# Acceleration - Torque vs Power

242 replies to this topic

### Poll: Max accelleration at max torque or max power (52 member(s) have cast votes)

#### Assume we have a car traveling at speed and a gearing so we can select gear to have the engine running at a rpm where it's either produces max torque or max power. What gear would achieve maximum acceleration, the one that put engine rpm at max torque or at max power?

1. I am sure it is at max torque and know the formula to prove it (6 votes [11.54%])

Percentage of vote: 11.54%

2. I know it is at max torque as I read it in books (2 votes [3.85%])

Percentage of vote: 3.85%

3. I believe it is at max torque by reading forums like this (0 votes [0.00%])

Percentage of vote: 0.00%

4. I think it is at max torque (4 votes [7.69%])

Percentage of vote: 7.69%

5. I have no idea what rpm would generate max acceleration (4 votes [7.69%])

Percentage of vote: 7.69%

6. I think it is at max power (11 votes [21.15%])

Percentage of vote: 21.15%

7. I believe it is at max power by reading forums like this (1 votes [1.92%])

Percentage of vote: 1.92%

8. I know it is at max power as I read it in books (3 votes [5.77%])

Percentage of vote: 5.77%

9. I am sure it is at max power and know the formula to prove it (21 votes [40.38%])

Percentage of vote: 40.38%

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### #201 Rasputin

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Posted 29 June 2013 - 14:42

Good Lord, I left this thread on April 8th and it's been going on until recently?

At least two thirds of the voters got it right.

Edited by Rasputin, 29 June 2013 - 16:09.

### #202 WhiteBlue

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Posted 01 July 2013 - 07:43

Given the vague description of the poll one has to assume that we are close to top speed and any kind of torque curve is applicable. That leads to the assumption that a local maximum of the product of torque and angular speed can be off the peak torque point or plateau. Hence logic tells you the last option must be correct.

It would obviously be a bit different if we were discussing acceleration out of corners. In cars that are not traction limited with say AWD torque will play a bigger role. In that situation you are usually not at peak power and torque is your trump card. We will see this in LMP1 next year when they will have monster torque from the turbo engines and the AWD electric hybrid drives. LMP1s will be even more superior in performance over GTs and LMP2s than they are today in the cornering sections.

Edited by WhiteBlue, 01 July 2013 - 07:50.

### #203 Canuck

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Posted 01 July 2013 - 13:03

Perhaps you're missing the trick here. No one is debating that the greatest available thrust comes from the greatest torqe at the wheels - torque moves the pork. At any given road speed, the greatest available thrust is in the lowest gear that permits (accceleration at) that speed. That will invariably put the engine speed above the torque peak.

Clearly, if we were constraining with a fixed gear rather than fixed speed, the torque peak would provide the greatest acceleration.

### #204 Rasputin

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Posted 01 July 2013 - 17:30

But at the end of the day, acceleration is about wheel-torque, or shear-force at the tire's contact patch, torque and rpm before the clutch is irrelevant.

### #205 meb58

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Posted 01 July 2013 - 18:54

I've always rationalized torque and HP, internally this way; torque is a function of engine displacement/stroke (mechanical leverage?) and horsepower is a function of engine breathing (atmospheric leverage?)

Large displacement engines don't need great atmospheric leverage and small ones do. Recently I've become much more interested in gearing and how gearing leverages a particular engine's characteristics/output and tire traction and chassis balance...and I know next to nothing about gearing!

Edited by meb58, 01 July 2013 - 18:55.

### #206 CSquared

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Posted 01 July 2013 - 19:18

At any given road speed, the greatest available thrust is in the lowest gear that permits (accceleration at) that speed.

Not true for all engines/vehicles. See, for example, this graph.

### #207 meb58

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Posted 01 July 2013 - 19:49

...so given my quest in post #209, how do torque/hp curves change with gear ratios? I assume that gearing affects both power curves?

quote name='CSquared' date='Jul 1 2013, 15:18' post='6338748']
Not true for all engines/vehicles. See, for example, this graph.

