RWHP vs. BHP, emails w/ Harold Bettes, Superflow VP 2

[RGRv6guy]

I'd like further comments from the experts on this, then I
will post some dynosheets and get your input on a related
question.

**************************************************************************
> > *****
> > mailto: rgrcfm@hotmail.com (OLD Email, do not use )
> > Date: 5/8/2001
> > Time: 10:49:58 PM
> >
> > question:
> >
> > Dear Ask Harold:
> > First of all, thanks for this service! I hope to get an answer to a
> > long-time question that I have had about Brake Horsepower and Rear Wheel
> > Horsepower. I have maintained for a long time that when someone modifies a
> > car, and adds a large amount of RWHP they always cite the so-called 20%
> > rule as if getting an 80 HP gain at the rear wheels is automatically a 100
> > BHP gain. I seriously have my doubts, especially if the gain occurs at the
> > same HP peak as before (which few do) but for example, if I use an
> > underdrive pulley setup and gain 8 RWHP at the same HP peak as before, the
> > "20% RULE" would say
> > I added 10 HP at the crank. I do not see a driveline sapping out an extra
> > 2 HP just because an engine is producing a few more HP.
> > Perhaps that is not the best example, but one that is commonly used. I
> > submit to you, that a transmission and rear end will have the same amount
> > of drag or torque requirement to turn a given RPM regardless of the amount
> > of torque that is put in, at least within the design limits of the pieces
> > in question. I believe the
> > "20% Rule" is basically an OEM benchmark for
> > efficient driveline losses for overall fuel economy and such. It seems to
> > be pretty close on most stock setups. Or am I all wet, and the "20% Rule"
> > is an absolute the way nearly ALL car magazines use it??? I have only seen
> > 1 article that claimed what I am saying, and that was many years ago.
> > Thank you for your time!

His Response:
>From: harold bettes <hbettes@superflow.com>
>To: "'rgrcfm@hotmail.com'" <rgrcfm@hotmail.com>
>CC: customer service <servicedept@superflow.com>
>Subject: RE: Data posted to form 1 of

http://www.superflow.com/support/ask

harold.html
>Date: Mon, 14 May 2001 17:03:23 -0600
>
>You are spot on with your assessment that a strict percentage loss is not a
>good way to evaluate drivetrain losses.
>
>Indeed, when a gearbox or drive axle is subjected to more power, there is
>little to indicate that the losses go up in percentage with higher power
>application. There is much more evidence that the losses in power remain
>fixed.
>
>For an example, we tested an engine on the engine dynamometer (brake power)
>and then tested it when it was installed in the racecar where it normally
>resides.
>The losses were 27% in this example (202hp)! The racecar drivetrain used a
>Ford 9" and a Power Glide trans (with a high stall - 5100rpm- converter).
>Other racecars that we tested on the AutoDyn product varied from 12% to 15%
>losses in their drivetrains. We saw that the loss associated with using 3d
>gear Vs high gear (4th) was about 12 to 20 horsepower. In over the road
>tractors (trucks) the power loss through the drive train is from 50 to 60
>hp, regardless if it is being powered by a 400hp or 750hp diesel engine.
>
>So, the percentage loss method is fundamentally flawed. Each drivetrain has
>a specific loss in hp. Now, having said that, it is necessary to consider
>that at some level of power increase that gears bend more, bearings are
>loaded more, and so forth - but that does not mean that the drivetrain
>losses stay at a fixed percentage of power output.
>
>I hope that this has helped and not confused the issues you asked about.
>
>Rgds,
>HB2


My follow up (any problems with this part?)

Thank you very much, Mr. Bettes. This is very enlightening, and by no means confusing. It is almost as if you read my mind and answered additional questions that I did not ask. Especially the part about exceeding the capacity of bearings and gears. When those types of parts are loaded past their ability to transmit power, they will sap out extra HP, but not for long! Then they have a catastrophic failure, usually.
(is this part relatively close or
do the losses progress more linearally
as the power input progresses past the
design limits of the driveline? My guess
is an exponential loss curve...)

Thanx again!
Robert Greene

 

[JayMaechtlen]

Most of the bearings in a driveline are rolling-element type, either ball bearings, roller bearings, or needle bearings.
Deflection won't be much of an issue there. I don't know the shape of the power loss curve on those, but I'll bet that it doesn't increase much until way past design loads. Rolling-element life is limited mostly by surface fatigue, which is function of (contact loading) * (time under load).
Gear life is similar, but torque appled will deflect shafts and teeth, leading to non-optimum tooth contact patterns.
High loadings on teeth can cause fracture prior to surface failure. (broken teeth prior to wearout or surface failure)
I think gears stay fairly efficient until they get way out of mesh or badly damaged.

The "percentage loss" number can still be useful, as you should size the powertrain according to the power and torqe it will transmit and the desired lifespan. Longer life requirement means you must generally accept a higher energy loss transmitting the poower.
cheers
Jay

Jay Maechtlen

 

[RGRv6guy]

Thanks Jay!

The good news is that the V6 Mustangs that I play with
use a V8 transmission, very few load bearing parts are
downgraded for the lower torque, (if any)
In fact there is not much less torque produced on a V6,
as compared to a 4.6 liter V8 anyway.
So we are using a "Heavy Duty" tranny in essence, with
decently low torque loss figures. They do use a smaller rear
axle, but it is adequate for nearly all built up V6's so far.

