I believe drivetrain loss is caused by several things. Friction of the gear-to-gear interfaces, twisting of various components (driveshaft, axels, etc.), gears moving through fluids, bearing friction, torqueing of the frame, etc. I believe there will be little change, in power absorption, for some of these items with greater engine power applied. For example, the gears moving through differential fluid and bearing friction. There will be some increase because of the greater rate of acceleration, but certainly not in the same ratio as the power increase. I can understand, and believe, that gearing friction will increase as power, and therefore, rate of acceleration, is increased, but certainly don't understand how it will increase in the same ratio as engine power is increased. That also follows for twisting and torquing. As driveshafts twist or torque, they stiffen, which would resist further deflection and therefore, become somewhat constant as a power transmitter. I just don't understand how loss percentages would follow, or increase at a faster rate, than a power increase. Not being a smart *** in this thread, but if you could explain how you perceive that loss percentage actually increases at the same rate, or greater rate, than a power increase, I would appreciate it. I may be missing something here and would love to understand.
Ed

 

I agree.
We went thru this same question a few weeks ago, with no real concensus.
Many of us have differing opinions of what the results will be, and the ZR1 rwhp results will help lead us all to a better understanding of what is really going on as hp levels increase.
Those SAE certified fwhp results are a real blessing in this discussion.

 

I look at the drive train losses as two independant components:

1) Losses due to components that have a constant frictional factor. These are the components that really don't contribute any less or more HP loss as the input power increases. Their losses are only a function of the rotational speed and temperature of the components. This would include such components as shaft and bearing seals, moving parts (gears, etc) in fluid (assuming the fluid viscosity is constant). Note that if the fluid viscosity changes, the frictional force will also change. For example, the differential and tranny will lose much more HP if you dynoed at -10 deg F fluid temp vs. 200 deg F fluid temp. It simply takes more HP to move those parts at the same velocity in much thicker fluid.

The HP lost to these components is defined by HP = (T X RPM)/5252. The friction torque (T) is basically a constant (if the drivetrain temps/fluid viscosities are constant), so as rotational velocity (RPM) increases the HP lost is directly proportional and linear to the rotational speed. For these components, the HP lost will basically remain at a constant loss percentage. Example - lets say that the overall rotational drag force (T) of the drivetrain is 100 ft-lbs. This could actually be measured if you put a complete drivetrain in the air and put a torque measuring device on the drivetrain input shaft and slowly applied torque until it started to rotate. This would be the "rotational torque" required to overcome all the friction in the system.


2) The second component (and largest factor IMO) of the drivetrain HP loss is due to the components transmits the HP loads through the system. These are the components that contribute a variable amount of HP loss as a function of the load on and the velocity of the components. This is the HP loss factor that I believe basically remains a constant percentage loss of the input HP, or may even exhibits a slight increase in percentage loss as the input HP levels increase due to additional stress on the components. IMO, from a mechanical system viewpoint, there is no reason that the efficiency of the drivetrain should get better (ie, friction goes down) with increased loads and stress. It's just not a trait of mechanical systems in a case like this.

The HP lost to these components is a function of the frictional loads on the components. The frictional loads will basically follow the dynamic friction relationship of F = u x N, where friction load (F) is equal to the coefficient of friction (u) times the normal load force (N) which is directly proportional (ie, meaning a linear relationship) to the torque input to the system. If u increases with increased load, then this component will not be linear but slightly non-linear with the F loads and hence lost HP increasing as the torque loads increases.

If both components that contribute the drivetrain HP loss are essentially linear with rotational velocity and with torque input, then my theory is that the overall HP loss will essentially be a fixed, linear percentage of input power. If the loss percentage does change I do not think it is drastic, and I think the percentage lost would increase slightly with higher and higher HP inputs to the driveline due to the added loads and stress on the mechanical components. That's my $0.02 on the physics involve.

 

... I have confidence that the SAE rated HP numbers from GM on the Vette is an accurate number. Then if good avg rwhp dyno data can be gleaned for all the Z06s one should be able to put together some data that tells the story.

 

The only people that ever care about engine hp #s...are car sellers/manufacturers...and people wanting to boast about a big hp#.

It doesn't much matter when your "calculated" 550fwhp car gets beat down by a 420rwhp car out there in the real world.

The loss percentage is pretty close to a constant. People tend to think of torque when arguing this...it takes X amount of force to turn the gears, axles and tires. Yes..
What is not, is the time factor (the difference between torque and power). You can turn a hand crank with little effort at 1rpm. Now try to turn it at 100 rpm, and tell me it isn't tougher. You'll probably even blister your hands trying because of the increased friction. That friction didn't come from increased load resistance, it came from you trying to turn it faster. The car's frictional components feel the same thing...trying to spool the drivetrain up quicker always saps more power.

