Quote from tristancliffe :

it's perfectly possible that the optimum gear shift point of a car is BEFORE peak power. It just depends on your gearing.

I agree - and that point can also come after your peak power output just as it can come before (granted not as likely as before depending on what gear your're in, but still...) It's about maximizing area under the curve.

But: Which curve?

Quote :

It's a simple F = M x a problem. For a fixed mass, the greatest force will result in the greatest acceleration.

Only for that moment in time, which is somewhat not relevant, or at least doesn't relate to what my argument is.

You still need to explain to me why you maintain that absolute force has more to do with acceleration over time than the rate that the force is applied does. Again, I feel that it's about work - it takes more work to move an object from here to there in less time, and that takes more power, not necessarily more force. Take any given force. Apply that same force twice to the same body under the same conditions, but at different rates. The quicker you apply the force, the more work has been done, and more POWER has been utilized. That example is a bit fantastic and impossible but it illustrates my point about the factor of time, and the potential to do work (which is power not force).

I could be dense, but it still seems like you're comparing apples to oranges. I am speaking about getting from A to B as fast as possible - not about any specific moment within the time it takes to get from A to B.



Edit: here's a question, and I've seen it before on here: Will a CVT equipped car accelerate most rapidly from A to B by holding RPM at peak power or peak torque?

Think about top fuel drag cars..... The whole objective of such beasts is to perform the most work on that car as possible to get it from here to there like a whirlwind on crack. They want the most force in the least TIME as possible, which takes power.... I think around 6000 horses of it or so.....

edited for clarity again.

I am taking bets ladies and gentlemen. I have my favorite . . .

Quote from Ball Bearing Turbo :

I agree - and that point can also come after your peak power output just as it can come before (granted not as likely as before depending on what gear your're in, but still...) It's about maximizing area under the curve.

But: Which curve?

Indeed, maximising the area under the tractive effort chart (torque) will yeild stronger performance, either through tuning, more revs (longer in lower gears) or less transmission losses. And maximising the area under the tractive effort chart will result in more area under the power curve, as the power curve is derived from the torque curve.

Quote from Ball Bearing Turbo :

Only for that moment in time, which is somewhat not relevant, or at least doesn't relate to what my argument is.

Eh? Okay, let mass change as fuel changes or whatever. You still know the force and the mass at any one time and can therefore workout the speed, acceleration and velocity if you know the starting conditions. Thus starting from rest and assuming no wheel spin or clutch slip it's possible to work out how long it'll take to do a 1/4 mile or reach 200mph. Track work is different as braking and turning come into it, but for pure acceleration tractive effort and mass is all you need.

Quote from Ball Bearing Turbo :

You still need to explain to me why you maintain that absolute force has more to do with acceleration over time than the rate that the force is applied does. Again, I feel that it's about work - it takes work to move an object from here to there, and that takes more power, not more force. Take any given force. Apply that same force twice to the same body under the same conditions, but at different rates. The quicker you apply the force, the more work has been done, ore POWER has been utilized. I could be dense, but it still seems like you're comparing apples to oranges. I am speaking about getting from A to B as fast as possible - not about any specific moment within the time it takes to get from A to B.

Power ISN'T the rate at which force is applied. 200N applied is 200N whether you do it for 1ms or 1 year. And the rate of acceleration will be the same if the mass doesn't vary. Power is the measure of work done on an object.
At it's simplest, lots of work doesn't necessarily mean lots of acceleration, just as lots of acceleration doesn't necessarily mean lots of power. The two are connected, but not interchangable. Torque is what drives a vehicle. Torque is what you (should) tune in an engine. You might tune for high end torque or low end torque, or mid range torque, but it's that that you want to alter. I'm not quite sure what you're saying still. It's probably me being Mr Thicko, but I'm sure we'll sort it out sooner or later...

Quote from Ball Bearing Turbo :

Edit: here's a question, and I've seen it before on here: Will a CVT equipped car accelerate most rapidly from A to B by holding RPM at peak power or peak torque?

Think about top fuel drag cars..... The whole objective of such beasts is to perform the most work on that car as possible to get it from here to there like a whirlwind on crack. They want the most force in the least TIME as possible, which takes power.... I think around 6000 horses of it or so.....

