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Acceleration - Torque vs Power


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Poll: Max accelleration at max torque or max power (51 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 [11.76%])

    Percentage of vote: 11.76%

  2. I know it is at max torque as I read it in books (2 votes [3.92%])

    Percentage of vote: 3.92%

  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 [7.84%])

    Percentage of vote: 7.84%

  5. I have no idea what rpm would generate max acceleration (4 votes [7.84%])

    Percentage of vote: 7.84%

  6. I think it is at max power (11 votes [21.57%])

    Percentage of vote: 21.57%

  7. I believe it is at max power by reading forums like this (1 votes [1.96%])

    Percentage of vote: 1.96%

  8. I know it is at max power as I read it in books (3 votes [5.88%])

    Percentage of vote: 5.88%

  9. I am sure it is at max power and know the formula to prove it (20 votes [39.22%])

    Percentage of vote: 39.22%

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#1 Hoax

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Posted 05 April 2013 - 07:48

It's been a long time since the last torque vs power discussion. I'm amused/surprised/chocked/sad that engine designers/tuners, race engineers, and other technical people fail to agree on one of the easier questions to answer theoretically in car dynamics. Let's check up on how the current situation is with a poll!

Hopefully there will be an answer close enough for everyone to participate regardless of technical knowledge. There is no need to start the discussion all over. This is just a poll to check where current forum members stand on this issue.


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#2 Gonzo

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Posted 05 April 2013 - 08:36

Here we go...

#3 saudoso

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Posted 05 April 2013 - 18:26

Once bitten twice shy...

#4 Magoo

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Posted 05 April 2013 - 18:59

From Wikipedia http://en.wikipedia....roll_(Internet)


In Chinese, trolling is referred to as bái mù (Chinese: 白目; literally "white eye"), which can be straightforwardly explained as "eyes without pupils", in the sense that whilst the pupil of the eye is used for vision, the white section of the eye cannot see, and trolling involves blindly talking nonsense over the internet, having total disregard to sensitivities or being oblivious to the situation at hand, akin to having eyes without pupils. An alternative term is bái làn (Chinese: 白爛; literally "white rot"), which describes a post completely nonsensical and full of folly made to upset others, and derives from a Taiwanese slang term for the male genitalia, where genitalia that is pale white in colour represents that someone is young, and thus foolish. Both terms originate from Taiwan, and are also used in Hong Kong and mainland China. Another term, xiǎo bái (Chinese: 小白; literally "little white") is a derogatory term that refers to both bái mù and bái làn that is used on anonymous posting internet forums. Another common term for a troll used in mainland China is pēn zi (Chinese: 噴子; literally "sprayer, spurter").

In Japanese, tsuri (釣り?) means "fishing" and refers to intentionally misleading posts whose only purpose is to get the readers to react, i.e. get trolled. arashi (荒らし?) means "laying waste" and can also be used to refer to simple spamming.

In Icelandic, þurs (a thurs) or tröll (a troll) may refer to trolls, the verbs þursa (to troll) or þursast (to be trolling, to troll about) may be used.

In Korean, nak-si (낚시) means "fishing", and is used to refer to Internet trolling attempts, as well as purposefully misleading post titles. A person who recognizes the troll after having responded (or, in case of a post title nak-si, having read the actual post) would often refer to himself as a caught fish.[citation needed]

In Portuguese, more commonly in its Brazilian variant, troll (produced [ˈtɾɔw] in most of Brazil as spelling pronunciation) is the usual term to denote internet trolls (examples of common derivate terms are trollismo or trollagem, "trolling", and the verb trollar, "to troll", which entered popular use), but an older expression, used by those which want to avoid anglicisms or slangs, is complexo do pombo enxadrista to denote trolling behavior, and pombos enxadristas (literally, "chessplayer pigeons") or simply pombos are the terms used to name the trolls. The terms are explained by an adage or popular saying: "Arguing with fulano (i.e. John Doe) is the same as playing chess with a pigeon: the pigeon defecates on the table, drop the pieces and simply fly, claiming victory."

