850 replies to this topic

[WPT]

A paper calculation comparing two different shift strategies was performed. For one shift sequence, the time required, distance travelled, and the change in speed for a bike shifted at peak power rpm was compared to one shifted at redline rpm. The method used to calculate time to speed and time to distance follows.
Suppose the drive wheel hp vs. road speed data exists for the speed range Va to Vb. The mass of the vehicle and drag forces are known. Divide the speed range into an appropriate number of speed intervals. These speed intervals need not be equal in width to one another. For an interval, say V2 to V3, perform the following:

• determine Hp2 and Hp3
  
• from p=Ft*V calculate the thrust forces Ft2 andFt3
  
• subtract from Ft2 and Ft3 the appropriate drag forces. The result, Fa2 and Fa3, are the thrust forces available for acceleration.
  
• divide Fa2 and Fa3 by the vehicle mass to obtain the acceleration rates a2 and a3
  
• calculate the time, t, to go from V2 to V3 by: t=2(V3-V2)/(a2+a3)
  
• calculate the distance, s, travelled by: s=(V2+V3)(t/2)
  
• repeat steps 1 to 6 for the other intervals and add all the t's and s's. The time and distance to go from Va to Vb is now known.

Of course drive wheel torque could be used to determine the thrust force Ft. Then follow steps 3 to 6 to obtain t and s. The very same answers will be obtained. Proof: where p is drive wheel power, T is drive wheel torque, Ft is the thrust force, V is vehicle speed, r is the rolling radius of the drive wheel, and w is the angular velocity of the drive wheel in radians per unit time.

• p=Ft*V; but V=r*w
  
• p=Ft*r*w; but Ft*r=T
  
• p=T*w; the correct equation for p in terms of t and w

The above assumes the drive wheel does not break traction and the clutch does not slip.

Now, for two identicle bikes, B1 and B2:

• 2nd gear ratio is 10.71 and gives 70 mph at 10,500 engine rpm
  
• 3rd gear ratio is 8.16 and gives 70 mph at 8000 engine rpm
  
• mass per bike is 20.8 slugs
  
• drag force is given from: Fd=(V^2)/111.524, where Fd is in lb and V is in mph
  
• shift time is 0.1 seconds, bike coasts at constant speed during the shift

For the engine of both bikes:

• ___RPM_______HP_______Torque(lb-ft)
  
• ___8000______67.07_______44.03____
  
• ___8500______70.90_______43.81____
  
• ___9000______73.89_______43.12____
  
• ___9500______78.42_______43.35____
  
• __10000______81.66_______42.89____
  
• __10500______82.55_______41.27____
  
• __11000______81.17_______38.71____
  
• __11500______77.00_______35.17____

Scenario: Both bikes are dead even accelerating in 2nd gear. B1 will shift at max power rpm (10,500), B2 at redline (11,500). The stop watch starts when 70 mph reached and B1 initiates its shift to 3rd gear.
Results: After 0.76 seconds both bikes have completed their shifts and are accelerating in 3rd gear. B1 has travelled 81.21 ft and is travelling at 76.67 mph. B2 has travelled 82.87 ft and is travelling at 77.65 mph. B2 leads B1 by 1.66 ft and is travelling 0.98 mph faster (i.e. is increasing its lead).
Conclusion: Shifting at max power rpm does not maximize acceleration.

[Jodum5]

I wish you'd take my calc class for me

[wegmann]

Originally posted by WPT Conclusion: Shifting at max power rpm does not maximize acceleration.

Why would it?

I'm not saying the redline strategy would always be best either. It might be, for this particular bike. Assuming you can't change your gear ratios, and that you know the torque curve, wouldn't a better strategy be optimizing the area under the curve?

[Engineguy]

Originally posted by WPT
Conclusion: Shifting at max power rpm does not maximize acceleration.

Train A is traveling east at 57 MPH. Train B is traveling west at 63 MPH. If Train A started 342 miles west of...

Seriously, it's all about maximizing thrust. At the speed that the multiplied torque in 3rd gear will exceed the multiplied torque in 2nd gear is the optimal time to upshift. This is assuming, of course, that you are beyond the torque peak, and torque is dropping with further increase in RPMs. So it is dependant on the interaction between the torque curve and the gear ratio spread involved. Note that I did not, and do not need to, mention horsepower.

[Wolf]

WPT, I think things are a bit more complicated than that- I believe one should take into account the tyre slip*.

* I don't know how things work in real life but GPL models it as v=vt/(1+k*F) , where vt is theoretical speed

I think the best way to accelerate would be to shift at redline in 1st and 2nd, and at intersection of remaining curves...

[Fat Boy]

Originally posted by Engineguy
Note that I did not, and do not need to, mention horsepower.

Let's try your game with a little twist. You don't know the tire size or the gear ratios. All I give you is the rev drop between gears. Now where do we shift? Do you want a horsepower curve now, or can you do it all based on engine torque?

[Engineguy]

Originally posted by Fat Boy


You don't know the tire size or the gear ratios.

I'm unlikely to be shifting gears in a vehicle that has neither tires nor a transmission.

