# 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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### #101 johnny yuma

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Posted 23 April 2013 - 05:20

I would say that "lesser amount of torque" as you would say, is what is actually generating the engine peak power at 6300 rpm and that's why we are shifting from a lower gear at that point for max acceleration!

Well its good to see you acknowledge torque generates power ! Thats why race engine builders tune for more TORQUE at higher rpm,because it will generate more power.But the peak torque is always below the peak power rpm,so in any given gear your period of best performance will be between the torque rpm peak and the power rpm peak. After the power peaks, cylinder filling etc starts to decay quickly,so you need that higher gear because...you have no chioce !

### #102 bigleagueslider

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Posted 23 April 2013 - 06:17

When it comes to acceleration of a vehicle, torque is irrelevant, and energy transfer (or power) is the only concern.

Consider this: How does a rocket accelerate to over 17,000 mph during launch without any "torque" being applied by the engine?

### #103 JimboJones

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Posted 23 April 2013 - 07:11

So in any given gear your period of best performance will be between the torque rpm peak and the power rpm peak. After the power peaks, cylinder filling etc starts to decay quickly,so you need that higher gear because...you have no chioce !

No, in any given gear best performance will come from maximizing the area under the torque-to-wheels curve. It so happens that if you shift down, you get even more performance as you are nearer the power peak. I think we all know torque generates power... the reason engine tuners tune for torque is because that's what a dyno measures!

### #104 gruntguru

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Posted 23 April 2013 - 07:21

Yes. You left out one important bit (probably because it seems obvious).

Shift points should "maximise average power" ie you should shift "above the power peak" (if possible) at rpms that will produce the same "power" in the next gear. So as you go through the gears, the rpm will be oscillating in a band centred on the power peak (assuming a symmetrical peak.

The location of the torque peak is totally irrelevant to the question of "shift points".

### #105 johnny yuma

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Posted 23 April 2013 - 08:47

When it comes to acceleration of a vehicle, torque is irrelevant, and energy transfer (or power) is the only concern.

Consider this: How does a rocket accelerate to over 17,000 mph during launch without any "torque" being applied by the engine?

It doesn't have any wheels to use torque,the force ,or thrust (but not power)if you like,goes straight from the combustion chambers
to the jet exit zone creating a region of high pressure and the rocket nose is attracted to the low pressure zone ahead. It is a very
powerful attraction..if you know what I mean.

### #106 johnny yuma

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Posted 23 April 2013 - 09:03

No, in any given gear best performance will come from maximizing the area under the torque-to-wheels curve. It so happens that if you shift down, you get even more performance as you are nearer the power peak. I think we all know torque generates power... the reason engine tuners tune for torque is because that's what a dyno measures!

Your first sentence fails on so many levels...no actually only from mixed metaphors and ..something else. Sentence 2,no,you get more performance because you shifted down,and multiplied your torque available at the
higher rpm area you entered,and even though the torque was less AT THE FLYWHEEL in the lower gear,what was available at the flywheel multiplied by the better gear ratio,gave you better torque at the axles,more force at
the tyre/road interface,and accelerated you quicker.A DYNO measures torque at the rollers which have an axle so you haven't reached the road yet...otherwise your ride would get shot into the wall of the dyno shop.
A mathematical formula converts your dyno torque to Power so the punters have something to wave at their mates/buddies down at the Horsepower Hotel.

### #107 johnny yuma

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Posted 23 April 2013 - 09:30

Yes. You left out one important bit (probably because it seems obvious).

Shift points should "maximise average power" ie you should shift "above the power peak" (if possible) at rpms that will produce the same "power" in the next gear. So as you go through the gears, the rpm will be oscillating in a band centred on the power peak (assuming a symmetrical peak.

The location of the torque peak is totally irrelevant to the question of "shift points".

