fasteddyss

iTrader: (1)

Does anyone have a handle on what is lost in the drivetrain?? For instance, if the factory advertised 320 HP what would the RWHP be with a 6-speed??

1BadAzzWS6

iTrader: (11)

It's about a 15% or 16% loss in the M6. So thats about 270rwhp. However GM underated the engine...so I think stock 01-02 LS1's are dynoing around 290-300rwhp.
Don't quote me, but thats what I've heard.

Jay02LS1ws6

Well, here's my question...

Sure, we say that 15-16% is an acceptable figure for drivetrain loss a stock f-body. Then we modify our engine with bolt-ons... in my case, headers, intake, cat-back, and tune. So now my engine is producing more horsepower. Is it still acceptable to use a 15% correction when I see that I dyno'ed 345 HP at the rear wheels? I have a difficult time understanding why 15% should still be used... the only reason we say that we use it in the first place is because the transmission, driveshaft, rear-end, axle all take some of the engine's energy to spin. Well... after the bolt-ons, what has changed? All of those parts are still the same, so if we still used a 15% correction, that would assume that those parts all of a sudden require more energy to spin. Why? Just because the engine is now making more power, we can assume that the driveline suddenly requires more energy to spin? I definitely don't follow here... although I know 15% is a pretty standard number that everyone goes by. Can anyone clear this up?

Jay02LS1ws6

01-02s are a little higher than that, I think. 295-310... some even in the 315's (manuals for the most part).

89FormulaLS2

iTrader: (1)

Anyone know what the loss is for an A4? I've got a 2001.

fasteddyss

iTrader: (1)

Jay is right. 15% of what horespower? Is 50 horsepower an acceptable figure for loss in a stick drivetrain, regardless of what your hp figure is??

parish8

iTrader: (6)

every car will be diferent, there are many variables, how many miles on the car, what oils you are using and how warm are they, how heavy are the tires, what gears, there is no way to pin point what your loss will be.

to make things even a little more complicated a dyno jet has a drum you have to accelerate, more hp means you will accelerate it faster, it isn't hard to see how it would take more hp to acelerate something quicker and that includes your driveline so in that case i think the hp loss due to your driveline does go up with more hp.

on a mustang dyno they keep the rpm rise/time constant so you wont be acelerating the parts any faster and i have to agree, the drive line loss will be prety close to a constant hp number no matter how much hp you are making.

bottom line is who cares, 15% is prety close for figuring things like fuel system requirements, just go with that.

jRaskell

iTrader: (1)

The horsepower lost through the drivetrain is a result of having to accelerate all the drivetrain components (it takes power to get everything rotating), and what is lost due to friction. These are NOT static power figures. It takes more horsepower to accelerate your car forward faster, so why would you think it doesn't take more horsepower to accelerate your drivetrain components faster?

Jay02LS1ws6

Good stuff.

Merlin

I think you may have a point on the ratio of power loss to power production. I was thinking of Newton's 2nd Law, F=ma. From that we can also get the equation for rotational dynamics which is T=Ia, where T=torque, I=moment of inertia and a(alpha)=angular acceleration. As you add more power to your engine through mods, you increase the generated torque. The moment of inertia is based strictly on the mass and geometric properties of the drivetrain in this case and, hence, does not change unless you change something in the drivetrain. We wouldn't worry about the (I) of the engine in this case because we are only interested in the power lost from the flywheel to the wheels. We would just say that the engine produces a cerain amount of torque and go from there. But this equation must still hold true and since (T) has increased and (I) remains the same, then alpha must increase proportionally to make the equation balance.

So we say Torque and angular acceleration (alpha) are proportional. Let's just say that for an example, (T)=300, (I)=10 so that (alpha)=30. In this case (I) is ~3.33% of (T). Now if we increase (T) to 400 and (I) still has to be 10, then (alpha) increases to 40. (I) is now 2.5% of (T). This would lead you to believe that as torque increases, the amount of drivetrain loss as a percentage of total power would decrease. That would be true if, of course, it weren't for friction. And the actual drivetrain loss you are talking about is due to friciton.

