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Acceleration - Torque vs Power


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Poll: Max accelleration at max torque or max power (65 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 (9 votes [13.85%])

    Percentage of vote: 13.85%

  2. I know it is at max torque as I read it in books (3 votes [4.62%])

    Percentage of vote: 4.62%

  3. I believe it is at max torque by reading forums like this (1 votes [1.54%])

    Percentage of vote: 1.54%

  4. I think it is at max torque (5 votes [7.69%])

    Percentage of vote: 7.69%

  5. I have no idea what rpm would generate max acceleration (4 votes [6.15%])

    Percentage of vote: 6.15%

  6. I think it is at max power (13 votes [20.00%])

    Percentage of vote: 20.00%

  7. I believe it is at max power by reading forums like this (1 votes [1.54%])

    Percentage of vote: 1.54%

  8. I know it is at max power as I read it in books (3 votes [4.62%])

    Percentage of vote: 4.62%

  9. I am sure it is at max power and know the formula to prove it (26 votes [40.00%])

    Percentage of vote: 40.00%

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#251 Charlieman

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Posted 04 August 2017 - 14:53

Hello all.

 

Here is a graph made with the DOS/QuicBasic RoadLoad program (at http://www.pattakon....attakonEduc.htm near the bottom of the web page):

I don't think that it shows anything. If it shows anything, it is that manufacturers try to make petrol cars perform like diesels, and vice versa.

 

Laurence Pomeroy's charts and speculation about engines in "The Grand Prix Car", or about frontal area and power, are more illuminating.



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#252 manolis

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Posted 04 August 2017 - 17:10

Hello Charlieman

 

You write:

"I don't think that it shows anything."

 

 

Yet, you can see a lot.

 

 

It shows the force that at most can apply the tires to the road (and to the car) at the various speeds, which is the most important info in calculating the accelerations, the times to cover some specific distances, the velocities achieved in specific time interval etc.

 

 

The same plot (with fewer horizontal and vertical lines) is here:

 

RoadLoad_Mercedes_Diesel_Gasoline.png

 

 

Here is the plot the RoadLoad program (in the "expert mode") gives for the Diesel Mercedes E-250:

 

RoadLoad_Mercedes_Diesel.png

 

 

And here is the plot the RoadLoad program (in the "expert mode") gives for the Gasoline Mercedes E-250:

 

RoadLoad_Mercedes_Gasoline.png

 

 

Lots of information.

 

If anything is confusing, please let me know to further explain.

 

 

 

PS.

 

Here is the same plot for the Gasoline E-250 with the maximum speed set at 253Km/h (i.e. 3Km/h above that of the Diesel, think why), and the final transmission ratio properly aligned:

 

RoadLoad_Mercedes_Gasoline2.png

 

Thanks

Manolis Pattakos


Edited by manolis, 04 August 2017 - 18:20.


#253 Kelpiecross

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Posted 05 August 2017 - 04:44

I don't think that it shows anything. If it shows anything, it is that manufacturers try to make petrol cars perform like diesels, and vice versa.
 
Laurence Pomeroy's charts and speculation about engines in "The Grand Prix Car", or about frontal area and power, are more illuminating.


If I remember correctly - didn't Pomeroy think that torque was far more important than horsepower when it came to acceleration?

#254 Greg Locock

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Posted 05 August 2017 - 05:50

Well it's certainly a red herring with long legs.



#255 manolis

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Posted 05 August 2017 - 10:51

Hello all.

 

Here is the “colored” graph for the Mercedes E-250 CDi Diesel:

 

RoadLoad_Mercedes_Diesel.png

 

The yellow curve shows the force the engine provides for the acceleration of the car. Its initial horizontal part at left is where the clutch is slipping (by slipping, it provides to the gearbox the peak torque until the revs to rise to the rpm of the peak torque).

Then the yellow curve follows the “force vs velocity” curve of the 1st gear.

At the velocity wherein the first gear gives less force at the wheels than the second gear, the yellow curve starts following the second gear “force vs velocity” curve.

Similarly for the rest gears.

I.e. the yellow curve is the maximum possible force the engine can apply to the road by the tires at a specific velocity (speed) of the car..

 

The red curve results from the yellow curve after the subtraction of the rolling and aerodynamic resistance (the purple curve). At the top speed is where the yellow curve intersects with the purple curve (and the red curve becomes zero: the force provided by the engine equals to the total resistance in the motion of the car).

 

The green curve is the distance covered versus the velocity of the car.

 

The blue curve is the time versus the velocity of the car.

