That's a good point. The actual HP that is required to operate the drivetrain should remain fairly constant with rpms. So the "percentage" of loss should be a little lower as the crank HP is increased. 20% loss is average for a stick shift and 30% is average for a low powered stock auto. As the engine combo gets more powerful the % of drivetrain loss will be less. But don't take my word for it, put it on a chassis dyno and see for yourself.

_________________
1966 Mustang, Custom glass hood, 331 stroker, 5.4 H-beams, Probe pistons, TrickFlow heads, ported Stealth 8020, XER solid roller, Holley 750 HP, Hedmans, 4spd, 9"/3.50, BFG DRs

"12.10 @ 114 mph on the street with Grandma driving!"

<font size=-1>[ This Message was edited by: bluestreek on 4/6/03 12:08pm ]</font>

 

20% to 30% is a little large. The Nova I'm working on now has a basically stock powerglide. It made about 100 less on a chassis dyno than on an engine dyno. I would figure about 8 to 10% on a stick & about 10-15% on an auto.

 

Are you sure? I've seen stock 5.0 Mustangs dyno at around 170-180 HP at the rear wheels. That seems like more than a 10-15% loss. Like I said in my earlier post, the percentage number should go down as the engine HP increases so your numbers may be correct for higher powered combos.

 

It's hard to say. The last place we went to was a little weird. He put a strip on the damper for an accurate tach reading. His reading was 5500 max rmp/hp but the tach read 7800. I've since re-wired the car & went to a different shop & got something all together different. Basically in my opinion, dyno numbers are just a small part of the equation, it's still only going to do what it does. Also an engine dyno uses the shops fuel, ignition, & exhaust systems so those numbers are just a ball park number for tuning. I just do mine the old fashoned way, race it like I stole it

 

I guess technically all that matters is rear-wheel, because that's what sees the road. Just curious. Thanks guys.

 

A constant percentage should be what happens. That percentage is determined by every moving interface that the power is transmitted through. For example, if the bevel gears in your diff are 98% effecient than no matter if you send 100 or 1000 hp through them you're only getting 98% out(98hp or 980hp in this case). What you do is multiply all the efficiencies together to get the overall net.

Trans input shaft -> output shaft (could vary depending on gear selected)

output shaft -> driveshaft (universal joint eff)

driveshaft -> diff input (universal joint eff)

diff input -> axles (bevel gear eff)

And then there are all the losses associated with seals, bearings and friction, which are also proportional. I think the bulk of it is all the gear to gear interfaces. So just using complete guesses for numbers this might be 95x99x99x95x99 etc and you get (in this made up example) 87% eff, a 13% loss. There also might be some items which are more static than proportional also.

That's what the textbooks say at least, who knows how close real life is to this.

Eric

 

Really interesting. I hadn't thought about it that way, but it makes sense from a scientific standpoint.

Thanks alot for your input.

 

If you say percentage then understand it only applies to *that* exact combination.
The parasitic losses are constant for a given combination.

Thus -- you make more FWHP you get more RWHP. End of story.

I have engine dyno'd and rw dyno'd multiple combinations. The numbers are not percent but fixed.

 

Thats the Joey!!!!

 

Hmmm...is it so that it takes a certain/same amount of power to rotate moving parts and to overcome friction? I mean it takes a certain amount of power to rotate, say a 9" rearend at 6000 rpm.But, the needed power is only certain/same without a load.When the load increases so does the friction.And the load increases with power.The harder the car accelerates the more friction is created in the rearend and thus more power is wasted. Without the load the rearend does not care whether the power comes from a 100 hp 4 cylinder or a 400 hp V8. If it takes 40 horses, that is 40 % from the 4 cyl output( 60 hp still available from rear wheels) but only 10 % from the V8 output( still 360 from rear wheels).With a load more power means more friction/heat buildup between the gears and in bearings more power loss/ heat build up.
The simple(?) conclusion is that power loss through the drivetrain is a combination of stationary loss of the rearend / trans + proportional to the power the input shaft of the trans rearend will see.

