20% Drivetrain loss doesn't make sense to my brain.

I know that somewhere between 15-20% drivetrain loss in the general rule of thumb. But that just doesn't make any sense to me.

Take, for example, any car you choose. Just for grins, let's make it a 4 speed with a 9" rearend.

Put a flathead in it that makes exactly 100 hp at the crank. Assume the standard 20% and you get 80 hp at the rear wheels with a 20hp loss.

Now, take that same car and put in a blown big block fire breather that makes 1000 hp at the crank. Now, all of a sudden, that 4 speed and 9" is losing 200 hp. to the back wheels?

 

That's why it's a rule of thumb and not an actual measurement. The only real way to tell is have the engine dyno'd then after it's in do a chassis dyno run. But you'll be surprised it'll still take a decent piece of power to turn that stuff. It does seem that the percentage tends to go down as the power goes up, but not as much as you'd think. Just a few percent in my experience.

 

Good question.

I'm certainly no expert, but I wouldn't figure the horsepower loss would be linear. There is loss of horsepower due to all the mechanical interaction of the various seals, bearings, gears, torque convertors, etc. that exert load that has to be overcome. Wouldn't that loss of power through the transmission and rear end remain constant? To my way of thinking, if the loss is 20 horsepower (just for a number), the loss would remain more or less constant or at least increase at a much lower percentage if nothing else is changed.

Am I screwed up? How about it engine builders and dyno-guys, set us both straight ?????

 

I think part of the reason that the 15-20% number works well is because if parts are sized correctly, the loss will be about the same. What I mean by this is, if you've got a flathead you can run a C4 without risk of blowing it up which is pretty efficient. if you've got a blown big block in there, you have to have a much bigger beefier trans. that takes much more hp to spin.

 

seems crazy, doesn't it. But it is true that you lose a considerable amount of power through the drive train. Every time you have to transfer power through gears or change the direction of power there is a loss.

Even driving a V-belt pulley over alternator or water pump can take up to 2-3 HP, not the load of the item just the pulley! Then factor in power loss through the bearings and from friction it's not hard to come up with 10-20% loss. This is why it is so important to get drive line angles correct, proper lubrication and adjustments.

 

Yes it seems the stronger you make something, i.e. the tranny or rear, the more power it consumes. There are differences in the power consumption of different rears for instance even though they are both built strong. A 12 bolt takes less power than a 9" mainly due to the location of the pinion.

 

C4 uses less than a FMX & that uses less than a C6. i suspect a 9" uses more than an 8" rear, so on & so on.

 

old HP rating with no tranny, nothing hanging on the front, vs 72 & up.

 

exactly even a banjo rear uses less power.

 

i always thought that 15-20% power loss was for automatics that manuals were more like 10-15%......


but if you think about all the "stuff" your engine has to turn before it evens starts to spin those rear wheels is what causes the power loss, if you think about like electronics the more "things" you have in a single circuit the more resistances there will be to power flow...

 

I've read somewhere that a 9" uses more power than the 8.8 Ford or 12 bolt Chevy.
They said it was from pinion altitude in relation to the ring gear centerline. Lower on a 9".

 

The 9" is about the most extreme offset pinion rear around...Banjo is efficient because of no offset, straight cut gears would be even more efficient but would lose the strength that is a byproduct of greater and smmother tooth contact in the spiral cut rears! I think a factor in the equation is that gear LOAD, that is more power going through, presses the gears together harder and so loss goes up with power, though probably not in a neat linear way.
Will post Ford engineering rules-of-thumb on power loss in gear train from the 1930's...

 

The total drivetrain losses will go down as a percentage as the power being used increases, as some of the friction sources are not affected. Much of the loss increases directly with applied power, however, as friction increases with force.

Using a non-drivetrain example (because I already typed it up): During the power stroke a piston in an engine making 500 horsepower is pushing much harder (5x) against the crankshaft than the same piston when that engine is making 100 horsepower (at the same rpm), so there is more friction (5x) between the rod bearing and the crank. The crankshaft is also pushed harder against the main bearings, so there is more friction there (and so on). The friction increase for the same piston during the other strokes will be less, as the load increase during those strokes is also less.

