Friction Loss Question
 

For example, let's use 10% as the arbitrary power loss somewhere between the engine transmission on a 100 HP vehicle.
That should read about 90 Hp on the dyno, subtracting the 10 hp figure correct?
Now, lets add a few parts to make it a 200 hp -
The deduction baffles me because now there is a subtraction of 10% which means there is a loss of 20 HP.
However, this is how its done in the industry and i think it might be wrong
Why would there be another 10 hp loss?

AFAIK, the frictional losses from the drivetrain are proportional to the power (load) being transmitted through them. There are also "fixed" losses that remain constant, no matter what the power/torque levels are. That said, there is no way a standard percentage value can be used for an accurate calculation, especially across different vehicles/engines/drivetrains. I think it's more of a rough guideline at best, and I've heard anywhere from 10-20% used as the "percentage of loss".

Like I said, just because the engine jumped a 100 HP why the extra loss?
One could say "the New CR did it", but a turbo would not necessarily have to be bumped a couple points.
Confused.

Why do high performance cars need a tranny cooler whereas commuter cars do not? Assuming both have the same parasitic losses in the trans...

As force is increased, so is friction.

Think of sliding a 200lb object across ice, now try sliding a 400lb object across the ice. Even if you have the object on a low friction surface, the 400lb object will have more drag on the ground.

You increase power, either by increasing speed(RPM), or Torque. The faster the transmission moves, the faster the frictional parts come apart and together, increasing friction.

There are reasons several forms of US racing use systems that do not change gears, more power to the wheels this way with less moving parts.

also consider(typical gear box) that involute gear faces roll (not slide) from one face/tooth to the next. But, as loads (and especially over-loads) are applied the gear, teeth deflect causing the profile to no longer be that great mathematically perfect involute rolling/meshing interface. IOW, the gear teeth start to slide past each other at the contact(mesh)

All this means that gear friction is a function of torque. And also, that the doubling of the load could make for more than doubling of the friction.

Nah...I see what you are saying...but,...nah
Jump from 100 HP to 200 HP .
There will be no added 10 hp added loss.
SEE?
That's what I am saying....these arbitrary numbers floating around....
Just changing HP can not cause more drag.

There will. Just using your example... doubling the horsepower of an engine. More power can be achieved with higher rpm. Same parts now moving greater distances in the same amount of time, equals more friction. In order to make those parts live at the higher rpm means for one, greater valve spring pressures, this puts more stress on rocker arms, pushrods, lifters, cam lobes, etc. more pressure equals more friction. Now you have Pistons rocking in their bores at a higher rate yep, more friction. Now the crank is moving around in its journals more creating more friction.

Or another way to take your engine and make it double its Hp. Make it larger. Now the larger pistons, possibly traveling further in their bores, actuating larger valves, etc. More surface area, more friction.

Forced induction? More pressure on the same components, more friction. The internal combustion engine is very inefficient because it looses so much to friction. The heat generated in an engine isn't from combustion alone. It's friction. Work it harder i.e. more hp, that means more heat, from more friction.

Losses might be 10% or they may be more or less. The SAE has concluded that every drivetrain is different and assuming arbitrary % loss is not accurate.

In real life drivers have to let their 500 hp engines loaf along (for traffic or whatever) most of the time. Drivetrain loss is a function of the power the engine is actually producing at any given time, not it's potential. Doubling the HP capacity of the engine won't materially affect transmission losses if you maintain the same driving style. Is that what you mean?

Yes it can. More power creates more friction which creates more heat = more wasted power. Faster rotational speeds increase windage losses. This assumes you are actually using the horsepower.

"In any drivetrain component with meshing gearsets, heat generated by contact friction between the gears is a significant contributor to drivetrain loss. This is true during steady-state driving, but is far more of an issue when the throttle is mashed to the floor and the resulting thrust force and angular acceleration builds up in these drivetrain components. The heat generated by this dynamic friction is absorbed by the transmission and differential fluid as well as radiated to the atmosphere through the transmission and differential housing(s), and in some cases, via a heat exchanger or oil cooler. This absorbed and radiated heat is literally the conversion of engine torque into thermal energy because you can't technically "lose" power, but can only convert it into other things (some of our favorites being forward motion and tire smoke)."

Drivetrain Power Loss - The 15% "Rule"- Modified Magazine

You guys I think are missing the point or, I have some density .
( which is possible )
Either way let's try this example as it is real EZ to repeat what we have been told for a forever about drive train loss.
Lets take the 100 Hp engine , add a turbo and dial in 200 HP.
The HP / TQ will rise to 200 but the RPM will be the same in this scenario .
The only added drag will be the exhaust pushing the spool, is that going to eat 10 HP?
If it is , it is not going to consume the 10 HP under all conditions .
So it would not be a strict rule ( the loss)
Just thinking here chewing the cudd .