 

 

Just a thought for you. Why not think of the drive train loss as a constant value like 45 hp. That number is the frictional loses to turn it. The frictional force loses does not increase just because you increase the input hp. So if you start with say 350 hp and you lose 45 hp in the drive train, you end up with 305 hp to the wheels. So if you increase the engine hp to 400 and the drive train is still 45 hp in losses, you end up with 355 hp to the wheels. That is assuming that the speed of the dyno is the same. Force will increase and decrease as a result of the speed. However, I think that if we are comparing apples to apples then the drive train losses will be constant. What do you think about that idea?

 

Bingo...

That's a more much accurate statement than a percentage loss (although every driveline has some given percentage loss but if you change the engine the percentage changes and the constant remains very close to the same assuming the new powerplant isnt completely overwhelming and flexing the driveline (not likely unless your at retarded power levels with the new combo).

I have both engine dyno'ed and chassis dyno'ed both of my C5 engine combinations and will be doing the same with the next engine combination I am currently working on. The bottom line is my 346 made 550 HP (flywheel) and on average generated 475 RWHP.....my 383 made 615 (flywheel) and it generated on average 540 RWHP (note that both made around 5 more on their better runs but the average runs of both show my driveline eating exactly 75 RWHP with both combo's). My gut is if I can get the next combo to generate around 675 HP at the crank, I will likely produce right at 600 to the ground (the same 75 delta). Note that this "Delta" I am referring to takes into account many variables but none of them have changed in my particular combination.

The reality is if I had a heavy LS7 clutch and heavy 60 pound rear tires my car would see easily 20-25 less RWHP. So whats changed with the output of my engine?....absolutely nothing but if you plug in the new numbers your "percentage loss" certainly has changed and that's why you really cant use this 15% Internet accepted factor to figure flywheel/crank numbers....its just something that gets you in the ballpark but it is subjective to many different variables in each of our combinations. What everyone does have is there own "Delta" or constant that their particular driveline, clutch, converter, rear end gears, etc., will eat up. If that's not changed and you increase your engine output by 40 HP you will see 99+% of the 40 HP at your rear wheels as well (changing your percentage but the constant "Delta" remains the same).

To summarize, from what I have seen from the numerous dyno results (both chassis and flywheel), an M6 car with a lightweight clutch and reasonably lightweight rims and tires will lose 75-80 HP from the crank to the rear wheel.....if you have a heavy stock clutch and aftermarket heavy rims you might lose a 100 to the tire but most M6 combo's will be in the 75-100 HP category regardless of how much power is being made at the crank (make 400 and you will see 300....make 700 and you will see 600).

Now if your running an A4 with an aftermarket street converter you will see an additional 20-30 less at the tire than the M6 scenarios I just highlighted (over 100 - 130 HP loss), but once we are discussing A4's with various designed converters and all the different stall speeds (with their varying efficiencies) its really a crap shoot. A really high stall could throw another 30 on top of that....(you could make 600 HP at the flywheel and only see 440-450 RWHP on a chassis dyno with a stalled auto combination....or possibly closer to 500 if its a more efficient lower stalled street converter....thats just the way it is).

Anyway....I'm sure the die hard "percentage" loss crowd will jump in to argue but I'm extremely confident that its much closer to a fixed loss (based on your individual combination) and I have empirical evidence to back my theory with my personal dyno experiences as well as other similar data from Westech in Mira Loma, California, a very popular testing facility here in So Cal that has two engine dyno cells as well as a chassis dyno and has done similar testing that also backs the "Delta" or fixed loss theory (try to get that from the percentage loss crowd).



-Tony

 

Hey, thanks for the time to put that together. I appreciate it. I just want to make clear that I never meant for you to think, that I think drivetrain losses decrease with added power, just that the rate of loss increase, decreases. Which, if true, would equal a lower percentage of loss (but higher hp number) with the more powerful engine. I see your points about linear relationships. The key here, as you allude to, is if they are of great enough effect to overcome the effect of those losses that are not related to increased torque, on the overall loss factor. Thanks again...good conversation. I'm also anxious to see the results of the various dynos put together. Should be interesting. I hope they all have factory tunes, and the same overall gearing.
Ed

 

Im with tony, I dont think the percentage to turn those same parts gets harder With HP. I think the resistance stays the same at all levels from stock to fully built.

Just think about it, the parts will require effort to turn but how could it get harder? Its seems like a set rate to me.

Whatever the driveline eats up, is what it eats up, as a set rate depending on your clutch, differential gear setup etc

 

Now lets add a 3200 stall and 3.42 differential on an auto

 

This theory has been shot down numerously and with some logic to support that the drivetrain loss is not a constant amount of HP. There is no way it could be from an engineering viewpoint.

