I have seen many estimates of hp loss through the transmission for a chassis dyno. The type of estimate I have the most trouble with is estimating the loss by percentage.
If I take for instance my 77 L-48 with it's TH350. Some claim a 25% loss through the trans would not be unreasonable.
My 77 came stock with 180hp installed. This would mean on a chassis dyno I would expect to see a 45 hp loss, or 135 hp RWHP.
Now take this 180hp motor and build it up to a 450hp 383. If the 25% rule is followed I would now have a 112.5 hp loss through that same TH350 giving me 337.5 RWHP. Nothing else changed so how can the same trans still use 25% of the available hp? To keep the variables limited we'll assume that both hp levels were reached at the same rpm.
Wouldn't it be more accurate to estimate loss through the transmission with a fixed hp loss?
I found some numbers for various transmissions. Can anybody confirm or deny these numbers.
None of this factors in the type or stall of the installed torque converter which I'm sure also has to factor into the RWHP results as well.
GM
PG 18 hp
TH-350 36 hp
TH-400 44 hp

FORD
C 6 55-60 hp
C 4 28 hp
FMX 25 hp

Chrysler
A904 25 hp
727 45 hp

 

The percentage calculation is more near the truth than an absolute value.
If you say your TH350 has always a drop of 45HP - well then there would nothing come out if you had a 45HP engine. But we all know thats not true, you can turn it by hand, and the output shaft moves.

Maybe it´s not a streight line, but again: percentage is the most accurate.

 

As stated above, percentage is a more accurate way to go. But that could change as well depending on the engine out front. The more horsepower you have in the engine - the more drag the transmission/drivetrain/exhaust creates.

I have a 70 L-46 that I have engine dynoed and chassis dynoed before and after a rebuild. The engine was detuned in a prior rebuild in the 80's with flat top pistons and a slug of a cam. It was rebuilt back to "stock" L-46 configuration, but with ported heads, a 10.4:1 CR and a slighly bigger cam. It looks like a bone stock L-46 from the outside.The car was run on the same engine and chassis dynos, the drivetrain remained the same and untouched - M21/3.70:1, same with the exhaust -stock exhaust manifolds, 2.5 exhaust pipes, 2.0 mufflers.

Pre-rebuild
285hp@4900/336tq@3400 Gross HP on engine dyno.
217hp@5000/282tq@3250 RWHP on chassis dyno. This was a 23.85% loss.

Post rebuild
341hp@5600/387tq@3400 Gross HP on engine dyno.
244hp@5000/304tq@3600 RWHP on a chassis dyno. This was a 28.44% loss.

The stock exhaust manifolds are one of the the limiting factors. You can only push so much air thru them. If I had built a 1000hp engine, the percentage of loss would be even higher.

 

Exactly how does that happen?

 

The faster you try to accelerate it the harded it is to turn. Like pushing a car. Its not that hard to move it but when you try to push it quickly its a lot harder. Turning the output shaft of a transmission is easy buy when you try and spin it quickly it is much harder.

 

Physics.

 

That doesn't mean that the transmission is consuming more hp. In order for the trans to consume more power, friction would need to increase, and it doesn't.

 

Is that the best you got?

 

F=MA
The more powerful engine is accelerating faster, more force is required.
Also, the gears accelerating faster through the trans fluid will create more friction and generate heat and heat is energy

Dont ask me to explain that any more in depth, thats the best I got

 

Yes, physics is the best answer. http://en.wikipedia.org/wiki/Drag_(physics)

Why don't you explain how friction doesn't increase?

 

How do you explain something that doesn't happen?

 

Well, you can't argue with science like that. You've out smarted physics. Drag/friction won't increase.

Now that you've proven drag/friction won't increase, could you provide us (and the OP) with the tranmission loss numbers as requested?

While you're at it could you explain why my rebuilt (more powerful) engine lost a larger percentage of HP using the same drivetrain & exhaust?