 

Here is the performance table calculated by the RoadLoad program:

 

RoadLoad_Mercedes_Diesel_perf.png

 

And here is, for comparison, the respective performance table the RoalLoad program calculated for the Petrol Mercedes E250:

 

RoadLoad_Mercedes_Gasoline2_perf.png

 

Here is the “colored” graph for the Mercedes E-250 Petrol:

 

RoadLoad_Mercedes_Gasoline2.png

 

As before, the yellow curve is the maximum possible force the engine can apply, through the tires, to the road to accelerate the car.

 

The red curve is the yellow after subtracting the resistance force (purple curve).

 

Etc.

 

If anything is confusing, let me know to further explain.

 

Thanks

Manolis Pattakos



#256 Kelpiecross

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Posted 06 August 2017 - 04:11

Well it's certainly a red herring with long legs.

Eh?

#257 manolis

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Posted 06 August 2017 - 04:42

Hello all.

 

In order the top speed of the Mercedes E-250 CDi Diesel to be maximized, a longer final transmission ratio (as compared to that used in the previous posts) is required to shift the peak power (provided at 3,700rpm) at the top speed.

 

Engine_comp.jpg

 
With the Diesel car at its top speed, its engine must operate at 3,700rpm. 
 
Side effect?
Even longer transmission ratios for the Diesel and so, even worse performance as shown in the following graph and performance table:

 

RoadLoad_Mercedes_Diesel_top_speed.png

 

RoadLoad_Mercedes_Diesel_perf_top_speed.

 

 

Thanks

Manolis Pattakos


Edited by manolis, 06 August 2017 - 04:45.


#258 manolis

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Posted 06 August 2017 - 05:41

Hello all.

 

 

The guys at http://www.competiti...com/PvTquiz.htm (those with the QUIZ) use wrong power-torque curves for the Diesel.

 

 

If the torque curve of the Diesel does increase linearly from 0rpm(!) to 1,500rpm, then the power curve cannot increase linearly in the same interval because the power is the torque multiplied by the revs.

 

The power curve of the Diesel will start, more or less, as the yellow curve shows.

 

Mercedes_E250_Diesel_Petrol_Engines_Comp

 

 

As for the 0rpm the curves start, this is wrong, too. 

 

Thanks

Manolis Pattakos  


Edited by manolis, 06 August 2017 - 05:41.


#259 Greg Locock

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Posted 06 August 2017 - 06:23

kelpiecross - Pomeroy was establishing/reinforcing a red herring that has persisted for decades.



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#260 Tenmantaylor

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Posted 06 August 2017 - 15:32

Torque: Lazy Efficiency
Power: Peak Potential

As I was driving a rather puny diesel on the targa florio route the other day I thought how I maybe would have prefered a puny petrol as the experience of finding the elusive peak power seems a more authetntic pastime. At the same time when climbing a steep hill out of a tight corner at suboptimal revs or in the wrong gear altogether due to not knowing the roads, a diesel is actually preferable due to the greater peak and earliness of torque. In this situation the ability to accelerate towards the (lower) peak power is greater hence more efficient and usable and probably faster. If driven properly with purpose I'm sure (things on the whole beig equal) greater power at the expense of torque wins every time should you have the knowledge and ability to keep it in the window of revs and gear.

But that's all in the past of the IC engine. What about hybrid and electric drivetrains? I'm sure the defiencies of both will be merely ironed out or in electric's case, obliterated;

https://youtu.be/eT7KKxoAvvk

That boat has already sailed. The IC engi e is toast. It is inefficient, out of date technology. The only saving grace is that it runs on a high energy density source vs cheaply available current alternatives. Ignore this at your peril. By all means enjoy being the horse racer, the steam engine tinkerer. Fill your boots. But you'll be an antiquity before you know it.

But what about experience? What about chasing the revs, matching the revs to the gearing on the downchange? The vibrations, the sonorous howl, the bark, the beating heart, the soul? I'm sure there were those who thought this on horse back seeing the first IC vehicles fly by.

Know your context and your desires. In the electric future I don't think we'll have these arguments; such is the superiority in every respect of the electric motor compared to the IC engine you'll be able to have more power and torque than is usable at pretty much all times and the entire discussion will shift to electronics, mappings, algorithms and control. The limitations (ie characteristics) of differing IC engines will become irrelevant.

The next thing I started thinking was how electric could replicate these sensations. A suitably engineered electric motor could easily produce the low torque high peak power characteristics of a large displacement NA IC engine ascribed to it electronically. It could then be mate to a manual gearbox. But people won't do this. It would be the hand built automatic Swiss watch mated to an LCD screen. Electric will be objectively superior in every way; torque and power. You will only be faced with experience or performance. As long as it's a legal choice anyway. Electric has it's own characteristics right now But as the battery technologies that are in the pipeline come to fruition the power/torque world will be your oyster.