Another doozie gg,think I see what you mean.All I can add is even though the torque peak can be irrelevant,it seems to turn out it WAS usually in the right place if an engine was flogged to power peak
plus a bit,then an upshift was made. My hobby car a 1950 Holden with 3 speed gearbox makes 60 bhp@3800 rpm,100 lbs/ft Torque@2000 rpm. Second gear ratio 1.59:1. You were way better off wringing it's neck
in second gear to keep that ratio as long as you could rather than changing into top gear to be at the engine's best torque rpm of 2000,because that would mean you were stuck in top at 40mph with an hourglass
as a stopwatch.However, as newer and newer cars got 4,5,6 and 7 speed boxes ,such thrashing became unnecessary... and counterproductive.

I wonder,has the "gap" between Torque peak and Power peak,expressed in rpm,changed much in the average car since say 1950? The number of gearbox ratios certainly has !

### #108 saudoso

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Posted 23 April 2013 - 13:03

We just won't learn, right? We will keep discussing this. And now its even worse.

In previous interactions we didn't agree about high rev/low torque or low rev/high torque engines being best.

Now we just can't get the concepts of force, power, speed and leverage settled. And that's not even high school physics level.

### #109 Tony Matthews

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Posted 23 April 2013 - 13:20

the force ,or thrust (but not power)if you like,goes straight from the combustion chambers
to the jet exit zone creating a region of high pressure and the rocket nose is attracted to the low pressure zone ahead.

Very low pressure zones infront and behind in the cruel, cold vaccuum of space. The thrust comes from the forward inner area of the combustion chamber, not at the exit.

### #110 CSquared

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Posted 23 April 2013 - 14:00

It doesn't have any wheels to use torque,the force ,or thrust (but not power)if you like,goes straight from the combustion chambers
to the jet exit zone creating a region of high pressure and the rocket nose is attracted to the low pressure zone ahead. It is a very
powerful attraction..if you know what I mean.

I don't know what you mean because that's not at all how rockets work.

### #111 Dmitriy_Guller

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Posted 23 April 2013 - 14:19

It doesn't have any wheels to use torque,the force ,or thrust (but not power)if you like,goes straight from the combustion chambers
to the jet exit zone creating a region of high pressure and the rocket nose is attracted to the low pressure zone ahead. It is a very
powerful attraction..if you know what I mean.

How does it accelerate in space?

### #112 saudoso

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Posted 23 April 2013 - 14:35

See, middle/high school level discussion. Better left alone.

### #113 JimboJones

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Posted 23 April 2013 - 21:57

Your first sentence fails on so many levels...no actually only from mixed metaphors and ..something else. Sentence 2,no,you get more performance because you shifted down,and multiplied your torque available at the
higher rpm area you entered,and even though the torque was less AT THE FLYWHEEL in the lower gear,what was available at the flywheel multiplied by the better gear ratio,gave you better torque at the axles,more force at
the tyre/road interface,and accelerated you quicker.A DYNO measures torque at the rollers which have an axle so you haven't reached the road yet...otherwise your ride would get shot into the wall of the dyno shop.
A mathematical formula converts your dyno torque to Power so the punters have something to wave at their mates/buddies down at the Horsepower Hotel.

I don't understand why you are arguing, you're explaining the same theory in a different way! Yes, the lower ratio scales up the torque, but it also scales up the engine speed, and torque x speed is power, so you are maximizing the power! You obviously didnt read the post I pointed to, or this would have been clear. Don't appreciate the wit either, especially when I'm a qualified engineer working on race engines every day, and you clearly are not. Oh, and engine tuners develop engines on engine dynos, not car dynos!

See, middle/high school level discussion. Better left alone.

haha, so true, I've given up. Anyone reading this thread actually seeking to understand the answer to the original question should be able to, because it has been made perfectly clear (in a couple of posts). It doesn't help when people come along and pose some confusing scenario, as if that will somehow prove physics wrong.

Loved the rocket analogy though!

### #114 johnny yuma

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Posted 24 April 2013 - 00:24

Free at last ! Now I know how a rocket really works ! I shall leave you rocket scientists to your work.

### #115 gruntguru

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Posted 24 April 2013 - 00:50

Rockets. The change in momentum of the gases being accelerated through the nozzle must be matched by a change in momentum of the space vehicle in the opposite direction. A guy standing at the back of the vehicle throwing tennis balls could achieve the same effect. No pressure required.