[WPT]

Optimum shift points for maximum acceleration occur at the hp cross-over points or redline if no cross-over. If one determined the drive wheel torque cross-over points, the very same shift points as the hp curves gave, would be noted.
One can see this from the above graph. Here the rear wheel hp vs road speed for each gear in a six-speed motorcycle is shown. The tick marks on each power curve represents engine rpm, 3500 at the beginning of each curve, and 11,500 at the end of each curve (so 500 rpm between marks on a given curve). Road speed is just rear wheel rpm times some constant, so that axis could just as well represent rear wheel rpm. One has a power vs rpm graph, where p=T*w.
Ask what shape would the power curve take for a constant torque engine. It would be a straight line running through the origin (0,0 point), and the greater the constant torque the steeper the slope. One now can visualize rear wheel torque on the above graph. Print out the graph and pin a piece of thread at the origin. The steepest slope that is just tangent to each power curve is tangent at the engine's torque peak (~8000 rpm). Check to see this is so. One can now easily see that both approaches yeild the same results. This is just a "visual proof" of the mathematical proof I gave above.

Posted by Fat Boy
Let's try your game with a little twist. You don't know the tire size or the gear ratios. All I give you is the rev drop between gears. Now where do we shift? Do you want a horsepower curve now, or can you do it all based on engine torque?

For best acceleration shift according to the hp curve, not the engine torque curve. Power is the time rate of doing work, it takes work to accelerate a vehicle. The more power used to accomplish a given amount of work, the quicker the work gets done.

Posted by wegmann
Why would it?

I never thought it would. I'm trying to correct mis-information from some time ago. See http://forums.atlasf...ghlight=Fourier for the off-topic discussion. WPT

[WPT]

Posted by Wolf
I think the best way to accelerate would be to shift at redline in 1st and 2nd, and at intersection of remaining curves...

Exactly right. I agree with you 100%. WPT

[Fat Boy]

Originally posted by Engineguy


I'm unlikely to be shifting gears in a vehicle that has neither tires nor a transmission.

OK, wise ass. I didn't say they didn't have tires or a gearbox. I said you didn't know what they were. A prime example of this is anything from a shifter kart to a stock car where you may not change the internal gears of the box, but you often change the final drive.

By working through the torque multiplication and calculating the thrust at the tire you are figuring out how much work that can be accomplished by the engine over a certain time period. Congrats, you're working with horsepower.

The shift will occur after the horsepower peak (on the descending side of the curve) when you are 'X' horsepower and when the next gear will put you to 'X' horsepower on the ascending side of the horsepower curve.

It's all about maintaining the highest average horsepower or, as Engineguy showed, the highest average forward thrust at the contact patch. They're the same thing.

[Todd]

I was taught to shift at the point where torque peaks in the next higher gear.

[McGuire]

Originally posted by Wolf
I think the best way to accelerate would be to shift at redline in 1st and 2nd, and at intersection of remaining curves...

I think that is a great idea, when restricted to video games and theoretical discussions.

We seem to have lost track of one of the key terms in this otherwise also futile discussion. What is a "redline"? The engine speed beyond which must not be exceeded on the insistence of its builder, boys and girls. What happens to an engine already running at redline when (NOT if) the driver misses an upshift? What happens on a downshift with the engine at redline?

In the race engine business there is an old term for those who operate their equipment this way: they're called "rod stretchers." It's a love/hate thing...it's painful to see fine machinery abused this way, but on the other hand, these folks are a key source of revenue. Meanwhile, why do we suppose engine builders go to all that trouble to issue dyno charts with their engines in the first place? In the words of Carroll Smith, "In the rpm department, don't confuse noise with power."

[Fat Boy]

Todd,

That's not necessarily right. Torque itself doesn't make the car go. Torque as it is applied through the gears of the car is what makes it go. Remember, we are looking for the maximum thrust at the contact patch of the tire or maximum average horsepower. They are 2 ways expressing the same thing. Looking at the problem as simply an issue of engine torque is an oversimplification.

[McGuire]

Originally posted by WPT
I never thought it would. I'm trying to correct mis-information from some time ago. See http://forums.atlasf...ghlight=Fourier for the off-topic discussion. WPT

Oh-oh. I had a feeling I was the inspiration for all this.

WPT's extensive pencil pushing above was all designed to refute a statement I made which has apparently had him stewing for awhile. What I said: "In any gear, top speed occurs at the rpm of peak hp."

He didn't understand what I meant by that and has latched onto an oblique meaning. At root this is all a semantical dispute that is in no danger of ever being resolved. Now it has started again with additional participants, woo-hoo. From here on I am going to just sit back and watch, and amuse myself watching others argue, saying essentially the same things from mutually inverted perspectives in comically different ways. Carry on!

[Todd]

Originally posted by Fat Boy
Todd,

That's not necessarily right. Torque itself doesn't make the car go. Torque as it is applied through the gears of the car is what makes it go. Remember, we are looking for the maximum thrust at the contact patch of the tire or maximum average horsepower. They are 2 ways expressing the same thing. Looking at the problem as simply an issue of engine torque is an oversimplification.

Torque is measured. Horsepower is calculated. What makes the car go? A gear ratio is just that. It is a constant relationship. If the engine is making maximum torque at 4k rpm, then it is getting more torque to the wheel at 4k in first gear than at any other engine speed in first gear, and it is getting more torque to the wheel at 4k in 4th gear than at any other speed in 4th gear.