In Thai, the term "krean" (เกรียน) has been adopted to address Internet trolls. The term literally refers to a closely cropped hairstyle worn by most school boys in Thailand, thus equating Internet trolls to school boys. The term "tob krean" (ตบเกรียน), or "slapping a cropped head", refers to the act of posting intellectual replies to refute and cause the messages of Internet trolls to be perceived as unintelligent.[citation needed]

#5 carlt

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Posted 05 April 2013 - 23:30

you forgot the answer category :

I have no idea , but can feel it in my arse

#6 Tony Matthews

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Posted 06 April 2013 - 07:13

In Thai, the term "krean" (เกรียน) has been adopted to address Internet trolls. The term literally refers to a closely cropped hairstyle worn by most school boys in Thailand, thus equating Internet trolls to school boys. The term "tob krean" (ตบเกรียน), or "slapping a cropped head", refers to the act of posting intellectual replies to refute and cause the messages of Internet trolls to be perceived as unintelligent.[citation needed]

The very worst trolls are the "north kreans".

#7 Hoax

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Posted 06 April 2013 - 08:13

Believe it or not this is not intended to be a troll post. Neither is it a thread to start another discussion about torque vs power. If it were it would have been enough to give you two options in the poll.

In the last week I must have read thousands of torque vs power posts. I'm already firmly planted at the very end in one of the sides on this issue even before starting to read all these posts, so why did I carry on reading? (I must confess that I still do read them.) I'm really not sure myself but one reason is that there is a plenty of knowledge on either side and in every discussion there is always a lot of interesting stuff coming out when the discussion starts to track off in another direction. Clearly there are people here with extreme experience in engines and cars that are still *wrong* in this relatively simple matter. How come?

I am interested in trying to take this to another level. This thread is actually about why do people that have the knowledge still get this wrong when they shouldn't? People are able to prove to themselves (and others) with physics formulas of their liking that they are correct even when they are on the wrong side. If you really want to know the theory on how fast a car accelerate it shouldn't take more than a hour to get the formulas right. If you are a little bit mathematically challenged you could spend a day and that’s it. Seems like a fair investment if your business is engines, cars or racing.

#8 Magoo

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Posted 06 April 2013 - 17:14

I am interested in trying to take this to another level. This thread is actually about why do people that have the knowledge still get this wrong when they shouldn't?


Maybe you just have poor mechanical aptitude. Not everyone has a natural grasp of these things.

#9 JimboJones

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Posted 06 April 2013 - 20:24

I am interested in trying to take this to another level. This thread is actually about why do people that have the knowledge still get this wrong when they shouldn't? People are able to prove to themselves (and others) with physics formulas of their liking that they are correct even when they are on the wrong side. If you really want to know the theory on how fast a car accelerate it shouldn't take more than a hour to get the formulas right. If you are a little bit mathematically challenged you could spend a day and that’s it. Seems like a fair investment if your business is engines, cars or racing.


Totally agree. In answer to your question, some people are just set in their ways, and unwilling to listen. They have a theory, quote some formulas that they do not truly understand (because as you suggest they haven't actually worked through the maths themselves), and no sound engineering explanation will convince them otherwise. Technical forums are full of these flawed personalities, but rest assured none of them would last in a job that depends on such knowledge...

#10 Canuck

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Posted 06 April 2013 - 21:59

This annoys me. You can't say highest acceleration is at peak torque and highest acceleration is at peak power without one of those statements being false (unless peak torque and power happen to be the same rpm obviously). I look forward to learning when I'm here and am disappointed that minds who's input I tend to value equally when it comes to a technical topic can be so protractedly stubborn in defending their ego when the math proves their position wrong.

To paraphrase someone somewhere - science doesn't give a sh!t what you believe.

#11 Magoo

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Posted 06 April 2013 - 22:10

All the choices in the survey fall short of being technically correct. There is a little or a lot wrong with every one of them.

Folks need to go back and actually read the original power/torque threads. I doubt if the participants are interested in typing all that again.

#12 ray b

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Posted 06 April 2013 - 23:29

quibble

if the car red line is 6500 and power peak is 6000
T peak is 4000

would not the accl be greatest 4000 to 6500 vs the short 500 rpm gain then shift and repeat

#13 manolis

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Posted 07 April 2013 - 04:21

Look at the given problem from the energy viewpoint.

In the question: Where the energy provided by the engine goes?

The answer is: One part is consumed to overcome the overall resistance (mechanical, aerodynamic etc) of the vehicle to move with the given speed (so, this part is the same in either case), the rest part is added to the kinetic energy of the vehicle and increases vehicle’s speed.

So, the more the energy provided per second by the engine (i.e. the more the power), the more the increase of the speed of the vehicle per second (i.e. the higher the acceleration).