Another thing I wanted to post (pic) hope it works, but
it's the Torque and HP curves of a cammed V6 Mustang,
the specs are unimportant, but the angle here is that
you can see what the BTQ curve might look like if you
"add in" and exponential TQ loss starting from the torque
peak and go up to HP peak. I added some extrapolated
lines in there to 6kRPM but the cam actually shows a
downturn in torque @ 5700-5800 RPM, however the HP goes
flat, hold on to 6K in bigger engines (strokers) that use
this same basic cam.

http://www.mustangmods.com/data/3523/rudedynorunto6krpm.jpg

It's full screen so I left it as a link. Sorry


My premise here is that the driveline losses are basically
an exponential curve (of some shape) and if you can
visualize that being about an additional 25#-feet total
loss from 3K to 6K then the torque curve is as flat as
Bonneville!

Props to FORD for their design, it's my (old) cam but that
was the only mod other than boltons, dual exhaust and tune.
Does anyone else agree that the torque curve is probably
nearly flat as a board from 3K to 6k??? Or is my assumption
bad?

 

[JayMaechtlen]

Looks like an amazingly flat torque curve!
Normally I think of an "exponential curve" as having an exponent => 2
The "percentage loss" formula would have an exponent of one, right?
I've never seen a perfectly flat torque curve, but my experience is limited.
Now, if you could just get numbers from that motor on an engine dyno, you would know for sure!
-hmm- can you get an idea of how much heat is getting dissipated by the trans and diff, in terms of temp rise and/or air being heated up?



Jay Maechtlen

 

[RGRv6guy]

I have no heat loss figures, but I see how that could be used
to calculate torque loss though!

Here is another graph, a stroker V6 using the same cam
and it has an even nastier torque curve... the graph is
superimposed on a 4.6 V8 dyno sheet, the V6 eats the V8
for lunch and spits out the bones

http://www.mustangmods.com/data/3523/265280_vs._bolton_00_gt.jpg

It peaks low and holds on until 6K (at least HP does)
perhaps with more cam, the one used was mighty lame as
far as ramp speed, lift and overall aggressiveness on the
lobes, but the durations are well matched. But hey, this
is a driveline forum, not engine!

 

[crysta1c1ear]

My premise here is that the driveline losses are basically an exponential curve (of some shape) and if you can visualize that being about an additional 25#-feet total loss from 3K to 6K then the torque curve is as flat as Bonneville!

Looks like an amazingly flat torque curve!
Normally I think of an "exponential curve" as having an exponent => 2

Saying "an exponential curve of some shape" is like saying 'a circle of some shape'. You have defined the shape: exponential or circular.

If the highest exponent is 2 (eg y=x^2 + x+1) you have a quadratic. If it is three (eg y=x^3 + x^2+x+1) you have a cubic. If it is 4 you have a quartic.

An example of an exponential curve is 10^x. The exponent is not a number but your variable, x. An increasing exponential curve will overtake any polynomial. Imagine y=x^1000. No matter how big x gets, you are still only taking things to the power of 1000. In contrast, 2^x means you are taking things to the power of thousands, then millions, and then billions, etc., as x goes through its range of values.

If something is a polynomial function, you'll have second or third order (or whatever order) derivatives that are zero. Exponential functions aren't like that. The derivatives have the exponential shape as the original curve.

Sorry to be so petty. I just have a background where pi doesn't equal 22/7. Maybe I'm not a real engineer!

 

[RGRv6guy]

Apparently I said that wrong, but I've always heard
"losses increase exponentially" meaning that at some
point the driveline loss curve would exceed a slope of
one and the HP/TQ curves will be diving the opposite
direction. What exactly do I want to call this then?

I am thinking that the driveline loss curve would be
more like a parabola, at zero RPM, no losses and it
increases with RPM. Set me straight here, sometimes
there is great wisdom in the proper terminology, many
engineering and mathematical terms are highly descriptive
and educational to the open mind...

My greatest claim to fame (other than the V6 arena) is
the fact that I was turned down by one of the best
engineering schools in the nation, Rose-Hulman Institute
of Technology. They loved my SAT's but said I needed a
higher GPA to get in

Make it CRYSTAL CLEAR please!


Thanks!

 

[crashbox455]

i'm with you, bud. i use "exponential" the same way- when the numbers start to run away from themselves up the graph.

i guess we'll have to straighten up, be serious, and start calling it an "increasing slope" situation.... or the "inverse property of diminishing returns". that has a nice sound to it

 

[JayMaechtlen]

oops-dang!
I did have a nagging feeling about that.
I was thinking x^n, with "x" the variable and "n" the exponant.
(he says, making excuses...)

Anyway- I'd expect the losses to be proportional to torque and speed, increasing with either.

If anyone has real data to plug in, that would be great!

BTW- power losses in drivetrain have nothing to do with torque and power curves over an rpm range, do they?


Jay Maechtlen

 

[EdDanzer]

From the information I have reviewed on vehicle drive trains, the power loss varies with both RPM and HP. At lower RPM and light loads hypoid gearing can be very inefficient. Because of the many variables, only detailed testing would provide actual results. The other variable that you have to consider is tire slip that will effect actual power to the ground.