Acceleration - Torque vs Power

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

>1000 posts and if this explanation was posted amongst that mess, it was lost in the noise. This simple, straight-forward explanation is all that I needed to sort me out. Thank you.

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.

If you put the 3.9 diff in the Ford ,and the 3.0 diff in the Holden,I'm guessing the Ford would win a s.s.400m
drag race quite easily,but the Holden would eventually have a higher top speed. You have only changed the TORQUE at the back axles,the power of the engines is unchanged.
My question is,if you put the two cars on the dyno again with their new diffs,does the dyno recalibrate itself and give you a greater Kw reading because it detects more torque to put back into it's black box,or what?

If you put the 3.9 diff in the Ford ,and the 3.0 diff in the Holden,I'm guessing the Ford would win a s.s.400m
drag race quite easily,but the Holden would eventually have a higher top speed. You have only changed the TORQUE at the back axles,the power of the engines is unchanged.
My question is,if you put the two cars on the dyno again with their new diffs,does the dyno recalibrate itself and give you a greater Kw reading because it detects more torque to put back into it's black box,or what?

As I said the Ford motor in the Holden would generally be slower as the engine does not rev very hard. Though pushing around a ton less would make it accelerate a good deal faster. IF the Ford had the useable 6000rpm it would be faster.[It may turn 6000 but not make any power over about4800]
All these 300kw road cars that don't rev very hard [max power 4500] are still slower than the Torana! Even with 6 speed boxes and 3.9 rear ends. The 700kilo weight penalty is part of the equasion. But they push the big heavy tanks around quite well as a road car.Wheras the little 3.3 would really struggle pushing around 1700+ kilo

Putting a 3.9 in the Galaxie however would increase the power output a bit on the rollers. Though I think that was only dynoed in second gear [of3]where the Holden was in 4th of 4. Same dyno, same operator and probably about the same road speed.
An engine dyno is a fairer measure ofcourse. But I do not ready access to one.

Several years ago the Galaxie had the original shorter 3.25 diff ratio which I swapped with the centre from a Ford Ranchero with a 4V 351 that I owned. The ratio change was beneficial for both vehicles as the torquey and lazy 400 appreciated the taller gears. As the lumpy grumpy 3514v appreciated the shorter gears. The 351 would turn 6000 very quickly and easily.
And these are engines of the same engine family. But 351 with 3.5 stroke, shortish conrods, large flowing cylinder heads v the 400 with 4" stroke, conrods nearly an inch longer and far smaller intake runners which makes far better velocity at low rpm, though ofcourse less volume at high rpm

Edited by Lee Nicolle, 12 April 2013 - 10:43.

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 reason is that if you stay in one gear, you're going faster when you hit peak power than you were going when you hit peak torque, so it's not an apples to apples comparison. The faster you go, the lower your maximum possible acceleration is.

Edited by Dmitriy_Guller, 12 April 2013 - 02:03.

The Ford has 3.00 gearing and 84" dia tyres where the Holden has 3.9 and 77" dia tyres.

Monster trucks?

Power is ultimately the only factor which controls both acceleration and top speed. It is an illusion (sometimes a seemingly very real one) brought about by gearing, engine characteristics etc. that torque is the main factor in acceleration. If cars had a infinitely variable ratio gearbox (of perfectly positive engagement etc.) from zero speed (which is sometimes referred to as an "engaged neutral") to 1:1, there would be little argument about whether torque or horsepower was the controlling factor in acceleration.

Recently I was trying to break a stuck, rusty clutch free by using the starter motor and holding the car with the brakes on hard in 5th gear etc. (it worked very effectively) - it is hard to find an accurate figure but the starter motor applies about 150/200 pounds-feet to the input shaft of the gearbox - as much as a pretty healthy engine. But although the torque at the flywheel is quite strong clearly the car is only going to accelerate to one or two mph - the HP at the flywheel only being 2 or 3 (or whatever) HP.

