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Discussion Starter · #1 ·
Got some very interesting information from Jim Bell tonight. He works very closely with Ford engineers and got this data from them:

Drivetrain efficiency

C6 tranny - 89%
AODE unlocked - 86%
AODE locked - 93%
T5 manual - 97%
E40D unlocked - 84%
E40D locked - 91%

Ford 9" rear - 89%
12-bold Chevy rear - 93%

here's the kicker:

Ford IRS - 91%


Jim Bell didn't have the Ford 8.8" rear efficiency numbers handy but he said that he'll get them for me this coming week.

We know that the drivetrain efficiency of the 03 Cobra is about 85% total (15% loss from crank to rear wheels). Could the IRS really account for 5% of that loss??

If the T5 tranny has a 97% drivetrain efficiency then the T-56 can't be much different. Although he didn't have efficiency numbers for the T-56, let's assume that it's about the same as the T-5.

If the IRS efficiency is 91% and the tranny is 97% then doesn't that mean that 6% is being lost in the IRS?

I am not sure how this all plays out but, again, Jim Bell says these numbers are directly from Ford and very accurate.

This may be another big reason to go to a live axle arrangement. Can we really pick up 35 RWHP from the swap??

I know that some guys have done this and reported very little pickup on power. But I also know of different, conflicting info.

Input would be appreciated!
 

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Hammer: You might be looking at this from the wrong perspective.

Assuming the T-56 has the same efficiency as the T-5 (97%), the tranny and IRS (together) would have an overall efficiency of 97%x91%=88%.

We can safely assume that add'l losses are present (engine/tranny bushings, u-joints, etc), so the figure that is typically thrown about for converting rwhp to flywheel for the '03 (85%) is probably pretty close.

For the sake of comparison, if you were to swap to a 9", you would lose an add'l 2% (or roughly 8-9rwhp on a typically modded '03).
 

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Man the IRS is more efficient then I expected. I doubt that an 8.8 is much better maybe 1 or 2% given a 9" is 3% less efficient then the IRS. What mosconiac said is correct.
 

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Discussion Starter · #5 ·
Thanks, I understand what you guys are saying and you are probably right.

But wouldn't an 8.8 be more efficient than a heavy-duty 9"?
 

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HAMMER said:
Thanks, I understand what you guys are saying and you are probably right.

But wouldn't an 8.8 be more efficient than a heavy-duty 9"?
Yes it would. I would expect the 8.8 to be around 92-93% efficient given the efficiency of the other numbers.
 

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I think Hammer just wants a solid rear lol and so do I :D Honesly my gt has the solid axle and i think it rides and drives just as good as the cobra...
I dont think the effiency is gonna be that big of a difference. What I feel would make a difference is the weight loss....If im thinking correctly its like 70 pounds lighter....NICE
 

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9"efficiency......

The 9" loss in efficiency is due to the low placement of the pinion gear in relation to the ring gear.....more "climb" is the term.......the 8.8 has almost exactly the same relationship location wise as the 12 bolt chevy .......hope this helps.....hermann
 

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One common mistake everyone makes is using a percentage figure for driveline loss. This is not correct unless my math is wrong. Its more along the lines of a fixed maximum hp loss similar to driving the engine accessories.

If you have a 390 bhp engine delivering 360 rwhp, its takes approx. 30 hp to drive the tranny. If you double the HP, it doesn't take any additional power to make the drivetrain spin. Its still 30 hp (for the sake of argument). Using the % calc., your driveline loss would continue to increase with increasing HP. Its a fixed figure unless you make a change to the driveline.

John
 

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Maynor: you have a very good point. I think you are right on. Its 30 Hp wether your making 900 hp or 600, it still takes the same horse to turn the tranny/drivetrain. Not a percentage thing I dont think.
Andrew
 

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I posted this in the "new information about the KB tune" thread, but it seems applicable here:

As I see it there are three main components to drivetrain loss with a stick car:
1) Friction
2) Loss thorugh gears
3) enertia of the driveline components.

Of these three, the first one changes little as the hp goes up. the second two appear to me to be approximatly a percentage.

Loss through gears goes up with the power transferred through them. In a modern mustang in 4th, you aren't going through any gears in the transmission (input shaft is coupled to the output shaft), but you do have the loss through the ring and pinion.

While it's true that the moment of enertia of the driveline does not go up with more power, BUT you are accelerating those components faster, which takes more power. i.e. if you double the HP,you shorten the time of your dyno pull in half, but you still have to accelerate all those components from the start speed to the end speed, but in half the time.