So, with the engine operating at its max power, the maximum acceleration is achieved.

Thanks
Manolis Pattakos

#14 gruntguru

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Posted 07 April 2013 - 07:36

Hi Hoax
There are at least two reasons for the ". . but torque wins races" myth.
1. Two engines of similar design produce the same maximum power but one of them has a lot more torque. This torquier engine will have a larger displacement and produce its peak power and peak torque at lower rpm. It will have a "milder" state of tune and will maintain close to peak power over a larger percentage of its rpm band. Consequently, over a given acceleration test (eg SS 1/4 mile) this engine can be operated at a higher average power output and will have superior acceleration. The "less torquey" engine could be maintained at the same average output by means of a close ratio gearbox but then more gearshifts would be required.

2. "Max power" is the correct answer - assuming the rotational inertia of the engine can be neglected. In practice this assumption is only valid if the engine speed is constant ie the vehicle has a CVT. For conventional transmissions the contribution of engine inertia to the total inertia of the vehicle is proportional to the gear ratio squared. This means that using a taller gear will reduce the effect of engine inertia and improve the acceleration - until the trade-off in power from reducing rpm begins to exceed any gains made. Thus maximum acceleration will occur by gearing for less than peak-power-rpm (but never as low as peak-torque-rpm). The effect is more pronounced at lower speeds.

As suggested by Magoo, the correct answer to your poll is "none of the above" - unless of course the vehicle happens to be a heavily loaded diesel-electric train which of course has a CVT and negligible engine inertia (expressed as a percentage of total inertia).

#15 manolis

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Posted 07 April 2013 - 09:08

Hi Hoax
There are at least two reasons for the ". . but torque wins races" myth.
. . .
As suggested by Magoo, the correct answer to your poll is "none of the above" - unless of course the vehicle happens to be a heavily loaded diesel-electric train which of course has a CVT and negligible engine inertia (expressed as a percentage of total inertia).


Hello Gruntguru

I think you make the simple complicate.

The problem is for a car moving at a specific speed, and asks for the maximum INSTANT acceleration at that speed.

This makes meaningless the shape of the torque or power curves.

For instance, take the case of a car having a total mass of, say, 1000Kg, moving with, say, 108 Km/h (i.e. the speed v is 30m/sec), with the overall resistance at this speed being, say, 20KW.

If the engine of the car provides a peak power of 80 KW to the car, and the overall resistance consumes, at the above speed, 20KW, the remaining 60 KW are used to accelerate the car.

So, 60KW= F*v, i.e. F=60KW / (30m/sec) = 2 KW * sec / m = 2 * (1000 Nt*m /sec) * sec / m = 2000 Nt, wherein F is the force applied on the car.

A 2000 Nt force acting on a 1000 Kg mass accelerates it INSTANTLY with 2 m/sec^2 (nearly 0,2g), no matter how peaky the power curve is.

At the maximum torque the engine cannot provide more power than the maximum power, so the simple answer is that with the engine providing its maximum power it results the maximum acceleration at a given speed, provided the gear ratios are the ideal ones (no need for CVT gearbox).

PS. Did you see the tilting valves?

Thanks
Manolis Pattakos

#16 Hoax

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Posted 07 April 2013 - 10:15

This annoys me. You can't say highest acceleration is at peak torque and highest acceleration is at peak power without one of those statements being false (unless peak torque and power happen to be the same rpm obviously).

I'm not sure what cause you to react. Of course one of the statements is false.

Most followers of the wrong side lack the math skill to derive the acceleration correctly from formulas. What I find strange is that there are also people with technical know-how that is so high that they should be able to see that they are wrong whenever the other side is presenting valid arguments. However, for some reason they still refuse to accept them. Engineers are supposed to be analytical and curious. Testing the validity of other sides arguments shouldn't take long and they should also be able to find them correct.

assuming the rotational inertia of the engine can be neglected.

Taking the inertia into consideration will put the question at a higher level of (unnecessary) complexity. I think you would agree that a correct understanding of the torque/gear ratio/power concepts is first required for a basic understanding of acceleration. Engineers in one of the sides is failing to fully grasp that even with huge experience in engines and cars.


#17 saudoso

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Posted 07 April 2013 - 11:18

Maximum area under the power curve. Maximum energy transfer. Maximum final speed. Maximum acceleration.

#18 Canuck

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Posted 07 April 2013 - 14:13

I'm not sure what cause you to react.