>1000 posts and if this explanation was posted amongst that mess, it was lost in the noise. This simple, straight-forward explanation is all that I needed to sort me out. Thank you.

I take it back.

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

Re-reading this sentence confused me. How can T.gr not be maximum when gr (in a given gear) is fixed and T is highest at peak torque? Or - what have I missed?

Power is ultimately the only factor which controls both acceleration and top speed. It is an illusion (sometimes a seemingly very real one) brought about by gearing, engine characteristics etc. that torque is the main factor in acceleration. If cars had a infinitely variable ratio gearbox (of perfectly positive engagement etc.) from zero speed (which is sometimes referred to as an "engaged neutral") to 1:1, there would be little argument about whether torque or horsepower was the controlling factor in acceleration.

Recently I was trying to break a stuck, rusty clutch free by using the starter motor and holding the car with the brakes on hard in 5th gear etc. (it worked very effectively) - it is hard to find an accurate figure but the starter motor applies about 150/200 pounds-feet to the input shaft of the gearbox - as much as a pretty healthy engine. But although the torque at the flywheel is quite strong clearly the car is only going to accelerate to one or two mph - the HP at the flywheel only being 2 or 3 (or whatever) HP.

If we take the following: a "low speed" engine that runs from 1000 to 3000 rpm with a torque peak at 2000 rpm and a 2nd engine engine that runs from 1000 to 6000 rpm with a torque peak at 3000 rpm, where each engine makes identical torque (it's a thought experiment) at 1000 rpm, peak torque and max rpm (so in essence the low speed engine's torque curve is simply compressed). Install into otherwise identical cars but the "low speed" engine has a 4:1 rear diff and an 8-speed transmission with a .5:1 final while the other has a 2:1 diff and a 4-speed transmission and 1:!1 final thus allowing both cars to be travelling the same terminal velocity at max rpm.
-Would both cars accelerate at the same rate (assuming both engines can get out of their own way, both setups have identical drag and truly seamless shifts)?

There is nothing in the original question that talks about the gearing being fixed, and in fact if you specify that you can select max P for a given v then you are changing the gearing. If you maximise T.gr then you will gear for max P AT THAT SPEED. There is no contradiction. As a real life example, if you were interested in max instantaneous acceleration at a given speed, would you be better off in third gear at Tmax, or second gear at Pmax?

Look at the plots of accceleration vs time as you go through the gears- you always want to be in the numerically higher gear, at any given speed, if the resulting rpm is between Tmax and Pmax.

here's an example of the general idea.

http://i1227.photobu...in/GvsSpeed.jpg

Edited by Greg Locock, 15 April 2013 - 22:41.

here's an example of the general idea.

http://i1227.photobu...in/GvsSpeed.jpg

A good example of an engine/transmission that needs to be shifted at the redline to achieve max acceleration. (Assuming the ends of the curves shown represent max rpm. Also assuming negligible engine inertia )

There is nothing in the original question that talks about the gearing being fixed, and in fact if you specify that you can select max P for a given v then you are changing the gearing. If you maximise T.gr then you will gear for max P AT THAT SPEED.

That's where I fell down - I failed to shift gears. Why do I find this subject so amorphous?

That's where I fell down - I failed to shift gears. Why do I find this subject so amorphous?

Since the torque max and powers max are at different rpm, then the gearing must be changed or the vehicle will be operating at a different (ground) speed.

Re-reading this sentence confused me. How can T.gr not be maximum when gr (in a given gear) is fixed and T is highest at peak torque? Or - what have I missed?

If you fix a gear ratio, you kind of render the question of which gear you should select moot.

Thank you Captain Obvious ;-)

Power is ultimately the only factor which controls both acceleration and top speed. It is an illusion (sometimes a seemingly very real one) brought about by gearing, engine characteristics etc. that torque is the main factor in acceleration. If cars had a infinitely variable ratio gearbox (of perfectly positive engagement etc.) from zero speed (which is sometimes referred to as an "engaged neutral") to 1:1, there would be little argument about whether torque or horsepower was the controlling factor in acceleration.