This is my humble theory...does it match with real world experiences from dynos?

 

..perhaps my theory will explain the different values quoted here: anything from 8% to 30+ % .
Cannot imagine it would take 300 hp to drive a C6 behind a 600 stroker ( 1000 flywheel hp and 700 rearwheel).That would be 30 % loss.
But, if it was physically possible to bolt a C6 behind a 100 hp 4 cylinder, it would be easy to imagine it would take 30 hp ( =30 %) to drive the C6.In that case 30% loss would sound reasonable.
Reverse that example : Lets imagine it would be possible to run the gearbox from the 100 hp 4 cylinder behind the 1000 hp bigblock.
If that case the loss would be very small, perhaps something like 1-2 %.
...and if you bolted that 4 cylinder to a Sherman tank the drivetrain loss would be 100 % since that motor could not move the tank at all

 

Good examples!!

Makes much better sense now.

 

On 2003-04-03 14:45, ville wrote:
Hmmm...is it so that it takes a certain/same amount of power to rotate moving parts and to overcome friction? I mean it takes a certain amount of power to rotate, say a 9" rearend at 6000 rpm.But, the needed power is only certain/same without a load.When the load increases so does the friction.And the load increases with power.The harder the car accelerates the more friction is created in the rearend and thus more power is wasted. Without the load the rearend does not care whether the power comes from a 100 hp 4 cylinder or a 400 hp V8. If it takes 40 horses, that is 40 % from the 4 cyl output( 60 hp still available from rear wheels) but only 10 % from the V8 output( still 360 from rear wheels).With a load more power means more friction/heat buildup between the gears and in bearings more power loss/ heat build up.
The simple(?) conclusion is that power loss through the drivetrain is a combination of stationary loss of the rearend / trans + proportional to the power the input shaft of the trans rearend will see.

This is my humble theory...does it match with real world experiences from dynos?

Sounds logical to me. Very well thought out. I guess unless someone has actually done a before/after comparison, it is pretty much a best-guess...

 

Some parts are constants (like seals, ring gear turning in oil etc.) and most are proportional to the load (gears etc).
If you where to have car with 500 flywheel/400 rear wheel hp, then it wouldn't be possible for a 100 hp 2 liter engine to move it an inch if it were a constant. A 302 from the seventies would have only 30 hp left at the wheels.... That is of course not the case.

 

It is my impression drivetrain losses are relatively "fixed" as some of the posts suggest.

I know of a very well funded team with reources sufficient to do the appropriate testing and I believe the differences between crank and wheel HP are well under 10% for them.

However, they deal with relatively high HP and very finely tuned components. I think you could interporlate that losses for a 400hp setup would be 15%-20% based on these results.

 

I believe the percent losses for a stick car are going to be far more constant than that of an automatic car. The losses associated with an automatic car can change greatly due to the torque converter. The stall of the torque converter alone can change the drivetrain efficiency. This is why late model vehicles come with a lock-up converter. The locking action of the converter makes the drivetrain more efficient leading to better gas mileage/more power to the ground.

 

Here is a very good website that really explains the difference between flywheet power and what gets to the rear wheels. There is an article on there from I think car craft and they took a 351 winsor that was just mildly built and put it on the engine dyno then the put it in the car. On the engine dyno it made 371hp. Then when they put it in the car with an AOD tranny and ford 9inch rear it lost 24%. The rear end makes a difference too. An 8inch will not soak up as much hp that I 9inch will. Also the AOD is most inefficient of all the automatics. I have seen time and time again stick cars lose about 15 to 17 percent and automatics always lose at least 20 percent or more. This is just the brutal truth. Here is the website that really explains it all.

http://www.superstang.com/horsepower.htm#Proof1


_________________
1968 protouring Mustang 342 stroker TFS heads, soon to be 185cc CNC ported TEA Trick Flows

AOD/3000 stall 4.11 9inch posi.
333rwhp 333 rwtq naturally aspirated
431rwhp 463 rwtq on Nitrous

<font size=-1>[ This Message was edited by: EricButler on 8/3/05 10:46am ]</font>

 

I have found 20-30 percent is lost on a dyno. My Mustang made 480hp on a dyno and 388hp on a chassis dyno.