Think of this. If the drivetrain loss was always 45 HP as you say, you would need more than 45 HP to even make it move. A C5 Vette only needs about 20 rwhp to go 60 mph down the road. Do the math.

 

That’s kind of hard to believe. From a conservation of energy viewpoint, how can a drivetrain become more efficient with higher and higher HP inputs to it? That’s like saying cold fusion works.

If what you are saying is correct, then the drivetrain losses are 25% for a car making 300 fwhp and having a constant 100 HP drivetrain loss. Also, on the same line of thinking, if an engine made 1500 fwhp and had only 100 HP drivetrain loss then the drivetrain would only have a 6.6% loss. I'm sure everyone would love a driveline that effiecient.

Even though you have supposed accurate engine and chassis dyno data, something is missing and just doesn’t add up in the logic department. When you look at extreme examples it seems the “constant HP loss in the drivetrain” theory falls apart

I will say that I don’t think the loss is a hard constant %, but I’m sure it’s not a constant amount of HP loss like “75 ~ 100 HP” regardless of fwhp being input to the drivetrain. If it does change with increased HP levels then it's not huge like a swing from 6% to 25% like your example.

 

I argued the position of a NEAR-constant driveline loss just last week, and was fairly trounced by the percentage guys. I still didn't believe their position, which was based on mathematical formula, and not something a business education/college dropout 30 yrs. ago, would understand.


Then my son-in-law explained it in terms I could visualize. Take your hands, palms in, and rub them slowly together....This isn't going to get dirty, so bear with.... Slowly increase the speed of this motion, which emulates adding power to the scenario. What do you observe? Added heat, and increase of friction! Same thing happens to engines. Increase heat and friction, and you increase the drag in NEAR equal proportions. Some would say equal, but I left myself some wiggle room.

 

Not a very parallel situation Zee.....

Dont lose the faith because near constant driveline loss (with each individuals set-up slightly different based on driveline components, clearances, oil type, etc.) is exactly what you would experience in real world situations....real world meaning a Vette with 300 HP up to say 800 HP. Pick up a hundred HP at the crank and MOST of that power gets to the rear wheel (very close to a one to one relationship). A driveline has lubricated bearings and parts and is meant to be rotated....it also has some fixed mass and rotating inertia. If you are always testing the same RPM, the amount of energy needed to spin it there remains the same....it doesn't change based on the engines power output that its bolted to....that would defy physics and the conservation of energy principle. An engine with more power (power equals twisting force over time) will spin a fixed mass quicker from a given RPM to a higher given RPM....thats the nature of what more power does. The amount lost in heat and "friction" is minimal (with lubricated bearings, trans gears, and differentials) if we are talking about sane levels of power that the driveline was designed to withstand.

Think about a given driveline accelerating from hypothetically 3000 - 6000 RPM and taking 10 seconds to do so based on an engine of some given power output, the drivelines mass and rotating inertia. Now swap engines and accelerate the same driveline from 3 - 6K in four seconds which btw would require a huge amount of additional power to make that happen. If you had a heat gun and could take a reading of the driveline before and after each run do you really think it would be significantly hotter?

The percentage theory throws away 20, 40, 50+ HP....where did that power (energy) go. Think about how large even a 15 HP electric motor is (thats a BIG air compressor motor for instance) and how much work and heat it is capable of generating....that type of power doesn't easily get misplaced or mysteriously "absorbed" in the driveline.

If your not changing the total mass of the driveline or the distance the mass is from centerline (rotating inertia), and not changing the test parameter speeds of that mass you will see just about all the extra power you apply at the crankshaft make it to the back wheel....and IMO that is a good thing because I work to damn hard to get that extra power at the flywheel.

Harold Bettes of SuperFlow fame....a veteran in this industry btw, who has forgotten more than most forum members will ever know (concerning dyno's, engine dynamics, physics related to this hobby, etc.) also agrees with the "near constant" theory as I had the opportunity to discuss this very topic with him at length a year or so ago when a similar thread was brewing elsewhere and I thought he would be a great guy to give "expert" opinion on concerning this.

One last piece of REAL world information pertains once again to some chassis dyno time I spent on my former 346 combo. I decided to spray a small shot of nitrous on the dyno to check out the A/F ratio and just how "happy" everything was with the nitrous engaged. With only an .053 single nitrous jet that Nitrous express called out to be a 100 shot (about right based on many other single nozzle deployment systems I had seen...in fact on the light side in orifice size if anything) my 346 combo jumped from 480 RWHP to 606 RWHP, a gain of 126 RWHP. I was elated to see that much power from such a tiny jet. Whats my point you ask....if I was doing the same test on an engine dyno guess how much you think I would have gained? (about 126 HP is my guess....give or take a pony or two). The percentage crowd would tell me I would have had to see over 145 HP on the engine dyno because certainly my driveline would develop a larger appetite for power and 20 HP would magically disappear. No way in hell would a single orifice .053 nitrous jet make close to 150 HP and even the 126 it made (26 more than the manufacturer claimed) was extremely impressive, some of which likely related to the fact of how dialed in and optimized that combo actually was.