Recently I was trying to break a stuck, rusty clutch free by using the starter motor and holding the car with the brakes on hard in 5th gear etc. (it worked very effectively) - it is hard to find an accurate figure but the starter motor applies about 150/200 pounds-feet to the input shaft of the gearbox - as much as a pretty healthy engine. But although the torque at the flywheel is quite strong clearly the car is only going to accelerate to one or two mph - the HP at the flywheel only being 2 or 3 (or whatever) HP.

The high torque at the gearbox input shaft is largely provided by the gearing resulting from the flywheel's radius.The output shaft of the starter motor alone would not be anywhere near 150lbs/ft,which is why it only makes 2 or 3 HP.If there was 150lbs/ft at the starter motor,it could indeed accelerate the car very well...but not if still geared by the flywheel as it would demolish the ring gear on the flywheel, or the gearbox, if you tried to move the car from rest. The analogy breaks down because the characteristics ofthe starter motor are very different to an IC engine.
A Toyota Prius evidently has an epicyclic gearbox so power to the driven wheels can accept torque from 2 shafts turning at different speeds ,from the electric motor which has max torque from zero rpm up to a tail off point ,while the IC motor does it's thing at other rpms.

Edited by johnny yuma, 17 April 2013 - 02:51.

The high torque at the gearbox input shaft is largely provided by the gearing resulting from the flywheel's radius.The output shaft of the starter motor alone would not be anywhere near 150lbs/ft,which is why it only makes 2 or 3 HP.If there was 150lbs/ft at the starter motor,it could indeed accelerate the car very well...but not if still geared by the flywheel as it would demolish the ring gear on the flywheel, or the gearbox, if you tried to move the car from rest. The analogy breaks down because the characteristics ofthe starter motor are very different to an IC engine.
A Toyota Prius evidently has an epicyclic gearbox so power to the driven wheels can accept torque from 2 shafts turning at different speeds ,from the electric motor which has max torque from zero rpm up to a tail off point ,while the IC motor does it's thing at other rpms.

The point I was trying to make is that the overall effect of the starter motor is to produce an engine that apparently has 150lbs/ft of torque and 2 or 3 horsepower at the flywheel. Clearly it is not going to go anywhere very quickly - I think this is one way of illustrating the fact that torque at the flywheel is not the controlling factor in acceleration. You could actually build a steam engine wth the odd 150/3 combination of torque/horsepower.

On the other hand the opposite combination of 150 horsepower and 3lbs/ft of torque could produce an engine that would cause plenty of acceleration (when suitably geared) of a car.
A small steam or air turbine might have a combination of torque/horsepower like this - 150HP at 260,000RPM.

The point I was trying to make is that the overall effect of the starter motor is to produce an engine that apparently has 150lbs/ft of torque and 2 or 3 horsepower at the flywheel. Clearly it is not going to go anywhere very quickly - I think this is one way of illustrating the fact that torque at the flywheel is not the controlling factor in acceleration. You could actually build a steam engine wth the odd 150/3 combination of torque/horsepower.

On the other hand the opposite combination of 150 horsepower and 3lbs/ft of torque could produce an engine that would cause plenty of acceleration (when suitably geared) of a car.
A small steam or air turbine might have a combination of torque/horsepower like this - 150HP at 260,000RPM.

The thing is the 2-3 HP from the starter motor exists at the starter motor shaft,not at the centre of the flywheel.Whatever torque the starter motor has at it's shaft is multiplied by the extreme gearing involved to have
150 lb ft. at the centre of the flywheel,not at the circumference of the flywheel where the starter motor is engaging.

If the flywheel was supported by a method which allowed the engine components to not be engaged,all the torque would be freed up to accelerate the car,which would be a jack rabbit start then ..no more as starter motor going flat out already. If you use the starter to move a manual transmission car in some sort of emergency,a good battery will move the car even while having to crank the engine as well.