Similarly, friction between the ring & pinion in the rear increases with the applied power. Sliding friction increases linearly with the force applied, so the amount of friction under 5 times the load will be 5 times as much. The percentage stays the same.

The same rules would apply throughout the drivetrain. More load means more friction, and for sliding friction the friction increase is linear with the increase in the load.

 

Rather than using the losses as the metric for this it makes more sense to use the efficiency. Simple conversion, 20% loss is 80% efficient.

Efficiency is just power out divided by power in. (400HP out)/(500HP in) = 80%

Here is a headache for you: If you jack up the rear end and let the tires spin, the power out is zero, and your drivetrain is therefor 0% efficient. Put a slight load on the tires and it becomes 70% or better.

It is important to realize that the power loss is dependent on the amount power transmitted, and that yes, putting 1000HP through the system will produce a higher loss than 100HP. This is due largely to the higher force, which means higher friction, as the gears slide over one another and the bearings support larger forces with more friction.

 

DIN vs SAE, yeah but for the most part when dyno-ing an engine the transmission and accessories are not installed. In fact I cannot recall seeing a modern engine dyno that uses a trans.

Bruce, you're spot on as always. The "hypoid" gearset is designed to be quiet and to be stronger because of the the increased gear tooth contact and fit into modern chassis designs, low floor tunnel and low ride height.

 

I read a fantastic article about this, recently.
I'll try to remember where it was, but it was pointing out exactly the same thing you mentioned...the percentage of power needed to accomplish a certain task, through a certain system configuration, should not automatically go up, because more power is applied.
Something that takes 20 horsepower to turn...takes 20 horsepower to turn.
It was a good read, and it really, quite well ( and in much better terms than my non-scientific ass can muster...) tore the "rule of thumb" a new one...
I'll try to find the article...It makes no sense to my brain either...
Also, I would think that dependent upon the ease in which the higher power input can overcome friction and get it turning, the losses ( percent-wise ) would drop off faster, as well, than say, that 100 horse Flattie...

 

Thats about what I would think...definately increased load, on gears, bearings, thrust surfaces...etc, but I don't see it being linear, either...
But, ???

 

Think of HP as a force. If you put a 50 pound kid on a sled on the ground and pull him. Now put a 300 pound guy on the same sled and pull across the same surface, do you really think you will use the same forces to pull?

It is the same for HP losses through gears and bearings. The more HP being transferred the more losses you will observe due to frictional forces. If you want to overcome the losses you need focus on eliminating the friction through different lubrication or remove the components that are consuming the HP.

 

So weight (static) and exerted force are equivalent???
Not being sarcastic...I'm actually asking...
AND, once that fat kid gets a-rolling, won't he tend to want to stay in motion, thus lessening the exerted force needed?
Someone blind me with science...

 

A thousand horse engine will by nature have more gear than the 100 hp package ,try just pushing a car in netural that has more gear and tire ,then push a low hp car with a proper gear package and the todler will move much easyer . if both cars are set up corectly then both would see about the same loss in drive line ,not exact but close .c

 

Thats what I was thinking about, as well...the gear reduction, or "work advantage"...


Too many variables for a "rule of thumb", it seems...

 

Some Ford numbers, from an intro engineering textbook put out by Ford England in 1935:
Gear box, in HIGH: 2% or less
In lower gears...8% for each pair of gears in action
speedo, u-joint, minor junk 1%
Rear end...assume banjo...10%

Remember each % is applied to total power arriving at that point after losing power at previous stations.

I have not followed through this math, but thet seem to say that in high, wheel power is ~87.32%, in indirect gears ~83.79%(should be 75.4% in lower gears...sorry)

Head hurts. Add in own guesstimates for QC spur gears, torque converter, 97 seals in auto trans, etc.

 

Please do. Because it's wrong.

As an engine's output increases, the load from all components driven by it increase as well. That's because of their inherent resistance to turn.