### #116 saudoso

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Posted 24 April 2013 - 01:20

Or a monkey loose in space with a machine gun.

### #117 gruntguru

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Posted 24 April 2013 - 06:10

Another doozie gg,think I see what you mean.All I can add is even though the torque peak can be irrelevant,it seems to turn out it WAS usually in the right place if an engine was flogged to power peak
plus a bit,then an upshift was made. My hobby car a 1950 Holden with 3 speed gearbox makes 60 bhp@3800 rpm,100 lbs/ft Torque@2000 rpm. Second gear ratio 1.59:1. You were way better off wringing it's neck
in second gear to keep that ratio as long as you could rather than changing into top gear to be at the engine's best torque rpm of 2000,because that would mean you were stuck in top at 40mph with an hourglass
as a stopwatch.However, as newer and newer cars got 4,5,6 and 7 speed boxes ,such thrashing became unnecessary... and counterproductive.

I wonder,has the "gap" between Torque peak and Power peak,expressed in rpm,changed much in the average car since say 1950? The number of gearbox ratios certainly has !

This illustrates the irrelevance of the torque peak. A cyclist generates peak torque at zero rpm but gearing for zero rpm would produce zero acceleration. To maximise acceleration, top speed, hill-climbing speed, whatever - the gearing should be adjusted to keep this cyclist as close to peak power (115 rpm) as possible.

### #118 Kelpiecross

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Posted 24 April 2013 - 06:50

This illustrates the irrelevance of the torque peak. A cyclist generates peak torque at zero rpm but gearing for zero rpm would produce zero acceleration. To maximise acceleration, top speed, hill-climbing speed, whatever - the gearing should be adjusted to keep this cyclist as close to peak power (115 rpm) as possible.

What are the dotted lines? Girly-Men?

### #119 Tony Matthews

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Posted 24 April 2013 - 09:30

Men who are drug-free.

### #120 gruntguru

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Posted 24 April 2013 - 22:50

At 700W they are probably the average guys that zoom past me every time I ride.

### #121 Dmitriy_Guller

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Posted 25 April 2013 - 00:14

Men who are drug-free.

### #122 rgsuspsa

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Posted 25 April 2013 - 18:22

Very low pressure zones infront and behind in the cruel, cold vaccuum of space. The thrust comes from the forward inner area of the combustion chamber, not at the exit.

No, a rocket's thrust does not result from pressure on the forward inner area of the combustion chamber. The rearward facing area and pressure acting over it produces a rearward force which is equal in magnitude, and opposite in direction to that force produced by the forward area and pressure acting over it. These forces cancel each other.

Rocket thrust is produced by acceleration of gases passing thru the convergent then diverent sections of the exhaust nozzle. The sole function of the combustion chamber is to supply pressurized gases for expansion and acceleration by the exhaust nozzle.
From Newton's laws of motion the thrust is proportional to the magnitude of the mass flow rate of propellant gases, and the exhaust velocity of propellant gases exiting the nozzle.

Ron Sparks

Edited by rgsuspsa, 25 April 2013 - 19:02.

### #123 Tony Matthews

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Posted 25 April 2013 - 18:59

Oh well - thanks. This is what comes of taking notice of what you are taught at school.

### #124 Tony Matthews

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Posted 25 April 2013 - 19:05

The idea behind a simple solid-fuel rocket is simple. What you want to do is create something that burns very quickly but does not explode. As you are probably aware, gunpowder explodes...

You see what the problem is. I just came across this in a search for what might have been my misinformation. Gunpowder does not explode unless it is in a sealed container and/or detonated. I wonder what the writer thinks propells Guy Fawkes rockets.

### #125 saudoso

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Posted 25 April 2013 - 20:48

I've made rockets out of black shotgun powder. Best 'engine ecnlosure' was the body of Bic pens.

Edited by saudoso, 25 April 2013 - 20:49.