A CVT car will accelerate quickest with the engine at peak torque. edit: hmmm, I have a feeling the extra variable (the gear ratio) means this is wrong, but for fixed geared cars it's 100% correct. *goes to read about cvt's and do some maths*

The drag car does indeed what the most force in the least time. Clearly the faster it accelerates the quicker it's 1/4 mile time will be. And to get the quickest acceleration at any given point or differentiated over the whole time period, it needs the most force at the wheels it can acheive. This is, for any given gear ratio, varying or not, at the point of peak torque, when the brake efficiency is at it highest.

Edit2: phew, my edit was before Colcobs edit, so I correctly deduced the CVT trick question just in time

OMG, KILL ME NOW.

Not this again, please for the love of all that is holy.

*col goes searching for the TORQUE VS HORSEPOWER THREAD OF DOOM.....*

(..and you're still wrong Tristan, a CVT car will accelerate quickest at peak power because by definition, the CVT will adjust the transmission to give greatest torque AT THE WHEELS. You're confusing engine torque with wheel torque again)

:monkey:

Hey keep it going . . . I never caught the last Power v torque thread of doom so this is really interesting . . . . . You guys know your stuff.

Quote from tristancliffe :

The drag car does indeed what the most force in the least time. Clearly the faster it accelerates the quicker it's 1/4 mile time will be. And to get the quickest acceleration at any given point or differentiated over the whole time period, it needs the most force at the wheels it can acheive. This is, for any given gear ratio, varying or not, at the point of peak torque, when the brake efficiency is at it highest.

Cars that want to get from point a to point b in the minimum amount of time always do so with the engine running at its maximum power, not its maximum torque. These cars have appropriate gearing.

This is commonly accomplished in drag cars by the use of various forms of loose gearing. Drag racing torque converters for automatic transmissions seek to bring the engine to peak power as quickly as possible. In true drag only applications, the point is to have flash stall at or very near peak power, not peak torque. The drag racers then further make use of the change in tire diameter that comes along with thin sidewall slicks, minimizing the number of ratio changes necessary and thereby maximizing time spent at peak power.

Really fast drag applications simply run the engine up to peak power and then slip the clutches just enough to keep it there all the way down the track. The power lost via this inefficient method of delivering energy to the road is far less than that lost by having fixed gearing that requires time consuming shifting and running the engine at points other than peak power.

BBT is correct that power is what determines the time from a to b, and given the appropriate gearing torque is completely meaningless. This is why all rudimentary estimates of time to distance and top speeds utilize power figures, not torque figures or curves.

To the CVT - think about it, for any given speed, the extra wheel torque given from the increased gearing reduction at peak power rpm outweighs the lower torque output at the engine.

The whole cofusion is caused because you keep thinking about ONE gear. You have to remember, you'll never have the as revs low as peak torque, because you'll have changed down, reduced engine output but increased wheel torque. It's something that took me a while to click from the last thread but it finally makes this weird kind of clarity in your head.

See attached GRC sceeny that the maximum wheel torque can be kept by keeping the revs above peak torque (ignoring first gear of course). OK it's not always the case but this proves that peak wheel torque coincides with peak power.

I hear Tristan getting his towel out....

And warming up his throwing arm....

I'm going to ignore CVT's for the time being until I can come up with or be shown a graph that actually proves anything... I'm reading a few things, and I'll see what I can come up with.

But for manual transmissions, I still can't see the peak wheel torque coninciding with maximum power. That screenshot of the GRC shows the the wheel torque is falling away after peak engine torque. A gear is a torque multiplier, and as such the peak wheel torque (or force at the wheels) will have to be at the peak engine revs in any given gear, although the speed at which this peak arrives will obviously change.

I have taken the GRC sceenshot and added the points where peak wheel torque in any of the gears occurs. In this example it is just after the shift point because of the gearing. To the left of the red blob you would be better to change down a gear due to torque multiplication, and anything to the right of the blobs is less that optimum for that gear, but better than the next gear until it crosses over. Thus peak torque in this case isn't at the engines peak torque because the gearing suggests you would change differently.

Perhaps we are trying to explain different things?

