My guess would be the reason Zero_to_69 mentioned, "stradling the curve." What are you trying to argue? That they have better instantaneous acceleration at that point 5% over the HP peak? Or that to overall do the best they can in that race, it is advantageous for them to run 5% over their HP peak? These two thoughts are not the same, and saying that "highend racers" run their RPMs over the HP peak has nothing to do with the best instantaneous acceleration at a given speed.

Then again, drivetrain losses skew the engine's torque curve. So even if an engine's HP peaked at 6000rpm, the most wheel torque might be obtained a little above or below that. But if one of us is arguing "in a vacuum, the theory is this" and the other is arguing "but in the real world, it's like this" then I think we're wasting our time, even though both have their merits.

IMO, I don't even see how it's up for debate. I like to think of horsepower as units of torque per time. In other words, it's a measure of how fast an engine can apply the torque it's generating (rpm). Gear the drivetrain so the engine is at this peak and you SHOULD have the most available torque at the drive wheels. Not at the torque peak.

 

Can we elaborate on this please? This is the basis for posting originally.
(Forget acceleration, or g force for the moment)

EDIT - Thinking about the amount of time which the torque has to complete a
cycle.

I'm also going to use my 3.8L V6 Camaro to test the data given in this post.
Since the 3.8L is a manual transmission, we can eliminate any sort of torque
multiplication via converter slip.

My test will be a sprint from idle to redline (above HP peak) in 1st gear, and
2nd gear.

By theory, the car will accelerate as the torque curve increases, which will
show an incline in the 'g' graph relative to the torque curve.

Does anyone see flaws with this test (Jon - OldSStroker)?

 

Interesting concept.

Does Jim's answer give anyone (T and/or K) an Aha Moment?

One can hope...

 

No, it would be at the torque peak in the lower gear.

Rich

 

Summary to date:

Questions at the bottom. Please try them out.

Arnold Schwarzenegger has a 30 pound rock. He throws it ten feet and it takes 6 seconds to complete the task.

Hulk Hogan has a 28 pound rock. He throws it ten feet and it takes 5.45 seconds
to complete the task.

If Arnold throws ten , 30 pound rocks per minute, he will have effectively thrown
300 pounds, 100 feet.

If Hulk Hogan throws eleven, 28 pound rocks per minute, he will have effectively thrown 305.5 pounds, 100 feet.

HP formula says it takes less time to complete the cycle, therefore I have
applied less force over less time.

At the next point in time, HP formula says it takes even less time to complete
the cycle. I have applied less force, in a lesser amount of time.

This continues until I reach HP peak. After the HP peak, the HP formula says,
my engine cannot breathe well enough and begins to incur pumping losses.

From HP peak onward, power is diminishing, and torque is diminishing.

G force:

A car cruising at a steady speed with no change in opposing forces will require
a constant force to maintain speed.

There is no change in speed, therefore there is no change in g force.

At the moment the accelerator is pressed, the g meter sees a change in force
according to the rate of change.

Questions:

#1. At what RPM would you shift this engine for a 1/4 mile race and why?:
T-56 transmission (stock ratios), 3.73 gear, 28" tire. RPM limit: 7000 RPM

http://gmthunder.com/tino/c292h1.jpg
http://gmthunder.com/tino/c292h2.jpg

#2. Does anyone here believe that shifting this motor at ~ 4500 RPM with the above parameters
is BETTER than shifting at ~ 6500 RPM?

#3. Does anyone here believe that lauching this motor 4500 RPM will have more
snap than launching at 5500 RPM?

#4. If you had a 2000 RPM shift window into 2nd gear, shifting from 1st,
where would you want your 1st gear shift point?

 

It wasn't meant to prove your statement correct. It was meant to point out how unnecessary the statement was in the first place, because it was what would commonly be called a no-brainer...

You can't, and thanks for playing.

The only people I've seen clinging so doggedly to horsepower as the determining factor of acceleration are rotary fanatics who don't have any torque to speak of.

 

The ideal shift point is always the rpm at which torque at the axles in the current gear falls below torque at the axles in the next gear higher, or redline, whichever occurs first.

If you plot torque at the axles over the engine's rpm range for all gears, it's very easy to determine the optimum shift points from the points at which the curves cross. If they don't cross, the optimum shiftpoint is redline.

I think you're making this more difficult than it really is.

 

Certainly.

Let's say our engine's torque peak is 560 lb-ft. @ 5,000 rpm and the horsepower peak is 640 @ 7,000 rpm. First gear is 2.97:1, second gear is 2.07:1, and we've got a 3.55:1 ring and pinion. Gee, this sounds familiar.

Maximum acceleration overall will be achieved at torque peak in 1st gear, with a potential of a little over 5,000 lb-ft. of torque at the axles. (560 lb-ft. * 2.97 * 3.55 * 0.85 = 5,019 lb-ft.)

Maximum acceleration in 2nd gear will also be at the torque peak, with a potential of about 3,500 lb-ft. of torque at the axles. (560 lb-ft. * 2.07 * 3.55 * 0.85 = 3,498 lb-ft.)

However, torque at the axles in 1st gear at the horsepower peak is still higher than the maximum torque at the axles in 2nd gear at torque peak that we just calculated. Flywheel torque at 7,000 rpm would be 480 lb-ft. That would give us about 4,300 lb-ft. at the axles (480 * 2.97 * 3.55 * 0.85 = 4,302 lb-ft.) in first gear, which is obviously more than the maximum possible in 2nd.