This:

Of course one of the statements is false.

... there are also people with technical know-how that is so high that they should be able to see that they are wrong whenever the other side is presenting valid arguments. However, for some reason they still refuse to accept them. Engineers are supposed to be analytical and curious.

As obviously both statements cannot be true, why are otherwise intelligent intellectuals refusing to accept the mathematical proof?

#19 johnny yuma

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Posted 08 April 2013 - 01:03

There seem to be too many variables in the assumptions to allow any particular box to be ticked.In general ,you need to be in as low a gear as possible to multiply the torque at the axles,be at or above the maximum torque rpm of the engine, but with some useful amount of revs left in the engine so you can accelerate for a useful amount of time before you must change up to a higher gear.

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#20 Greg Locock

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Posted 08 April 2013 - 01:52

As obviously both statements cannot be true, why are otherwise intelligent intellectuals refusing to accept the mathematical proof?


Because the accurate provable answer for the simplified case does not translate especially well to real world experience, for reasons that can be calculated and explained. These are then used to throw doubt and confusion over the original simplified case, which whatever its faults is a good place to start.

#21 saudoso

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Posted 08 April 2013 - 02:23

If we are to take a practical example, let's say F1, how are those carsa geared? Where're the max torque and power points?

Because AFAIK they'll spin it up to the rev limiter every time if the pit radio allows it. Prety much since the beggining of times.

#22 munks

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Posted 08 April 2013 - 02:44

quibble

if the car red line is 6500 and power peak is 6000
T peak is 4000

would not the accl be greatest 4000 to 6500 vs the short 500 rpm gain then shift and repeat


Yes, this highlights one thing that is imprecise about the original question. Perhaps Hoax should have asked for the maximum *instantaneous* acceleration. But then again, that would have produced other hysterics (well does that mean the moment I stepped on the gas pedal? what about driveline elasticities, blah blah blah). In a normal conversation, this is where one would say, "you KNOW what I MEAN".

#23 munks

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Posted 08 April 2013 - 02:46

The very worst trolls are the "north kreans".


By the way, bravo on this post. It's great on at least three different levels.

#24 Tony Matthews

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Posted 08 April 2013 - 07:29

Pun, silly pun and very silly pun? :)

#25 malbear

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Posted 08 April 2013 - 12:23

Yes, this highlights one thing that is imprecise about the original question. Perhaps Hoax should have asked for the maximum *instantaneous* acceleration. But then again, that would have produced other hysterics (well does that mean the moment I stepped on the gas pedal? what about driveline elasticities, blah blah blah). In a normal conversation, this is where one would say, "you KNOW what I MEAN".

Back many years ago when I was young and needed to impress girls . I owned a Valiant AP5 push button auto, so at the lights I was pretty much unbeatable by other cars of the earer. The Cam and Carby were standard, only extractors and a two inch exhaust pollished ports standard valves.
slightly lowered and pump up Munro shockers. when the lights green the other car would light up smoke and spinning wheels . I just held about half throttle and let the brake off, I always gained a couple of meters not much spin and floored it a few secconds later those torque flight gearboxes were just great. If I floored it from a standing start there would allways be two black marks . dont know why as it was a standard diff and not limmited slip.
maybe the shocks stopped any wheel hop.
I love a tourque big motor.


#26 Rasputin

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Posted 08 April 2013 - 13:44

There's hardly any subject that has generated so many chuckles and frustrations between myself and other mechanical engineers than this one,
but let's try it once again, from the very beginning, even if I have learned that engineers are few and far between on forums such as this;

- What the car reacts to when you press the accelerator is the propulsion-force generated between tire and road, nothing else.

- Between engine and road is gearbox, final drive and wheel radius, why there is no given relation between engine-torque and propulsion-force.

- But power is always force times speed, why there is a direct relation between applied power, the car's speed and propulsion force.

Say you're driving a 2000 kg car with 200 kW at 72 km/h (20 m/s) and floor it at peak-power, you get 200 000/20 = 10 000 N of propulsion-force.

Disregarding the air-resistance, as force is always mass times acceleration, your car will get an instant 10 000/2000 = 5 /s^2 (0.5 g) acceleration.

That's all there's to it really, many standard road-cars have more torque than an F1 car these days anyway, about 300 Nm.

Edited by Rasputin, 08 April 2013 - 13:46.


