Now, using my own derived equations.
196ss 3m ago
Where did you get this formula for calculating Pf? It is absolutely wrong. In the physical sense of your equation, you find the ratio of the torque at peak power to the maximum torque of the engine. This ratio will be higher for high revving engines (NA with short stroke and lightweight flywheel), and lower for low revving engines (gasoline OHV, diesel). But it has nothing to do with the fraction of power transferred from the engine to the wheels (which is the physical meaning of your "powertrain factor").
In addition, the coefficient that you use (5252) introduces some error (insignificant in fact). I recommend you to convert power to torque (and vice versa) for a given engine speed by using the following formula with the parameters expressed in SI units:
P – engine power, W
T – torque, Nm
R – engine speed, revolution per second (RPM/60)
and the inverse expression:
Please take this post not so much as criticism, but as a desire to help. Sorry for the clumsy English.
NatsM 3m ago
Here I go again...hahaha. I have to explain again to you and BTICronox the "evolution" of this "powertrain factor" that I concocted. I have already explained it to Nimrod on my reply to his query. Sorry, but I don't have the time and mood to write a long explanation again here. If you really want to know what it is all about, I suggest you read my reply to Nimrod here. It's okay, I expected skepticism and criticism. My Pf equation works hand in hand using the Engineering ToolBox's Cr equation as far as I can tell. Regarding your formula, it's another problem since we talk in different languages — you guys in Metric and I'm with the English or Imperial. I always have to convert. Anyway, it is just the same using this:
H = R T/5,252
H = Horsepower
R = engine speed or revolution per minute
T = Torque in lb-ft
Now, can you now see how my "Pf" factor evolved? Transpose the variables. I explained it to Nimrod already how I came up with that factor. Thanks for your time.
196ss 3m ago @NatsM
I read your explanation and it didn't change anything. Your calculation of the "powertrain factor" in the equation above is incorrect.
NatsM 3m ago @196ss
As I said, I'm not forcing anyone to trust and use it. Besides, I'm still experimenting, remember? To each his own. Cheers!