### #126 gruntguru

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Posted 25 April 2013 - 22:57

No, a rocket's thrust does not result from pressure on the forward inner area of the combustion chamber. The rearward facing area and pressure acting over it produces a rearward force which is equal in magnitude, and opposite in direction to that force produced by the forward area and pressure acting over it. These forces cancel each other.

I will see your "No" and raise you to a "No No".

If you draw a free body diagram of the rocket - not including the exhaust gases, there must be a resultant force because it is accelerating. The area of application of this force is mostly the inner surfaces of the divergent nozzle, but also the front inner face of the combustion chamber which has a greater area than the rear inner face (the rear inner face has a hole to let the gas out).

Edited by gruntguru, 26 April 2013 - 04:55.

### #127 Magoo

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Posted 26 April 2013 - 04:41

"That Professor Goddard with his 'chair' in Clark College and the countenancing of the Smithsonian Institution does not know the relation of action to reaction, and of the need to have something better than a vacuum against which to react--to say that would be absurd. Of course, he only seems to lack the knowledge ladled out daily in high schools."

-New York Times, January 13, 1920

"Further investigation and experimentation have confirmed the findings of Isaac Newton in the 17th Century and it is now definitely established that a rocket can function in a vacuum as well as in an atmosphere. The Times regrets the error."

-New York Times, July 17, 1969

### #128 gruntguru

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Posted 26 April 2013 - 05:00

"Further investigation and experimentation have confirmed the findings of Isaac Newton in the 17th Century and it is now definitely established that a rocket can function in a vacuum as well as in an atmosphere. The Times regrets the error."

-New York Times, July 17, 1969

Classic!

Luckily they confirmed Newton in time for Apollo 11.

### #129 Joe Bosworth

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Posted 26 April 2013 - 11:07

This illustrates the irrelevance of the torque peak. A cyclist generates peak torque at zero rpm but gearing for zero rpm would produce zero acceleration. To maximise acceleration, top speed, hill-climbing speed, whatever - the gearing should be adjusted to keep this cyclist as close to peak power (115 rpm) as possible.

Grunt

I normally stay far away from threads such as this that finds much discussion of low technical merit. But this time I am fascinated by your message found in #117.

In over 40 years of cliosely following cycling technology I have never seen anybody publishing data similar to that shown in your graph of crank speed versus torque and power. I wonder where you got the data as it looks very dodgy and counter intuitive.

Given that acceleration is the result of forces on the subject body divided by the mass of the body.

Unravelling the power/torque curves shown in the graph into forward motion forces at the wheel using any set of ratios normal to cycling we find dramatically greater accelerative forces at low crank speeds than we do at high crank speeds.

I am used to the power provided to the cranks by the rider, (and therefore availavble to the driven wheel), to be more nearly flat over a wide range of pedalling speeds.

Any in-sights??

Regards

### #130 Greg Locock

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Posted 27 April 2013 - 05:46

Great stuff. Interesting to see that elite cyclists weren't pulling up. Incidentally I was cycling with a knee surgeon once and he noticed I was pulling up. He said "that's how I got my kids through uni". Your knee is designed to push not pull.

### #131 manolis

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Posted 27 April 2013 - 06:01

"That Professor Goddard with his 'chair' in Clark College and the countenancing of the Smithsonian Institution does not know the relation of action to reaction, and of the need to have something better than a vacuum against which to react--to say that would be absurd. Of course, he only seems to lack the knowledge ladled out daily in high schools."
-New York Times, January 13, 1920
"Further investigation and experimentation have confirmed the findings of Isaac Newton in the 17th Century and it is now definitely established that a rocket can function in a vacuum as well as in an atmosphere. The Times regrets the error."
-New York Times, July 17, 1969

Manolis Pattakos

### #132 Magoo

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Posted 27 April 2013 - 06:19

Classic!

Luckily they confirmed Newton in time for Apollo 11.

Exactly. There's a good deal of tongue-in-cheek here. Note the date is one day after Apollo 11's launch.