All that said, I intend to make the most of the IC engine during these last echoes of the 20th Century and buy the most powerful least efficient version of the car I currently own; the 4 litre V8 E92 M3. It produces 80% more peak power than my current 3 litre turbo diesel E92. But it can only achieve 50% of the peak efficiency and will only achieve 30-40% comparable efficiency on average. I dont even want to contemplate what efficency it would be achieving at it's peak power. I just want to experience it while I can.

#261 Bob Riebe

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Posted 22 September 2017 - 01:40

From way back when I used to know more than I forgot, is torque lb/ft and work ft/lb?

Torque does not mean something is moving but work does?

Power determines how fast something is done, I tink.

 

ON a bicycle, from when I rode such things, big sprocket little sprocket easy pedaling but get no where fast; little sprocket and little sprocket bitch to pedal but got you where you wanted to be a lot faster and quicker, it you had the muscle power.


Edited by Bob Riebe, 22 September 2017 - 01:44.


#262 Greg Locock

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Posted 22 September 2017 - 04:54

There is a formal approach to equations called Dimensional Aanalysis. It is a powerful method of checking whether an equation is plausible. Strictly speaking the units for work and torque are dimensionally identical and are the product of force and length. At first sight that is odd. However, the work done by a torque is the product of the torque and the angle in radians by which it rotates. Angles in radians are pure numbers and have no dimensions. So in order for work=torque*angle to be dimensionally consistent the units for work and torque must be the same.

#263 munks

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Posted 25 September 2017 - 14:56

Thank you Greg for pointing out this anomaly and for clarifying the connection between work and torque.

 

Occasionally one might see angles as a unit, for example the torsional stiffness/rigidity of a car body might be expressed as Nm/rad or lb-ft/deg (degrees are of course nothing more than a constant multiple of radians). My addled brain has trouble with the concept that angles are dimensionless and yet have a unit? Do I need to reconcile this before I die, and if so, how?



#264 Greg Locock

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Posted 28 September 2017 - 12:11

Let's agree that numbers in themselves don't have dimensions. So degrees which are just 180/pi * radians, have the same dimensions as radians. And now you are going to hate me. The value of an angle in radians is the ratio of the length of the subtended arc to the radius of the circle, ie a length divided by a length, which therefore has no dimensions.  DA is a formal field, I don't know what their word is for a unit like degrees, and i slipped up before confusing units and dimensions.

 

the wiki article may help https://en.wikipedia...sional_analysis



#265 munks

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Posted 29 September 2017 - 14:20

Actually, the wiki article touches on that exactly. There are a couple of extensions to DA described there, and one of them (Siano's orientational analysis) specifically addresses angles. The math is a bit too much for me, but it appears that extension can help determine whether a solution involving angles is acceptable.



#266 Dmitriy_Guller

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Posted 04 October 2017 - 00:48

Back in college, when taking the dynamics course, I used dimensional analysis as a sanity check for my calculations without knowing it was a formal field.  I was a math major, so it seemed obvious that treating units as constants should work.  It appeared to over-complicate things by doing extra math when I didn't need to, but it helped a great deal making everything click together. 

 

That worked really well, until the professor on the final exams kept asking questions like "The acceleration of the object at time x is x^2 + x + 3.  Calculate the ...".  She was probably the worst professor I ever had.



#267 Greg Locock

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Posted 04 October 2017 - 05:40

Mathcad can be used to enforce units in equations, so you can't add apples to pears. I must confess that that is one feature I don't use, but some people swear by it.



#268 Lee Nicolle

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Posted 06 October 2017 - 23:04

From way back when I used to know more than I forgot, is torque lb/ft and work ft/lb?

Torque does not mean something is moving but work does?

Power determines how fast something is done, I tink.

 

ON a bicycle, from when I rode such things, big sprocket little sprocket easy pedaling but get no where fast; little sprocket and little sprocket bitch to pedal but got you where you wanted to be a lot faster and quicker, it you had the muscle power.

In Oz foot lbs in the US pounds feet!



#269 Lee Nicolle

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Posted 06 October 2017 - 23:19

Torque in a roadcar is very important, but engine response and HP is as well.  Diesels are usually very torquey but have poor response so good for towing but poor for overtaking. Modern ones can be better though I dislike them and drive a petrol V8 Landcruiser. Auto as I am an old fart!

All the modern SUVs, 4WD and tradie utes are now being offered with the stinkers only, and auto only at least here in Oz. I did notice very different engine/ trans combos though in Europe where manuals are far more common. 