Edited by johnny yuma, 18 April 2013 - 00:55.

The thing is the 2-3 HP from the starter motor exists at the starter motor shaft,not at the centre of the flywheel.

Power is conserved through a gearset so 2-3 hp at the starter is still 2-3 hp at the flywheel centre. Torque is increased and rpm is reduced by the same factor so power stays the same.

Power is conserved through a gearset so 2-3 hp at the starter is still 2-3 hp at the flywheel centre. Torque is increased and rpm is reduced by the same factor so power stays the same.

One thing I have learnt,gg,is not to dismiss ANYTHING that you say.I stand corrected.By the same logic,it makes no difference what gear you are in,your car still has the same power at the rear wheels at it's power rpm peak...so it is the maximising of Torque which makes your car accelerate faster in lower gears than higher , not any increase in power.
That being the case,why is it not generally agreed you best acceleration available in any given gear will be at maximum Torque RPM,not maximum Power RPM ?

That being the case,why is it not generally agreed you best acceleration available in any given gear will be at maximum Torque RPM,not maximum Power RPM ?

No argument with that.

Yup, no argument with that. You'll see in those graphs I posted that in each gear the max accel was in the middle of the rpm range for each gear, where we'd expect max torque to be, not at the upper end of each gear, where we'd expect max power to be.

But a given roadspeed, it is better to be in a numerically higher gear ratio than it is to be at max torque, because more power is available, at that speed.

Yup, no argument with that. You'll see in those graphs I posted that in each gear the max accel was in the middle of the rpm range for each gear, where we'd expect max torque to be, not at the upper end of each gear, where we'd expect max power to be.

But a given roadspeed, it is better to be in a numerically higher gear ratio than it is to be at max torque, because more power is available, at that speed.

hmmm don't you mean "because more torque is available ,at that speed" ?

hmmm don't you mean "because more torque is available ,at that speed" ?

At the wheel there will be more tractive force available because of the higher gear ratio, even tho the engine torque is less than at max torque.

At the wheel there will be more tractive force available because of the higher gear ratio, even tho the engine torque is less than at max torque.

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.

This subject really brings the cream to the top...

Edited by saudoso, 18 April 2013 - 10:30.

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.

Oh sorry, I'd assumed from most of your posts that you weren't a troll. I won't waste your time any further.

Monster trucks?

Diameter of the circumference, not the height.

Oh sorry, I'd assumed from most of your posts that you weren't a troll. I won't waste your time any further.

No,I'm sorry,my attempt at levity has failed.Serves me right for posting while drinking.Now I am
the lowest of the low in 2013 social pecking order...A TROLL !!! ah well back under the bridge for me.

Ah well if you were posting while dring that is a whole separate thing, entirely understandable (not at 830 am though, unless you are having champagne for breakfast, a fine tradition) .

Ah well if you were posting while dring that is a whole separate thing, entirely understandable (not at 830 am though, unless you are having champagne for breakfast, a fine tradition) .

Certainly not old chap,it was an hour after sunset and I was enjoying a Corona with a twist of lemon.It was as every bit as refreshing as the previous three .

".......Trolling, Trolling, ..Trolling on the River... ... ... "

Diameter of the circumference of the tyre, not the height.

FTFY

Power is merely an attempt to quantify how much work the machinery in
question can perform over a period of time.

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

Hi everyone,

here's another way of thinking about this that helps me understand it. Hope it helps.

Maximum acceleration is achieved when you have the maximum torque - at the wheels.

Torque at the wheels is basically just the engine torque multiplied by the gear ratio. Therefore if you had an engine at 1000 RPM (peak torque), producing 100 torques with the wheels also rotating at 1000 RPM, then you would have a gear ratio of 1:1 and 100 torques at the wheels.