### #133 manolis

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Posted 27 April 2013 - 06:37

Grunt
I normally stay far away from threads such as this that finds much discussion of low technical merit. But this time I am fascinated by your message found in #117.
In over 40 years of cliosely following cycling technology I have never seen anybody publishing data similar to that shown in your graph of crank speed versus torque and power. I wonder where you got the data as it looks very dodgy and counter intuitive.
Given that acceleration is the result of forces on the subject body divided by the mass of the body.
Unravelling the power/torque curves shown in the graph into forward motion forces at the wheel using any set of ratios normal to cycling we find dramatically greater accelerative forces at low crank speeds than we do at high crank speeds.
I am used to the power provided to the cranks by the rider, (and therefore availavble to the driven wheel), to be more nearly flat over a wide range of pedalling speeds.
Any in-sights??
Regards

Joe Bosworth,

based on the Torque / Power data of Gruntguru's post #117:

The gear ratios are indicative, yet they show what really happens.

According this, for the maximum acceleration the first gear must be used until 120-130 rpm, making the shape of the torque vs rpm at low - medium revs so significant.

The shape of the torque curve in the case of a bicycle is "the opposite" of the torque curve of the old 2-stroke motorcycles, those with the peacky torque curves wherein at low rpms the engine could not even idle.
By the proper use of the clutch (controllable slipping), the old 2-stroke motorcycle can maintain (for a given gear ratio) the maximum force for any speed lower than the speed corresponding to the maximum torque.
A good part of the energy provided by the engine is consumed on the clutch overheating it and reducing its expected life, however the bike is like having a "flat" torque curve and achieves the best possible acceleration.

Thanks
Manolis Pattakos

### #134 gruntguru

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Posted 27 April 2013 - 07:51

In over 40 years of cliosely following cycling technology I have never seen anybody publishing data similar to that shown in your graph of crank speed versus torque and power. I wonder where you got the data as it looks very dodgy and counter intuitive.

Given that acceleration is the result of forces on the subject body divided by the mass of the body.

Unravelling the power/torque curves shown in the graph into forward motion forces at the wheel using any set of ratios normal to cycling we find dramatically greater accelerative forces at low crank speeds than we do at high crank speeds.

I am used to the power provided to the cranks by the rider, (and therefore availavble to the driven wheel), to be more nearly flat over a wide range of pedalling speeds.

Joe.
I found the chart via a Google search and chose it on the basis of the simple curves. The power shown is in fact relatively flat if you consider the rpm range covered by 80% max power and above.

I always expected max torque to occur at zero rpm, but I doubt that torque drops as quickly as shown, ie I would expect a rider to be able to maintain close to max torque up to at least 20 or 30 rpm.

### #135 kikiturbo2

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Posted 27 April 2013 - 22:15

Grunt

I normally stay far away from threads such as this that finds much discussion of low technical merit. But this time I am fascinated by your message found in #117.

In over 40 years of cliosely following cycling technology I have never seen anybody publishing data similar to that shown in your graph of crank speed versus torque and power. I wonder where you got the data as it looks very dodgy and counter intuitive.

Given that acceleration is the result of forces on the subject body divided by the mass of the body.

Unravelling the power/torque curves shown in the graph into forward motion forces at the wheel using any set of ratios normal to cycling we find dramatically greater accelerative forces at low crank speeds than we do at high crank speeds.

I am used to the power provided to the cranks by the rider, (and therefore availavble to the driven wheel), to be more nearly flat over a wide range of pedalling speeds.

Any in-sights??

Regards

If I remember correctly, and I will check tomorrow, that chart is from the book Bicycling science, by David Gordon Wilson, MIT press, http://mitpress.mit....cycling-science
I highly recommend it to anyone interested in cycling, and related tech... It is a great piece of literature..

### #136 RDV

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Posted 28 April 2013 - 13:13

...oh, no...deja vu all over again..

### #137 Joe Bosworth

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Posted 28 April 2013 - 13:45

I pulled Wilson's book out of my bookshelves before responding above. There is nothing there that justifies the graph reproduced in thread #117. The most sense seems to be stated under "Conclusions", page 100 at the end of Chapter 2. It states that,"A riders ability to produce power can not be described with a single number'; I MIGHT ADD ANY SINGLE SET OF NUMBERS. It goes on to describe how the highest levels of power come from anaerobic work capacity and aerobic capacity is virtually tied to steady state ability of muscles to oxidise fuel.

