Nacelle

some have asked... engine stock.

DeerHunter

Healthy!

lax01

To save everyone from downloading the PDF and then rotating it:

wooge

what does the CEng prefix mean?

shotor

Thank you.

Was this done in BC? If so, where?

DeerHunter

I can't tell you what they stand for, specifically, but here is a pretty good guess:

CEngPW = Crank Engine Power (estimated)
SAEPwr = SAE horsepower, as measured at the rear wheels (actual)
CEngTq = Crank Engine Torque (estimated)
WhlTqr = Wheel torque, as measured at the rear wheels (actual)

EsDTLs = Estimated Drive Train Loss (either a WAG or calculated from a coast-down measurement)

In general, the estimated crank power is derived from wheel (actual) plus the estimated drive train loss at that engine speed.

Nacelle

No, Ontario

Nacelle

DeerHunter covered it pretty well...

ChrisF

I think 19% drivetrain loss is HIGHLY optimistic. 2WD rear drive is likely closer to 15% max loss which equals (surprisingly) 434.x (435).

NAM VET

I have no doubt the operator of this dyno inputs accurate data into the computer. But dyno's can be programed to "fudge" the raw data very easily. Back when I had my Cobra, with a stroked and bored 351 Winsor, pushed out to 396 ci, with all the proper race parts, port matched,Trick Flow heads, Pro Holley, Ford Motorsports cam, etc, i had it on a rear wheel dyno several times. The race shop dyno asked me what numbers I wanted. He can input data like temp and humidity, and change the curves quite a bit.

quite a few of us ran our Cobras, and one question we never could resolve is the % of engine power lost to friction.

More later.....

NAM VET

well, as I was saying....

For instance, if you have 250 crank HP, and assume the standard RWD 15% loss to friction and exhaust, then you have about 215 or so at the wheels. Now, if you put a turbo on the motor, and put out say 400 HP, the same 15% means now the same drivetrain consumes 60 HP. So why did the same drivetrain now pull an extra 25 HP out of the engine. We could never figure it out.

Also, I have an engineer friend, who used to do gearing calculations for NASCAR road course cars. He taught me to look at gear splits, and where torque and HP curves begin to flatten out, and thus give a suggestion as to upshift RPM. I have no idea of the gear splits on our cars, nor could I really give an knowledgeable opinion as to shift points for maximal acceleration. But looking at where the above dyno curve begins to flatten out, it would seem to me that about 7K rpm, it makes no sense to continue to 8400 rpm, as an upshift at redline just drops the engine back down to a flattening part of the HP curve. So, I think, for our driving, we generally shift as the tach passes 7K rpm, as that keeps us off the rev limiter, and drops the next gear into a meatier part of the curve.

As I recall, my Cobra's motor, which could rev to 6500 rpm, was pretty much out of steam at about 5000 rpm, so that is where we shifted. On a cool day, it put out 495 horses, and about 550+ pounds of torque. Not bad for a car all up weighing about 2200 pounds.

But then, what would I know.

All the best...

DeerHunter

The debate about whether drivetrain losses are a straight percentage or a fixed number will rage on until every car has an electric motor powering each driven wheel. I believe that it's a combination of the two: A fixed loss (dependent on the drivetrain configuration, gearbox efficiency, etc, etc.) plus a small percentage (more power generates more heat, which is a reflection of inefficiency).

Regardless, the only sane way to use a chassis dyno is for comparison purposes. The actual crank horsepower/torque is fairly immaterial if all you're looking for is the change between stock and mod A, or mod A vs. mod B.

Jake951

To maximize acceleration you must think in terms of maximizing rear wheel torque, not engine torque. The difference is the torque multiplication you get through your gearbox. Lower gears give you more torque multiplication. Once you upshift to a higher gear, you get less multiplication.

The bottom line is that you almost always want to stay in the lowest gear possible, all the way to redline, to maximize acceleration. If you work out the math, this is exactly the same as staying as close as possible to the engine's horsepower peak, which is close to redline. Upshifting too early drops you back down the horsepower curve and costs you in horsepower.

DeerHunter

You got the first part right but are generalizing on the second ("...almost always want to stay in the lowest gear possible, all the way to redline..."). This is often the case with peaky motors (torque peak occurring higher-up in the rpm range), but often not the case with engines exhibiting lots and lots of low-end grunt. The only way to calculate the ideal shift points is to plot torque (at the wheels) for each gear and rpm range. Shift at redline or where torque available in that gear falls below the torque available once you shift to the next gear - whichever occurs first. Oftentimes it will be redline for the first few gears, but then successively lower shift points in the higher gears.

thekid96

The 15% number isn't an exact science but it works in a general sense because it represents an average loss due to parasitic drivetrain loss PLUS loss to centrifugal forces. The drivetrain loss is fairly static but the centrifugal losses are dynamic and increase with hp production.