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"Peak" acceleration

It looks to me like you might have been better off just leaving it in third and letting the engine rev past the horsepower peak to avoid that last 0.59 second shift time, if the engine can safely rev that high.
That pro stock racing transmission would make a lousy road racing transmission because it’s not designed to downshift, in fact, if you lift your foot off the gas the ramped engagement dogs would immediately cam out of engagement kicking the transmission into neutral, possibly with shifting linkage damage.
The whole operating principle is that the shift fork engages the next higher gear and then the dog clutch in the lower gear cams out of engagement, like the starting crank of an antique farm tractor or Model T car.

I had a 3.55 final drive in it. A 4.10 works better, tops out 4th through the traps. I couldn’t wind out 3rd that far. I was already past the power peak and pushing the RPM a bit for a cast crank (silly rules again!) motor. I was just trying to get a little data and see if I could create dyno curves in various gears. Data was a little noisy for that. It was still fun though… :wink:

So, you agree that the fastest way down the 1/4 mile is with a CVT that keeps the engine at max power? If not, please cite references.


I looked up the power and torque of a 2014 Corvette.
Power peak was 402.48 HP @ 5936 RPM, which comes out to 356 ft-lb of torque.

Torque peak was 400.02 ft-lb @ 4839 RPM which comes out to 368.6 HP.

Let’s assume we use this engine in a car with a CVT.
Driver A programs the CVT to hold the torque peak 4839 rpm.
Driver B programs the CVT to hold the power peak 5936 rpm.

At 10 mph, the rear axle is turning 144 rpm. Driver A’s transmission is set to a 33.6:1 reduction ratio, driver B’s transmission is set to a 41.2:1 reduction ratio.
400.02 ft lb X 33.6 = 13,441 ft lb of torque at the rear wheels.
356 ft lb X 41.2 =14,667 ft lb of torque at the rear wheels.

At 30 mph, the rear axle is turning 432 rpm.
Driver A’s transmission is set to a 11.2:1 gear reduction ratio.
Driver B’s transmission is set to a 13.7:1 gear reduction ratio.
400.02 ft lb X 11.2 = 4480 ft lb at the rear wheels.
356 ft lb X 13.7 = 4891 ft lb at the rear wheels.

It seems to me that driver B will win this drag race.

No matter what speed you choose, the CVT that is holding the power peak rpm is delivering more torque to the rear wheels than the transmission that is holding the torque peak. It’s lower gearing at that road speed more than compensates for the engine’s lower torque.

After careful consideration and @B.L.E 's terrific math example, yes I now agree! Thanks to all who participated in the discussion!

I had to stew on this a couple days to put it into words. B. L. E. summed it up real well. In a given gear, the max acceleration will be at the engine’s torque peak. At a given vehicle speed, a gear that puts the engine at its horsepower peak will provide the maximum thrust. In most car engines these days the torque peak is at a fairly high RPM, usually between 4,500 and 5,000. In B. L. E’s example above, the torque peak and the horsepower peak are only about 1,100 RPM apart…Not a whole lot when you’re over 4,500, So I’d have to guess that, given that the horsepower peak is way up there near 6,000, shifting at or near redline is gonna give you the fastest acceleration.