[Dmitriy_Guller]

Originally posted by McGuire


Oh-oh. I had a feeling I was the inspiration for all this.

WPT's extensive pencil pushing above was all designed to refute a statement I made which has apparently had him stewing for awhile. What I said: "In any gear, top speed occurs at the rpm of peak hp."

He didn't understand what I meant by that and has latched onto an oblique meaning. At root this is all a semantical dispute that is in no danger of ever being resolved. Now it has started again with additional participants, woo-hoo. From here on I am going to just sit back and watch, and amuse myself watching others argue, saying essentially the same things from mutually inverted perspectives in comically different ways. Carry on!

Actually, the reason he corrected you that you don't know the first thing about relationship between torque, horsepower, and acceleration. "In any gear, top speed occurs at RPM of horsepower peak" is an entirely wrong thing to say. This is not a matter of semantics, it's a matter of stating something completely false. The only relationship between horsepower peak and top speed is that if you pick the top gear just right so that the car tops out at HP peak, then the car would reach its maximum possible top speed. Now here is a good place to quote your pet theory which use this false bit to come up with absolutely nonsensical result.

Not at all. The principles of power, torque, and gearing are the same for all piston-engine road vehicles.

For example, take these two engines:

Engine A makes 40 hp at 2200 rpm. (95 lb ft of torque)

Engine B makes 40 hp at 4400 rpm. (47 lb ft of torque)

We will take these two engines to Bonneville, installed in identical vehicles with direct drive, 1:1 gearing. Since the engines make identical horsepower, we know that their top speeds will also be identical. (Theoretically speaking, of course.) However, engine B makes its 40 hp at twice the rpm, so we know it has exactly half the torque, by definition. The acceleration of vehicle B will be much poorer, possibly requiring a push start just to get away. How embarrassing.

Can we fix the acceleration problem with Vehicle B? Sure, we can install 2:1 gear reduction, which will double the net torque and make it equal to Vehicle A. Now their acceleration will be equal as well. However, now we have also cut the top speed of Vehicle B in half. The engine is still turning 4400 rpm, but its output shaft is turning only 2200 rpm. That is the part about gearing that keeps escaping you. If we expect Vehicle B to keep up with Vehicle A, we can't just double the gear ratio. We have to add change-gears too, and shift them. There is no free lunch here. Gearing can only multiply torque. Gearing cannot multiply power. Power is simply torque over time.

By the way, engine A above is not an F1 engine or a Chevy V8. It really is an "A," as in Model A Ford. Developed 40 hp at 2200 rpm. Manufactured from 1928 to 1931. Hell of an engine. -mcg

Your clearly nonsensical assertion that "in any gear, top speed occurs at RPM of horsepower peak" leads you to deduce that 1 = 2 in the next to last paragraph. I think that most of us can agree that one does not equal to two.

[Dmitriy_Guller]

Originally posted by Todd


Torque is measured. Horsepower is calculated. What makes the car go? A gear ratio is just that. It is a constant relationship. If the engine is making maximum torque at 4k rpm, then it is getting more torque to the wheel at 4k in first gear than at any other engine speed in first gear, and it is getting more torque to the wheel at 4k in 4th gear than at any other speed in 4th gear.

What makes the car go is torque at the wheels, which is a product of engine torque and gear multiplication. It is a very common fallacy to equate maximizing torque at the wheels to maximizing engine torque, that would only be correct if gear multiplication was constant. However, it's not. It would be constant only if you were stuck in one gear. However, you always have a choice of at least two, and usually three gears at any point on the track. Therefore, for maximum acceleration you have to choose the gear that maximizes the product of engine torque and gear mulitplication. Conveniently enough, that product is also known as horsepower (times a unit constant). Therefore, to maximize acceleration at any given speed, all you do is pick a gear that would make the engine run at the RPM which provides the highest horsepower. Engine torque has nothing to do with a decision of when you should shift.

[Fat Boy]

Originally posted by Dmitriy_Guller

What makes the car go is torque at the wheels.......Engine torque has nothing to do with a decision of when you should shift.

BINGO!!!!!!!!!!!!!!!!!!!!!!!!

[Dmitriy_Guller]

Originally posted by Fat Boy


BINGO!!!!!!!!!!!!!!!!!!!!!!!!

You think there is any chance that we can convince some "experts" of this once and for all? Nuh, me neither.

[Wolf]

Originally posted by McGuire


I think that is a great idea, when restricted to video games and theoretical discussions.

We seem to have lost track of one of the key terms in this otherwise also futile discussion. What is a "redline"? ...

McGuire- for me 'redline' is highest revs where it's safe to change a gear*...

* TBH, I'd say it's subjective 'redline'- to clarify: if mechanic told me not to exceed, say, 11,000 rpm (I'd reckon that engine can take a bit more, because he's impose a safety margin), I'd start shifting at 10,800-10,900 to prevent going over rev limit. Somebody else might have fancier footwork, and be able to change closer to 11,000 though.

[Fat Boy]

In road racing applications, an engine almost always sees it's highest RPM on downshifts, not during acceleration. Redline is generally where the limiter steps in and shuts down the party.