Anyway...I wont be wavering in my beliefs just as sure as some of you on the other side of the fence wont waiver on yours, but those of you out there not sure which way to go certainly have some additional things to ponder over.

Tony

PS....Anyone want to bet me that my next combo wont be close to a 75-80 HP loss from flywheel to RWHP #'s because thats my guess of exactly what it will be....again. How could that be possible if the percentage theory holds any water?

 

This I agree with, and I think is basically the point your making. You are taking a peak hp number (at a certain rpm), and comparing it to that same peak reading of an engine dyno.

That's where the conflict of opinion comes in.

According to what you just claimed , those parts just go along for the ride on the quicker accelerating car, and the hp-maker in front of those parts is the only part seeing the additional load needed to do so.

If that were true, then the driveline didn't endure any additional stresses while accelerating the car that much quicker.
And, if that's the case, then we certainly wouldn't ever have a need for things like better clutches, better output shafts, better rearend gears, etc.

The fact is, those parts can only absorb so much power being transferred through them, and yes, that's exactly what they do...absorb energy...right up to the point of breakage (or slippage, in the case of a clutch).


So yes, if you measure the driveline loss factor while the car is on jackstands, it probably will remain the same in the case of a 300 or 3000 hp engine.
But put the hp to the ground and accelerate the car that much quicker, I'm reasonably sure the drivetrain fights back a little more too...and if it does, then how can you say it didn't consume any additional power?

 

Your last post was long, but the part above caught my eye and I had to reply.

If you apply thermodynamics to the issue, you would clearly see that the increased friction and HP lost is turned into heat, and that the more HP you put into the drivetrain the more heat (ie, lost HP there is). This shoots down the "constant HP loss" theory, as there is no way thermodynamically that the drivetrain loses the same amount of HP regardless of how much you put into it. The following example might put some perspective towards the thermal physics involved and open some eyes.

First some baseline info. 95+ % of all the HP lost by the drivetrain is turned into heat. X amount of HP = Y amount of heat. As HP is turned to heat, the temperature of mass increases. In the drivetrain, most of the HP lost, and heat generated, is in the transmission and differential. Other components also increase in temperature, but the tranny and diff comprise the largest amount.

So, lets say we have a truck that makes 500 rwhp (peak) just below redline. Let's also say that we have a 50 mile long hill with a steep and constant grade. Instrument the tranny and diff with thermocouples to measure their temperatures.

Drive the truck up the hill at a constant 70 mph -- let's say the required amount of rwhp was 100 to achieve this. Measure the tranny and diff temps at the top of the hill.

The next day, hook-up a 15,000 trailer to the truck and drive it up the hill again at 70 mph. Let's say that in order to achieve this feat, it takes 450 rwhp. Measure the tranny and diff temps at the top of the hill again.

Which scenario do you think will produce higher tranny and diff temps? Do you know why?

If you assumed that the drivetrain lost a constant 75 HP in both senarios then wouldn't it make sense that the tranny and diff temps would be the same if the truck was driven up the hill without the trailer vs with the trailer? Do think that is really what happens and makes sense?

The same example could be used with a race car. Let's say a race car that makes 800 rwhp just tootles around the track at 20% throttle, so it's making way less than it's capable 800 rwhp -- let's say it's only putting down 150 rwhp at tootle speeds. Now drive the car at WOT and near its 800 rwhp level. Check the tranny and diff temps after 1 hour of continuous driveing at those two levels. Guess which one has much hotter drivetrain temps.

If more HP wasn't lost with increased drivetrain HP inputs, then the tranny and diff temps would always be the same no matter if you were driving around town or racing WOT for hours at a time. We all know that never happens and never will.

 

That's actually a pretty good example, and correlates along the lines of my truck pulling a heavy trailer uphill example. Also, try doing that but also press your hands tighter and tighter together at the same time - that would basically simulate higher HP and loads into the drivetrain. This causes even more friction and heat of course.

A convert is always welcome.

 

Good points, and again this correlates with the truck pulling a heavy trailer up the hill example.

 

In the end...this will all simply boil down to the type of dyno differences, and why a load-bearing chassis dyno is more accurate (even though they give you lower readings).

Spooling up a Dynojet drum takes a lot less effort in comparison, so in turn, the drivetrain loss experienced is that much less too.

.