And it seems the US market is different again though stinkers are being pushed down their throats as well it seems. Though it has cost VW billions!

In reality you need both decent power with low end torque. In a racecar? I wonder as the Falcon I sporadically play with probably has more torque than power and getting it too the ground is a real problem. Especially getting moving with traction issues.

The V8 Stupidcar engines however have very good hp and very little torque. The last supposedly real figures I have heard [Ford] was 628hp and 415 foot pounds. And they struggle to get moving through lack of torque but move fairly well once launched. The 415 foot pounds seems a hangover from GpA. It has not seemed to change.



#270 rms

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Posted 06 October 2017 - 23:55

Lee ..... do you really  believe the "real" bhp and torqe figures ?  If so ..... you need to brush up on your maths !



#271 DeKnyff

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Posted 07 October 2017 - 14:45

Something in between peak torque and peak power, depending basically on speed.

 

There are two main forces with act on an accelerating car (for the sake of simplicity, let's assume it's on level ground and we'll don't take into account losses and rolling resistance, which are usually small in comparison): the advance force on the traction wheels and the aero drag.

 

Let's divide the moving car into two different ideal half-cars:

 

1) a car with the original car's mass and moving in vaccum.

2) a car devoid of mass but having a body moving in a draggy atmosphere

 

For the first car, the maximum acceleration is obtained with the maximum forward push (2nd law of Newton). This forward push is proportional to the torque applied to the wheels, itself proportional to the engine's torque (constant of proportionality depending on the kinematic chain) and therefore, the maximum acceleration is obtained at peak torque. Peak power is irrelevant.

 

For the second car, there is no mass to move and therefore torque applied to the wheels is irrelevant. In that case, the more power you can get (which is, by definition, the produce of drag force by speed), the more you can overcome the drag and the quicker you can go.

 

Real world cars are a mix of both ideal cars. At low speeds, aero drag is of little importance, so cars will look ideal car 1 and will have the the maximum acceleration at or near the torque power. At high speeds, considering that drag is proportional to the square of speed, it will be the dominant factor (like in ideal car 2) and power will become essential for accelerating.

 

Which is consistent with the well-known fact that torquey cars accelerate well from traffic lights, but only powerful cars keep pushing above 200 kph.



#272 Dmitriy_Guller

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Posted 07 October 2017 - 17:53

For the first car, the maximum acceleration is obtained with the maximum forward push (2nd law of Newton). This forward push is proportional to the torque applied to the wheels, itself proportional to the engine's torque (constant of proportionality depending on the kinematic chain) and therefore, the maximum acceleration is obtained at peak torque. Peak power is irrelevant.

 

That "constant" of proportionality is only constant if you're stuck in gear.  Every pro-torque argument ultimately rests on this fallacy of keeping a non-constant constant.

 

Which is consistent with the well-known fact that torquey cars accelerate well from traffic lights, but only powerful cars keep pushing above 200 kph.

 

It's a widely shared belief, but it's not a fact.  If you compare 0-60 times for different cars, you will find that power to weight ratio is a much better predictor than torque to weight ratio.


Edited by Dmitriy_Guller, 07 October 2017 - 17:53.


#273 DeKnyff

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Posted 07 October 2017 - 20:46

That "constant" of proportionality is only constant if you're stuck in gear.  Every pro-torque argument ultimately rests on this fallacy of keeping a non-constant constant.

 

I know, but the thread opener says "revs at a given gear" and I'm answering to that.


It's a widely shared belief, but it's not a fact.  If you compare 0-60 times for different cars, you will find that power to weight ratio is a much better predictor than torque to weight ratio.

 

 

When I said "starting from the traffic lights" I was referring to speeds lower than 60 mph. Anyway, in the absence of drag, at low speeds, the force which accelerates the car is only the torque coming from the engine, through the kinematic chain to the wheels, so max acceleration must necessarily happen at max torque revs. Of course, as you said, this forward push will be different in different gears, but for a given gear max acceleration = max torque revs. I insist, I'm talking of low speeds, before aero drag stars to be significative. It's not a belief, it's physics.


Edited by DeKnyff, 07 October 2017 - 20:46.


#274 Dmitriy_Guller

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Posted 07 October 2017 - 21:18

Aero drag does not factor in anywhere.  The answer is the same whether you have aero drag or not.  "Torque is for acceleration, horsepower is for speed" is another one of those widely shared beliefs that are not in fact true.