Putting this all into perspective, if you shift a couple hundred RPM too early, you may lose 0.002 seconds down a particular straight. If you shift a tick too late, after the engine has already touched the limiter, you will most likely lose a tenth or 2. While you always want to use all the HP available to you, it is much better to shift a little early and give up the thou or two rather to bump the limiter and give up a position.

[McGuire]

Originally posted by Dmitriy_Guller

Therefore, for maximum acceleration you have to choose the gear that maximizes the product of engine torque and gear mulitplication. Conveniently enough, that product is also known as horsepower (times a unit constant).

The statement in the first sentence above is correct enough as far as it goes, but beyond that you are totally lost.

The product of torque and torque multiplication is not "horsepower". Sorry, but that term has been taken. Horsepower has already had a recognized meaning for over two hundred years, and that's not it.

The product of torque and torque multiplication (gear reduction) is simply more torque, not horsepower.

[McGuire]

Originally posted by Dmitriy_Guller

You think there is any chance that we can convince some "experts" of this once and for all? Nuh, me neither.

I don't think so either. Before I'd listen to you, first you would have to be able to describe horsepower.

[McGuire]

Originally posted by Fat Boy
In road racing applications, an engine almost always sees it's highest RPM on downshifts, not during acceleration. Redline is generally where the limiter steps in and shuts down the party.

Putting this all into perspective, if you shift a couple hundred RPM too early, you may lose 0.002 seconds down a particular straight. If you shift a tick too late, after the engine has already touched the limiter, you will most likely lose a tenth or 2. While you always want to use all the HP available to you, it is much better to shift a little early and give up the thou or two rather to bump the limiter and give up a position.

Exactly right. In road racing, more engines are damaged by clueless downshifting than by anything else, by far. And if the engine is at redline when you downshift, you WILL overspeed the engine. (Unless the limiter can catch it.)

[Dmitriy_Guller]

Originally posted by McGuire


The statement in the first sentence above is correct enough as far as it goes, but beyond that you are totally lost.

The product of torque and torque multiplication is not "horsepower". Sorry, but that term has been taken. Horsepower has already had a recognized meaning for over two hundred years, and that's not it.

The product of torque and torque multiplication (gear reduction) is simply more torque, not horsepower.

In cars where engine is mechanically connected to the drive wheels (in other words, any normal car), the product of engine torque and gear multiplication at any fixed speed is perfectly proportional to horsepower (which is identical to saying that torque to the wheels at any given speed is horsepower times a constant, in case you weren't aware). If for example at 100 MPH you change a gear and double your torque at drive wheels, you'll find that the engine would be outputting exactly twice the horsepower at that point in the new gear. This is trivial to show mathematically.

I don't even have to make this argument. Horsepower by definition is the amount of work the engine does (EDIT: over time)*. Since the engine only accelerates the car, it would seem quite obvious that more work from the engine would equal more acceleration. Oh, and my first sentence is not "correct enough": it's perfectly correct, as is every other sentence in my post, which is more than I can say for the enormous volume of nonsense that you keep posting on this subject.

[Wuzak]

Torque is perfectly proportional to power at all times.....not just when put through a reduction gearbox.

Maximum thrust in any given gear is at the point of maximum torque at the engine.

The maximum thrust overall, therefore, is when the gear ratio is the lowest possible (1st) and the engine is at the point of maximum torque.

So one would think the best way to get maximum acceleration would be to have the gears arranged so that when you shift into the next highest gear you would be shifting from maximum hp rpm and dropping back to maximum torque rpm.

[McGuire]

Originally posted by Dmitriy_Guller

In cars where engine is mechanically connected to the drive wheels (in other words, any normal car), the product of engine torque and gear multiplication at any fixed speed is perfectly proportional to horsepower (which is identical to saying that torque to the wheels at any given speed is horsepower times a constant, in case you weren't aware). If for example at 100 MPH you change a gear and double your torque at drive wheels, you'll find that the engine would be outputting exactly twice the horsepower at that point in the new gear. This is trivial to show mathematically.

I don't even have to make this argument. Horsepower by definition is the amount of work the engine does. Since the engine only accelerates the car, it would seem quite obvious that more work from the engine would equal more acceleration. Oh, and my first sentence is not "correct enough": it's perfectly correct, as is every other sentence in my post, which is more than I can say for the enormous volume of nonsense that you keep posting on this subject.

You can't be serious. Sometimes people look at things through the wrong end of the telescope, but you are looking through yours through the wrong orifice.

Acceleration is the resultant of instantaneous wheel thrust, which is solely a function of torque. Horsepower is work over elapsed time, and has no independent meaning removed from time. By definition, horsepower removed from its time element is torque. The only way instantantaneous wheel thrust can be calculated from horsepower is to extract the equivalent torque value from it, and then call THAT "horsepower." How transparent is that?

Your misconception is repeated and illustrated in your defintion of horsepower. You wrote above that horsepower is "the amount of work the engine does." That is EXACTLY wrong here. Work is force-displacement, irrespective of time. A horsepower is 550 pounds per feet per second of work. What you conceive is closer to torque; however: torque is not really work either but a special class of rotational force (not joules but Newton-meters, not ft lb but lb ft).