Having said all that, I think the '03 stock trans, stock rear has a driveline loss of 12-13%.

JT
 

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JT,

Not quite true. 1, 2 and 3 have a maximum "peak" power value neccassary to drive the compenents. When cold, the peak value may be higher than when warm due to fluid viscosity reduction. eg, when cold, it may take 30 hp to turn the drivetrain versus 27 hp when warm (example only) due to viscosity reduction. You could consider that the PEAK level neccassary to drive the drivetrain.

The drivetrain is just another engine accessory only instead of being hung off the front of the engine with the water pump, alternator and A/C condesor, its hung off the back using a shaft. The A/C condensor and alternator share this same principle. They take XX amount of HP to drive regardless of engine size or HP levels.

If you double you engine power, the peak power values neccassary to power the drivetrain and accessories does not vary. If it did, we'd be rating the alternator, water pump and other compenenents the same way and we don't.


John
 

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Discussion Starter · #15 ·
Good points, guys. Maybe those percentages given to me were at a specific power level and not to be interpreted as a static formula across varying power.

And I was under the impression that the drivetrain efficiency of the 03 Cobra, as a complete system, is 85% (meaning 15% loss from crank to rear wheels).

I do agree with you guys that loss is a constant and the percentage would go down as HP goes up. I guess that the percentage is based upon stock HP numbers.

Let's take a look at these figures. Ford rates the stock 03 Cobra at 390 HP. Given 15% loss through drivetrain, bone stock 03 Cobras would dyno at 332 RWHP. Well, I can't think of anyone that was that low. Does that mean that the drivetrain efficiency is higher than claimed? Or does it mean that the HP rating is off. I believe it's the latter.

I made about 350 RWHP bone stock with a brand new 03 Cobra.

http://mywebpages.comcast.net/richardjuliano/Bone Stock Dyno Run.jpg

If we work the formula in reverse and plug 15% drivetrain loss into my RWHP, we have 403 RWHP. That is probably accurate. And that means that I lose 50 RWHP from drivetrain.

Last dyno pull, my car made about 620 RWHP. If we assume that drivetrain loss is a constant, then the percentage is not applicable. Instead, we add the 50 HP loss from drivetrain and come up with 670 engine ponies.

And that would mean that the drivetrain loss, in this particular application, is about 7% or 93% drivetrain efficiency.

Could that be accurate?

I don't think so. I think that there are other variables not taken into consideration here, for example the holding power of the clutch. There may be no loss with stock power but I have a stock clutch and I am sure that I am losing power with it now through slippage.
 

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Maynor said:
JT,

Not quite true. 1, 2 and 3 have a maximum "peak" power value neccassary to drive the compenents. When cold, the peak value may be higher than when warm due to fluid viscosity reduction. eg, when cold, it may take 30 hp to turn the drivetrain versus 27 hp when warm (example only) due to viscosity reduction. You could consider that the PEAK level neccassary to drive the drivetrain.

The drivetrain is just another engine accessory only instead of being hung off the front of the engine with the water pump, alternator and A/C condesor, its hung off the back using a shaft. The A/C condensor and alternator share this same principle. They take XX amount of HP to drive regardless of engine size or HP levels.

If you double you engine power, the peak power values neccassary to power the drivetrain and accessories does not vary. If it did, we'd be rating the alternator, water pump and other compenenents the same way and we don't.

John
The drive train is not just "another engine accessory". You are not transmitting 480 ft lbs of torque through the alternator or the A/C compressor. You are transmitting it through rotating shafts ,and gear faces of the driveline. When the power goes up, those shafts want to seperate from each other rather than transmit the extra Torque, which causes extra load on the bearings. The teeth on the gears deflect causing extra power to be used.

Using an enertial load dyno, like a dynojet, the biggest factor is the moment of enertia of the drive train. If car A has 200 hp and car B has 400 hp, both using the same drivetrain, a dyno pull with car b will take half as long as the dyno pull with car A. Both engines have to accelerate the transmission input shaft, the countershaft, the output shaft, the driveshaft, the pinion, the differential, two hafshafts and two wheels from the same starting speed to the same ending speed, but car B has to do it in half the time. This takes more power. The power that it takes to accelerate these components in half the time, is not available to accelerate the dyno rollers.

JT
 

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Good explanation JT.

Here's a simple question to ask yourself before you come to the conclusion that driveline losses are fixed amounts....

Why would Ford quote efficiencies for their components in percentages if they WERE fixed amounts? Why wouldn't they then say that the IRS comsumes "X" horsepower and be done with it?

There is much more here than meets the eye.
 
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