Remember that all power measured from an engine is based on TIME. A 3cyl Geo engine will put out more power than any top fuel dragster engine. But of course it will take years to do it.

When you have your car on a dyno, you are measuring power based on output over a period of time. If you want those gears, pulleys, belts, bearings, and tires to turn over sooner, you have to feed them more power. This is the power that is robbed from your rear wheel power measurement.

You are right that "something that takes 20 horsepower to turn...takes 20 horsepower to turn" if you are talking about a steady speed. The power it takes to get the mass of that "something" accelerated - the force that is measured on a dyno - will vary greatly depending on how fast it happens. The faster the drivetrain is accelerated, the more force the drivetrain eats up.

 

I think we tend to get confused over the terms in Newtonian mechanics. If you want to learn the real complete answers its not going to happen in a few posts on the forum. These are not trivial problems and the answers are not simple. It is, however, understandable with some reading from the right places. Wikipedia pages on basic physics are not the best, but they are a start.

Force, F, measured in Newtons (N) is the quantity that makes mass change acceleration, as in F=MA.
http://en.wikipedia.org/wiki/Force
http://en.wikipedia.org/wiki/Newtons

Energy, E, is a measure of work, which is measured in Newton*Meters, (N*m), sometimes called a Joule (J).
http://en.wikipedia.org/wiki/Energy
http://en.wikipedia.org/wiki/Joule

Power is a measurement of work over time, often given in Joules/Seconds, or Watts (W). There are 745.7 Watts in one horsepower.
http://en.wikipedia.org/wiki/Power_(physics)
http://en.wikipedia.org/wiki/Horsepower

These things work just the way the units and math does, where an engine making 1HP makes 745.7 Watts of power. Over 1 second the motor made 745.7 Joules of energy, which is the same as 1 Newtons of force pushing over 745.7 meters, or 745.7 Newtons over 1 meter.

Imagine our 1HP engine spinning at 1000 RPM. If it had a wheel on it with a radius of 1 meter, the edge would be moving at 104.7 meters per second (104.7 m/s). Dividing the 745.7 watts by 104.7 meters per second gives 7.12 Newtons at the edge of the wheel. This is the way dynomometer work, by measuring the force at some radius from the centerline and the rotational speed, and calculating the power from this.

I could go on, and this stuff is awesome to learn, and it really works. I can only hope to kind of get some curiosity going, to help connect terms like torque, power, and those numbers written on lightbulbs and the boxes on stereo equipment.

 

This math is making my head hurt. Once you get thru all the numbers then look at the lubrication factor of each lube you use.

Do a coast down test with petroleum base verses synthetic. This will add another factor to the equation. It makes a difference worth looking at when you want to get all you can out of a vehicle.

 

Both the chassis dyno and the engine dyno interpet (do not directly measure from zero) the HP measurement. As mentioned, they do this by loading and measuring acceleration over time. Even tire pressure can affect a chassis dyno, so while they try to make them accurate, the main thing is to to make them repeatable so changes can be quantified.
Another way to hurt your head is something brought up in a similar discussion: Where does that 20 percent go? If you are losing 200 HP in the drivetrain, why doesn't the transmission and rear get very hot?

 

I'll try to find it, I'm sure 'ol google should be able to help...
It WAS, in any case, just more fuel for a very interesting fire.
I'm no math/physics whizz, so my head hurts faster than some of yours...
If I can locate it, I'll post the link...

 

They certainly DO get pretty hot, though...that power is lost through the trans. cooler, and the lube in the rear, no? As well as through the housing?
I guess it makes sense, to, then, that by the time it gets to the rear, many of the losses are already accounted for, and perhaps thats why the rear doesn't quite need the cooling care of the trans...Also, I imagine its general shape and size help to dissipate heat better on its own than the confinement of a transmission...
Gonna have to do some physics reading...although I hear scientists arguing all the time too, and they know all this math...

 

Not to be off topic but a snowmobile uses almost 50% of it's HP to make the clutch work properly & turn the track....food for thought.