Maybe you can provide numbers for the "why peak torque can never appear after peak HP" question. I get it, as RPM is always increasing at the moment torque peaks (and therefore more power is made at least for a short part of the curve past peak torque), but like to see it written mathematically. I plugged some random numbers in to find limits, but that's a pretty backwards approach.
I think you have the right idea. On the dyno graphs, RPM is always plotted on the horizontal x-axis (and torque or hp on the vertical y-axis). The 'after' in the question "why peak torque can never appear
after peak HP" means higher RPM. Okay. It's late, and I can tell the more I try to explain it in words, the more confusing it'll sound (perhaps I'll try to explain it in words tomorrow).
But anyway, here's the math...
First a little calculus tutorial:
A
derivative is a measure of the rate of change of a quantity. The derivative of a function
f with respect to some quantity
x is a measure of the rate that the quantity
f changes with respect to the quantity
x, and is written as df/dx.
Here is the important part for this discussion; the derivative is either positive (increasing), negative (decreasing), or zero (not changing at all). (We will use this fact later.)
All right. We have a function (equation) for the horsepower:
hp = torque*RPM/5252
To make this function looks 'nicer,' let me write it as
h = T*x/c
where h = horsepower,
x = RPM,
T = torque, and
c = 5252.
Now, bear with me...
let's take the derivative of h with respect to x (in other words, we ask, how does hp change with respect to RPM):
dh/dx = (x*dT/dx + T)/c
To get that, I had to use calculus, and something called a "product rule." (Anyone interested can find these differentiation rules in any first-semester calculus text.)
Okay... remember when I said that a derivative is either positive, negative, or zero...
it turns out that when it is 0, it is at either its maximum or its minimum. One quick look at myhondas' graphs (or anyone else's) and we see that it would be a maximum (i.e., where it rises, just levels off, and starts to turn down). We could justify that it's the maximum (as opposed to the minimum) with a little more math, but we might as well use common sense to spare us the work.
So when the derivative is zero, it's at its max (another important fact we'll use later).
All right. Either I'm getting rusty, or the above equation, when equated to zero, becomes a complicated differential equation that seems hard to solve. My hunch is that I'm rusty. But it turns out with a little cleverness we can avoid that. Let's look at the equation again:
dh/dx = (x*dT/dx + T)/c
What happens when the torque (T) is at its max. Remember, when it's at its max, its derivative is zero (dT/dx=0). So let's plug that in the equation above and see what we get:
dh/dx = (x*0 + T)/c = (0 + T)/c = T/c, or simply
dh/dx = T/c.
This tells us what the horsepower (h) is doing with respect to the RPM (x) when the torque (T) is at its max (and remember 'c' is just 5252).
Ahah!
Remember what I said about derivatives?
Is dh/dx positive (increasing), negative (decreasing), or zero (not changing)?
The torque (T) is at its max (hence, a positive number), and c is just 5252 (also a positive number), and thus T/c is positive, which means dh/dx is positive (and remember, dh/dx is the rate of change of horsepower with respect to RPM).
So when the torque is max, the horsepower is
increasing (its derivative is positive), which means that it must peak
after the torque peaks.
Whew!
