There is one fundamental difference in an oil thread when compared to this thread. An oil thread is founded on opinions. This thread is founded on engineering facts.

There can be no dispute with the facts as Duke has provided them in such an eloquent fashion. Basic physics is indisputable.

Correct! Unless a force acts through a distance - either linearly or an arc of a circle - no work is done, nor is any power expended.

Brian pointed out the inconsistencies in your example. It's just not a valid exercise, even for academic purposes.

You still haven't anwered the question in post #35, yet. Do you want to give it another try? Anyone else?

Duke

Obviously the car making the lower torque will win, because the 1000 ft-lb machine will have blown itself up at 2000 rpm, since you stated it can only deliver 200 hp. At 2000 rpm, it would be producing 380 hp, which of course it can't do.

Big cigar for Tom! You caught it, nobody else did!
As far as post 35, I am not sure I got it right. First half I get. We get 500 lbf at each wheel. Second part I am not sure. If the 250 lb-ft is the torque at the wheel (as it appears) then the force on each wheel will be 250 lbf. As I see it, the first car (assuming the same weight) will accelerate faster because of a larger driving force, regardless of the rpms. If the 250 lb-ft are measured somewhere else let me know and I will try to understand.

JL

From the description of the engines, the first is a low revving engine, say 500 CID. The second is a high revving engine, say 250 CID. The key is the following statement.

"Now substitute an engine that makes half the torque, 250 lb-ft, but at twice the speed, and we shorten the axle ratio from 2.5:1 to 5:1, so that peak torque occurs at the same road speed."

Both engines are making the same RWHP at road speed. Though the "small" engine makes only half the torque, with suitable gearing it delivers the same RWHP, so both will accelerate at the same rate if the throttle is floored at this road speed.

Again, I stated this example to illustrate that POWER is what counts to accelerate a vehicle.

If a one-liter motorcycle engine rated at 120 HP @ 12000 was substituted in a small sedan for the original 2-liter engine rated at 120 HP @ 6000, acceleration performance and top speed would be the same provided gearing was shortened accordingly. If the two-liter was geared to yield 3000 revs at 60, the one-liter would have to be geared to yield 6000 @ 60.

The same argument can be made for substitution of a low revving four-liter industrial engine that is rated at 120 HP @ 3000 in which case you would gear it to yield 1500 @ 60.

Assuming the car required 120 HP to achieve 120 MPH, all three combinations would achieve 120 MPH at engine revs corresponding to their peak power.

Duke

My wrx can have multiple 'maps' run.

a)One map is set for 91 octane fuel....280HP 275T
b)Another map for 93 octane 280Hp 280 T

The peaks occur at identical rpms and the dyno graphs almost overlap.
One car so gearing, drag etc. constant.

Do you think there is a diff in accel. and top speed?

__________________
82 300D....went to MB heaven
90 350 SDL....excercising con rods

All that may very well be true, but no matter how you cut it, torque at the wheel uniquely determines acceleration (given fixed mass, tire size, total resistance). We do not need to know the power or rpms. If you know the velocity and overall drag coefficient (including rolling resistance) you can calculate the power required (as stated in previous posts). If in addition you know the gearing, you can get the rpms.
The physical quantity that propels the car is force, that force is created directly by the torque, and it (the force) does not depend directly on rpms. or power. This is true for the wheel. For a particular engine, torque, power and rpms are uniquely tied together. That does not change the previosly stated, it will just determine the torque available and the power being used.

JL

Top speed should be the same (same max power)
Acceleration cannot be predicted unless you know the details of the torque vs. rpm curve. Max torque by itself is not enough to predict average acceleration. We need two know how remapping alters the shape of the curve.

JL

The bottom line here is that to really understand acceleration and power curves, you need to sit on a 1975 Kawasaki 2-stroke-triple (H1, H2, or S3) with expansion chambers and just keep the throttle wide open as long as you can while you shift through the gears.
Remember to hold on tight and to avoid policemen.

I remember when Kawasaki first came out with them. Unbelievably fast......would beat any other bike hands down for quite awhile.

The exhaust arrangement was certainly kind of funky.

I've changed my mind. It's hopeless. I don't think there's one in a hundred in the US population that understand 400 year old physics.

Our education system has just gone to the dogs. I was lucky to have gone through high school and college in the sixties when math and science were actually taught.

English, too!

Duke

Don't think so. The problem is not physics. The problem is usage of physics formulae without an understanding what the underlaying principles are. Putting numbers into a formula (or to make it worse, into a computer program) is no guarantee of an answer with a real meaning. Garbage in, garbage out. It is possible to get so wrapped up with anecdotes and cases that basic concepts blur and magical forces appear out of nowhere. Basic laws apply wether we have a gas engine or a diesel or an electric motor or a hand-crank.

JL

I'm not clear in my head about #35 so lets step back to #38...
148hp @ 4800 rpm's
143 ft-lbs @ 5800
(the work being done increases from 81,400 lbs/sec to 86,900 lbs/sec with the rpm's while the torque declines from 162 to 143)
I'm not sure what determins torque but if I were to guess it would be the mechanical limitations of the engine air-flow/exaust applied to the connecting rods...

Torque wrench is 1' + 80lbs weight applied = 80 ft-lbs.
Work = 0 movement

How am I doing so far?

"If ...500/2.5 = 200 lb-ft.

F = TxN/5252V (we're still talking about torque and power at the rear wheels)

Now substitute an engine that makes half the torque, 250 lb-ft, but at twice the speed, and we shorten the axle ratio from 2.5:1 to 5:1, so that peak torque occurs at the same road speed.

At this road speed which setup provides the greatest drive thrust? Which one provides the greatest RWHP?"


When you say "twice the speed" you mean twice the rpm's or road speed?

__________________
-Marty

1986 300E 220,000 miles+ transmission impossible
(Now waiting under a bridge in order to become one)

Reading your M103 duty cycle:
http://www.peachparts.com/shopforum/831799-post13.html
http://www.peachparts.com/shopforum/831807-post14.html

"Twice the speed" means twice the rpm's.

The relationship P=FV still holds, but we can rewrite it as:
F=P/V or F = TxN/5252V (we're still talking about torque and power at the rear wheels)
I have the feeling I'm still missing something but:

250 lb-ft/5 = 50 lb-ft...at wheels

83.77 hp = 200 lb-ft X (2200 rpm/5252) = 46,073.5 lbs/sec.
41.88 hp = 50 lb-ft X (4400 rpm/5252) = 23,034.0 lbs/sec.

__________________
-Marty

1986 300E 220,000 miles+ transmission impossible
(Now waiting under a bridge in order to become one)

Reading your M103 duty cycle:
http://www.peachparts.com/shopforum/831799-post13.html
http://www.peachparts.com/shopforum/831807-post14.html