[/quote]

### #208 Rasputin

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Posted 01 July 2013 - 20:09

It could be explained much simpler, I picked up this piece from another site;

What should be interesting to Pirelli here is obviously the wheel-torque, or rather the shear-force on the contact patch.

This can be calculated in two ways:

A) With 588 kW (800 Hp), the wheel-torque will of course taper-off with the speed, but at 30 m/s (108 km/h), two 660 mm dia wheel will spin with 14.5 Rps (868 Rpm) and 6470 Nm,
resulting in a shear-force of 6470/0.330 =19.6 kN on the two wheels.

B) With 588 kW at 30 m/s, the total shear-force will be 588 000/30 = 19.6 kN as Power is Force times speed.

Quite brilliantly expressed me thinks.

Edited by Rasputin, 01 July 2013 - 20:12.

### #209 gruntguru

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Posted 02 July 2013 - 03:55

Not true for all engines/vehicles. See, for example, this graph.

Nice graph what is it - 200 hp turbo thingy geared for 165 mph?

Edited by gruntguru, 02 July 2013 - 06:09.

### #210 Greg Locock

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Posted 02 July 2013 - 04:31

Csquared- what is that graph supposed to be proving? Surely having a red line much higher than max power rpm is pretty much irrelevant to this discussion?

### #211 Rasputin

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Posted 02 July 2013 - 05:03

Csquared- what is that graph supposed to be proving? Surely having a red line much higher than max power rpm is pretty much irrelevant to this discussion?

I think it pretty much proves that Power is always Force (Thrust) times Speed?

2500 lb at 30 mph in first gear equals 500 lb at 150 in 5th, doesn't it?

### #212 gruntguru

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Posted 02 July 2013 - 06:33

Csquared- what is that graph supposed to be proving? Surely having a red line much higher than max power rpm is pretty much irrelevant to this discussion?

At least it does disprove Canuck's statement . . . . .

Perhaps you're missing the trick here. No one is debating that the greatest available thrust comes from the greatest torqe at the wheels - torque moves the pork. At any given road speed, the greatest available thrust is in the lowest gear that permits (accceleration at) that speed. That will invariably put the engine speed above the torque peak.

Clearly, if we were constraining with a fixed gear rather than fixed speed, the torque peak would provide the greatest acceleration.

Not true for all engines/vehicles. See, for example, this graph.

. . . . eg at 100 mph Canuck's statement suggests you should be in third gear when in fact there is better acceleratiopn in fourth.

Note to all: These "thrust" curves mimic the engine's torque curve not the power curve. For example in 4th gear (RED curve), max torque occurs somewhere between 85 and 100 mph and max power occurs at about 115 mph (where an imaginary smooth curve drawn to "touch" each thrust curve - touches the red curve).

It is interesting to note that each gearchange hits the next gear at obout max torque (except 1 - 2 shift which hits below max torque rpm). This only coincidence. If the ratios were closer, the upshifts would hit the next gear above the max torque rpm.

Max torque rpm is irrelevent to shift point selection.

### #213 CSquared

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Posted 02 July 2013 - 06:46

At least it does disprove Canuck's statement . . . . .
. . . . eg at 100 mph Canuck's statement suggests you should be in third gear when in fact there is better acceleratiopn in fourth.

That's exactly what I was trying to say. If my post or the graph were unclear, I apologize. I searched something like "torque gear curve" and picked the first image that came up where the lines crossed. From the image's URL, I'm assuming the car is a 3rd-gen RX-7.

I don't see why the redline makes the graph or the engine irrelevant to the discussion, though.

### #214 Canuck

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Posted 02 July 2013 - 11:43

Urgh. I stand by what I meant if not what I typed. If you're being a technical pedant, then my statement is incorrect. Clearly if you're operating the engine at such a high (relative) RPM that the engine torque multiplied by drivetrain ratios produce less torque at the wheels than the next higher gear selection, you will find greater acceleration in the next gear at that road speed.

