Let's do some calculations...

According to ifidonlike, the higher the hp, the more the force at wheels, and you’ll always shift at the peak-hp rpm. This is wrong.

Let’s look at the following engine, with 200 lb-ft at 1500-4500 rpm and 200 hp at 7500 rpm. I list rpm, torque and hp in a table:

1000____150______29
1500____200______57
2000____200______76
2500____200______95
3000____200_____114
3500____200_____133
4000____200_____152
4500____200_____158
5000____175_____167
5500____166_____174
6000____158_____181
6500____152_____188
7000____146_____195
7500____140_____200
8000____118_____180

Let’s say you are driving in 2nd gear, with a ratio of 4:1. 3rd gear has a ratio of 3:1. Where do you shift?

If you shift at 7500 rpm, before the shift:

Engine torque = 140, wheel torque = 4*140 = 560, wheel force = 560 / R.

After the shift, rpm = 5625, let’s round to a bigger torque:

Engine torque = 166, wheel torque = 3*166 = 498, wheel force = 498 / R.

But if you shifted at 5000 rpm, before the shift:

Engine torque = 175, wheel torque = 4*175 = 700, wheel force = 700 / R > 560 / R.

After the shift, rpm = 3750:

Engine torque = 200, wheel torque = 3*200 = 600, wheel force = 600 / R > 498 / R.

So both before and after the shift, you get more force at the wheel if you shift at 5000 than at 7500. The hp figures are 167 @ 5000 and 200 @ 7500.

Generally, if Q(x) is the torque curve, you are in a gear with ratio K1 and need to up-shift into a gear with ratio K2, the optimal shift point x satisfies the equation:

K1 * Q(x) = K2 * Q(x * K2 / K1).

As you can see from this, x is always greater than the peak-torque rpm, but can be substantially less than the peak-hp rpm. In fact, the equation doesn’t involve the hp curve at all.

If Honda comes in and holds the torque curve flat at 140 lb-ft way beyond 7500 rpm, they can keep making more and more power. But does this change the optimal shifting point a single bit? Nope.