 

On 2005-08-05 14:53, 472Mach1 wrote:
I have found 20-30 percent is lost on a dyno. My Mustang made 480hp on a dyno and 388hp on a chassis dyno.

It's not correct to state that as a percentage. You probably had power steering and a fan on the engine during the RWHP test that you did not have on during the engine test.

Regardless...if you were to bump the horsepower production of your motor up your transmission is unable to resist *more* as a percentage indicates. It's lossy, yes, but fixed.

One of my engines made 455 FWHP and 392 RWHP with a C6 and the addition of the fan and P/S. With a 125 Shot, well, it made 125 more HP not 70% of 125HP.

 

I'm thinking that losses are fairly constant in a given situation, but that is not to say that putting 1000 HP through the same transmission as 100HP would give you the same loss.... The 100 HP engine would indeed move the car because the 100 HP isn't gathering the amount of friction as the first because a.) it isn't accelerating the rotational assembly nearly as fast, and b.) the force generated on the meshing surfaces isn't nearly as high as a higher HP motor - the fluid can lubricate much better and you are losing much less due to metal contact.

I think that once the HP levels reach a certain point, the transmission becomes "saturated" (you're pushing all the fluid out that you can and all the metal that will touch is touching. Once you get there, in higher HP motors, you lose about a given amount of power, cause you've got the transmission losing the most that it can.

Now, in an auto tranny, you lose more because you have a torque convertor, as has been said. You have to rely on a fluid to transmit force, and since the input fins on the engine side are moving much faster than that of the transmission side, you're going to generate heat simply due to friction - any heat generated is a direct loss of power. That, in addition to the fact that the sun and planetary gears on an auto generate a lot more friction than their manual counterparts.

I'm curious to see if someone could study the loss due to friction on the main bearings as a function of horsepower.

<font size=-1>[ This Message was edited by: thekingofazle on 8/23/05 2:40am ]</font>

 

The loss is a combination of frictional loss, efficiency losses, and inertia loss needed to accelerate rotational masses including the trans and rear end.

I like to think of things in extremes to help to understand them. Imagine a 4 ft diameter 250 lb flywheel. The amount of inertia required to accelerate this mass is going to be significant. The faster the flywheel is accelerated, the more power is needed to rotate the flywheel. The same principal holds for rotating the torque converter or flywheel/clutch assy, trans internals, driveshaft, ring and pinion, axles, and wheels/tires. So you end up with frictional heat losses, gear efficiency loss, and inertia loss. Inertia loss is one reason why a 9" uses more power than an 8". 100 HP loss through a drivetrain is not due to just friction. If it was, the heat generated would be tremendous.

A friend who has a built 289 with a toploader and built 8" behind it lost 15.5% from flywheel to chassis. 375 FWHP and 317 RWHP.

_________________
Tracy Blackford: Corona, Ca
'65 FB Mustang 331 with H-beam 289 rods, KB 322's, 282S cam, fully preped 351W heads. Built C4 and 3.50 9" posi

<font size=-1>[ This Message was edited by: blkfrd on 10/19/05 11:31pm ]</font>

 

FWIW, the below numbers are an example the differences of an engine dyno and a chassis dyno on the same engine.

Engine dyno:
415HP@6200RPM
408TQ@4500RPM


Chassis dyno:
323HP@6000RPM
299TQ@4750RPM

I'm not sure of the make of the engine dyno, but the chassis dyno was a Mustang Dyno which will typically show lower numbers than say a Dynojet chassis dyno. You can count on about 5-10% in most cases I've seen.

This particular 331 stroker engine was installed in a 1966 Shelby Mustang with a manual transmission.