soslo

iTrader: (12)

Thanks guys (and gals, if any)! This has been fun - a good intellectual thread.

What are we agreeing to...
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by soslo:
<strong>
If there is a relationship, it is probably something like: L = X + Y , where
L is Total Loss in HP
X is HP it takes to turn the drivetrain
Y is 1% of actual engine output

so our engines in question may pull something like
100 HP engine - 49 HP
2000 HP engine - 1930 HP

That seems reasonable to me.

</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">

XAntivenomX

iTrader: (5)

No.

If you have some knowledge of physics, you could, theoretically find the sum of the frictional losses in the drivetrain and therefore have a coefficient of friction for the entire drivetrain. You can write this as a percentage.

Another way to look at it is to say it is like drag force, the faster you go, the more force is exerted on the object by resistance. This is why a 2000HP engine does not lose just 50hp through the drivetrain. The loss is proportional to the power, which gives you a frictional coefficient(or percentage).

Fenris Ulf

Drivetrain loss is dynamic, it is not directly a percentage but that approximation works quite well. It takes a certain amount of torque to accelerate the drivetrain and increase the rotational inertia, and this relation is based on input torque, not a fixed value.


Your L=X+Y thing is pretty far off. So you think a 50 hp motor would not even be able to turn the wheels?

WhynotSS

Let's sum things up.

L = X + Y
L is Total Loss in HP
X is HP it takes to turn the drivetrain
Y is 1% of actual engine output

In the beginning, soslo was assuming that the X was constant for any given hp. For example, when X = 50, a 100 hp engine will make 49 rwhp and a 2000 hp engine will make 1930.

After reading all of the information on this page, we've seen that the hp loss inreases as the rate of acceleration increases.

Horsepower is the amount of work performed over the time needed to complete the work. We know that you can accelerate the same drivertrain from idle to redline quicker with 1000 hp than 50 hp. In other words, the same weight is accelerated in less time with 1000 hp than with 50 hp. Soslo argued that the drivetrain weight doesn't change from the 1000 hp engine to the 50 hp engine, so the loss should be the same.

However, trackbird pointed out that it takes a lot more work to accelerate something quickly than it does to accelerate something slowly! In other words, more work needs to be done in less time. The key is to focus on work--not weight.

From what we've seen, engines with high hp have more of a drivetrain loss because more work is needed to accelerate the drivetrain quickly. Low hp engines do not have as much drivetrain loss because less work is needed to accelerate the drivetrain at such a slow rate.
This is why people use percentages (15%) to predict flywheel hp.

As we've seen, the straight percentage concept doesn't seem to hold true. Cars with 2000 hp lose about 25% while cars with 100 hp lose about 15%. Of course, when comparing different cars and engines, the percentages will differ. The question is, does this variance in percentages occur only because the cars/engines/drivetrains are so different?

I believe the variance exists mostly because the amount of work needed is an exponential equation. Thus, high hp cars use unproportionally more hp to accelerate the drivetrain than low hp cars.

To sum it up, cars with 50-400 hp should be around a 15% loss. Cars with 750 might be around 17%. Cars with 6000 hp might be around 30%. I doubt we'll be able to figure out exact percentages in this thread, but you get the idea.

As for the equation L = X + Y:
Although this might fit the data, it does not give a causal explanation. X and Y are actually the same thing--part of the percentage of hp that is lost in accelerating the drivetrain. The other part is friction.

A better equation would be L = W + F.
L is the drivetrain loss.
W is the loss from more work being performed in less time.
F is the loss from friction that increases with rate of acceleration (contributed by XAntivenomX).

soslo

iTrader: (12)

Thanks guys, I think I finally get it.

XAntivenomX flicked the first half of my light switch with </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif"> So you think a 50 hp motor would not even be able to turn the wheels?
</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">I thought to myself, "Self, the 50 HP engine would spin it, but it would be really hard to do...(slow)"

trackbird has been trying to tell me this since the beginning, but for some reason I couldn't get it.

Finally WhynotSS comes along and finishes the flicking of the light switch, and it is clear again.

The reason why high HP cars lose more HP through the drivetrain is because they are working harder and revving the motor quicker.

Bingo.

It is the revving of the motor quicker than I was missing, and makes sense now. That, and the graduated percent loss comment </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif"> To sum it up, cars with 50-400 hp should be around a 15% loss. Cars with 750 might be around 17%. Cars with 6000 hp might be around 30%. </font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">So where are we?

I guess 15% drivetrain loss for manuals and 20% for autos <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" />

Fenris Ulf

</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by soslo:
<strong> ok, a couple of things:
1. We don't know the actual engine output like you said...it is entirely possible that the manufacturers just take the wheel HP and multiply it by 1.15 </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Just FYI, to find flywheel ratings assuming a 15 percent loss, divide by .85, don't multiply by 1.15 <img border="0" title="" alt="[Smile]" src="gr_stretch.gif" />

trackbird

iTrader: (4)

This all came together nicely.

One last thought.

Front wheel drive driveline losses seem to be less than rear drive (for a given hp). No driveshaft to turn (among other things). It seems to be a case of less weight. Fewer parts, less air drag (A driveshaft is seeing some air drag), etc. The front drive/front engine or mid or rear engine/rear drive seem to be a little bit more efficient than the front engine/rear drive setups. Is it a big difference....no....but the theory does seem to be supported in dyno testing. Anyway....it looks as if the big picture has finally started to show it's self....Thanks everyone for the assistance and information!!!! Enjoy.....

