TL, I think what you're missing in the equations isn't the transmission or engine, but the work being done. It's not just the engine side of the equation that's changing. The DYNO side it changing too, in that more power is being transferred to the dyno. THAT is what's pulling the extra power, expressed in a ball park percentage. What your quote above should say is:

If you want less power loss through the drivetrain, do less work. -in which case you'll only NEED a less powerfull engine.

Or you could just trust 100 years of engineers hot rodding cars. I'm not one for blind faith, but sometimes you can simply trust smarter people. Not being snotty, just sayin' you don't have to trust me.

 

 

I understand the varying efficiencies but what I was getting at relates to the first post.
I was postulating that the drivetrain loss is fixed and that 100 HP added is seen at the rear wheels too. Example: Transmission A takes 50 HP, rear gear B takes 20 HP and so on. All cumulative but still fixed. Instead, only a portion will be transmitted as the rest is eaten up by the accumulating losses. Output power equals input power times the drivetrain efficiency.
Certainly more to it than the obvious.

 

 

Carrying this to the extremes almost makes the results impossible.
Push an 8000 HP Top Fuel engine through a transmission (for these purposes let's not discuss what a Top Fuel transmission is) that eats at 20% and that's 1600 HP to drive it. Take that same trans and put 500 HP in front of it and and only 100 HP is lost.
If I hadn't seen it written in classical terms, I'd still be on the fixed part, fixed consumption fence.

 

This NASA white paper has some graphs in it that will help illustrate the torque element and speed element. Now if somebody tells you it's not rocket science just tell them, Bull****!

https://ntrs.nasa.gov/archive/nasa/c...9830011870.pdf

 

The drive train loss is NOT fixed. Not EVER.

Part of it is constant in the sense that it's proportional to SPEED, whether rotational or ground speed depending on the particular thing. Most of that of course is related to the RPM of the thing in question. So for example a transmission pump doesn't use ... 10 HP or whatever, it uses 3 HP at idle, 10 HP at 2000 RPM, 20 HP at 5000 RPM, and so on. Not necessarily a completely linear function of RPM, but related, just the same. Some other losses, gear oil windage for example, are pretty much totally RPM dependent, and totally independent of load.

Part of it is like that. Part is PROPORTIONAL TO THE FORCE APPLIED TO THE PARTS THAT SLIDE, like the R&P.

Part of it is proportional to BOTH applied force and RPMs, such as bearings... they obviously don't consume any power at all when the part they're holding is sitting still, and relatively little if the part is spinning with no load on it; but as load increases, which also increases the load on the bearings from having to hold the part aligned in spite of the load, the losses in the bearings increase as well.

The TOTAL LOSSES are the result of ADDING all those things together. None of the sources of loss is "fixed"; some are RPM-dependent; some are load (torque, power, whatever you want to think of it as) dependent; some are a more complex function of the 2.

The graph of loss vs HP therefore needs to be 3-dimensional: the independent variables are RPM/speed and applied torque (note that HP itself is readily calculated from those 2 quantities), and the dependent variable is loss in HP, NOT directly as a percentage. Then to get a percentage or fraction or whatever you want to call that, divide the loss by the applied HP at that speed.

Note also that we customarily use track mph to "calculate" RWHP; this is absolutely accurate in a way, since the car acquires kinetic energy as it goes down the track, and power (HP) is simply the time rate at which work is being done (kinetic energy is being added to the car). If the engine were able to be kept running 100% of the time at its max HP RPM, such as with a very well-matched racing torque converter, this number would be quite close to the engine's HP. Unfortunately it ignores the aerodynamic losses. But for a typical street car, and even for some much faster ones, that factor can be safely ignored... the car simply isn't going fast enough for it to be large enough to be the largest error involved.

Cmon guys, this isn't rocket surgery here or anything. It's just high-school physics. Which for me, was a good bit longer ago than 43 yrs; but fortunately, like many other things in HS such as Latin for example, it hasn't changed much since then.

 

Probably girls were more on my mind than physics. More hormones than brains back then.

 

So, at a fixed RPM, doubling the power will also increase the losses due to inefficiency.
Seems simple enough to grasp and yet counter-intuitive (despite my education). The physical properties of the drivetrain introduce aspects that I hadn't considered (but should have known all along).
That said, as I posted earlier, I can't expect the 25% bump in engine output to translate to a 25% bump in rear wheel output.
I need more power Scotty!