#27 NTSOS

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Posted 08 April 2013 - 16:15

There's hardly any subject that has generated so many chuckles and frustrations between myself and other mechanical engineers than this one,
but let's try it once again, from the very beginning, even if I have learned that engineers are few and far between on forums such as this;

- What the car reacts to when you press the accelerator is the propulsion-force generated between tire and road, nothing else.

- Between engine and road is gearbox, final drive and wheel radius, why there is no given relation between engine-torque and propulsion-force.

- But power is always force times speed, why there is a direct relation between applied power, the car's speed and propulsion force.

Say you're driving a 2000 kg car with 200 kW at 72 km/h (20 m/s) and floor it at peak-power, you get 200 000/20 = 10 000 N of propulsion-force.

Disregarding the air-resistance, as force is always mass times acceleration, your car will get an instant 10 000/2000 = 5 /s^2 (0.5 g) acceleration.

That's all there's to it really, many standard road-cars have more torque than an F1 car these days anyway, about 300 Nm.


Ok, let's see....so when you floor it at peak power, there is zero engine torque....now I get it!

#28 blkirk

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Posted 08 April 2013 - 17:11

Ok, let's see....so when you floor it at peak power, there is zero engine torque....now I get it!


If you want a TrollTreat ™, you'll have to do a better trick than that.

#29 NTSOS

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Posted 08 April 2013 - 17:29

If you want a TrollTreat ™, you'll have to do a better trick than that.


A TrollTreat ™ you say.......would that go good with a pint of Guinness?

Edited by NTSOS, 08 April 2013 - 17:29.


#30 JimboJones

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Posted 08 April 2013 - 21:45

Yes, this highlights one thing that is imprecise about the original question. Perhaps Hoax should have asked for the maximum *instantaneous* acceleration. But then again, that would have produced other hysterics (well does that mean the moment I stepped on the gas pedal? what about driveline elasticities, blah blah blah). In a normal conversation, this is where one would say, "you KNOW what I MEAN".


It's fairly obvious the OP is talking of instantaneous acceleration, why confuse it even more by mentioning shift points...

It's about power, power, POWER!
If you want to make a car go fast, you tune the engine for as much power as possible, over the rpm range it will be used. The torque is IRRELEVANT, as you simply gear it appropriately.
As Rasputin correctly mentions, F1 engines have a pathetic torque output, less than my diesel in fact, but it goes a hell of a lot faster.

Considering 13% of people here 'know' the answer is actually max torque, I'm surprised none of them are caring to venture why? funny that...


#31 Gold

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Posted 09 April 2013 - 00:00

I'm not an engineer but when I floor it in a turbo diesel Golf, I feel greater acceleration just after the rpm hits peak torque, this "kick in the backside" dies down as the rpms approach peak power. Until I shift up and again I feel greatest push after 3.5k rpm. Even on my gsxr which does 0-100 in 2.5 seconds the greatest feeling of acceleration is not at peak power but a few k rpm under it.

Why is that then?

Edited by Gold, 09 April 2013 - 00:01.


#32 johnny yuma

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Posted 09 April 2013 - 02:10

Can we agree acceleration is greatest in lowest gear,least in top gear? Why is this so ? I would say
because the gearing multiplies the torque.The Formula 1 engine is able to deliver it's limited torque at very high RPM,so it can be geared extremely low in all gears compared to a road car of similar capacity,thus giving impressive acceleration, and speed in gears, from a small engine. Power is only an expression of work done over time,derived from torque readings put through a formula involving time.Yes,we torque people have learnt nothing,I hear you say,but we remain calm and carry on.
It seems final drive ratio about 7.5:1 ??

Edited by johnny yuma, 09 April 2013 - 06:01.


#33 johnny yuma

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Posted 09 April 2013 - 06:04

I'm not an engineer but when I floor it in a turbo diesel Golf, I feel greater acceleration just after the rpm hits peak torque, this "kick in the backside" dies down as the rpms approach peak power. Until I shift up and again I feel greatest push after 3.5k rpm. Even on my gsxr which does 0-100 in 2.5 seconds the greatest feeling of acceleration is not at peak power but a few k rpm under it.

Why is that then?

Makes sense if acceleration rate decreases after peak power,does it not ? The motor's ability to do more is diminishing.

#34 NTSOS

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Posted 09 April 2013 - 10:30

I'm not an engineer but when I floor it in a turbo diesel Golf, I feel greater acceleration just after the rpm hits peak torque, this "kick in the backside" dies down as the rpms approach peak power. Until I shift up and again I feel greatest push after 3.5k rpm. Even on my gsxr which does 0-100 in 2.5 seconds the greatest feeling of acceleration is not at peak power but a few k rpm under it.