In the context of the question - peak power or peak torque at a given road speed - the lowest possible gearing at peak power will produce the greatest instantaneous acceleration at the given speed. People so busy looking to prove someone else wrong, they're happily missing the point.

### #215 Greg Locock

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Posted 02 July 2013 - 23:36

Correct. If you add dots on that graph to indicate max power and max torque in each gear then I strongly suspect all you discover is that the red line is very high compared with max power rpm.

### #216 johnny yuma

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Posted 03 July 2013 - 01:53

So racing,you wring it's neck in each gear till it sounds unhappy and acceleration is heading south,then change up,...agrees with Canuck's graph.
Looks like you would change up from third to fourth at 92 mph. Each gear will have a slightly different rpm change up point I think?

### #217 gruntguru

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Posted 03 July 2013 - 06:01

People so busy looking to prove someone else wrong, they're happily missing the point.

I am OK with Csquared's post, the point I think needs to be made is - for max acceleration you should be in the gear that sees the highest engine power. Any other suggestions about "higher revs", "lowest gear", "torque peak" etc are irrelevent.

When the power has dropped off to the point where an upshift will produce identical power - its time to upshift (ie the intersection of lines on the thrust chart above.)

### #218 Rasputin

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Posted 03 July 2013 - 06:45

I am OK with Csquared's post, the point I think needs to be made is - for max acceleration you should be in the gear that sees the highest engine power. Any other suggestions about "higher revs", "lowest gear", "torque peak" etc are irrelevent.

When the power has dropped off to the point where an upshift will produce identical power - its time to upshift (ie the intersection of lines on the thrust chart above.)

Finally some words of wisdom, thank you.

### #219 rgsuspsa

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Posted 03 July 2013 - 22:41

Finally some words of wisdom, thank you.

Gear ratios, number of ratios, contact patch and any number of other parameter are irrelevant to the central issue of how fast a mass can be accelerated. The rate of kinetic energy gain by the mass, (rate of gain of (1/2 X Mass X Velocity squared)) is the only issue. The rate at
which kinetic energy is gained by the vehicle is maximized by definition at maximum power of the power source. In an actual race car, gear ratios, contact patch and all other parameters relevant to operating at maximum power are selected such that this is achieved to
the greatest degree feasible within rule limitations and physical limits of materials.

Ron Sparks

Edited by rgsuspsa, 03 July 2013 - 23:07.

### #220 Dmitriy_Guller

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Posted 04 July 2013 - 01:35

I am OK with Csquared's post, the point I think needs to be made is - for max acceleration you should be in the gear that sees the highest engine power. Any other suggestions about "higher revs", "lowest gear", "torque peak" etc are irrelevent.

When the power has dropped off to the point where an upshift will produce identical power - its time to upshift (ie the intersection of lines on the thrust chart above.)

Seconded. People coming up with various rules of thumb is the #1 reason that this discussion can get so confusing. Those rules of thumb almost always correlate well with the correct things to do with the real engines we actually have, which is why they're hard to disprove, but they are not generally correct. Yes, in reality, engines that we all drive don't redline at RPM far above power peak, and in reality we don't have gearboxes with 100 gears, so waiting to upshift until you hit the rev limiter can't be that wrong, but that beats around the bush in a way that "always be in gear that gives you maximum power at that speed" doesn't.

Edited by Dmitriy_Guller, 04 July 2013 - 01:38.

### #221 Dmitriy_Guller

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Posted 04 July 2013 - 01:47

Urgh. I stand by what I meant if not what I typed. If you're being a technical pedant, then my statement is incorrect. Clearly if you're operating the engine at such a high (relative) RPM that the engine torque multiplied by drivetrain ratios produce less torque at the wheels than the next higher gear selection, you will find greater acceleration in the next gear at that road speed.

