, it was hard to push.
The power required to spin an object is proportional to
(Mass moment of inertia)RPM^2/time.
When you reduce the mass moment of inertia (lighter wheels, tires, flywheels) the power required is less. Notice that the power required is proportional to RPM squared and divided by time. This tells us that things that spin to high RPM and do it quickly have an even bigger effect. This is why spinning a flywheel in 1st gear takes up to 20 HP at high RPM, but that same flywheel might only take 1-2 HP to spin it in fifth gear because the rate of change of RPM^2 is much less.
This is why you can not quote (or even measure) the horsepower loss, because it changes at different speeds. Measuring the change on a chassis dyno is another impossiblity because the machine doesn't spin the wheels and tires at the same change in rate as an actual accelerating vehicle.
For the numbers. In 1st gear, the stock flywheel can require anwhere from 5 to 20 HP to spin it any different RPM's. Therefore, if the mass moment of inertia is reduced by 50%, you would save half this amount in 1st gear only. In higher gears, the flywheel requires 1-5 HP to spin it, therefore the lighter flywheel would only save a fraction of this amount.
Four tires (20,000 lbm in^2 total) require anywhere from 2.5 to 10 HP to accelerate them (the mass moment of inertia is high, but the RPM^2/time term is much smaller). Because most of the mass moment of inertia is in the tires, lighter wheels will only save a small fraction of the required power.
If you combine the effects of rotating mass and the regular (transational) force required to push a mass. 1 pound 8.5 inches from the center of rotation (for an NSX, it will vary with vehicle) is roughly similar to 1.5 pounds. Therefore saving 1 pound of wheel is like shaving 1.5 actual pounds. Because the tire is further from the center, saving 1 pound of tire (12.5" from center) is like saving 2 pounds.
Bob
[This message has been edited by 1BADNSX (edited 06 April 2002).]