#275 Greg Locock

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Posted 11 October 2017 - 23:23

...as i think has been demonstrated about 30 times in this thread. The exception is if you can't change the gearing, and you have limited options when tuning a NA SI engine, in which case the rpm of the torque peak is your only variable. Then it might be a slightly useful rule of thumb, in some cases.



#276 V8 Fireworks

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Posted 22 October 2017 - 16:59

Hi Hoax
There are at least two reasons for the ". . but torque wins races" myth.

 

Yes, what some people fail to understand is that statement is about comparing the power curves of two otherwise similar engines.  Two 2000cc class engines with a category limit of 8000rpm, one making 275bhp the other making 285bhp, but the former perhaps producing a greater cumulative total in the range 5500 to 7500rpm whereas the latter has been tuned to the nth-degree for a peak figure that arrives at 7950rpm but losing out to the former all the way up to 7650rpm.


Edited by V8 Fireworks, 22 October 2017 - 17:32.


#277 V8 Fireworks

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Posted 22 October 2017 - 17:05

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 feeling jerk, the rate of change of acceleration. Turbos, generally, have a more severe rate of change of acceleration than NA.



#278 V8 Fireworks

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Posted 22 October 2017 - 17:07

Thx Manolis,

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

Is it true or not?

 

Not true.


Edited by V8 Fireworks, 22 October 2017 - 17:08.


#279 V8 Fireworks

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Posted 22 October 2017 - 17:17

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

 

:up:

 

Power is related to the rate of energy put to the wheels. Watts/second.   Fundamentally related to effective energy rate the engine extracts from the combustion of petrol or diesel or ethanol, or that the electric motor draws from the batteries.

 

Torque is just the twisting force of the drivetrain.  You can have lots of torque with no power at all, it doesn't have any relationship to performance...


Edited by V8 Fireworks, 22 October 2017 - 17:18.


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#280 V8 Fireworks

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Posted 22 October 2017 - 17:25

The V8 Stupidcar engines however have very good hp and very little torque. The last supposedly real figures I have heard [Ford] was 628hp and 415 foot pounds. And they struggle to get moving through lack of torque but move fairly well once launched. 

 

:rolleyes:



#281 V8 Fireworks

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Posted 22 October 2017 - 17:28

I know, but the thread opener says "revs at a given gear" and I'm answering to that.


 

 

When I said "starting from the traffic lights" I was referring to speeds lower than 60 mph. Anyway, in the absence of drag, at low speeds, the force which accelerates the car is only the torque coming from the engine, through the kinematic chain to the wheels, so max acceleration must necessarily happen at max torque revs. 

 

 

This is completely wrong.

 

The force to accelerate the car is caused by power being extracted by the engine from petrol...

 

Power = force * velocity

Force = mass * acceleration

Therefore

Acceleration = Power / (mass * velocity)

 

Therefore maximum acceleration occurs at maximum power, as of course it should given that is when the maximum energy rate is being produced by the engine.


Edited by V8 Fireworks, 22 October 2017 - 17:32.


#282 gruntguru

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Posted 22 October 2017 - 22:13

You are feeling jerk, the rate of change of acceleration. Turbos, generally, have a more severe rate of change of acceleration than NA.

No he is feeling acceleration. If the entire rev range is explored in a particular gear, the highest acceleration will coincide with the torque peak. Of course if you wanted maximum acceleration at that road speed, you should have been in a lower gear.



#283 gruntguru

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Posted 22 October 2017 - 22:14

:up:

 

Power is related to the rate of energy put to the wheels. Watts/second.

Watts is a unit of power not energy, Perhaps you meant Joules/second?



#284 Canuck

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Posted 23 October 2017 - 00:28

No he is feeling acceleration.


I am tempted to say WRONG!! because we do not sense acceleration only a change in the rate of acceleration. If you were to stuff up your visual and auditory input, you would have no sense of difference between an acceleration of 2m/s and 20m/s under sustained conditions. Once the rate of change went to zero, you would not have the data to make a conclusion about your acceleration relative to a previous acceleration event. That is to say you would not know if you’d been accelerating at different rates.

I say tempted because I’ve been around here long enough to be wrong despite my absolute certainty at least once.

#285 Greg Locock

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Posted 23 October 2017 - 02:15

" you would have no sense of difference between an acceleration of 2m/s and 20m/s under sustained conditions." So those pictures of people with their faces pulled back and their eyeballs popping out on centrifuges are fakes?



#286 Kelpiecross

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Posted 23 October 2017 - 04:30

No he is feeling acceleration. If the entire rev range is explored in a particular gear, the highest acceleration will coincide with the torque peak. Of course if you wanted maximum acceleration at that road speed, you should have been in a lower gear.


The highest acceleration in a particular gear would be at the power peak surely?