Edit: forgot screenshot :S

Edit2: Right, I've done some calculations and reading, and I am 100% happy that CVT's will run at peak power for maximum tractive effort at the wheels. It took a bit of thinking, but I understand it now... As for Manual cars, you still get peak torque at the wheels when the engine is at maximum torque. The gearing you have defines when the crossover point is, which defines the gear change point. The more gears you have the closer you can get to the maximum area under the graph being a smooth hyperbola (i think it is), but as most cars only have 4, 5 or 6 gears this bit of the argument isn't brilliant. You either have (mostly) 6 or less, or continuously variable. For a manual transmission with definite, well spaced gearing for all conditions (acceleration, top speed, economy etc) as on road cars then the maximum wheel torque, as I have said, WILL occur at maximum engine torque NOT maximum power.

I am back

Mr. Tristan:
Well, you conceded that with a CVT - peak power determines maximum acceleration, but in doing so you also condeded the whole debate, because the fact that a CVT is infinitely variable is not relevant. If you consider this carefully you will see that if it's true for a CVT, it's true for any GIVEN gear ratio, if for no other reason than simply on the basis that CVTs "use" every gear ratio possible - including the ratios you would use in a regular manual. Put differently, if it's true for a CVT, it's true for ANY transmission known to man because a CVT is effectively incorporating every ratio you could ever have in any feasible transmission.

Which brings me to my second point, the death blow.

Recall how I've been trying to convey the importance of time in this whole debate? It FINALLY occured to me to define power in terms of it's components (yes, it took a long time for that revelation....) being force and time, and therefore I submit to you that 1HP = 550 lb/ft per second.

This is exactly why more power yeilds greater acceleration, not necessarily more force alone, but more force over time is the key factor. Thus, even an engine that has long past it's torque peak, but is at it's POWER peak has more actual output. Surely I don't even have to post the numbers for everyone to realize that HP*550 gives you the "amount of force per second" (vulgarized term for convenience) being generated.

However I will give an example:
in the attached graph, Note the following:

Peaks:
Torque: ~300lb/ft at ~4000RPM.
HP: ~290HP @ ~5500RPM

Note the force output at both peaks:
5500RPM = 290HP = 159500 Pound Feet of Torque per Second
4000RPM = ~200HP = 110000 Pound Feet of Torque per Second

Thus, 45% more work is being done at peak power than at peak torque.

Your need for force is now satified in the proper venue: time.

Quote from colcob :

OMG, KILL ME NOW.

Not this again, please for the love of all that is holy.

Forgot to mention: ROFL LMAO

:munching_

What is happening here is a clash of definitions.

There are two ways of defining acceleration in this debate.
The strictly scientifically correct definition, which is the rate of change of velocity at an instantaneous moment in time; and the more practical but semantically innacurate definition of 'how quickly can we go from point A to point B'.

In the first case, the highest instantaneous acceleration will always occur when the longitudinal force is highest, which is at the torque peak in any given gear. Of that there is no doubt.

But when people talk about the 'acceleration' of cars, they tend to be actually thinking of how long does it take to get from point A to point B, or even how long does it take to get from velocity A to velocity B.

In these cases we get back to thinking about the area underneath graphs, rather than the maxima of them.
If you take a graph of acceleration over time, the total area under that graph is equal to the average velocity over that time. So anything you do to increase the overall area under the tractive effort curve gives a faster average velocity over the course of the acceleration. Which means that your final velocity will be higher and your overall distance travelled will be greater.
Ergo, in laymens terms, the 'acceleration' is better.

So basically, the performance potential of an engine is dictated by the area under the torque curve. Now before someone jumps in and says 'but what about peak power', I should say that peak power is dictated indirectly by the area under the torque curve, the longer the torque curve stays above a certain point, the higher the peak power will be.

An excellent illustration of all these points is the restricted V10's that Torro Rosso are using in F1 this year. They have been rev and inlet restricted to give supposedly the same peak power as a V8, but they have a greater area under the torque curve due to those 10 cylinders pulling before the restrictor kicks in, which means they have a better performance potential than they should.

So in summary, acceleration performance is not dictated by peak power only, or peak torque only, but by the total area underneath the torque curve (within the rev range used by the current gearing setup).

I love mixing in enlightened company. Makes me feel clever.

I must say that I am reading this with interest and with only a slight fuzzing around the edges.