In the context of the question - peak power or peak torque at a given road speed - the lowest possible gearing at peak power will produce the greatest instantaneous acceleration at the given speed. People so busy looking to prove someone else wrong, they're happily missing the point.

I guess you're not going to like what I'm about to say next then. However, I must. When you are at peak power at a given speed, the lowest possible gearing or highest possible gearing won't matter, you will be accelerating at the same rate.

### #222 CSquared

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Posted 04 July 2013 - 02:05

If you're being a technical pedant ...

This is The Technical Forum, son!

### #223 johnny yuma

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Posted 04 July 2013 - 02:11

I guess you're not going to like what I'm about to say next then. However, I must. When you are at peak power at a given speed, the lowest possible gearing or highest possible gearing won't matter, you will be accelerating at the same rate.

If you wish to BE at peak power (I presume you mean engine rpm peak power)at any given speed,does not that automatically require you to be at a quite precise gearing !!! ? You appear to be saying acceleration in first gear has no potential to be superior to acceleration in top gear.

Edited by johnny yuma, 04 July 2013 - 02:11.

### #224 Dmitriy_Guller

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Posted 04 July 2013 - 02:21

I am saying that if several different gears all put you in peak power at the speed you're in, it doesn't matter which one of those gears you choose.

### #225 Rasputin

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Posted 04 July 2013 - 02:34

I am OK with Csquared's post, the point I think needs to be made is - for max acceleration you should be in the gear that sees the highest engine power. Any other suggestions about "higher revs", "lowest gear", "torque peak" etc are irrelevent.

When the power has dropped off to the point where an upshift will produce identical power - its time to upshift (ie the intersection of lines on the thrust chart above.)

This is it really, gruntguru hit the nail on the head, as acc is Force over Mass and Force is Power over Speed, it's all about keeping the engine at its powerwise sweetspot.

That's all there's to it.

### #226 johnny yuma

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Posted 04 July 2013 - 03:09

I am saying that if several different gears all put you in peak power at the speed you're in, it doesn't matter which one of those gears you choose.

How can several different gears all put you in the power peak at a given speed ???

### #227 Dmitriy_Guller

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Posted 04 July 2013 - 03:27

How can several different gears all put you in the power peak at a given speed ???

You can have a flat power peak (or plateau, if you will). Many new turbo engines at least on paper have those.

### #228 johnny yuma

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Posted 04 July 2013 - 03:57

You can have a flat power peak (or plateau, if you will). Many new turbo engines at least on paper have those.

Well good-oh , but you will still accelerate at a greater rate in a lower gear even if you do have a flat graph.Dare I add it's because ,mechanically,and in no way relating to what your flat graph engine might be doing,your Torque at the Axles is much improved by the lower gearing chosen..Thus,the Force at the tyre/road contact patch is greater,thus greater acceleration. In other words torque at the first motion shaft of the gearbox won't change with your flat graph engine,but it will change at exit shaft from gearbox in each gear,unless it's 1:1.

Edited by johnny yuma, 04 July 2013 - 04:03.

### #229 gruntguru

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Posted 04 July 2013 - 04:00

Well good-oh , but you will still accelerate at a greater rate in a lower gear even if you do have a flat graph.Dare I add it's because ,mechanically,and in no way relating to what your flat graph engine might be doing,your Torque at the Axles is much improved.Thus,the Force at the tyre/road contact patch is greater,thus greater acceleration.

No. If you have a flat POWER curve, it doesn't matter which gear you are in. A flat power curve will have a torque curve that rises sharply as revs reduce.

### #230 johnny yuma

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Posted 04 July 2013 - 04:08

This is The Technical Forum, son!

Yeah,only boffins and rocket scientists allowed.

### #231 johnny yuma

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Posted 04 July 2013 - 04:11

No. If you have a flat POWER curve, it doesn't matter which gear you are in. A flat power curve will have a torque curve that rises sharply as revs reduce.

yeah right,Hey I gotta get me some of these fancy curves,dammit !...(no harm in feeding the chooks is there?)