I have found the source of the graph which does not provide any real in-sight other than to say the curves are from less than 10 seconds of effort. Therefore fully anaerobic.

The real bicycling world is based on virtually 100% aerobic work except for track sprint work which might be half anaerobic.

The world 1 hour records are set by riders generating about 600 watts at a cadence of about 90 rpm or less. This sets a bit of a marker for output over an hour .

As nearly as I can tell the published graph is a test conducted on a dyno utilising a flywheel as a load and having the riders perdal at max force from a full stop to max with the load set so it can be accomplished in seconds. Definetly not the real cycling world where all work is done between with cadences between about 75 and 115 rpm. Also not real world as higher speeds are developed without the retarding factor of aero increasing exponentially.

The road racing guys virtually never go above about 105 rpm. The track sprinters go up to about 115, well above there max power because they have fixed gears and have to win/lose on jumping from quite low speeds, depending on how their opponent sets up the race.

I I have played around a bit with a rear wheel with a power meter installed. I have found that between about a cadence of 70 and 95 that pedall forces remain nearly constant and power goes up in direct relationship to cadence rpm. My finding is generally verified by Figure 2.16 data published in Wilson.

Therefor acceleration forces at the rear wheel stay constant at any road speed using gears in the range that I find. It takes a very accomplished rider to not lose rear wheel power at much above 100 rpm as steering and frame motions take power away from the pedalling action. I might add that I have determined that my aerobic sweet spot is between 86 and 93 rpm cadence. Everybody is different but I have never found a good cyclist that doesn't loose efficienct below 80 R=rpm, (though one TdF rider was known as being effective with gearings that found him at about 75rpm.

You can monitor all kinds of first class international riders and find that 95% of them stay between 85t and 100 wheter they are climbing, sprinting or cruising in the peloton.

Now that I have poo-pooed the curves from #117, I have unravelled the lower curve a bit. Say you are pedalling a 40/12 gear at 80 rpm you are developeng a rear wheel force of 8.5 Newtons at a speed of 33.5 kph. If you go to a 40/15 gear you will be cranking 100rpm at the same road speed and still be generating 8.5 Newtons at the back wheel. Therefore acceleration will be identical, or at least within the accuracy that you can pull data off of small graphs without the backup data. This conforms to my experience.

For running, swimming and cycling max power output is closely tied to output duration with any one individual's capability virtually a constant percentage of world record capability over a wide range of time spans between about 1 minute and 12 hours. This is because the transfer of oxygen across the lungs to blood and accross blood to muscle is the controlling actions. VOmax rules. Below one minute the anaerobic capacity is so high as to provide very non-specific results.

I guess all this says is that human power results should not be extrapolated to try to prove mechanical feats of physics and engineering.

Regards

Edited by Joe Bosworth, 28 April 2013 - 14:32.

### #138 kikiturbo2

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Posted 29 April 2013 - 05:38

I pulled Wilson's book out of my bookshelves before responding above.

you are right... found the surce..

http://sportsexercis...different-rpms/

### #139 gruntguru

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Posted 29 April 2013 - 07:59

Joe
Apologies but this thread was never about real world cycling. It is about acceleration and if you were to perform an acceleration test (drag race) using bicycles over say 200 metres, the maximum output, anaerobic power curve would be the starting point for selecting gears.

If the test demanded a single gear (as for a track sprint) the acceleration obtained would have the shape of the torque curve.

If multiple gears were allowed, the rider would select gears to keep his pedal velocity as close to the power peak as possible.

The point of using these graphs was to demonstrate the irrelevance of the torque peak (zero rpm in this case) when shifting for maximum acceleration. If you prefer, we could consider the toque and power curves as coming from a steam engine or electric motor where duration of effort can be more or less neglected.