Hopefully, when you have come to some kind of consensus I might know whether it's better to put the gear change of my FZR on the power peak or the Torque peak. TBH I'm going for the Torque atm. From what I know of engines and cars, torque does seem to be the more important of the two. But I do realise that you can't have one without the other . . .

No. You change gear when the red light comes on. Simple as that.

LFS calculates when the next gear will give you more acceleration than the current one, and thats when the light comes on.

Yea, ok. Smartarse. Maybe what I really meant was to . . Maybe what I meant was that if you . . . . Maybe . . . Damnit. I'll just shut up and read.

Yeah, the ideal shift point is very shortly after the red light comes on. ("Shortly after" because you lose a bit of speed in the shift process)

Let's return to the original question - why is the FZR faster. Someone said the XRR has more power, and I said that didn't matter one iota. Neither did the max torque figure. I stated that the torque curve was the important thing. This is because of the area beneath it.

Quote from tristancliffe :

More power at some specific point. Remember the peak power figure of a car is the most useless figure bandied about by anyone. Peak torque is much better, but still fairly useless. What you want is a torque curve - the only bit that matters.

For a given fixed gear ratio the acceleration in that gear WILL be greatest at peak torque output. For a CVT you can't state the same thing as you don't have a fixed gear to judge when 'best' acceleration takes place. The CVT alters it's ratio to ensure that the area under the tractive effort curve is at it's greatest.
With a manual it is neccessary to choose the gear ratios for a number of factors - low speed control (i.e. acceleration with wheelspin or boggin down), high speed atainability (reaching max speed on the straights), and drivability in between. For best performance you will, of course (and I've never stated differently) have to change gear around the peak power figure so that you 'use' as much of the area under the tractive effort curve as you can.
That's what the red light in LFS tells you. If you change when it comes on you use the area under the graph more effectively. Shift too early and you waste a gear, whift too late and you'll waste a gear. In wasting the gear you lose some potential (except when purposely stretching a gear before a corner, but thats different).

Hence power is only good for judging top speed, where you are at a fixed rpm (eventually) and only in one gear. I don't think any shortcut for estimating acceleration will ever be particularly accurate, there are just too many factors to consider - you need to sit down and simulate it. That's why I don't really like power-to-weight ratios, they don't tell you enough about acceleration to be that meaningful, and because the traction of the drive wheels needs to be known also, you don't know how difficult they'll be to get off the line either.

Quote from tristancliffe :

Remember the peak power figure of a car is the most useless figure bandied about by anyone. Peak torque is much better, but still fairly useless.

This is just woefully untrue.

Some rather obvious (tired, even) examples come to mind:

-I have an electric motor here capable of peak torque greater than an F1 engine. It produces less than 5HP.
-I have seen hydraulic wheel motors capable of similar feats.
-In terms of more plausable stuff in the automotive realm, the peak torque of the 3.0 liter Vulcan in my van is 165 ft-lbs. Same as the motor in my Mazda. The motor in my Mazda produces an extra 70 horsepower. It gets from A to B much quicker. Yes, in the same vehicle with the same gearing, the Vulcan is capable of the same peak instantaneous acceleration figure. Speaking of dumb figures, there is one right there. The only thing that matters is the average acceleration over time. High peak accelerations just break things and break loose tires.
-Importantly, the reverse of the non-auto examples above do not demonstrate a situation in which a power figure is "worthless". The torque output of the power turbine of a PT6A is rather tiny, but gear the thing from 20,000+ rpms to 2100 rpms at a propeller, and you've got more than a 1000ft-lbs and hundreds of horsepower. No matter what speed the engine moves at, you can find a way to make use of the power it creates.

In summary, a 300ft-lb torque electric motor might chirp the tires, but might not get you anywhere quickly, while a 40ft-lb turbine engine might get you there faster than the 200ft-lb car engine.

There are some really good reasons why you do, as you say, need a torque curve to look at. However, for a cursory inspection or simple matchup, the peak power figure will be vastly more meaningful than the peak torque figure. This is why nearly everything on earth used to provide motivation to twist anything has a power output specified and printed on it, not a torque output.

Maybe the reason why this isn't so obvious to you is that you haven't driven a 1970's Cadillac equipped with a emissions regged big block. Uh, 400ft-lbs of torque, check. Dead slow, check. Less power than a new Camry, check.