#287 V8 Fireworks

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Posted 23 October 2017 - 06:03

The highest acceleration in a particular gear would be at the power peak surely?

 

I agree.

 

Think of driving a properly tuned NA car, max power arrives just before redline, that is when max acceleration is.

 

 

No he is feeling acceleration. If the entire rev range is explored in a particular gear, the highest acceleration will coincide with the torque peak.

 

No.

 

Think of driving an old-fashioned turbo car.  The greatest sensation occurs when the turbo suddenly spools up and the car takes off *whoosh*.

 

This is  because the torque has suddenly increased so dramatically as the turbo has come on boost.  The slope (rate of change) of torque is the steepest at the point, i.e., the rate of change of acceleration is greatest at that point and that is what you feel -- jerk, rate of change of acceleration (giving the term "jerkiness").

 

For example, the Lancer Evolution 6 feels like "whoosh" at about 3000rpm when the rate of change of torque is the greatest.  [It so happens that is when torque is the greatest, but that is not the cause of the perceived sensation.]  This is why NA cars feel less dramatic as they are accelerating, even though they may as just as fast (or faster), since they typically don't have aggressive spikes in the power curve as old-fashioned turbo cars do.

 

Mitsubishi-Lancer-EVO-6-Vipec-Makinen.jp


Edited by V8 Fireworks, 23 October 2017 - 06:03.


#288 gruntguru

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Posted 23 October 2017 - 06:49

A graph of acceleration vs road-speed in a particular gear is similar to a thrust curve which is the leverage to the contact patch times the TORQUE curve. This is what the body feels.

 

Yes the body is also sensitive to jerk.

 

Acceleration determines how far into the seat-back your body is pressed.

 

Jerk determines how rapidly your body is slammed back or forward in the seat.



#289 gruntguru

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Posted 23 October 2017 - 06:53

. . . . an acceleration of 2m/s and 20m/s . . . 

Yes you can't sense the difference between constant velocities of 2 and 20 m/s.

 

No, you can sense the difference between constant accelerations of 2 and 20 m/s/s.

 

(m/s are units of velocity. You probably meant m/s/s?)


Edited by gruntguru, 23 October 2017 - 06:53.


#290 V8 Fireworks

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Posted 23 October 2017 - 09:56

A graph of acceleration vs road-speed in a particular gear is similar to a thrust curve which is the leverage to the contact patch times the TORQUE curve. This is what the body feels.

 

Yes the body is also sensitive to jerk.

 

Acceleration determines how far into the seat-back your body is pressed.

 

Jerk determines how rapidly your body is slammed back or forward in the seat.

 

Is this correct?  Care to elaborate further please?  :)


Edited by V8 Fireworks, 23 October 2017 - 09:57.


#291 gruntguru

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Posted 23 October 2017 - 21:53

Surely you can see what I am saying?

 

If you ride a rocket sled accelerating at a constant 10g you will "feel" the difference compared to say a 747 accelerating down the runway at less than 1g.

 

Compression of the seat back is also obvious. Lay the seat back through 90* with you sitting in it - that's 1g of compression of the cushion. Now increase your body mass x 10 - that's 10g of compression. The two scenarios feel different - both to you and the poor cushion.



#292 manolis

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Posted 24 October 2017 - 06:04

Hello Kelpiecross.

 

You write:

"The highest acceleration in a particular gear would be at the power peak surely?"

 

 

Here is modified the plot in the post #287 of V8 Fireworks:

 

Force_vs_Speed_vs_Power_vs_Torque_Mitsub

 

 

The green curve "Accelerating force on the vehicle" has identical shape with the original torque curve.

 

At 60Km/h vehicle speed (i.e. at 16.7m/sec), wherein the engine revs at 6,000rpm, the engine makes its peak power (282HP / 207kW). 

 

The relation between the power provided, the speed of the vehicle and the force acting on the vehicle is: Power = Speed * Force.

 

This means that at 60Km/h the force that accelerates the vehicle is 207kW / 16.7m/sec = 12,500N. It is the point P4 in the above plot.

 

Supposing a total vehicle weight of 10,000N (1,000Kg, 2,200lb), this gives an acceleration of 1.25g (g=9.81m/sec^2).

 

 

At the point P1 of the maximum torque, the vehicle speed is 37,5km/h (10,4m/sec) and the force acting on the vehicle is 14,600N. This gives an acceleration of 1.46g, which is 17% higher than the acceleration at the maximum power.

 

 

Worth to note: at the point P1, wherein the acceleration maximizes, the power provided by the engine (point P2) is: 207HP (152kW), i.e. it is more than 25% less than the peak power.