### #140 gruntguru

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Posted 30 April 2013 - 02:54

Tractive Force where the rubber meets the road is where forces which began in the combustion chambers,and were then converted
into torque by the crankshaft,get re-converted into force again. Power is merely an attempt to quantify how much work the machinery in
question can perform over a period of time. Power does not cause acceleration any more than the Candy Man causes love and makes
the world feel good.

How much torque should a good engine make from a gallon of gasoline?

Say its 1970 ,and you're cruising on a hilly highway in your Austin 1800,fully laden.You're sitting on 70 mph and a steepish hill begins.You don't feel like changing back a gear so you keep the throttle wide open ,the car slows to 50 mph and then cruises up the hill.If you were "making more power" at
65mph,why wouldn't the car climb the hill at 65mph,but could climb the hill at 50mph. Perhaps it was providing more torque at 50mph,and torque is what holds speed on a hill,or accelerates your car to a higher speed,not Power? Of course you could have gone back to Third and climbed the hill faster,but that would not have been because the engine made more power at higher revs,but because you were in a lower gear and multiplying the torque by 1.41 or whatever. I know I'm not answering your question ,and why would I,given the facetious nature of the question.However I thought a little "reverse engineering" question on hillclimbing/acceleration was interesting.

Agreed - the question was facetious and actually nonsensical however, the intention was to provoke thought as to whether it is power or torque that is the more fundamental property when analysing vehicle propulsion. The rocket example is a clue although it can still be argued whether power or "force" (thrust) is more fundamental. Here is how I see it.

There are typically three reasons a vehicle requires propulsion.
1. Acceleration. (to increase Kinetic Energy (KE))
3. Climbing. (to increase Potential Energy (PE))
3. Steady state. (Overcoming parasitic loads - friction, aero drag etc (losses)).

In each of these three cases, the one property that is absolutely defined is Energy, eg in the case of acceleration, the mass of the vehicle, the initial speed and the final speed combine to tell us how much energy is required to perform the acceleration - regardless of how quickly we accelerate or even how we might vary the acceleration rate during the test. The energy required is:

Delta KE = 1/2MV2^2 - 1/2MV1^2

Notice - no mention of torque or force. The time taken to complete the acceleration test will depend only on how quickly energy can be transferred or applied to the propulsion task and power is simply that - the rate of energy transfer. Consider also where the energy comes from - fuel. All liquid hydrocarbon fuels have a specified energy content eg gasolene about 43 MJ per kg. So we can also determine how much fuel is required to perform the acceleration test (providing we know the conversion efficiency of our propulsion system). Once again, no mention of torque or force, it doesn't even matter what type of propulsion system is used, the gasolene could be used in an IC engine, a jet engine, a steam recip engine, a steam turbine or even a steam-nozzle rocket! Most interesting of all - the amount of energy (or the quantity of fuel) required does not depend on how quickly the vehicle accelerates. So 0-60 in 10 sec requires the same amount of fuel as 0-60 in 5 seconds provided the car weighs the same.

Here's another way to look at it:

Q1. Vehicle mass = M. Max engine power = P. How long will it take to accelerate from 20 m/s to 30 m/s? (answer = 1/2M(20^2 - 30^2)/P)

Q2. Vehicle mass = M. Max engine torque = T. How long will it take to accelerate from 20 m/s to 30 m/s? (Can't be solved. Not enough information)

Edited by gruntguru, 30 April 2013 - 02:55.

### #141 desmo

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Posted 30 April 2013 - 14:09

Best exposition yet of the underlying logic. And thanks for keeping the math(s) simple.

### #142 NTSOS

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Posted 30 April 2013 - 15:32

Well, kind of interesting....an ICE consumes fuel and air by the pound and is measured in lbs/hr...if both examples consume the same amount of fuel and air during their acceleration runs, the quicker accelerating vehicle is obviously consuming twice the amount of fuel/air per second.....the slower accelerating/broken ass vehicle is simply out on the track longer!

OTOH, there is a symbolic world where you can snatch power figures out of thin air and simply ignore how they were generated and insert them into a convenient formula to support an argument! Then there are physical/real world companies like Detroit (formally Detroit Diesel), that put out pesky brochures that describes what happens in *their* physical world and how it supports *their* way of thinking.