Quote from Bob Smith :

I don't think any shortcut for estimating acceleration will ever be particularly accurate

I think you are quite correct, Bob, although the inaccurate shortcuts are quite useful when trying to determine what size motor you need to run the lawn mower, or roughly how many horsepower you'll need to drop in the 2800lb auto to run the quarter mile in 13's. For relatively slow cars, the simple three/four variable quarter mile calculators can be accurate within +/- 10%, which is damned impressive considering the small number of variables used. Stuff gets twisted when you get to cars with enough power (meh, torque over a bunch of time) to spin the rubber bits in bigger gears.

Quote from colcob :

What is happening here is a clash of definitions.

Ok...

Quote :

There are two ways of defining acceleration in this debate.
The strictly scientifically correct definition, which is the rate of change of velocity at an instantaneous moment in time; and the more practical but semantically innacurate definition of 'how quickly can we go from point A to point B'.

How can there be measure of rate of change at an instant? Change is measured over time, if no time elapses it's impossible for change to take place!

Quote :

In the first case, the highest instantaneous acceleration will always occur when the longitudinal force is highest, which is at the torque peak in any given gear. Of that there is no doubt.

I beleive I proved above that the the delivery over time of force is actually highest at peak power output.

Quote :

But when people talk about the 'acceleration' of cars, they tend to be actually thinking of how long does it take to get from point A to point B, or even how long does it take to get from velocity A to velocity B.

Again with my density, but I fail to see how one can be true but not the other. If "peak acceleration" occurs at "peak torque", then how can holding a CVT at the "peak torque" not yeild "peak acceleration" from point A to B? That's paradoxical to me! If the "rate of change for that instant" is the greatest it can be, how could it (the torque peak!) NOT produce the greatest rate of change over time? But it doesn't!


We've already established that CVTs will have maximum acceleration at peak engine POWER output (Power; BTW I've clearly also defined above, in terms of "force output" for lack of better ability to articulate.) That being true, and it is, how could the same possibly NOT be true for any given gear ratio - since by saying it's true for a CVT, we have said it's true for an infinite number of ratios! That includes the 3, 4, 5, 6 ratios in any automatic or manal transmission on the earth!

Quote :

In these cases we get back to thinking about the area underneath graphs, rather than the maxima of them.
If you take a graph of acceleration over time, the total area under that graph is equal to the average velocity over that time. So anything you do to increase the overall area under the tractive effort curve gives a faster average velocity over the course of the acceleration. Which means that your final velocity will be higher and your overall distance travelled will be greater.

Yes. We've all been saying this.

Quote :

Ergo, in laymens terms, the 'acceleration' is better.

The problem is if a particular condition (which some are saying is the peak torque output of the engine) is capable of producing the greatest acceleration even for a "moment", then holding that condition constant (as in the case of a CVT) would yeild the greatest acceleration over time as well (because as stated, "for any given gear ratio peak acceleration happens at peak torque" right?) , but it does not. Why? Because the overall output is still less than it is at peak power, and less tractive effort is available.

Quote :

So basically, the performance potential of an engine is dictated by the area under the torque curve. Now before someone jumps in and says 'but what about peak power', I should say that peak power is dictated indirectly by the area under the torque curve, the longer the torque curve stays above a certain point, the higher the peak power will be.

Peak power is not dictated indirectly, it's DIRECTLY influnced by the torque curve!

Analogy:
What's the point in measuring a room in 2 dimentions? It tells you nothing about the "potential" of the room (what you can fit) Torque = 2D, Power = 3D

Quote :

So in summary, acceleration performance is not dictated by peak power only, or peak torque only, but by the total area underneath the torque curve (within the rev range used by the current gearing setup).

Both curves are relevant. NOT just the torque curve!

Quote from Ball Bearing Turbo :

How can there be measure of rate of change at an instant? Change is measured over time, if no time elapses it's impossible for change to take place!

How can you state the momentary acceleration of an object if it's at a moment. Acceleration is a rate of change, yet you can state the acceleration at time x.

Quote from Ball Bearing Turbo :

I beleive I proved above that the the delivery over time of force is actually highest at peak power output.

Not really, you just stated some stuff. Some is correct, some isn't.