So since we have have friction, the equation becomes something like (T-Tf=Ia), where (Tf) is the torque loss or resistance due to friction. Now (Tf) is not very easy to calculate, but just as rubbing your hands together generates more heat the faster you do it, accelerating the engine at a faster rate also generates more heat or friction. And so as you increase the acceleration of your engine through mods, imagine that (Tf) also increases along with it and you lose a part of your new found power due to increased friction. At what rate it increases, I have no idea but I would bet that it is not linear. So I think it would be possible that as you increase power, there is a certain point at which the power loss would start to decrease as a percentage of total power, though it would always be there and would always increase as you increased power.

parish8

iTrader: (6)

z98

If you go from 300 hp to 600 hp, your driveline losses will be twice as much because you are spinning them (or attempting to spin them) twice as fast.

That's simple enough for our purposes

blueWS6

there was a thread on the other site that basically came up with a 12% loss with an M6 and about 18% loss with an A4, are these numbers anywhere close to correct?

TeeKay

I'd like to try to shed some light on the question that a couple of folks have posed. So here goes...

Engineering-wise, Efficiency is always expressed as a percent (e.g. 85% efficient). 85% efficient means that only 85% of the input is going to be the output. By definition, the efficiency of a given machine doesn't change even though the input power is changed (up or down). The remarks about inertia and accelerating a mass are good points. However the main villians are friction and gear pressure angles (along with a few other small contributors). Most manual trans drivetrains (flywheel to pavement) lose about 15% of the power put in by the engine and most auto trans drivetrains lose about 18%. That's 85% and 82% efficient respectively. The extra loss in the auto is mostly ATF oil pumping losses.

Read no further if you don't want some details.

Gear teeth don't mesh exactly straight on, they always meet at some angle called the pressure angle. So some of the power transmitted is lost as thrust against the bearings. As you apply power, the teeth mesh, the drive gear tries to push the driven gear harder against its shaft as well as turning it. The "push" power does NOT help turn the gear, so it's lost. From triginometry, the percentage lost is related directly to the pressure angle AND NOTHING ELSE. The more power you apply, the harder the push 'cause it's a percentage of power applied. The gears have to be strong enough to take the power transmitted without failing of course.

The other element is friction (yep, you guys are right on about that). Bearing friction is directly related to the force applied to the bearing, the "push". See where I'm going? Friction (and power loss) is the coefficient of friction (always the same for a given bearing) times the "normal" force (load applied straight into the bearing). As the input power increases, the "push" power increases proportionantly. So the friction and loss increase the same amount. That loss turns up as heat. Running a trans and diff hard generates a LOT of heat, compared to gentle driving, right?

So from the above facts, power loss through a drivetrain is always a percentage of the power applied. That percent number does not change when you change the input power.

This little piece is not complete by any means, but it's meant to show some basics. I hope it helps and doesn't sound uppity.

89FormulaLS2

iTrader: (1)

Nice post, TeeKay. That makes complete sense to me. Thanks.

Adam

yellowbird

so if you dynoed at 350 rwhp what would you say your engine hp would be .2002 a4.

Gev

You would multiply that number by 1.18.

TeeKay

You divide the whp by the estimated efficiency for crank hp. In your case 350/.82 (approx. for auto) = 427 crank hp. If it were an M6 with 350 whp then it would be 350/.85 (approx. manual) = 412 crank hp. Note that if the efficiencies are very much different than estimated, then the calculated hp is gonna change. The same 350 whp with an 80% efficiency in the drivetrain is 350/.8 = 438 chp. HTH

TeeKay

Yep, that's exactly right for 85%. 1/.85 = 1.18! IMO 85% is a pretty good number for ballparking.

blackws6

sounds like we have a few physicists (spelling) here they should condence this post down and make it a sticky in one of the sections.