Now, I know I keep telling you that you don't really understand the concepts of hp and torque, nor the difference between them, and you haven't wanted to hear it. Seriously, how much more evidence do you think you will need?

[WPT]

Posted by Wulzak
So one would think the best way to get maximum acceleration would be to have the gears arranged so that when you shift into the next highest gear you would be shifting from maximum hp rpm and dropping back to maximum torque rpm.

Not so, that is just what my worked example above showed is false. The torque peak rpm in that example is 8000 rpm and one is better off shifting at redline. If one had an infinitely varible computer controlled transmission, for best acceleration would one program it to hold peak torque rpm or peak power rpm during an acceleration run? Look at the above graph. At 108 mph the engine is at peak torque rpm in 6th gear, but one would have greater acceleration in either 5th gear or 4th gear. At a given speed for best acceleration choose the gear that delivers the most power to the drive wheels. WPT

[WPT]

Posted by McGuire
horsepower removed from its time element is torque.

Power is the time rate of doing work, p=(ft*lb)/t. Remove the time "element" and one is left with work, not torque. Another power equation is; p=f*v. Remove the time "element" from the velocity, v, and one is left again with the units of work.

Posted by McGuire
The only way instantantaneous wheel thrust can be calculated from horsepower is to extract the equivalent torque value from it, and then call THAT "horsepower

Not true. One can use p=f*v to calculate the instantaneous thrust. WPT

[McGuire]

Originally posted by Dmitriy_Guller
If for example at 100 MPH you change a gear and double your torque at drive wheels, you'll find that the engine would be outputting exactly twice the horsepower at that point in the new gear. This is trivial to show mathematically.

Maybe so with your math, but with all due respect you suck at story problems.

If you double the torque via the gearbox at 100 mph, that would be a downshift. When torque is doubled at the gearbox, the speed of the output shaft is reduced by precisely one-half, and the wheel speed too. Very quickly this vehicle would no longer be traveling at 100 mph. It's slowing down, fast.

Meanwhile the hp at the rear wheels has not changed, at least to the extent engine rpm remains constant. (I understand you posed this as a hypothetical, but it should be noted that in reality the drive wheels would almost certainly break loose.) Note that gearing cannot alter horsepower, only torque and speed relative to each other. If it multiplies one, it must always divide the other in exact proportion:

When you gear a machine 2:1 (MA = 2) you double the output shaft's torque, but also reduce its speed by half.

Conversely, if you gear the machine 1:2 (MA = .5) you double the output shaft's speed but also reduce its torque by half.

Thus horsepower remains constant. Which is only logical: A pair of dissimilar gears will multiply torque at a fixed rate proportional to their relative speed as determined by their tooth ratio, regardless of their combined operating speed. Therefore no net increase in the machine's horsepower will result.

The explanation that "torque times gearing is horsepower" etc. is an inadvertent but undeniable violation of thermodynamic law: gearing cannot increase the net energy in a system. It would be nice if we could create or increase multiply horsepower with gearing, but unfortunately it doesn't work that way. All gearing can do manipulate torque vs. speed -- there is no way gears can increase both at the same time.

Now in regard to your statement pasted just above, here you truly are looking at things perfectly backward. Please read the following very carefully, for it can help you. If you don't get it, try to avoid the usual pulling your hair and stomping your feet. Simply stop and read it again. Keep studying the sentence below until you fully grasp its meaning, and then you will be placed on the path of illumination:

The ground is not driving the wheels; the wheels are being driven by the engine.

[McGuire]

Originally posted by WPT

Power is the time rate of doing work, p=(ft*lb)/t. Remove the time "element" and one is left with work, not torque. Another power equation is; p=f*v. Remove the time "element" from the velocity, v, and one is left again with the units of work.

Sorry no, not in this case. Interestingly, there is nothing about an automotive engine which precisely meets the classical definition of work. Here we have only torque, rpm and horsepower.

Watt's basic definition of horsepower: 33,000 lb per feet per minute of work.

Here, translated to its equivalent in a 4-stroke automotive engine:

HP = torque x rpm / 5252 (5252 is simply 33,000/2pi)

Note that all the same elements exist in both cases, with the exception of work. And as you say, torque is not work. Technically it is a class of rotational force. Note that in order to avoid confusion, torque is measured in newton meters rather than joules (a representation of equal quanitity but not an equal property) and in lb ft rather than ft lb. We could say that 33,000/2pi is a "conversion" of torque into work, but that is a bit of a fudge. And in any case, acording to HP = torque x rpm/5252, when horsepower is removed from time interval torque remains. No mention of work, eh.

[McGuire]

Originally posted by WPT
Not true. One can use p=f*v to calculate the instantaneous thrust. WPT

Please go back and read what I wrote again.

[Dmitriy_Guller]

Here's a simple example:

The engine produces 50 lb-ft of torque at 2000 RPM, 100 lb-ft of torque at 4000 RPM , and 75 lb-ft of torque at 8000 RPM. Consequently, the horsepower figures are respectively 19.04, 76.16, and 114.24 . In first gear, the gear ratio is 4:1, in second gear it's 2:1, and third gear is 1:1. You're driving in second gear at 4000 RPM, and therefore your engine currently produces 76.16 HP. Torque to the wheels is 200 lb-ft. Right now your car is at peak engine torque, so it must be accelerating the best it can at this speed, right?