Quote from Ball Bearing Turbo :

Again with my density, but I fail to see how one can be true but not the other. If "peak acceleration" occurs at "peak torque", then how can holding a CVT at the "peak torque" not yeild "peak acceleration" from point A to B? That's paradoxical to me! If the "rate of change for that instant" is the greatest it can be, how could it (the torque peak!) NOT produce the greatest rate of change over time? But it doesn't!

This is what I was struggling with last night - how come in CVT's you keep them at peak power for best performance but with a MT you want max torque for best performance in any gear. This is because of several factors.
If you had a manual gearbox with an infinite number of gears then maximum acceleration in any one of those gears is still at peak torque. If the ratio is constant, best acceleration is at peak torque. Always. But the ratio in a CVT is an extra variable. What I did to understand it is take a tractive effort graph for a 6 speed gearbox, and then plot curves with the same torque curve for a gearbox with 100 ratios. The area under the tractive effort curves is maximised when you run the engine on the right hand side of the torque curve, which happens to be around peak power. I'm not sure if it is actually AT peak power though, but it certainly is close.

Quote from Ball Bearing Turbo :

We've already established that CVTs will have maximum acceleration at peak engine POWER output (Power; BTW I've clearly also defined above, in terms of "force output" for lack of better ability to articulate.) That being true, and it is, how could the same possibly NOT be true for any given gear ratio - since by saying it's true for a CVT, we have said it's true for an infinite number of ratios! That includes the 3, 4, 5, 6 ratios in any automatic or manal transmission on the earth!

Dealt with that above. You are trying a compare two systems with a different number of variables - one has fixed ratios the other are variable.

Quote from Ball Bearing Turbo :

The problem is if a particular condition (which some are saying is the peak torque output of the engine) is capable of producing the greatest acceleration even for a "moment", then holding that condition constant (as in the case of a CVT) would yeild the greatest acceleration over time as well (because as stated, "for any given gear ratio peak acceleration happens at peak torque" right?) , but it does not. Why? Because the overall output is still less than it is at peak power, and less tractive effort is available.

Still, comparing systems with different variables. If you could 'lock' a CVT at a given ratio (and you can on some modern CVT's) you will find the peak acceleration in that gear will occur at peak torque. If you are in your car and you want to overtake a tractor it's best to hold the car in a gear so that it's just below peak torque - because acceleration will be greatest whilst you overtake.
Another test - use LFS. On a flat road (dragstrip) accelerate in 4th gear and watch the g-meter. It should peak at peak torque (although I suspect that air resistance will skew the results too much anyway).

Quote from Ball Bearing Turbo :

Peak power is not dictated indirectly, it's DIRECTLY influnced by the torque curve!

Yes, but indirectly dictated by the AREA UNDER THE torque curve (sorry for caps, using it to exmphasise).

Quote from Ball Bearing Turbo :

Analogy:
What's the point in measuring a room in 2 dimentions? It tells you nothing about the "potential" of the room (what you can fit) Torque = 2D, Power = 3D

Sorry, thats rubbish. Torque is more relevant. We all, apart from you, say so. Read a few books and you'll see

Quote from Ball Bearing Turbo :

Both curves are relevant. NOT just the torque curve!

Yes, power is relevent, but a LOT less than the torque curve.

Quote from tristancliffe :

How can you state the momentary acceleration of an object if it's at a moment. Acceleration is a rate of change, yet you can state the acceleration at time x.

An instant is not relevant when you're concerned with time, over which acceleration happens.

Quote :

Not really, you just stated some stuff. Some is correct, some isn't.

Which is correct and which isn't and why? I noticed you conveniently skipped that post! I did in fact prove completely that the most force is delivered in the least time at peak power when compared with peak torque, and I even did it with numbers. You cannot deny that that is true, because it would be ludicrous. Plain and simple, and I don't even need a book!

Quote :

If you had a manual gearbox with an infinite number of gears then maximum acceleration in any one of those gears is still at peak torque.

Are you being serious? You just basically pulled a 180! A CVT IS a transmission with an infinite number of ratios!, and the max acceleration you stated yourself, using graphs and all, occurs at peak power. I don't know how you could say that with a straight face!

Quote :

If you could 'lock' a CVT at a given ratio (and you can on some modern CVT's) you will find the peak acceleration in that gear will occur at peak torque.