Why is that then?





The rate of acceleration follows the engine torque curve and is highest at the torque peak! The rate of acceleration decay is equal to the engine's torque decrease as it approaches the power peak......IOW, the vehicle is still accelerating past the torque peak, but at a slower rate. If the vehicle is equipped with a gear box, the shortest elapsed time over (x) amount of distance is achieved by changing gear at an RPM point where torque is decreasing at a faster rate than power is increasing......or slightly above the power peak!

#35 saudoso

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Posted 09 April 2013 - 11:44

The rate of acceleration follows the engine torque curve and is highest at the torque peak! The rate of acceleration decay is equal to the engine's torque decrease as it approaches the power peak......IOW, the vehicle is still accelerating past the torque peak, but at a slower rate. If the vehicle is equipped with a gear box, the shortest elapsed time over (x) amount of distance is achieved by changing gear at an RPM point where torque is decreasing at a faster rate than power is increasing......or slightly above the power peak!



Id est the maximum area under the power curve.

#36 Gold

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Posted 09 April 2013 - 16:53

The rate of acceleration follows the engine torque curve and is highest at the torque peak! The rate of acceleration decay is equal to the engine's torque decrease as it approaches the power peak......IOW, the vehicle is still accelerating past the torque peak, but at a slower rate. If the vehicle is equipped with a gear box, the shortest elapsed time over (x) amount of distance is achieved by changing gear at an RPM point where torque is decreasing at a faster rate than power is increasing......or slightly above the power peak!


Yeah in single seater schools they say you should change up somewhere between peak power and redline.

So to conclude I still don't get it, why the discrepancy between the mathematical formula (and the poll result) and the rpms when I feel the "kick in the backside"?

Edited by Gold, 09 April 2013 - 16:53.


#37 Ursus

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Posted 09 April 2013 - 19:38

Yeah in single seater schools they say you should change up somewhere between peak power and redline.

So to conclude I still don't get it, why the discrepancy between the mathematical formula (and the poll result) and the rpms when I feel the "kick in the backside"?


Because even if the acceleration drops past the tourque peak, it will drop even lower by shifting early.

By shifting at lower revs you will experience a new acceleration peak at the tourque peak. However in reality this peak will be lower in absolute numbers than the dimishing acceleration in the lower gear.

#38 JimboJones

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Posted 09 April 2013 - 21:40

I'm not an engineer but when I floor it in a turbo diesel Golf, I feel greater acceleration just after the rpm hits peak torque, this "kick in the backside" dies down as the rpms approach peak power. Until I shift up and again I feel greatest push after 3.5k rpm. Even on my gsxr which does 0-100 in 2.5 seconds the greatest feeling of acceleration is not at peak power but a few k rpm under it.

Why is that then?


You are right... Contradiction I hear you say? The acceleration will follow the torque curve in a GIVEN GEAR. But you would have accelerated faster if you were in the gear below, as it would have likely put you closer to peak power at that particular road speed, giving you even more torque at the wheels. The gearing scales the torque curve, so even though the lower gear ratio would put you way beyond peak torque at that road speed, you have more torque to ground thanks to the shorter ratio. And it all boils back to maximizing engine power, i.e. you should shift gear in order to maximize the area under the power curve. Doing this, you will always feel the sensation of decreasing acceleration, it's inevitable with increasing carspeed.

By shifting at lower revs you will experience a new acceleration peak at the torque peak. However in reality this peak will be lower in absolute numbers than the dimishing acceleration in the lower gear.


Bang on!

Another thing to aid understanding, is consider the ideal solution, where you have a CVT (continuously variable transmission). You would set the ratio to always keep the engine at maximum power, as for any given carspeed, this will give maximum torque to the wheels. If you set the ratio to operate the engine at peak torque, yes, you would have more engine torque, but the higher gear ratio would scale this down more than the shorter ratio at the slightly lower torque.

If you are still not convinced, create a torque curve with a different rpm for peak torque and peak power, and plot a graph of torque to ground vs. carspeed for each gear ratio. You will see the shape of the torque curve is simply stretched and scaled down for each gear, and that maximum torque to ground for any carspeed will be beyond the peak engine torque, in a lower gear.