WhynotSS

soslo, I'm glad I sent that private message to you on the other board asking you about this. This discussion has been fun and educational. Hopefully, we're right!

Camaroholic

iTrader: (1)

I hope to have an answer for myself soon. This pic was taken yesterday. The recipe used to build this motor should put it at about 460 rwhp +/- 10 (anywhere from 450 to 470 - assuming M6 with typical 'all the bolt ons' configuration). The motor dyno should be on Friday, assuming it all comes together.

I too am very curious about % losses.

-Andrew <img border="0" title="" alt="[Smile]" src="gr_stretch.gif" />

soslo

iTrader: (12)

please post the results!

PacerX

iTrader: (4)

I think this was missed in this entire discussion, although it was danced around a bit....

As horsepower increases, drivetrain loss will increase.

Why? Well, mostly because I'm loading the drivetrain up more. Gear teeth, even if they are set up perfectly, deform slightly as they mesh and thereby become less efficient. More torque on the gears equals more deformation. More deformation will cause the pitch diameters to shift relative to each other and cause a progressively increasing loss of efficiency.

The same holds true for bearings, as the normal force on a bearing increases the losses increase.

The loss is fairly linear in nature in bearings and gears until really nasty deformation starts. Then it increases exponentially and things start flying apart. Because of the linearity of loss prior to catastrophic failure, a percentage works out nicely.

Why did I bring this up? Well, mostly because this is not easily predictable. The physics surrounding the power required to accelerate a given mass at a given rate is pretty well known. Predicting frictional losses in a complex machine (like a drivetrain) is a bit more difficult... it can be done, but you'll want a computer for it.

Finally, EFFICIENCY is by definition expressed as a percentage (power out divided by power in).

Here's the final caveat, the drivetrain losses due to deformation aren't truly linear when caused by deformation in the geartrain. They increase exponentially.

trackbird

iTrader: (4)

And, 2000 hp puts lots of strain on the gears, and increases frictional losses (wasted as heat), etc.


That would be the interesting part of the study....

Cal

iTrader: (1)

Sorry to reserect this old thread, but I ran on to it doing a search and wanted to add something to it. The force required to overcome friction is given by F=KN, where K is a constant for the materials used and N is the Normal force. The Normal force is just the force between the two materials. As you apply more torque, this Normal force increases, increasing the F or Frictional force in direct proportion. This is why drivetrain loss is a percentage rather than a fixed amout. Once deformation starts, then it would be a non-linear proportional relationship.

trackbird

iTrader: (4)

Wow, I was sure this thing was ancient history. I was suprised to see it show up.

That would explain things up to the point of deformation pretty well. After that, it should get interesting in a hurry.

skippytheloon

iTrader: (14)

Holy Jupiter,
I always thought physic problems were supposed to be drawn out with the parts labeled? Makes perfect sense even without the illustrations.

Kevin,
I didn't know you were such a good physicist maybe you should consider a new career: Topless Tudors! Think of the possibilities!
maybe not

trackbird

iTrader: (4)

Me???

I'm just glad that I understood what I said.....

Kevin

z98

Flywheel HP =(RWHP X weight of drivetrain)(exponential equation where RWHP = X, includes the constant <weight of drivetrain>)

The exponential equation would be a representation of the increasing loads faced by friction and the entire weight of the drivetrain accelerated to any given horsepower. Without knowing the actual equation, it is easy to assume engine of approximately the same power level will have similar levels of loss. For example, 300-400 hp is directly comparable in the same drivetrain, just as 400-500 would be. But the jump from 300-600 would skew things just a bit.

or something

It only matters what you're getting to the wheels though, right? Right?

critter

iTrader: (9)

Good thread. The dyno issue was glossed over, though. The question at hand is the comparison of flywheel horsepower and rear wheel horsepower, or engine dyno vs. chassis dyno. A Dynojet does not measure horsepower. It does not even measure torque. I measures acceleration (well, measures speed and differentiates to get acceleration) of a drum of known weight. Given that, the acceleration of all rotating components (wheels, axles, posi, driveshaft, trans shafts, cankshaft, flywheel, clutch, etc) will consume HP and reduce the acceleration of the drum, and thus will not be measured. The dyno software then corrects its computed number to try to compensate, but as was pointed out earlier, the horsepower of the vehicle under test will skew this as more power will be "lost" in the faster acceleration of the non-drum rotating components. (And in fact, Dynojets seem to over correct, giving numbers too large) Given all this, there is simply no way you can write a simple formula that will correlate fw and rw hp. For a graphic demonstration of this, you might try dynoing in 2nd and 4th, or with 3.23 gears and then 4.56 gears and compare the readings. However, if you use a Mustang dyno (which does measure torque) and don't use the acceleration mode, you will get real RW HP and stand a much better chance of deriving a conversion formula.

jstpsi

as much as i enjoy math to answer a question on any given car it sounds easier to dyno the car, pull the motor and dyno it. Trackbird your points are valid, thanks for the logic. It just seems with so many combinations and variables with each specific drivetrain
you will need to spin it to find out the truth

CHRISPY

It has been verifed multiple times that the drivetrain loss through an M6 with stock rear is right around 50hp.

This is a constant value and doesn't fluctuate as a percentage.

ie: if my car dynos 450RWHP and I am running stock T56/rear/clutch/flywheel/DS then my engine horsepower is right around 500.

Cheers,
Chris