And that's why oil coolers and transmission coolers and differential coolers exist.
A lot of the extra hp gets wasted as friction. More power, more friction losses.
Best
Les

Maybe we should, for the purposes of discussion, assume that the losses are 15% at max power, which is the only part of the equation that ever gets addressed.

Perhaps they are less when running at 10% power, and perhaps they are more. Most people are only interested in what total HP numbers they can infer at max output.

Of course a lot of guys only quote wheel horsepower for accuracy. but I am still going to say that drive train losses are related to load on the system, and they would be much less in a no load situation, and much more reflective of what you are saying. If we are talking about a 75 HP loss at 4 or 500 HP, I can only say that that must reflect real world conditions, as engine builders and dyno operators always seem to quote the same loss figures. FWIW, they quote higher figures for AWD vehicles and fuel economy also drops, so I expect the losses are real, whether it has to do with friction or inertial losses or whatever.

They say it is mathematically impossible for bumble bees to fly, but they do anyway, so it must be real. Driveline losses at WOT are the same, for whatever the cause.

Simplify. Remember High School Physics? u= coefficient of friction. It was always expressed as a fraction or decimal, same as a percentage, and was constant. f=uF.

Logically, think of resistance in the gearbox as friction. Friction causes gearbox/fluids to heat up. More heat, more loss. If you put 10HP on a 915 gearbox, it will likely barely raise the temps. 500HP and it will get hot, cook fluid, and self-destruct. So yes, % loss may be similar, but with more HP there will be more (frictional)Loss, as noted in raised temps.

I'll chime in. I think for simplicity sake, I see it as this. Take an engine that makes 100 HP @ the crank. It is coupled to a conventional rear wheel drive setup sans vehicle for this discussion. If measured at the drive wheels the loss might be10% / 10 HP.

Now take the same engine which makes a verified 200 HP @ the crank, bolt to to the same existing drive train connected to the wheel dyno. Assuming the trans/ diff would tolerate a smooth application of twice the power - I don't believe, as a matter of fact some one would have to prove this in real world testing, we'd see the same % of loss.

Now in actual on the road testing, the same drive train rated for 100 HP would most likely fail at some point with shock loading, torque. I'm sure the friction would increase but I don't see how it would double prior to failure.

And thats my point..too.
It does not seem logical.

Search "gear friction loss as a function of load"

The first hit is http://www.geartechnology.com/issues/0905x/mba.pdf

Looks legit. Should be useful.

Drive train losses are a function of the load in the transmission, the normal force between the gear teeth increase at higher loads, and the friction loads at that interface are roughly proportional. Their are fixed losses as well which dont scale with speed.

Applying a gross percentage as an estimate of drive train losses is a rule of thumb at best, and for entertainment purposes at worst. There are 100 million related discussions on the internet. Nobody uses that number in the engineering context. The good thing for car dynos is that torque/power at the wheels is what matters for car performance, so the crank hp is just a curiosity anyways. If you really need to know it, better grab your wrenches and book an engine dyno


https://www.nap.edu/openbook/21744/x.../img-204-2.jpg

It is indeed a lot of loss, that is why F1 cars can't follow at distances of much closer than 60 feet for one or two laps, before the driver has to back off. Before the current electric Hybrid's F1, you'd hear radio communication if a driver didn't get a pass done in a lap or two, to back off, and let the transmission cool.

If you ever get the chance to see a fully assembled NASCAR, the amount of cooling systems they have is amazing.

The ducts on the upper rear fenders of the current Gen Corvette, feed dual radiators for the rear mounted gearbox and differential.

This. ^^^^ The difference between load losses and internal losses.

That's all you need to know. Your intuition is correct, not accounting for load.

http://www.mdpi.com/2075-4442/1/4/95/pdf
I thought Carbon60 was going to become the next miracle oil. https://en.wikipedia.org/wiki/Fullerene
It does prevent wear at high load, but no corrosion inhibitors, and not for Nikasil/Alusil.

Nanodiamonds polish surfaces well, but also cylinders.

Boric acid doesn't mix with some lubes or replace ZDDP.

PTFE clogs everything up.

+
Those two ^ together totally did it for me. I've often wondered. Assuming you have the same car, so a theoretical Veyron, and the "valet" key knocks the motor down to 100hp, so the frictional loss (15%) is 15hp, but then the owner gets in and has the full 1000hp, so 150hp loss. Yeah, sounds crazy and extreme, but it makes sense now. Granted, I'm assuming these are as stated, guidelines, and the graphs aren't a straight line, but still, pretty cool that I can visualize it now.

Yep.

Dragging a large steel cube over a large, lubricated steel sheet (MD Holloway is there). Together, the coefficient of friction is of steel on steel with lube is .16.

Ff = μ m g

First, a 100lb cube of steel.
Ff = .16 * 100lb = 16lb

Now, lets double the weight of the cube.
Ff = .16 * 200lb = 32lb

This is like doubling the horsepower, except I assume we would really be talking about the torque.

So the friction of 2 gears on each other with twice as much force applied would be double (you know, assuming very general, ideal type circumstances).

So, The guy in the Veyron with 1000hp at 35 mph is seeing the same losses as the guy with a Veyron using the valet key so he only has 100hp. Where their losses would be different (150hp vs 15hp assuming 15% loss) is if they were full throttle at the RPM where they are each making peak power.

Yeah, following your reason, this makes sense, but you also have to think about, the transmission for a 52hp Chevy chevette is probably, much, much smaller and lighter than the transmission for a 500hp Corvette, even if they were both 4 speeds. And when they take something like a corvette, the standard model has a tranny, but the Z06 model adds a tranny cooler. Some cars have added finned diff covers.

Some trannies will have more fluid, bigger, heavier gears which will withstand the power AND the heat better, or maybe more gears which add more mass to absorb the additional heat. Besides, I'm sure there are things that are calculated by the engineers like "Is the average corvette owner going to be at peak output all of the time or only for a few secs here and there most of the time. Do you think that the Corvettes at Le Mans use the same tranny with no additional cooling as a street 'vette. No, of course not, they use at least a cooler if not a different trans. And that's because they are always forcing max power through the trans. Since most vette owners are probably mostly loafing around at 1500 rpm using a tiny fraction of the power on tap, they aren't creating the heat or experiencing those losses