#39 Tony Matthews

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Posted 09 April 2013 - 22:19

Another thing to aid understanding, is consider the ideal solution, where you have a CVT (continuously variable transmission). You would set the ratio to always keep the engine at maximum power, as for any given carspeed, this will give maximum torque to the wheels. If you set the ratio to operate the engine at peak torque, yes, you would have more engine torque, but the higher gear ratio would scale this down more than the shorter ratio at the slightly lower torque.

I think your 'aid to understanding' is going to cause some confusion...

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#40 manolis

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Posted 10 April 2013 - 05:12

I'm not an engineer but when I floor it in a turbo diesel Golf, I feel greater acceleration just after the rpm hits peak torque, this "kick in the backside" dies down as the rpms approach peak power. Until I shift up and again I feel greatest push after 3.5k rpm. Even on my gsxr which does 0-100 in 2.5 seconds the greatest feeling of acceleration is not at peak power but a few k rpm under it.

Why is that then?


Gold,

what you feel is the true and agrees with the theory.

The plot below is the power and torque curve vs r.p.m for an engine. For simplicity the torque curve is sinusoidal.

Posted Image

The maximum torque is at 5000 rpm, while the maximum power is at 6300 rpm.

If the gearbox of the vehicle is such that at 100Km/h with the 2nd gear the engine operates at 6300 rpm (wherein it provides its maximum power) and the 3rd gear is such that at the same 100Km/h the engine operates at 5000 rpm (wherein it provides its maximum torque), the force from the road to the vehicle vs the speed of the vehicle is:

Posted Image

The "force vs speed" curve has the same form with the "torque curve vs rpm" curve (if you think how they measure the power and torque in an accelerating drum dynamometer, you will understand why: they measure the force the tyre(s) apply on the drum periphery vs the rpm of the engine).

By the circle it is marked the point of the maximum torque. By the rectangle it is marked the point of the maximum power.

If you accelerate with 2nd gear in the gearbox, the force increases until the 80 Km/h (wherein the engine provides its maximum torque) and then it decreases (at 80 Km/h and 2nd gear the force that accelerates the vehicle is 2500 Nt, while at 100Km/h and 2nd gear, wherein the maximum power - with 2nd gear - is provided, the accelerating force is only 2300Nt.)

But the problem of Hoax is different.
At 100Km/h how can you achieve the maximum possible acceleration of the vehicle?
By selecting the 2nd gear (the engine operates at its maximum power) or by selecting the 3rd gear (the engine operates at its maximum torque).

The 2nd gear is better because at 100Km/h the resulting force from the road to the vehicle is 2300 Nt.
With the 3rd gear and the engine operating at 5000rpm (manixum torque), the resulting force is only 2000Nt, which means that the acceleration at 100Km/h with 3rd gear (and the engine operating at its maximum torque) is only 87% of the acceleration at the same 100Km/h speed with 2nd gear (and the engine operating at its maximum power).

In the question why the rectangle can never be at a lower height than the circle, the answer is simple: if you multiply the speed by the force you take the power provided to the vehicle. And the rectangle corresponds to the maximum power, so the height of the rectangle from the horizontal axis can never be lower than the height of the circle.

Several years ago the RoadLoad program (in DOS) was written and is available at http://www.pattakon....attakonEduc.htm It explains all these in details.

If you are still confused, just let me know.

Thanks
Manolis Pattakos

#41 Hoax

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Posted 10 April 2013 - 11:39

As obviously both statements cannot be true, why are otherwise intelligent intellectuals refusing to accept the mathematical proof?

I don't know, I hope your answer is better than mine. So far 20 of 32 in this poll know or can prove that they are correct in this issue. The problem is that they are all not proving the same thing. Arguments are starting to fly around...


#42 Hoax

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Posted 10 April 2013 - 11:41

If the vehicle is equipped with a gear box, the shortest elapsed time over (x) amount of distance is achieved by changing gear at an RPM point where torque is decreasing at a faster rate than power is increasing......or slightly above the power peak!

Hmm... Isn't both torque and power decreasing above the power peak?


#43 Gold

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Posted 10 April 2013 - 12:38

Thx Manolis,

What about the old proverb "Torque gives 0-60, hp gives top speed".

Is it true or not?

#44 JimboJones

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Posted 10 April 2013 - 13:11

Thx Manolis,

What about the old proverb "Torque gives 0-60, hp gives top speed".

Is it true or not?


