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?

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

Bottom line
 

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?

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

Thanks
 

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.

When Duke switched to 5:1 gearing, the output is slower than the input........just like a real differential.......so you'll have 1250 ft.-lb. at the rear axles. You'd have 1250 lb. of thrust if the diameter of the tires was 24".

So you end up with 625 lbs with the 12" wheels?

I am now also confused about the 500 lb generating 2.5:1 geared car with 12" wheels laying down 200 lbs. Is this not derived by dividing the torque by the shaft input to the differential assuming the output aspect is 1?

I knew I was missing something on #35 but am eager to learn. :-)

With 12" diameter wheels, the torque is doubled (radius of .5 feet) so you'll get 2500 lb. of driving force. If you put larger tires on a vehicle, you'll notice a bit of a loss in acceleration for the first time you take it out.......and vice-versa.

500 lb.-ft. though a 2.5:1 differential gives 1250 lb.-ft. at the axles. With 24" tires, the driving force is 1250 lb. With 12" tires, the driving force is 2500 lb.

It may not have been a trick question but it sure tricked me.

So my M103 sohc...177hp@5700rpms
177hp X 500 lbs (is that right for hp?) = 88,500 lbs/sec = 5,310,000 lbs/min.
idealy.

177hp = 163 ft-lbs X 5700rpm/5252 (is that crank shaft torque output?)
through it's 3.07:1 differential at its 1:1 4th gear puts out 500 lbs through a 24" (1ft rad.) wheel 1000 lbs through a 12" wheel

and 875 lbs/sec through it's actual 15" wheels?

and if so does it generate:

4,987,500 lbs/min. = 875lbs X 5700 rpm's
or
52,500 lbs/min = 875 lbs. X 60 sec.

???

You've got that basically correct.

It puts 500 lb. driving force to the pavement through a 24" tire.

It can't have 15" "wheels."

The diameter that we are concerned with it the diameter of the tire.

We've confused you by using "wheel" when the "wheel" is the O.D. of the tire.

The tires are slightly larger than 24" and the force to the pavement is reduced very slightly because of this.

The answer is both setups produce the same RWHP and drive thrust at this road speed.

I'm done posting to this thread.

Duke

My opinion has always been that if one is a scientist, at least once a day that scientist should stand in front of a mirror and practice saying "I don't know.".
The calculations given for power or thrust or whatever it's being named are for a wheel that is fairly solid such as a round steel wheel. Given that a rubber tire, regardless of construction type, flexes, this results in a wheel diameter which is dependent upon where the diameter is measured. The steel wheel of comparable diameter has a single metric which describes the diameter. Is there a difference between the two wheels as regards power/thrust, all else being equal, I ask myself. At the moment I think a rubber tire is always going up hill functionally whether it is on a flat grade or a downward grade, but I think I'll stand in front of the mirror before I try to figure that one out. Being one of the Great Unwashed, I find genuflecting an "I don't know" is good for the Spirit. This is a good thread & I'll keep reading it & post when the urge strikes me. I think I'll start a thread regarding the fuel/air ignition in a diesel using a precombustion chamber. Speak of what happens after the initial fuel ignition which is a function of compression, then the burning of other areas in the fuel/air cloud that is a function not of compression, but of radiative heat from the initial compression ignition pattern. Perhaps we can be enlightened in both an intuitive fashion and a formal fashion using differential geometry or tensors or a high level calculus. I don't know.

And, after some reflection, you may have responded with "absolutely".

The rubber tire has a large flat spot on the bottom so the distance from the centerline of the axle to the pavement is less than the radius of the tire.

This will have the effect of providing more forward thrust from a given torque from the axles.

This was ignored in the previous calculations.

Amen
 

The inner core of science is always "I don't know it all" coupled with "I do not understand it all either". Time after time, even the the strongest scientific principles have proved to be incomplete, if not wrong.
Reading the threads in this forum for the last couple of years I have learned that every poster presents valuable information, though sometimes it may not be perfect information. I have learned a lot from what I read and I have also learned to respect the posters. Even if I do not agree with them I know there something that I may be missing, do not know or do not understand. The wide variety of knowledge and experience is what makes this forum great.

JL

I agree. But, there is also energy being lost because the tire is rolling and as it rolls it is forever creating a new flat spot using the front of the tire and the flat spot at the rear changes it's geometry to be part of the tire "circle". So, I wonder when one calculates the "effect of providing more forward thrust from a given torque from the axles" and the cancelling of some of the increased forward thrust with the energy lost by continually flexing the tire, is the net effect a loss or gain in expressed thrust. I'm thinking the net effect is a loss in expressed thrust/power. And, of course, we have the further confound that the length of the "flat spot" lessens with velocity and the diameter of the tire increases with velocity..
Duke's contributions were of great value in this thread, but I would say that Newton was/is one of the giants in basic science; but, one must also consider that Einstein's General Relativity appears largely correct and General Relativity denies there is such a thing as a gravitational force but only inertial forces whose force vectors are in the opposite direction of Newtons gravitational attractive force. Being fluent in Newtonian physics is a good and valuable thing, but it is not a substitute for thought. I don't direct that at you, Brian, but just at a few instances of the development of this thread that doesn't include you and I merely thought I'd throw it in this post.

Now, this sure is not Physics 101.
I love it!

JL

You're referring to the loss created by the frictional losses in the tires. These would definitely need to be addressed, as would the losses in the drivetrain, if you wanted to be precise.