Detroit DD16

"With a displacement of 952 cubic inches (15.6 liters) and up to
600 horsepower, the DD16 is the largest, most powerful engine
to ever roll off the Detroit assembly line. So go ahead. Take on
even the most challenging jobs with confidence.

Detroit offers an innovative ACRS that gives the DD16 superior
torque response for pulling heavy loads with confidence. As
much as 2050 pound-feet of torque is delivered at only 1100
RPM, giving you the launch and acceleration you need to get
the heaviest load moving. And, with a wide, flat torque curve,
DD16 lets you hang in top gear longer for more efficient,
quieter cruising."

### #143 Dmitriy_Guller

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Posted 30 April 2013 - 15:46

I guess we also needed the "I am sure it is at max torque and I've got the marketing pamphlet to prove it" option in this poll.

### #144 NTSOS

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Posted 30 April 2013 - 17:01

I guess we also needed the "I am sure it is at max torque and I've got the marketing pamphlet to prove it" option in this poll.

No, not at all.....this is not about max torque, that is simply a point on a curve, it is about the shape of the power and torque "curves", how they are generated, how they relate to one another and how they relate to max acceleration. Detroit is equally adept at using power or torque figures and jargon in their literature.....I simply posted one side for an opposing view!

### #145 Tony Matthews

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Posted 30 April 2013 - 17:07

Trouble maker.

### #146 NTSOS

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Posted 30 April 2013 - 17:43

Trouble maker.

Me.....well maybe just a little, but I used quotation marks.....and opposing views will have to discuss it with the boys from Detroit!

### #147 saudoso

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Posted 30 April 2013 - 21:46

Big engines and flat torque curves are great to haul big weights and to drive around confortably. Cross the RFID toll gate in top gear and just floor it back to cruising speed.

But what would be the optimal shifting point for maximum acceleration with a 952 ci Detroit engine?

### #148 NTSOS

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Posted 30 April 2013 - 22:28

Big engines and flat torque curves are great to haul big weights and to drive around confortably. Cross the RFID toll gate in top gear and just floor it back to cruising speed.

But what would be the optimal shifting point for maximum acceleration with a 952 ci Detroit engine?

Yes sir......I guess around the area of 1800 RPM or so where it peaks at 600 hp and is still generating 1751 lb-ft of torque. I think it has a 2000 rpm governor.

### #149 gruntguru

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Posted 30 April 2013 - 22:40

But what would be the optimal shifting point for maximum acceleration with a 952 ci Detroit engine?

Unfortunately, this is a "real world" question. If it is assumed that engine rotational inertia is negligible compared to vehicle mass, the answer is the same - "if possible shift at revs above the power peak such that power prior to the gearchange is equal to power after the gearchange." (Electronic truck engines typically have a flat-top power curve so it is easy to accelerate while keeping the engine constantly at peak power).

Unfortunately, in the "real world" a truck in low gear has significant engine inertia, so the rule does not hold. Max acceleration would be obtained by shifting somewhat "shorter" as the contribution of engine inertia is reduced significantly with higher gearing.

Edited by gruntguru, 30 April 2013 - 22:41.

### #150 saudoso

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Posted 01 May 2013 - 02:44

Unfortunately, this is a "real world" question. If it is assumed that engine rotational inertia is negligible compared to vehicle mass, the answer is the same - "if possible shift at revs above the power peak such that power prior to the gearchange is equal to power after the gearchange." (Electronic truck engines typically have a flat-top power curve so it is easy to accelerate while keeping the engine constantly at peak power).

Unfortunately, in the "real world" a truck in low gear has significant engine inertia, so the rule does not hold. Max acceleration would be obtained by shifting somewhat "shorter" as the contribution of engine inertia is reduced significantly with higher gearing.

But anyway, I'd assume you wouldn't be prescribing shifting around the peak torque RPM on the curve. And bear with me, flat top torque curve means a straight pointing up power curve, not a flat one.