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      12-17-2008, 02:44 AM   #30
doba_s
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here is some info i just found on m3forum ...

Originally Posted by Carver - m3forum.net

A member asked me to come and give my $.02 here so that hopefully some truth and some understanding could be gained.
Now, in physics, engineering, and design, as in many other fields, there are always multiple ways to arrive at the correct answer. Elegant explanations generally consist of the most basic examples which most clearly demonstrate the proof with a minimum of complexity.

Ok, so we can solve this problem in any one of these ways:

a) Area under the HP curve along with time
b) Area under the TQ curve along with time and gear ratio
c) Total work through integrated force and distance
d) Kinetic energy through fuel burn and time
e) Impulse through integrated force and time
f) Power through integrated force, distance and time
g) Force:time ratio with time in gear
h) A couple more which are really derivations of the above

So, which one of the above is the most elegant, meaning simple and easy to understand while yielding accurate results?
g is by far. Why? You use the fact that both cars have the same mass, burn the same fuel, have the same horsepower curves, have the same torque curves (redundant), use the same transmissions, have the same wheel/tire rolling diameters, have the same aerodynamic drag etc.
By cancelling these similarities out, you come up with a basic F=mA and time ratio along with speed which clearly tells the story.

Here we go.

F(t) = the time function of force at the crankshaft
since we're not interested in the actual time or distance of the race but a relative position of the two cars in relation to gears and each other, as we race through time, the functions will cancel out as each car runs through it's powerband. Remember, the only reason this works is that we are comparing the cars with identical tq/hp curves in a ratio and not looking for an absolute value. If we wanted actual time or distance separation, we would have to integrate the torque curves over time along with the gearing to get actual force derivatives. That's why this ratio is so powerful and easy. You don't need the actual integration of the area under the curve. With axle ratios of 3.620 and 4.100 and gear ratios of 4.230, 2.530, 1.670, 1.230, 1.000 and .8300 we have 11 unique force ratio stages yielding unitless distance units when combined with speed and time or acceleration in each gear. Call the distance factor d(g) or distance separation relative to gear with all of the distance units summed. We will make positive values the 4.100 car and negative values the 3.620 car. The distance units are a constant distance/time ratio derived from the force ratio ie. it is torque/mass independent. That's why there are no units. If you used actual derived integrated torque and mass, you could figure out the actual distances.

Car A = 4.100 car, Car B = 3.620 car

Stage 1-Both cars in first-ends at 36mph
(17.34-15.31)36/(15.31*17.34) = .2752 distance units total d(g) = .2752 du

What this means is, depending on the identical torque or HP curves of the cars (the same for each car) and their identical masses, there is a constant times the distance factor which will give you the car's relative positions in actual units ie feet, meters etc. Car A is .2752 distance units ahead.

Stage 2-Car A shifts to 2nd gear-ends at 40mph
(15.31-10.37)4/(10.37*15.31) = -.1245 du total d(g) = .1507 du

Continuing the process, Car B had an advantage of .1245 du from 36mph to 40 mph and closed the gap to .1507 du, but didn't make up for the advantage Car A had to 36mph. Why? Even though the percentage advantage of Car B was greater in Scenario 2 than the advantage Car A had in Scenario 1, the time available was much shorter from 36 mph to 40 mph than from 0-36 mph.

Stage 3-Car B shifts to 2nd gear-ends at 60mph
(10.37-9.159)20/(9.159*10.37) = .2550 du total d(g) = .4057 du

Car A is now .4057 du ahead. The extra time in identical gears for the 4.100 car is too much of an advantage compared to the short period of time the 3.620 car has in the lower gear.

Stage 4-Car A shifts to 3rd. gear-ends at 68mph
(9.159-6.847)8/(6.847*9.159) = -.2949 du total d(g) = .1108 du

Car B gets a break because of the big jump in the M3's 2nd.-3rd. gear ratios and closes the gap to .1108 du.....car A still leading.

Stage 5-Car B shifts to 3rd. gear-ends at 90mph.
(6.847-6.045)22/(6.045*6.847) = .4263 du total d(g) = .5371 du

Car B pays dearly for the combination of the 3.62 rear end and the M3's 2nd.-3rd. gear ratio split....Car A is now .5371 du out in front.

Stage 6-Car A shifts to 4th. gear-ends at 102mph.
(6.045-5.043)12/(5.043*6.045) = -.3944 du total d(g) = .1427 du

Car B takes advantage of the 12mph that it can stay in 3rd. Notice however that on each chance that Car B has to do this, Car A is still slowly increasing it's advantage.

Stage 7-Car B shifts to 4th. gear-ends at 123mph.
(5.043-4.453)21/(4.453*5.043) = .5517 du total d(g) = .6944 du

The extra time it takes to reach each speed is now starting to favor the 4.100 car's rear wheel force advantage even more. Notice each even and each odd stage gap is increasing.

Stage 8-Car A shifts to 5th. gear-ends at 139 mph.
(4.453-4.100)16/(4.100*4.453) = -.3094 du total d(g) = .3850 du

By looking at the jump from stage 6 to stage 8, you can see that the 4.100 car is taking advantage of the big 4.100-3.620 advantage while having closer gear ratios.

Stage 9-Car B shifts to 5th. gear-ends at 151 mph.
(4.100-3.620)12/(3.620*4.100) = .3881 du total d(g) = .7731 du

Notice that each advantage is greater for Car A than Car B as long as the gear ratio jump isn't too large.

Stage 10-Car A shifts to 6th. gear-ends at 171 mph.
(3.620-3.403)20/(3.403*3.620) = -.3523 du total d(g) = .4208 du

Car B gets a little help from the 5th.-6th. jump in ratios on Car A but it's not enough.

Stage 11-Car B shifts to 6th. gear-ends at 182mph.
(3.403-3.005)11/(3.005*3.403) = .4281 du total d(g) = .8489 du

Notice that the gap at the top of 4th in the 4.100 car at 123mph is 2.5 times the gap at the top of 1st. gear (a good rough 1/4 mile+ example) The main idea here is that the advantage and time the 4.100 car has in the same gear as the 3.620 overcomes the advantage the 3.620 has in a lower gear because of less time in that scenario. Another way to say it is when you integrate force at the rear wheels (F=mA), the sum of F(4.100) over time is greater than the sum of F(3.620) over time up to the top speed of the 4.100 car. As a result, the total integration of acceleration over time is greater in the 4.100 car. Notice also that the outcome was partially dependent on the M3's ability to hold a gear for a long period of time ie. flat torque curve and high redline. I could do a proof showing that for the M3's gearing, the optimum shift points are actually up in the 8400-8700 rpm range depending on the gear change. Obviously an added shift, corner, or turbo car spool time etc. between cars on a track could change this equation, but that is a different argument.

As a side point, what you feel as a driver/passenger isn't always what the clock shows. The human body is pretty good at feeling force but not very good at feeling the integration or time of force. That's why quite often a turbo car will "feel" faster than an NA car of equal acceleration because of the torque peak or "hit" of the turbo. Likewise, the added force in each gear of the 4.100 car will "feel" faster than it really is because of the force increase. You won't notice the reduced time of the force in each gear as much. It wouldn't be surprising to see someone reduce their 1/4 time by .3sec with a 4.100, but "feel" like the car is over .5 sec. quicker. Since most people get an M3 for how it "feels", this is even more of a reason to go with a 4.100 differential. A 400rpm increase on the highway, imho, is nothing with this car. The M's moderate torque and high redline along with fairly large weight make the argument for the added force "feel" even more weighted toward the 4.100 side.
My $.02