### #232 gruntguru

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Posted 04 July 2013 - 06:33

yeah right,Hey I gotta get me some of these fancy curves,dammit !...(no harm in feeding the chooks is there?)

### #233 Rasputin

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Posted 04 July 2013 - 08:08

All this reminds my of why I became an engineer, kudos en masse to gg!

Though I can add that as Force is Power over Speed, Force goes to the sky at zero speed, why we have whee-spin.

Edited by Rasputin, 04 July 2013 - 08:43.

### #234 Magoo

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Posted 04 July 2013 - 13:03

Though I can add that as Force is Power over Speed, Force goes to the sky at zero speed, why we have whee-spin.

Oh, so that's why.

Once again I would like to congratulate myself for not participating in this discussion.

### #235 Rasputin

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Posted 04 July 2013 - 13:56

Oh, so that's why.

Once again I would like to congratulate myself for not participating in this discussion.

Happy o enlighten you, in theory you have as much force as you want at take-off, funny isn't it?

### #236 Dmitriy_Guller

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Posted 04 July 2013 - 18:25

Happy o enlighten you, in theory you have as much force as you want at take-off, funny isn't it?

This is an area where our simplifying assumption really break apart. It's impossible to accelerate from zero without a slip somewhere, and once you have a slip somewhere, these simple equations which are still very hard for some people to fathom don't apply anymore. Usually one of the hardest racing physics things to program in racing simulations is the tire behavior at low speeds.

### #237 gruntguru

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Posted 04 July 2013 - 22:52

If only we could divide by zero.

### #238 johnny yuma

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Posted 04 July 2013 - 23:40

Oh, so that's why.

Once again I would like to congratulate myself for not participating in this discussion.

to quote John Cleese's customer, " I came here for an argument"

Cleese replies "no you didn't".........etc etc.

Edited by johnny yuma, 04 July 2013 - 23:41.

### #239 Greg Locock

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Posted 05 July 2013 - 02:04

It's impossible to accelerate from zero without a slip somewhere

Pedantic interlude...
It is not impossible, even in a normal car. Switch engine off. Engage first gear. Release clutch. Crank engine.

Now, that is a completely useless exercise 9 times out of 10, but it is SOP when starting up a steep slope in a 4wd diesel. If you slip the tires you will skid to the bottom of the hill, on typical clay dams for example.

### #240 Dmitriy_Guller

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Posted 05 July 2013 - 03:00

Pedantic interlude...
It is not impossible, even in a normal car. Switch engine off. Engage first gear. Release clutch. Crank engine.

Now, that is a completely useless exercise 9 times out of 10, but it is SOP when starting up a steep slope in a 4wd diesel. If you slip the tires you will skid to the bottom of the hill, on typical clay dams for example.

Point taken. In that case, at standstill, you're going to be dividing zero power by zero speed, which would reduce to some finite acceleration once you cancel out the terms that give you zeros at standstill.

### #241 johnny yuma

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Posted 05 July 2013 - 05:09

Pedantic interlude...
It is not impossible, even in a normal car. Switch engine off. Engage first gear. Release clutch. Crank engine.

Now, that is a completely useless exercise 9 times out of 10, but it is SOP when starting up a steep slope in a 4wd diesel. If you slip the tires you will skid to the bottom of the hill, on typical clay dams for example.

Electric motor has max torque from 0 rpm.Will spin wheels if insufficient traction just the same,low range or high.Is SOP in petrol 4WDs too(14 years with Parks and Wildlife Service)for starting movement on steep slippy slopes.But it can't extract you if you're bogged !! Best to keep off muddy roads in wet weather,farmers justifiably hate you when you rut their roads.

### #242 Rasputin

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Posted 06 July 2013 - 09:49

If only we could divide by zero.

Isn't that called an asymptote or something?

### #243 gruntguru

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Posted 07 July 2013 - 23:50

Very hyperbolic oh proboscotic one.