 

 

It seems strange: with substantially less power, the acceleration is higher!

 

 

If you look at it from the energy viewpoint, it gets simpler:

 

in order to gain 1Km/h speed when the vehicle moves with 30Km/h, it is required only the 1/2 of the energy required in order to gain the same 1Km/h speed when the same vehicle moves with 60Km/h (because the kinetic energy increases with speed square).

 

(31^2 - 30^2) / (61^2 - 60^2) =~ 0.5

 

 

If something is still confusing, please let me know to further explain.

 

Thanks

Manolis Pattakos  

 

 

 

 



#293 V8 Fireworks

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Posted 24 October 2017 - 06:56

 

 

At the point P1 of the maximum torque, the vehicle speed is 37,5km/h (10,4m/sec) and the force acting on the vehicle is 14,600N. 

 

Great work manolis!

 

I had to check this for myself, you are absolutely right, 14,600N force at about 38 km/hr, much less than the 12,500N force at 60 km/hr.  :eek: [A Lancer Evo which only weighed 1000kg and had zero drivetrain loss would be very fast indeed it seems.  :p ]

 

So maximum torque indeed produces maximum acceleration? Is this always true?  :confused:

 

Edit - Wait of course it is not true that maximum torque always produces maximum acceleration.  

 

As to the original question, higher power will produce more acceleration for a given road speed, given that Force = power/velocity...  That must be true.  

 

That is, it's better to be in the gear that gives engine speed closest to maximum power for a given road speed in order to achieve maximum acceleration when planting one's foot -- right? [I.e., for the Lancer Evo, better to be in 3rd gear near peak power, rather in 4th gear near peak torque, for instance?]


Edited by V8 Fireworks, 24 October 2017 - 07:24.


#294 V8 Fireworks

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Posted 24 October 2017 - 07:30

For a Honda 4-cylinder car engine with a power and torque curve like this, could be said that the best (or any(!)) acceleration can always be found by going to the lowest possible gear for the given road speed?  :)

 

Honda Civic

Gielisep3.jpg

With the stock tune (red), it was two torque peaks neatly corresponding to the two sets of (VTEC) cam profiles...

 

Point 1, 30 km/hr, 4000rpm:

F=P/v= 82 kW / 8.33 m/s = 9,840 N

 

Point 2, 60 km/hr, 8000rpm:

F=P/v= 156 kW / 16.67 m/s = 9,360 N (less!)

 

 

I would like to try this for another vehicle.

 

Ford Mustang Shelby

2016-ford-mustang-shelby-gt350-dyno-657x

 

Point 1: 5000rpm, 31 mph, 13.9 m/s, 360hp (peak torque)

 

F = P/v = 268 kW / 13.9 m/s = 19,300 N 

 

Point 2: 7000rpm, 44 mph, 22.3 m/s, 470hp (peak power)

 

F = P/v = 350 kW / 22.3 m/s = 15,700 N (much less )

 

 

]


Edited by V8 Fireworks, 24 October 2017 - 07:54.


#295 V8 Fireworks

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Posted 24 October 2017 - 08:10

Hmm, another way of looking at it:

 

Idealised naturally aspirated engine with linear power characteristic:

Torque should be constant at all rpms (nearly), thus Power is proportional to velocity only (since P = Torque * rotational velocity, where torque is constant), and thus propulsive force is constant.

 

Typical turbocharged engine, with torque peak at low to mid rpms, and power peak at mid rpms:

Power is a function of velocity (since torque is far from constant with velocity), thus the propulsive force is not constant.

 

Given the above, the maximum propulsive force (acceleration) within a given fixed gear ratio does indeed occur at the peak torque.  :eek:  (Because that peak torque, by definition, is the same location where the power curve deviates above the level of an idealised straight line power "curve" by the most. )

 

However, the opening question called for fixed speed and indeterminate gear ratio (i.e., a car with a CVT?) in which case maximum power is better than maximum torque, for maximum acceleration.

 

Phew.  :stoned:


Edited by V8 Fireworks, 24 October 2017 - 08:37.


#296 manolis

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Posted 25 October 2017 - 03:01

Hello V8 Fireworks.

 

You write:

"the maximum propulsive force (acceleration) within a given fixed gear ratio does indeed occur at the peak torque."

 

Yes, without exceptions.

 

It doesn't matter whether the engine is naturally aspirated or turbocharged, it doesn't matter whether the engine is a 2-stroke or a 4-stroke, it doesn't matter whether the engine has "flat" or "peaky" torque curve: in all cases the maximum acceleration is achieved at the maximum torque. 