The flat spot remains fairly constant for any of the speeds that we are discussing. The reason is that the flat spot must support the weight of the vehicle. If you have a flat spot of 28 square inches and an internal pressure of 32 psi, the tire will support 896 lb. which, coincidentally, is about the weight on one tire on a typical W126.

I hesitate to speak for Duke, however, I will say that he understands the concepts of horsepower, torque, and force better than anyone else on this forum, including myself, and he appears to be frustrated with those who wish to render opinions on a topic where the science is cast in stone.

Don't sell him short on the "thought" aspects as well. His position clearly requires some thinking outside the box on many levels. However, on the basic facts of this subject, I agree with him completely.........there is no room for "discussion".

Just to add a little levity...
 

to this otherwise "heady" thread:

"If I had only known, I would have been a locksmith."
-- Albert Einstein

Good day!

Jeff Pierce

It is not a problem of understanding the concepts of power, torque etc. That is easy. It is how they are put together through formulas. After you link them, they still have to hold together on their own. That is where this reasoning fails. That is why, if what you want to know if torque at the wheel is the one factor determining acceleration, you can concentrate on the wheel alone. Anybody who studied engineering went through something called Statics, and there they learned about something called "free body diagram". A free body diagram of a 1 foot radius wheel subject to X foot-pound of torque at the axle results in X pounds of force backwards applied on the ground, and X pounds of force applied on the axle forward. Period. You not need anything else. There cannot be any other force or it would not be in equilibrium. I do not think you can get any simpler than that. If after going through all your formulae, equilbrium at the wheel is not satisfied, it is all wrong. By this I mean, if the torque at the 1 foot radius wheel is specified at 250 ft-lbf, the thrusting force you calculate cannot be 2500 lbf. That would give you 2500 ft-lbf of torque.

JL

A torque applied to a wheel bolt that does not move is statics. Power delivered through a rotating shaft is dynamics.

That's the next course in an engineering curriculum, and one much understand dynamics to analyze vehicle performance.

Duke

No, not the frictional losses, if I understand you correctly. I'm obviously having difficulty in being clear, which is my fault. Imagine that one has a steel wheel which has a flat spot similar to the radial where wheel meets flat road, level grade. But, the steel wheels geometry doesn't change from flat surface to circle instantaneously. So, we begin to roll this steel wheel forward. We first have some difficulty (power = X), then as we hit pure circle geometry we only have the rolling resistance, with all the variables that contribute to that rolling resistance (here Power to roll the wheel = X-a, where "a" is some constant). In the rubber radial, I think, we have the same basic phenomenon going. Though, with the caveat that the first non-flat spot is also circular (for some small distance) but that arc belongs to a circle of much greater diameter than our 20" diameter (for example) wheel. That arc always just ahead of the flat spot belonging to a circle diameter of >20", which in a continuous fashion moves towards the arc belonging to the 20" diameter circle uses some of the power put to the wheel by the axel. This is the basis by my saying the tire of relatively non-rigid geometry is essentially always going "uphill" in the same manner as if the relatively rigid steel wheel "experiences" going uphill, though the radial does it all of the time and the steel wheel only goe's uphill when the road grade is uphill. I do understand that even in my analogy that the flexing rubber has it's own intrinsic frictional loss as it rolls. apologize for moving towards a point, correct or not, that moves perhaps too far away from Post #1. I agree, Duke has provided us with the mathematical versus the intuitive physics, which is invaluable.

Absolutely, and I believe there are free body diagrams and equilibrium in dynamics too. I am interested in knowing where is that extra force applied, and how it is transmitted to the wheel. If you are right, you are right. I want to learn.

JL

There is no "uphill". The steel wheel has a single contact point and there is no deflection in the wheel. There are no losses associated with the deflection.

The rubber tire must be deflected as it moves forward. The losses are not great but they are more than the steel wheel.

If there was an "uphill", you would never be able to push an automobile yourself........the slightest "uphill" and you're toast.......as you probably are aware.

I frequently push the SD in and out of the garage because I don't want to bother to start it.........it usually takes most of my capabilities to do that...........and I'm not going uphill.

Yes, the "uphill" I was referring to in the rigid wheel is an actual uphill gradient of the road surface. The single point contact of the rigid wheel is what makes it easier to push your "car". So, I agree with you on that.
I do disagree if there was an "uphill" (the uphill instrinsic to the tire, but with 0% gradient) in the case of the radial, one would still be able to push the car on level gradient. It is more difficult, by far, than the rigid wheel, but that is the difference created by the non-rigid wheel, exclusive of frictional losses versus the rigid wheel. Yes, the difficulty in pushing your car out of the garage on flat gradient, versus what work you would experience with a rigid wheel is the "uphill" (poorly named, I know) created by the radial. It doesn't approach trying to push your car up a 6% gradient, but phenomenologically it is similar and a portion of that difficulty is the "uphill" intrinsic to the non-rigid wheel. The arc length of the non-rigid wheel just ahead of where it going to next meet the pavement is greater than the arc length of the non-rigid wheel at the top of the tire. It is the change in arc length which is the metric associated with what I've referred to as "uphill", in the mathematical sense.

I agree, but, I don't advocate the term "uphill" to describe the phenomenon.

The effort is expended to effect the angular change of the tire contact patch.

Yes, "uphill" was a bad term. At the end of the post I awoke enough to say "arc length". Arc length directly yields the tangent vectors, which is, I think, synonymous with "angular change".

Thanks for helping me get this. And thanks to all the contributors to this thread... I appreciate the chance to eavesdrop on your insights.

Anytime.

Help is what we do here.