Well, let's try downshifting. In first gear, your engine would do 8000 RPM, and torque to the wheels would be 300 lb-ft! After downshifting, your torque to the wheels (and therefore acceleration) is increased by a factor of 1.5! In fact, your engine now outputs 114.24 HP, exactly 1.5 times the 76.16 HP it put out in second gear. Wait a minute, I put my engine into worse torque range, how did my torque at the wheels (and therefore acceleration) suddenly increase by a factor of 1.5? Is it a miracle? Nuh, just mathematical tautology. Your acceleration would always increase by the same factor that your horsepower increases, try it with upshifting to 3rd gear instead.

Engine torque is not proportional to wheel torque, because you can't change engine torque without also changing the gear multiplication while staying at the same speed. However, the product of engine torque and gear multiplication is always truly proportional to horsepower at any given speed, and therefore that is why horsepower is relevant to this thread, and engine torque isn't.

[Dmitriy_Guller]

Originally posted by Wuzak
The maximum thrust overall, therefore, is when the gear ratio is the lowest possible (1st) and the engine is at the point of maximum torque.

Once again, you're doing the common fallacy of assuming that gear ratio stays fixed. It doesn't as long as you have the ability to change gears. The problem with keeping the engine at torque peak instead of horsepower peak is that to accomplish that you have to reduce torque multiplication to get the engine down to RPM of torque peak. What you gain from higher engine torque you more than lose by lowering gear multiplication.

[Dmitriy_Guller]

My mistake, I meant to write that horsepower is the amount of work done over time. My mistake came from being absent-minded when I posted, not from trying to redefine physics and arithmetic to make my theory work.

[Dmitriy_Guller]

Originally posted by McGuire


Maybe so with your math, but with all due respect you suck at story problems.

If you double the torque via the gearbox at 100 mph, that would be a downshift. When torque is doubled at the gearbox, the speed of the output shaft is reduced by precisely one-half, and the wheel speed too. Very quickly this vehicle would no longer be traveling at 100 mph. It's slowing down, fast.

Notice I said "double wheel torque", I did not say "double gear multiplication". There is an enormous difference, even if you fail to see it.

[Dmitriy_Guller]

Just to make sure we're all on the same page regarding the concept of "proportional". Let's take two functions, f(x) and g(x). Let f(x) = 5x. In this case f(x) is directly proportional to x. If you double x, you also double f(x). Now let's look at g(x) = 5*x *(1-x) . Is g(x) proportional to x? No!!! It is not proportional to x because x multiplies another funciton of x, you can't change x without also changing (1-x). If you double x, you would most certainly not double g(x). However, g(x) is proportional to x*(1-x). If you double x*(1-x) by picking the x's that would work, then g(x) would also double.

Now let's tie it to this discussion. Wheel torque = engine torque * gearing multiplication . The only way to change engine torque at given speed (assuming full throttle acceleration, of course) is to change the gear. Now you should see where I'm going with this: you can't change one term without also changing another term. Therefore, engine torque really is not proportional of wheel torque, and wheel torque is the thing that accelerates the car.

[Big Block 8]

Too busy at the moment to gather what you guys are actually arguing about, but to put it quickly:

P (power) = M (torque at wheels) x w (angular velocity at the wheel)

Shortly, this means that a car always accelerates the fastest through the gears at the peak power rpm, despite the fact that the maximum acceleration at one gear is highest at peak torque rpm. Gear ratios and gear changes must be selected in a way, that we get the maximum area under the power graph. Usually this means, that to get the maximum acceleration, we have to rev the engine beyond peak power rpm, then change and go past peak power rpm again and repeat until we are on the last gear.

[Dmitriy_Guller]

Originally posted by Big Block 8
Too busy at the moment to gather what you guys are actually arguing about, but to put it quickly:

P (power) = M (torque at wheels) x w (angular velocity at the wheel)

Shortly, this means that a car always accelerates the fastest through the gears at the peak power rpm, despite the fact that the maximum acceleration at one gear is highest at peak torque rpm. Gear ratios and gear changes must be selected in a way, that we get the maximum area under the power graph. Usually this means, that to get the maximum acceleration, we have to rev the engine beyond peak power rpm, then change and go past peak power rpm again and repeat until we are on the last gear.

Yep, I agree with you 100% percent.

[McGuire]

Originally posted by Dmitriy_Guller

Notice I said "double wheel torque", I did not say "double gear multiplication". There is an enormous difference, even if you fail to see it.

Huh? What are you talking about? Actually, there is no such thing as "gear multiplication." Torque multiplication is accomplished through gear reduction. You are simply unfamiliar with common mechanical terms. That's fine and understandable, but unfortunately the concepts behind them are equally foreign to you. I'm sure you don't quite know what you mean at this point.

And what I told you is true either way, whether you meant gearing up or down. With 1:2 gearing, output speed is doubled while torque is halved. With 2:1 gearing, output torque is doubled while speed is halved. That's why it's called gear reduction, eh.

However, in this case you did say and then repeat "double wheel torque," which means doubling the mechanical advantage of the gears, which also reduces the wheel speed by half. Just exactly as I said before.

So where were we? Oh yes...keep on bailing, your boat is sinking faster now.

