270HP at 7100 RPM, HP at 8000 rpm?

[Gokwi]

ok we know the early year nsx hits 270 hp at 7100 rpm, if we try to redline it, would the HP numbers raise, or still stay at 270 HP?

Thanks

 

[Periokid]

it drops. not sure to what value.

 

[nsxtasy]

 

[Ojas]

Maybe 250 HP?

From the FAQ:

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ojaspatel.com/nsx

 

[NetViper]

So is it faster to shift at 7100 RPM or go all the way to redline? Say on a 0-120 run.

 

[MvM]

I think shifting at 7100 rpm will not make the car go faster. When shifting, the gear ratios will become longer, slowing down your acceleration. And don't forget, after shifting, the rpm drops to a lot lower level, giving you even less HP. Also, shifting at 7100 will drop you out of VTEC.

Looking to the graph standard engine with OEM ECU and comparing them to the graph of the Dali Hotchip it seems that this power decrease is ECU-induced. With the Dali chip the numbers keep going up after 7100 rpm almost to 8000rpm. Maybe the ECU is programmed in such a way to protect your engine. Reliability was after all a very important design aspect for Honda.

 

[nsxtasy]

Shift the NSX at redline for maximum acceleration.

With most other cars, you want to shift when the horsepower in the lower gear has fallen to the point where it's equal to the horsepower in the higher gear at the point where your revs end up after the upshift. With the NSX, you never reach that point, so you shift at redline.

 

[sjs]

Originally posted by nsxtasy:
Shift the NSX at redline for maximum acceleration.

With most other cars, you want to shift when the horsepower in the lower gear has fallen to the point where it's equal to the horsepower in the higher gear at the point where your revs end up after the upshift. With the NSX, you never reach that point, so you shift at redline.

That and "shift to land at max torque in the next gear" are two very common old generalities I have heard quoted forever. But they both seem to disregard the loss of torque multiplication as already mentioned by MvM. I think that both are flawed and wrong as often as right, but in this case the answer is definitely to use all the revs in each gear and everyone here agrees.

 

[nsxtasy]

Originally posted by sjs:
That and "shift to land at max torque in the next gear" are two very common old generalities I have heard quoted forever. But they both seem to disregard the loss of torque multiplication as already mentioned by MvM. I think that both are flawed and wrong as often as right

I don't agree with shifting to land at max torque, but shifting where the horsepower is the same should work - and does include the loss of torque multiplication. Here's why.

You would want to shift where your acceleration in the higher gear after the upshift is equal to your acceleration in the lower gear before the upshift.

Acceleration is a function of torque at the wheels, so you would want to shift if you reach the point where your torque at the wheels in the higher gear after the upshift is equal to that in the lower gear before the upshift.

Torque at the wheels equals torque at the crank times gearing. Crank horsepower is proportional to crank torque times revs.

For torque at the wheels to be equal in both gears, the torque at the crank needs to be proportional to gearing (higher torque in the higher gear compensating for the gearing loss). But the two horsepower figures are equal at that point because when you multiply the crank horsepower numbers times the revs, you are reversing the gearing factor between the two crank torque numbers.

Hope that makes sense.

However, as you and I both note, you never reach this point in the NSX, so your best bet is to shift at redline. (I seem to recall someone doing the math and finding that at the highest gearshifts, it actually does make sense to shift before redline, but the revs were so close - 7700 to 7900 - that it wasn't worth making a distinction between that and shifting at redline.)

[This message has been edited by nsxtasy (edited 11 March 2003).]

 

[sjs]

Originally posted by nsxtasy:
...For torque at the wheels to be equal in both gears, the torque at the crank needs to be proportional to gearing (higher torque in the higher gear compensating for the gearing loss). But the two horsepower figures are equal at that point because when you multiply the crank horsepower numbers times the revs, you are reversing the gearing factor between the two crank torque numbers.

Hope that makes sense.
...

Yup, that seems to work. I've never done the math on it, but then I never had power curves for my cars in the old days so it didn't really matter. Everything came down to the butt dyno and occasional stopwatch test.

Thanks for laying it out.