Then why wouldn't it happen (peak acc @ peak tq) if you didn't lock it? It still has to "pass" that particular ratio as it modifies the ratio! There are no more variables, all there is is a greater resolution of ratios, which if broken down in into any subset of ratios be it 1000, 100, or 6, all yield the same result I'm afraid.

Quote :

If you are in your car and you want to overtake a tractor it's best to hold the car in a gear so that it's just below peak torque - because acceleration will be greatest whilst you overtake.

That's not true for that reason. Ideally, at any given speed, in any given ratio, you would want peak power output to perform the most work possible in the least time possible, regardless of available force. You gear there so that you can move into the max area under the curve, which moves you towards peak power, and that's how they relate. Passing a tractor takes time, not an instant.

Quote :

Yes, but indirectly dictated by the AREA UNDER THE torque curve (sorry for caps, using it to exmphasise).

same with all my caps, just for emphasis. Not yelling

Quote :

Sorry, thats rubbish. Torque is more relevant. We all, apart from you, say so. Read a few books and you'll see

Actually, not everyone agrees with you, I'm sure you've read all the posts. I think I have at least one supporter and a few wishwashers lol.

And I don't read. Too much bogus information out there.

Oh I give up.

BBO, I was with you up to a point, but obviously you havent studied maths or physics to a high enough level because although your instincts are good, you're struggling with the maths.

The concept of rate of change in an instant is fundamental to the entirety of calculus. In a sense you've been quite astute because the instant is actually defined as an infintestimal quantity of time. But if you take a graph, the gradient of a graph at any point is its rate of change.

Clearly you can point to a discrete spot on a curve and say that it has a gradient. This is what calculus is all about in a way, its a mathematical method of determining what the gradient is at a given point on a curve (and loads of other things as well, but lets not complicate it).

In the question of the CVT transmission, you are making the same mistake as Tristan in confusing the force from the engine with the force applied at the tyres.
A car accelerates ONLY by applying force to it. Acceleration is directly proportional to force (a = F/mass). So the torque applied at the wheels is the only determining factor (ignoring mass) as to how fast the car will accelerate at a given moment.

The torque generated by the engine is multiplied by the gearing factor to give the torque at the wheels. So if an engine puts 100Nm at 1000RPM into a gearbox with a ratio of 1:2 (or 2:1, I get confused) the wheels will apply 200Nm of torque and turn at 500RPM.

So your CVT gearbox basically calculates the correct ratio to ensure that for the current road speed, the engine will be turning at maximum power RPM, and this will result in a higher torque at the wheels than would be the case at peak torque, because the gear is lower.

An example.

the FXO gives 225lb ft at 4338RPM, and 234bhp @ 6365.

Using the formula BHP = torque * rpm / 5252 (yes, torque and bhp are directly related)

Torque at peak BHP = 234 * 5252 / 6365 = 193 lb ft.

Lets say we are travelling at 25m/s (about 50mph) and the circumference of the wheel is 1.7m, that means the wheel is rotating at 15rps = 900 RPM.

If the CVT matches our speed to peak power revs, that means the gear ratio will be 6365/900 = 1: 7.07.

So if we multiply our torque at peak power (193 lb ft) by our gear ratio, we get:

torque at wheels = 1365 lb ft

If our CVT matches the speed to our peak torque rpm, the gear ration will be 4338/900 = 4.82.

So the torque at the wheels will be 4.82 * 225 = 1084 lb ft

Voila. The reason a CVT produces greater acceleration at peak power is because it lets you use a lower gear which gives greater torque at the wheels. This is fundamental even ignoring CVT's. The reason more power gives you better acceleration over time is because it lets you stay in a lower gear for longer, thus giving you more torque at the wheels for longer.


And finally, as the astute will have notices above, torque and power are directly mathematically related to eachother. power = torque * rpm*constant depending on units used.

So all this nonsense about how power is more relavant than the torque curve, or peak torque is more relavant than peak power, is all just tripe.

In the beginning there was the torque curve. A dyno measures torque. From the torque at each given RPM we mathematically derive the power at each given RPM using the simple formula above. The point at which peak power occurs is a product of the torque curve, or the other way round depending on how you look at it.

The point is that the two curves are not independent, their shapes are inexorably bound to eachother by maths.