 

If the torque is perfectly flat (the same at all revs), then the maximum acceleration will be achieved at the lower speeds due to the increased aerodynamic resistance at the higher speeds.

 

 

 

 

In the case the gear ratio is allowed to be optimized (say by a CVT having negligible friction loss), then, for the maximum acceleration at all speeds you should keep permanently the engine at its peak power.

 

Thanks

Manolis Pattakos 



#297 Kelpiecross

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Posted 25 October 2017 - 03:54


Thank you Manny - for the first time in about 50 years I think I understand this seemingly paradoxical situation.

I imagine that you can also say that, in a way, that the "great unwashed's" claim that torque causes maximum acceleration is true. It may be that an older car with three gears and a big V8 may give the impression that torque is more important etc.

#298 manolis

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Posted 27 October 2017 - 06:03

Hello Kelpiecross.

 

 

In the following plot:

 

Force_vs_Speed_vs_Power_vs_Torque_Mitsub

 

 

the instant acceleration maximizes at 37.5Km/h vehicle speed. This is an instant peak which has only a little effect on the actual acceleration of the vehicle (before 32Km/h and after 40Km/sec the acceleration of the car is substantially lower than at the 37.5Km/h of the acceleration peak).

 

 

So, for the performance (i.e. how fast the car can go) more important than a local peak of the acceleration is its distribution along the speed range.

 

 

From the original dyno plot (post #287 of V8 Fireworks), the following plot was made: 

 

Force_vs_Speed_vs_Power_vs_Torque_Mitsub

 

As said, the shape of the torque curve and of the acceleration curve (case of zero aerodynamic and rolling friction) are identical.

 

 

Think what does it means "acceleration".

It is the rate of speed change:

 

a=dv/dt=(v2-v1)/(t2-t1)

 

wherein v2 is the final speed of the vehicle, v1 is the initial speed of the vehicle, and dt=t2-t1 is the time required for the specific "small" speed change to happen.

 

 

So, dt=(v2-v1)/a.

 

Taking a constant speed change (or speed step) v2-v1 (say 1Km/sec, or 5Km/sec, or 10Km/sec . . )  , the time required for this speed change is linearly proportional to inverse acceleration; this is the rule used to calculate / to draw the red curve.

 

 

The above plot "says" that in order to go from 22.5Km/sec to 27.5Km/sec (which is a 5Km/h speed change), they are required 0.290 seconds (it is the rev range / speed range wherein the engine torque is too low).

 

The above plot also "says" that the time required to go from 32.5Km/h to 37.5Km/h (which is again a 5Km/h speed change) is 0.160 seconds, i.e. almost half than before (22.5Km/h to 27.5Km/h).

 

A rough  calculation of the time required in order the vehicle to go, say, from 30 to 65 Km/h is to estimate the mean time per 5Km/h speed increase in the 30 to 65 Km/h speed range (it is about 190ms) and multiply it by 7 (7=(65-30)/5), which is ~1.5sec.

 

 

Please let me know if something is confusing to further explain.

 

Thanks

Manolis Pattakos



#299 RogerGraham

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Posted 03 November 2017 - 12:51

I've been reading this thread on and off for ages now, and the bit I keep tripping over was summarised by Manolis above:
 

1. maximum acceleration in a given gear occurs at peak torque

2. if the gear ratio can be optimized (e.g. with a CVT) then maximum acceleration occurs at peak power

 

At face value, statement 1 implies that for statement 2, at any instant (i.e. when the CVT is at a "fixed" ratio at that instant), the maximum acceleration will occur at peak torque.  But that's not the case.

 

Can someone help a poor village idiot understand (if possible, using hand-waving arguments instead of (or, as well as) charts)?


Edited by RogerGraham, 03 November 2017 - 12:53.


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#300 V8 Fireworks

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Posted 03 November 2017 - 14:25

 

At face value, statement 1 implies that for statement 2, at any instant (i.e. when the CVT is at a "fixed" ratio at that instant), the maximum acceleration will occur at peak torque.  But that's not the case.

 

No it doesn't mean that.  :)  The maximum acceleration will only occur at peak torque within a fixed gear, however a CVT is not a fixed gear.

 

Acceleration = power / velocity

 

The reason is that within a fixed gear, while power increases with engine speed (usually), velocity (road speed) also increases with engine speed.  The increase the increase in velocity (road speed) cancels out the increase in power.

 

Since a CVT is not a fixed gear, this does not apply, the CVT can put the engine to maximum power regardless of the road velocity.

 

I have a mathematical proof of these statements if it helps, just need to fetch it.  :)


Edited by V8 Fireworks, 03 November 2017 - 14:29.