Indeed, great post Manolis, that's the exact illustration I was referring to...

No it isn't. Both acceleration and top speed are defined by engine power (assuming you shift gear at the right rpms).
Knowledge of torque is useless without knowing the engine rpm and gear ratio, so you can translate it to the wheels. If you just deal with power, you needn't be concerned with the engine torque.

When Clarkson bangs on about all that torque an AMG V8 is putting out, it's actually the power that's relevant. But it's the delivery of that power that's surprising, because it comes at such low rpm (high torque). An engine with a lot more torque but the same power will generally have a flatter power curve, as it's less dependent on engine rpm, which means you don't have to rev the hell out of it to get that power (like an F1 car). The F1 car will destroy my diesel as it has more power, but only if it operates in a narrow rpm band (16-18k). My diesel won't be anywhere near as fast, but at low rpm, I will still get a good kick in the bum as it delivers plenty of torque.

#45 saudoso

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Posted 10 April 2013 - 13:23

:up: There is a huge confusion between how nice is to drive a torquey engine and the effects of that on the stop watch.

I do love them. The torquey engines. Until I turned 40 and became and old man I wouldn't hear off an auto transmission. And I loved being able to cruise the electronic toll booth at 50km/h in 5th gear and just floor it back to 120km/h. Or come uphill back from the beach in 5th gear.

But when I was up to no good shifting to the redline always gave the nastier results.

On the other hand, it's open to discussion if replacing a heavy car's engine with some high revn, low displacement bike engine will be the best solution. The narrow usefull RPM band and the number of gears needed might not be practical ot effective.

Edited by saudoso, 10 April 2013 - 13:27.


#46 NTSOS

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Posted 10 April 2013 - 14:11

Hmm... Isn't both torque and power decreasing above the power peak?

Hmm....OR, slightly above the power peak!!

#47 NTSOS

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Posted 10 April 2013 - 15:39

On the other hand, it's open to discussion if replacing a heavy car's engine with some high revn, low displacement bike engine will be the best solution. The narrow usefull RPM band and the number of gears needed might not be practical ot effective.


Kinda like my Kawasaki Centurion MX bike I used to race and then out of boredom, licensed it to ride on the street for the commute to work.
18.5 Hp @ 10,500 rpm from 100 cc's.......in stop and go traffic, people were visably upset having to listen to the exhaust note blaring next to their side windows as I was trying to keep it up on the pipe, so I wouldn't get run over in traffic. :)

Kawasaki Centurion

John

#48 Canuck

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Posted 10 April 2013 - 18:41

I don't know, I hope your answer is better than mine. So far 20 of 32 in this poll know or can prove that they are correct in this issue. The problem is that they are all not proving the same thing. Arguments are starting to fly around...

I have no answer to offer. I'm not one of the 20 by any means.

#49 Greg Locock

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Posted 10 April 2013 - 22:55

I have no answer to offer. I'm not one of the 20 by any means.


In a nutshell
f=m.a+drags
And p=f.v
so a=(p/v-drags)/m

so for a given m and v and drags, to maximise a at that exact situation you need to maximise p. drags is an all encompassing term to cover losses (aero, rr, driveline) at that speed.

The confusing argument is as follows.

f=m.a+drags
f=T.gr

where T is the instantaneous engine torque and gr is the overall gear ratio and effect of rolling radius

So a=(T.gr-drags)/m

In order to maximise a we have to maximise T.gr, which is not the same as maximising T For a given torque curve and speed T.gr is a maximum at peak power, not peak torque

Edited by Greg Locock, 11 April 2013 - 00:22.


#50 Lee Nicolle

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Posted 11 April 2013 - 00:00

I had a car dynoed yesterday. 3.3 litre Holden race engine. Comparing old dyno sheets and the total KW is about the same as another engine. My 400ci Ford Galaxie engine. Very different engines! Both deliver nearly the same number, but one is around 6000rpm, the other about 3800.Now the Ford has a LOT of torque whereas the Holden does not have much. But one engine pulls around 2 tonne of Yank Tank and the other 1 tonne of Holden. The Ford has 3.00 gearing and 84" dia tyres where the Holden has 3.9 and 77" dia tyres. Both have about the same top speed though the Holden will get there a lot earlier. But put the Ford motor in the Holden and it will probably be slower. [and squash the car!!] Or put the Holden in the Ford and top speed eventually would be greater but quite slow getting there.
I feel this may differentiate horsepower and torque.