98BlownCobra

Okay, I've always had some difficulty understanding how dynos generate the rear-wheel torque numbers that they do. Let me explain how I THINK it works, and then I'd be greatly appreciative is someone more knowledeable could confirm or deny it.

I'm gonna use a stock '97 Cobra as an example, because I happen to have the specs for it in front of me. The max torque at the flywheel is listed as 300@4800rpm. The first gear ratio is 3.37:1 and the rear end is 3.27:1. When I multiply those ratios out, I get 11.01 for my final drive. I then mulitply the engine torque (300) by the final drive ratio to get effective (multiplied) torque. This works out to 3303, not counting for driveline loss. The rear wheel RPM must, of course, be obtained to use the HP formula. So dividing flywheel RPM through the gearing, I get a wheel RPM of approximately 435.6. When I run these numbers through the formula (HP=(TQ*RPM)/5252), they work out to 273.9 HP. This is a realistic horsepower number for that car, especially when you factor in driveline loss.

So my question is, Why does the dyno not show 3303 lb-ft of torque at the wheel, since that's what it has to really be measuring? My theory is that it takes the calculated horsepower value and the known engine RPM and uses the HP formula to calculate a new rear wheel torque value. By doing this, you can show a more useable value on the dyno chart, and get some idea of what you're engine is doing for you. If this theory of mine is true, then the only real effect gears will have on a dyno is that the vehicle will lose a larger percentage of power through the driveline as the torque multiplication diminishes.

Like I said, the last paragraph was total guesswork on my part. It seems to make sense, but I'd love to hear from someone who really knows. Thanks.
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