Originally Posted by Petza914

This thread was a very interesting read.

So it sounds like the accessories on an engine consume so little power that they're not the things being compensated for proportionally. It's all the other major stuff.

That makes sense.

Originally Posted by GregBBRD

I'm certainly fine with dividing by .85 to calculate flywheel horsepower from rear wheel horsepower, when using a 5 speed.

That "adds" 18 horsepower to my most recent engine build/dyno test, which is the easiest 18 horsepower I've ever found!

What's the calculation for an automatic?

Originally Posted by ptuomov

Accessories can draw a lot of power, like an air conditioning compressor does. The key difference is that accessory power consumption is not dependent on the engine torque, whereas the transmission loss in the gearbox is dependent on the torque.

Originally Posted by Petza914

Yep, got it. So on a before and after run on the same car with the same accessories installed, the load from them is the same. It's the other stuff that changes and takes more power to overcome. Are those linear functions where a straight 15% adjustment is accurate?

Originally Posted by ptuomov

I don't know. My guess is that the 928 S4 five speed driveline consumes 10hp and 12% of the torque. So the crank hp I'd use is 10hp+rwhp/0.88.

Originally Posted by Ian928

Or is 10hp+rwhp/0.89 more correct for 12% loss?

Originally Posted by ptuomov

What sort of rwhp numbers do you see for completely five-speed S4s and GTs in nice mechanical condition?

Originally Posted by SwayBar

What did your stroker dyno at - rwhp?

Originally Posted by GregBBRD

GTS: 287.4 to 303.7. Minimum and maximum.....with assortment of numbers between those two. Your formula would result in 338.0 to 356.5 minimum and maximum flywheel horsepower.

Originally Posted by daveo90s4

In other words, gearbox transmission loss is - I believe - independent of engine power (for a any given nominated rotational speed). Every force (slave unit friction) is met by an equal and opposite force (engine power loss). Wasn't that Newton's 3rd law of motion?).

Now - to labour the point - why would the alternator or water pump or other peripherals differ from my hypothetical slave unit discussed above (for a given rotational speed etc)? [Note - I am not suggesting that these units do not consume power - I am only suggesting that the amount of power they consume is independent of the inherent power of the motor driving them (at a given rotational speed).

Originally Posted by Bulvot

The thing that you may be overlooking here is that force required to overcome driveline loss is not just a function of angular speed (RPM) and unchanging resistance. When you measure the ability of the motor to produce power, you are measuring it by applying a known opposing force and measuring the motor's ability to overcome that opposing force. When we're talking about a dyno, the opposing force is applied at the tires. This is a very important point because the driveline loss is not a static number, it is a function of the coefficient of friction times the force applied against the surfaces that are in friction with each other.

As the force of the motor increases, it applies greater force to overcome the opposing force faster than before. The two opposing forces (motor and dyno rollers) result in applying greater force against the surfaces in friction with each other in the driveline. Things like bearings, gear teeth, etc. Therefor, yes, it is impossible to avoid losing more power in the drive line if you are applying more power to it.

Originally Posted by ptuomov

What do you mean by "bigger" and "smaller" engine? If you mean displacement being different but power and torque per rpm being the same, it doesn't matter to the transmission losses. If you however mean more powerful with higher torque per rpm, then the more powerful engine will result in higher driveline losses when run at that higher power. This is because the load between the gears grows proportionally to the torque. If the friction coefficient is constant, the friction force and losses are proportional to torque.

The story is completely different from accessories. The accessories draw the same amount of power at a given rpm regardless of the engine torque. Therefore the friction losses in the gears, pulleys, and belts is independent of the engine torque. They are however dependent on the torque that the accessory device consumes.

Originally Posted by daveo90s4

Hi Ptuomov. By 'bigger' and 'smaller' I am meaning 'more powerful and less powerful at a given nominated rpm'.

Maybe I'm not putting forward my my proposition as clearly as I should be.

If I have some sort of driven device that requires (say) 20 HP to spin it at 2,000 rpm with a 100 HP 4 cylinder engine spinning it, then what I'm suggesting is that that same device will still consume the same 20 HP when spun at the same 2,000 rpm with a 400HP V8 engine.. How can it be otherwise? The device that consumes the HP has not changed one iota.

I really am trying to see / understand how it could be otherwise (note in all case I am talking about losses at the same RPM between scenarios).

Happy to continue this interesting discussion off-line or on a new thread - am mindful this somewhat esoteric discussion may not sit well within Carl's original thread. Bed time here in Oz. :-)

Cheers

DaveO