LongBeachCoupe wrote:Theres quite a few guys on here than have mechanical engineering degrees I believe... someone get the numbers together and prove this out on paper!
Guess no one wants to jump on this one, so I'll give it a go. Unfortunately, these are all estimates and will only give a rough idea. Been a while since I've had to do this, so someone correct me if I'm wrong.
First, the assumptions:
1. Let's call the pulley an open cylinder with all of its mass on the outer edge (think of a piece of paper rolled into a cylinder), as this is a close estimate to the pulley (which has most of its mass near the outer edge).
2. Angular acceleration is constant (IE, the time it takes for the engine to go from 2,000 RPMs to 4,000 RPMs is the same as from 4,000 RPMs to 6,000 RPMs).
3. Since I don't have a pulley (and don't really want to do the work with the actual dimensions), let's say it has a radius of 2.5 inches (we'll use the Stillen-measured weight of 4.71/1.10 lb respectively).
Equations:
I = m * r^2where 'I' is moment of inertia (lb*ft^2 - I'll use the notation Io for stock and Is for the Stillen pulley), 'm' is mass (lb) and 'r' is the radius of the pulley (ft)
T = I * awhere 'T' is torque (lb*m^2*sec^-2 - again using To and Ts) and a is angular acceleration (radians*sec^-2)
OK, the fun part:
Io = 4.71*(2.5/12)^2 = 0.2044 lb*ft^2Is = 1.10*(2.5/12)^2 = 0.0477 lb*ft^2
Now, say we want to run the engine from 1,000 RPM (roughly idle) up to 6,000 RPM (roughly redline) in 2 seconds. That is a difference in velocity of 5,000 RPMs. converting this to seconds gives us 83.33 rotations per second difference. Next, we need to convert to the units above to be able to use the equations. 1 rotation = 360 degrees = 2*pi radians, so an increase of83.33 rotations per second is an increase of 523.598 radians per second. Now, divide this by the time consumed in the process to get the angular acceleration:
523.598 / 2 = 261.799 radians per second squared (rad*sec^-2)
Ok, put the two together:
To = 0.2044 * 261.799 = 53.51 lb*ft^2*sec^-2Ts = 0.0477 * 261.799 = 12.49 lb*ft^2*sec^-2
Now, keep in mind the pound-foot of torque measured for the car is equal to the force a one pound object would exert at a distance of one foot from the pivot point. The estimate for gravitational force is 32.2 ft/sec^2 so:
1 lb-ft = 1lb * 32.2ft/sec^2 *1ft = 32.2 lb*ft^2*sec^-2
Using the torque determined above and this conversion value:
To = 1.66 lb-ftTs = 0.39 lb-ft
There you have the difference in torque (roughly) that is being consumed by the pulley. It may look minimal in comparison to the rated torque output at the crank, but think of how much is lost before hitting the pavement (not to mention how much force is needed to overcome rolling friction and wind resistance), and this becomes more critical. Also, realize that at 1,000 RPMs the engine is certainly not putting out peak torque, so percentage-wise this becomes more important.
Again, keep in mind the assumptions that I made (to make calculations a bit simpler). The numbers may look trivial, but if you factor everything in, they begin to appear more substantial.
My head hurts...