Horsepower % lost
 

Does anyone know the exact % lost between the crank and to the wheels.

about %15..........

On a GTO it's more like 27%!!!! :eek:

Blk Boxster's about right. It's in the neighborhood of 15% for a manual and 20% for an automatic. It varies from car to car and a lot of factors play into the final percentage lost.

This 15% number is always thrown around.

Let's think about it for a second. If I have a 100hp car with a 15% drive train loss. That means 15hp is lost to the rotation of components between the flywheel and the tires (friction heat etc), fine. Now I take the same car and drop in a 300hp engine, by this flat 15% loss method the components between the flywheel somehow now consume 45hp. Doesn't make much sense does it. I am open to suggestions as to how this is but it seems to me that the parasitic loss from the flywheel should be constant for a particular car and not a function of the amount of horsepower being produced.

My take is that the trend of comparing runs on different dynos is worthless, also trying to extract accurate flywheel hp numbers from chassis dyno numbers is also a pointless pursuit.

Todd

The drivetrain you need to handle 300hp (really the torque is the determinant, but I'll go with hp since that is your example) will have about 3 times higher loss (versus the 100hp drivetrain) due to higher rotational inertia (heavier gears) and more friction (larger surface area on gears and bearings). Will it be a perfectly linear 15% at all hp/torque levels? No, but it is a pretty good rule of thumb that matches nicely with real-world data.

Lucky, I follow your example, and would agree, but in Tholyoak's example the car is the same (without a beefed-up tranny). So he does pose an interesting question to ponder.

Guess I should read better... :cool:

Even so, I believe this is easily explained by the following High School Physics formula for the friction force observed between sliding surfaces:

F = mu * Fn

Where:
F = the force due to friction (loss in a tranny, for example)
mu = the coefficient of friction between two moving surfaces
Fn (as in "normal force") = the force applied perpendicular to the two moving surfaces

The key parameter above is "Fn", which inside a tranny occurs where gear surfaces touch and shafts push sideways against bearings (to counteract the torque that is spinning the shaft). As this "normal" force goes up, friction loss (F) goes up in a linear manner. So yes, the climbing loss as more power is put through a tranny makes perfect sense.

Yes, rotational inertia remains the same, but as friction loss climbs the inertia becomes less of a factor.

Yup, the amount of friction is directly proportional to the amount of force. Rotational inertia is a very minor contributor to drivetrain losses.

If that small engine in the example is at full throttle making all of it's 100hp and the bigger engine is at part throttle, making the same 100hp, the losses in the system will be the same, but add more throttle to the big engine and friction goes up in the pistons, crank, gears, axle joints and tires.

Yes the frictional loss will increase, but do you really think it would account for a 30 hp loss in the two examples, I for one don't think so, as the amount of heat generated would be pretty extreme and lead to failure of parts in pretty short order.

Anyway, I think trying to extract flywheel numbers from rwhp numbers is pointless.

Todd

Yes, it will increase in direct proportion of the power (actually torque, but whatever) increase.

Believe it. That is exactly why there are trannies desiged for 300 HP and different ones for 100 HP.

Every system in the transmission is built with what's called a factor of safety, which means they are overdesigned for purposes of functionality under extreme conditions, accounting for any manufacturing variations/defects, accounting for any misjudgement in engineering or inaccuracy in calculation, and longevity. When a more powerful engine is placed in a car that wasn't designed to accept it, this margin for safety is reduced, but it still may be enough for the car to work, at least for a while.

So that's why I go through so many AODs. ;)

This is what I think Tholyoak is saying:

Let's say you have a 986 Boxster with 240 HP. If it looses 15% HP (or 36 hp) you have about 204 HP at the wheels.

Now you do a engine swap with a 3.6L engine that has maybe 330 HP. It will be in the same car with the same exact tranny, tires, everything. So, how much HP to the wheels? 330- 36= 294 HP or 330-15% (50 hp) = 280??

Like Lucky and I said, 15%.

I think 280, but of course, I didn't finish engineering school. I went a different route.

No, it's your right foot that's to blame. :D

OK well I really didn't believe that in transplanting my 2.5L engine for my 3.6L X51 motor that it takes 15 more hp to get the power from the crank to the wheels.

I think I found the answer. It turns out that the 15% loss has to do with the rate of acceleration of the large drums on an inertial dyno. On a load control dyno, you assume a constant amount of parasitic drive train loss.

A good explanation can be found here for those that are interested.

http://home.earthlink.net/~spchurch/churchautomotivetesting/id12.html

Todd

That's accounted for and subtracted out when you get the dyno sheet so it's not part of the loss we're talking about. You're stuck with 15% man.