I beleive OP is talking about drivetrain loss, not total fuel efficiency per HP.

Just by adding 100 HP can not double the loss in the rear end(s)
That would double with every 100 hp in my example .
It would not take long everything would be glowing red.

In the OP you were talking about drivetrain loss. The power a turbo uses is accounted for at the crankshaft.
No, a drivetrain is not always going to consume exactly 10% of an engine's output. It could be more, it could be less. Depends on the drivetrain.

I was referring to the engine in this instance as it is part of the equation when accounting for loss.
Drive train loss is figured from total HP .

Think of the bearing and cylinder walls. More force against these surfaces means more friction, heat etc. That is what is eating up your power.
No free lunch.
Best
Les

In a no load system, there won't be a whole lot of loss, and there won't be a lot of increase of loss. With a dyno or a car on the road, when you use that power, you are loading the system.

There is a reason why you can only run so much power through a 915 gearbox, and that is because the load keeps increasing with the power. If you are just using the engine to spin the gears in the air, not much load, and an engine with twice as much power won't do much to the losses, but you won't need to use all that power to spin the gears to red line either, so it isn't really a test of the system.

The only way to find the loss of power in the drive train is to take the engine to an engine dyno shop and have them run it. Then put the engine in and take it to a rolling dyno and make some runs with no engine changes. EngHP - RWHP = loss and then you can figure percent.

UncleBilly asked about coolers in race cars and many do not need them but those that are very high speed, high rev or ancient mechanical (901 trans) need a cooler for longevity. In the 914-6 I used to race we had a cooler/pump system to make the ring and pinion last longer and also the same with those hard to find gears. Most serious vintage Porsche racers I knew had coolers in their cars and tranny failure was rare.

Aren't their road dynos that measure the to slow down after the power pull to help calculate friction loss of the engine and drivetrain?

Generally, for car's, HP is measured by OEM's as "Crank Horse Power", this is the horse power measured at the crankshaft.

Since taking the engine out to measure isn't often practical, most owners, and tuners, measure Wheel Horse Power, this is the measurement taken from the driven wheels driving the dyno. The 10% estimated loss figure, occurs from the engine at full power in top gear, and getting that power to the dyno at the wheels. You are confusing the number used for transmission loss - the difference between Crank HP and Wheel HP, for the loss in the engine system itself.

Losses involved in the engine itself, is much much much higher than 10%!!!!

When it comes to modification to an engine itself, you can increase crank HP, by decreasing the loss you are confused about. Reducing the restriction of the exhaust, by changing from crimp bends, to rolled bends, increases horse power, and decreases loss.

In take side modifications, are a little different, in that they allow more air flow in, but not always at an increase in efficiency, if the exhaust side can't match the new input, you probably still get more power, but drop efficiency.

For my 944's, I prefer to modify in ways that focus on lowering losses, reducing crank case windage, removing exhaust restriction, etc. These have minor gains, and the gains, are from making the system more efficient.

So the reason you think we are dense, is that the 10% number you pulled, is attributed as a normal loss for a manual transmission car from the crankshaft, to the wheels, when under full power in top gear. We see that number, and translate your first post into talking about that, as you mention "engine transmission loss".

What you seem to be speaking of, now that you've made a few more posts, is actually a loss of about 70-80% of the power the fuel is capable of.

When you add the Turbo, more air and fuel is rammed into the engine, so both the fuel and air used, and the horse power, are going up together. It may not always be a direct ratio, but you are increasing the fuel used in order to gain that horsepower, so the loss is increasing with the power.

First - I said I was dense .
I never said you, or anyone else in the topic is /was.
It is truly quite EZ to repeat what we have been told.
I can do that too.
Which again is why I drool along here.......
I do understand that number thrown out for drive train loss is the 15% number.
I was using the 10%number for simplicity, as stated.
My argument (again) is-
Using the 100 hp @ crankshaft figure, it is expected to see 85HP at the wheels, if we insist on using the 15% rule.
Now we pump up the engine to 200 HP it is said we lose15% which will now equal 170 HP , linear, with 30 HP loss.
To me, going linear up the scale, -it gets ridiculous.
A 500 hp will now see 425 @ the rear. with 75 HP missing in losses.
So with the 100 HP we lose 15HP
At 500, we see 425, 75 HP now consumed in the mass somewhere.
That is one HUGE number to eat!
Stuff is going to be glowing red RED HOT right now.

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