[Dmitriy_Guller]

Originally posted by McGuire
However, in this case you did say and then repeat "double wheel torque," which means doubling the mechanical advantage of the gears, which also reduces the wheel speed by half. Just exactly as I said before.

So where were we? Oh yes...keep on bailing, your boat is sinking faster now.

No, simply doubling mechanical advantage through gears would almost certainly not double wheel torque, for the same reasons I alluded to earlier. While torque multiplication is doubled, engine torque is definitely changed, as RPM changes. Whatever the RPM, engine torque, or torque multiplication you wind up with to achive doubling the torque to the wheels isn't important, as long as you achieve it in some way.

Once you do double the torque to the wheels, you would find that your engine in its new RPM is putting out exactly double the horsepower that it did before the gear change. Don't even focus on "double", any factor would do as long as you keep the engine out of rev limiter. My boat is staying afloat just fine, thank you, it can withstand strong gusts of hot air.

[Wuzak]

Originally posted by Dmitriy_Guller
Just to make sure we're all on the same page regarding the concept of "proportional". Let's take two functions, f(x) and g(x). Let f(x) = 5x. In this case f(x) is directly proportional to x. If you double x, you also double f(x). Now let's look at g(x) = 5*x *(1-x) . Is g(x) proportional to x? No!!! It is not proportional to x because x multiplies another funciton of x, you can't change x without also changing (1-x). If you double x, you would most certainly not double g(x). However, g(x) is proportional to x*(1-x). If you double x*(1-x) by picking the x's that would work, then g(x) would also double.

Now let's tie it to this discussion. Wheel torque = engine torque * gearing multiplication . The only way to change engine torque at given speed (assuming full throttle acceleration, of course) is to change the gear. Now you should see where I'm going with this: you can't change one term without also changing another term. Therefore, engine torque really is not proportional of wheel torque, and wheel torque is the thing that accelerates the car.

Gear ratio is not a function of engine torque. Not in a manual gearbox, anyway.

The gear ratio is T2/T1 where T1 and T2 are the numbers of teeth on the driving and driven gears.

It is also equivalent to N1/N2, where N1 and N2 are the input and output speeds of the gearbox.

The output torque of a gearbox is given by:

OT = IT x N1/N2,

or OT = IT x T2/T1,

where IT = input torque.

Thus output torque is proportional to input torque.

Ouput torque is also proportional to the speed ratio/gear ratio.

[pio!pio!]

I am not mathematically or technically inclined to the level that you fellows are but would this be a good strategy?

1) Get the vehicle on a dyno and do several pulls in all the gears (assuming no wheel spin etc etc)
2) Graph the hp vs speed of all the dyno runs in the different gears

From that point there will be times where, hypothetically, the hp in 3rd gear starts to decrease and intersects with the hp in 4th gear on the rise..at the point is the place to shift. There may also be hypothetically, the hp in 2nd gear decreases to redline but never intersects the hp in 2rd gear on the increase. In that case you shift at redline...

Does that make sense? that maximizes the area under the overall hp curve when rowing through all the gears

[Dmitriy_Guller]

Originally posted by Wuzak


Gear ratio is not a function of engine torque. Not in a manual gearbox, anyway.

The gear ratio is T2/T1 where T1 and T2 are the numbers of teeth on the driving and driven gears.

It is also equivalent to N1/N2, where N1 and N2 are the input and output speeds of the gearbox.

The output torque of a gearbox is given by:

OT = IT x N1/N2,

or OT = IT x T2/T1,

where IT = input torque.

Thus output torque is proportional to input torque.

Ouput torque is also proportional to the speed ratio/gear ratio.

Engine torque is certainly a function of a gear at any given speed. You have to appreciate the constraints that "at any given speed" implies. If you change a gear, then you necesserily change the RPM of the engine (assuming no slip), because the engine is mechanically connected to drive wheels. If you change the RPM of the engine, then you're changing the engine's torque output.

Same principle works backwards: the only way to change engine's torque (if you keep it at full throttle) is to change its RPM, and the only way to change the engine's RPM is to choose another gear. However, then you're changing the gear ratio (torque multiplication) as well. Since you can't change one quantity without changing the other at any given speed, neither enginer torque nor gear ratio is directly proportional to torque at the wheels, only their product is.

[Big Block 8]

Originally posted by Dmitriy_Guller

Yep, I agree with you 100% percent.

Still no time to meddle into the argument, but a couple of words more - generally speaking, it's amazing how hard for some people is to grasp the meaning of torque, crank torque and horsepower.

Lots of "experts" all around, even in reputable car magazines, declare that "torque" moves the car, not horsepower - statement which is exactly 1/2 wrong and totally wrong, if they mean by the "torque" the crank torque, which they most often do. Another trivialized chant is the "(crank) torque means acceleration, hp means the top speed". Crank torque does not mean acceleration and hp means the top speed only, if coincidentally the vehicle is geared to reach equivalent rolling & drag resistance power at that given speed, which actually never happens outside a math exam.

Horsepower from the engine's output shaft at given speed of the vehicle exert a torque to *wheels* and that is what moves the car - horsepower, period. Mathematically and for pretty long in practice as well, crank torque means exactly dick when all-out performance is considered. But flexibility, longevity, driving comfort, manufacturing costs - here are some aspects where changes in the crank torque number become interesting.