 

I can see how you computed your loss here but I think you're failing to factor in RPM. Your 341HP was reached at 5600 RPM on the engine dyno. Your 244 HP was reached at 5000 RPM. So not a direct comparison exactly.
If I was to guess it looks to me that the losses in that engine were due to a restrictive exhaust system. your peak HP should have been the same RPM whether in chassis or not, shouldn't it? if not I would suspect something in the installation is not correct.
Often the way to prove or disprove a theory is to take it to the extreme.
If I install a 50 HP motor then 25% would only be 12.5 HP to spin the drive train. So we now are assuming that this it is even possible to spin the drive train to 5500 RPM with 50 HP, because after all as the HP goes down the transmission consumes less power according to what I'm hearing.
Now let's install 1000 HP in front of that trans and suddenly it consumes 250 HP? That would be 5 times the HP the previous engine was even capable of. Yet both can spin this drive train to 5500 RPM?
This is where this method of computation will not hold up.
I find it easier to swallow that at X rpm with all other factors being equal that X drive train will consume X amount of power. How can it be any different?
If you could hook up a 10 HP engine to a corvette drive train it wouldn't likely even be able to turn it 500 RPM, much less 5500.

I noticed your engine dynos were gross HP. So none of the accessories were hooked up? That could explain a percentage of your losses as well.

 

All I can say is good luck with your search.

 

That article on drag does not apply in this situation. I can see what you're trying to illustrate but we are not changing the speed of any operating component here just the power being applied to said component.
If what I'm hearing we're true and I was to apply it to that Wikipedia example then a Cessna 150 with the standard 100 HP would cruise at the same speed as one with 200 HP because the drag increased due to the increase in HP. Obviously that is not true. The 200 Hp Cessna would cruise faster, although not anywhere close to twice as fast since drag increases with the square of the speed. Ie twice the speed is equal to four times the drag.
What else you got?

 

The more power you transmit through a set of gears, the higher the absolute losses. It's not a perfectly stable percentage, but fairly close to it.

That's why you can spin a transmission input shaft by hand all day and the gear oil never gets warm. Yet make it move the car and it gets plenty hot.

If you do not understand/accept this part of basic physics, I don't think there's anything else we can do you.

 

Any gearbox (automotive or industrial) has a power loss (or "driveline loss"). Key word here is power, as in HP, not torque. As horsepower is increased, torque or rpm has to increase. An increase in torque will increase the normal load on the gear teeth. An increase in rpm will increase the sliding velocity on the gear teeth. Both of these results in heat generation and power loss. Therefore, the more HP an engine makes, the more HP the gearbox and differential will loose. That's why the "driveline" loss stays approximately at a constant percent (not HP) . There are other factors that will affect this power loss like gear teeth design, type of oil, the amount of oil, types of bearings, etc.

Think of it this way. You can rotate a transmission by hand since your producing very little torque at a very slow speed (i.e. almost no HP). If the transmission took say a constant 20 HP to rotate, an avg. human wouldn't be able to rotate the gears. When in fact, its quite easy to rotate a transmission.

Oh by the way a side note on the dyno plot I presented above. The transmission losses could have been a bit higher than normal because I wasn't in high gear with the transmission in a 1:1 gear ratio as is typically used for dyno runs. I was only in 4th gear which has a 1.24 ratio. This means that the input and output shafts were not locked together and the power was going thru an additional set of gears.This is how it was explained to me

 

I provided data for the same engine (in two different configurations) in the same car. Following your logic the percentage of loss should have been the same. It wasn't.

Do your own experimant and provide us with your data. Again, good luck with your search.

 

If you could spin it by hand to 5500 RPM it would get plenty hot. What you're talking about is friction due to speed and applied force to get to a certain speed. Since you cannot apply enough force to get it to 5500 RPM then your not applying the same force and will not see same heat of friction or losses due to lack of sufficient force to get it to 5500 RPM.
I think what you're trying to compare it to is the rate of acceleration and therefore the additional force applied to the rotating components and additional friction that force and rate of acceleration will cause.
That is not what I'm talking about here. The rate of acceleration is not relavent nor is the applied force and additional friction due to those two. Constant speed, 5500 RPM, irregardless of HP being developed to create that RPM, why would the consumed HP increase simply because the HP of the driving component increased?
If you understand, it then explain it.

 

The percentage lost in the drivetrain is measured at full power. The more power applied to the gearing, the more the gears/shafts will 'twist and bend' which is this lost power we are speaking of that makes those gears/shafts twist and bend. At idle, it would be 0% because there is no load applied. At 5 mph it can't be 45HP, using zuendler's example, because the car only requires maybe 2 HP to maintain a speed of 5 mph. So, from idle to max power can be anywhere in between - 0-~25% as an example. My .02.