Maximum acceleration is achieved when torque at the axles is maximized, therefore we'd hold 1st gear to redline (400 lb-ft. = 3,585 lb-ft. at the axles) before shifting to 2nd. We'd also downshift to 1st gear to accelerate faster than we possibly could in 2nd gear, assuming that we weren't traveling at a speed which would place us too high in the rpm range of 1st for the gear change to be worthwhile. There's no point in downshifting if you have to shift right back up again. You'd be faster just burying the pedal in the current gear.

Torque at the axles plotted for all gears would look something like this. You can see that the torque curve for 1st gear doesn't overlap 2nd gear at all. Maximum acceleration would therefore be achieved by holding 1st gear to redline before shifting. However, the curve for 4th gear (light blue) does overlap 5th. You'd be faster shifting into 5th at about 7,200-7,300 rpm than holding 4th to redline. Clear as mud?

 

Here's the same graph showing the path of maximum acceleration (red line). I've had to explain this concept to the rotards before, obviously.

http://home.gci.net/~jimlab/images/M...eleration2.jpg

The blue circles show the engine's torque peak, or maximum acceleration in each gear. The yellow circles show the engine's horsepower peak, just for reference. This illustrates that you don't shift at either the torque peak or the horsepower peak for maximum acceleration. As I stated above, you shift at the point at which torque at the axles in the current gear falls below maximum torque at the axles in the next gear higher.

The next graph shows acceleration in g's over time through 4 shifts. You can see that maximum acceleration is obviously in 1st gear, at peak torque, and acceleration falls off from there, and in each subsequent gear.

http://home.gci.net/~jimlab/images/M...ation%20gs.jpg

The last graph shows drive power (thrust at the wheels) for a different car (a stock 3rd Gen. RX-7). You can see that the curve for each gear does overlap the curve for the next gear higher, so you wouldn't hold any gear, even first, all the way to redline for maximum acceleration. You can also see the increasing losses associated with rolling and air resistance plotted on the purple line labeled "Losses". The point at which the losses line and 5th gear curve cross is "terminal velocity", or drag limited top speed. In this case, right around 167-168 mph.

http://home.gci.net/~jimlab/images/M...ve%20Power.jpg

Is the effect of an engine's torque curve on acceleration clearer now?

 

Jim: nice graphs! If someone doesn't get it now, I am not sure how you could possibly explain it better.

Rich

 

Jimlab,

Thank you for the elaborate post. I'm about to post my thoughts, but to
clarify, I need to know your thoughts on the following:

edit - hang on...

 

Tino,

Instead of overthinking this and wasting time....

http://www.maxracesoftware.com/products.htm

ET analyist will make your job a whole lot easier.

BTW why don't you pull out the forumla for Work since that's what HP is. I never think about HP and TQ seperately maybe that's why this whole arguement has little interest to me.

Bret

 

Bret /Jim,

I don't feel this is a waste of time because we're now dealing with peak numbers
as opposed to a graph.

I was hoping everyone would reference the graph of TQ./HP so we could
get on the same page.

According to Larry (Maxracesoftware), he doesn't use the torque curve, he uses the horsepower
curve to calculate shift points.

According to the math, holding all the gears to the HP peak, or just above
always produces more torque than the next gear starting RPM point.

Again, I agree with you and Jim that the most torque to the wheels is best,
however, I'm having a difficult time understanding Jim's shift points if he's
not holding each gear to HP peak.

 

Here's an Excel spreadsheet

http://gmthunder.com/tino/excel.jpg

of the shift points for this dyno chart:

http://gmthunder.com/tino/c292h1.jpg

If you take the time to read the following links, you will see why I'm pulling my
hair out over this.

And for the record, I just haven't sat back and picked out the stories I like best.

I've tried this with two cars, on a dyno, on a drag strip, on the street and
with a G-tech meter.

With all due respect to everyone in this forum, I think I have some valid points.
Maybe I don't have the lingo to communicate my thoughts.

Larry Meaux (Maxracesoftware):

http://web.camaross.com/forums/showp...58&postcount=8
http://web.camaross.com/forums/showp...4&postcount=25
http://web.camaross.com/forums/showp...0&postcount=73
http://web.camaross.com/forums/showp...7&postcount=86


SteveinSeattle:

http://web.camaross.com/forums/showp...9&postcount=29
http://web.camaross.com/forums/showp...6&postcount=38
http://web.camaross.com/forums/showp...3&postcount=47


Erik Koenig:

http://web.camaross.com/forums/showp...8&postcount=40
http://web.camaross.com/forums/showp...4&postcount=82

 

For shift point calculation you can use either one, since horsepower and torque are inextricably linked. They'll cross (or not) at the same rpm as shown in this graph of horsepower vs. speed plotted for the same engine as above. Note that the curve for 4th gear crosses the curve for 5th gear at the same rpm as it does in the chart showing torque at the axles.

http://home.gci.net/~jimlab/images/M...vs%20speed.jpg

However, the torque curve is what ultimately determines rate of acceleration. Acceleration falls off after the torque peak which can easily be demonstrated by accelerating to redline in 4th gear starting from below the torque peak and noting what happens after you pass it. You will accelerate more and more slowly until you finally reach redline. If acceleration followed the horsepower curve, you'd just keep accelerating faster and faster (rate of acceleration, not vehicle speed) as you approached peak horsepower, which is obviously not the case.

Refer to the first graph in my last post and note where the peak horsepower point (yellow circle) occurs. The ideal shift points are all above the horsepower peak. The math proves that torque is still higher in the current gear above the horsepower peak than it would be in the next gear higher.