Usually I give many things a benefit of doubt, but this is very simple business, which I've witnessed both in theory and in practice and hence I must say it is true.

[Big Block 8]

Originally posted by pio!pio!
I am not mathematically or technically inclined to the level that you fellows are but would this be a good strategy?

1) Get the vehicle on a dyno and do several pulls in all the gears (assuming no wheel spin etc etc)
2) Graph the hp vs speed of all the dyno runs in the different gears

From that point there will be times where, hypothetically, the hp in 3rd gear starts to decrease and intersects with the hp in 4th gear on the rise..at the point is the place to shift. There may also be hypothetically, the hp in 2nd gear decreases to redline but never intersects the hp in 2rd gear on the increase. In that case you shift at redline...

Does that make sense? that maximizes the area under the overall hp curve when rowing through all the gears

Yes, that's exactly the way to do it.

[Big Block 8]

Originally posted by Dmitriy_Guller

Engine torque is certainly a function of a gear at any given speed. You have to appreciate the constraints that "at any given speed" implies. If you change a gear, then you necesserily change the RPM of the engine (assuming no slip), because the engine is mechanically connected to drive wheels. If you change the RPM of the engine, then you're changing the engine's torque output.

Same principle works backwards: the only way to change engine's torque (if you keep it at full throttle) is to change its RPM, and the only way to change the engine's RPM is to choose another gear. However, then you're changing the gear ratio (torque multiplication) as well. Since you can't change one quantity without changing the other at any given speed, neither enginer torque nor gear ratio is directly proportional to torque at the wheels, only their product is.

Minor correction - you are not changing the engine's (crank) torque output by changing it's rpm (assuming relatively flat torque curve). Changing the engine RPM at WOT will only change it's horsepower output. Then, if the vehicle's speed remains constant, torque at wheels will change accordingly.

[Dmitriy_Guller]

Originally posted by Big Block 8


Minor correction - you are not changing the engine's (crank) torque output by changing it's rpm (assuming relatively flat torque curve). Changing the engine RPM at WOT will only change it's horsepower output. Then, if the vehicle's speed remains constant, torque at wheels will change accordingly.

I don't want to make any more assumptions than necessary so that I would not lose generality, and I don't see the need to assume flat torque curve. What I'm talking about applies even to the bottom end of the engine, where torque curve still rises sharply. In any way, what's important is that crank torque still changes as RPM changes, however much or little is up to specific engine, and my statement applies equally well to any engine.

As for your previous post, I agree completely that this is an extremely simple issue. I think the difficulty lies entirely with the fact that crank torque fallacy is widely accepted, and that many of the people who push the theory don't exactly have the mental rigor of mathematicians and therefore do not accept the debunking. I guess there is also desire to think that you as "expert" know more than the general public, which only knows about peak HP figure. There is indeed more to know, but that hidden information is horsepower curve, not the peak torque figure.

[Big Block 8]

Originally posted by Dmitriy_Guller

I don't want to make any more assumptions than necessary so that I would not lose generality, and I don't see the need to assume flat torque curve. What I'm talking about applies even to the bottom end of the engine, where torque curve still rises sharply. In any way, what's important is that crank torque still changes as RPM changes, however much or little is up to specific engine, and my statement applies equally well to any engine.

Ok, my advice then would be to mention, that changing the engine rpm will change the engine's torque output to corresponding point in the torque/rpm curve. Flat curves are often both a design goal and the ideal status, so assuming them relatively "flat" in above example would perhaps make things more clear. Then the only variable would be the engines horsepower output. Just a hint for the future lectures.;)

Originally posted by Dmitriy_Guller

As for your previous post, I agree completely that this is an extremely simple issue. I think the difficulty lies entirely with the fact that crank torque fallacy is widely accepted, and that many of the people who push the theory don't exactly have the mental rigor of mathematicians and therefore do not accept the debunking. I guess there is also desire to think that you as "expert" know more than the general public, which only knows about peak HP figure. There is indeed more to know, but that hidden information is horsepower curve, not the peak torque figure.

Yes, I could pretty much agree with that. Peak torque number is a nice little piece of data though - if it's understood and interpreted correctly, that is.

One more of my favorite *OMG* chants is the "engines don't create horsepower, they create torque". I'd really like to see these "engines". What are they exactly? Crowbars?

[Big Block 8]

Finally had time to read this thread more thoroughly...

Originally posted by McGuire

For example, take these two engines:

Engine A makes 40 hp at 2200 rpm. (95 lb ft of torque)

Engine B makes 40 hp at 4400 rpm. (47 lb ft of torque)

We will take these two engines to Bonneville, installed in identical vehicles with direct drive, 1:1 gearing. Since the engines make identical horsepower, we know that their top speeds will also be identical. (Theoretically speaking, of course.)

I think I can now see what's the argument here. No offense McGuire, but the only practical theory I can think of that would make these two vehicles have identical topspeeds, would be to set them to start running against a large stone, which would give both a topspeed of zero. The other would be that the A would coincidentally reach it's geared limit just exactly when the B reaches drag/roll resistance limit, which would definitely be just a math exam anomaly.