If peak power was at 1500 RPM, then you could use a gear ration of 1:1.5 to travel at the same speed. So as long as the torque at 1500 RPM is greater than 2/3 or the torque at 1000 RPM, there will be more torque at the wheels. This must be the case, because that is the definition of peak power.

Why so much confusion? Because if you stay in the same gear, then you will have greater acceleration at peak torque than you will at peak power. The torque at the wheels is independent of the engine RPM if the gearing is fixed. However, if you were accelerating and reached peak power, then shifted in to a higher gear that dropped you to peak torque, your acceleration would decrease.

This is what the question asked, so the correct answer peak power in this case. However, you could ask a question where the correct answer was peak torque. So I think the people who have got this wrong are probably just answering the wrong question.

Cheers

This is what the question asked, so the correct answer peak power in this case. However, you could ask a question where the correct answer was peak torque. So I think the people who have got this wrong are probably just answering the wrong question.

I doubt they're answering the wrong question. They're just holding the wrong variable as fixed. There are three variables in this question: engine torque, RPM, and the speed of the car. The correct solution requires holding the speed of the car fixed, and then realizing that engine torque and RPM are not independent, and thus can't be maximized in isolation. Once you realize that fact, you're only one algebraic manipulation away from realizing that you should be maximizing power.

Edited by Dmitriy_Guller, 19 April 2013 - 18:16.

If engine torque was the thing to have..

Look at it! it want to eat the operator!

Edited by MatsNorway, 19 April 2013 - 21:18.

If engine torque was the thing to have..
Look at it! it want to eat the operator!

One of my favourite videos. A model for green energy and workplace health and safety. Just count the hazards, risks, near-misses.

Additional info. This is from Norway and the engine is a norwegian made engine. My father had this type or similar engine. With the period correct "gearbox" so to go backwards they had to get the engine to go the other way. It could go horribly wrong and it often did.
Some picture from the factory: http://www.sverrep.c...v/rappmotor.pdf

Yet another Koenigsegg video...

If engine torque was the thing to have..

Look at it! it want to eat the operator!

it's a wonderful video, but your observation about torque makes no sense.

However, it is indeed amusing and amazing to watch the engine hop around on the floor in spite of its great weight.

Yet another Koenigsegg video...

your observation about torque makes no sense.

its 24hp.. and around 400Nm.. given its 450rpm max rating.

Lets ignore the likely high engine volume and its potential in the modern world.

Edited by MatsNorway, 21 April 2013 - 08:23.

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.

As you slow the car some of the 'drags' drop, in particular, the aero drag. I agree in the real world most of the effect is due to what the tractor guys call torque backup, that is the ratio of torque at max torque to the torque at max power. When you are ploughing or subsoiling you set the tractor up in the gear at which it will just pull max power (ie it is bouncing off the red line), and then hope that the torque backup is enough to get you through the gnarly bits without having to fiddle with the depth of cut or angle..

As you slow the car some of the 'drags' drop, in particular, the aero drag. I agree in the real world most of the effect is due to what the tractor guys call torque backup, that is the ratio of torque at max torque to the torque at max power. When you are ploughing or subsoiling you set the tractor up in the gear at which it will just pull max power (ie it is bouncing off the red line), and then hope that the torque backup is enough to get you through the gnarly bits without having to fiddle with the depth of cut or angle..

So after you get slowed a bit by the gnarlies,there is enough torque liberated as you get back
into the friable loamies to accelerate back up to Redline 2000 !!

Edited by johnny yuma, 22 April 2013 - 00:12.

The rate at which things are raised against the force of gravity is the actual definition of horsepower. So the power needed by a car going up a hill is directly related to its speed and thus how fast it is raising its own weight and its load as it climbs the hill. Presumably at 65mph the rate of vertical movement is greater than the power the 1800 develops at that speed in top gear. Presumably again the power developed at 50mph by the 1800 matches the power needed to raise its total weight vertically at that speed (plus the lesser aero drag and rolling resistance at that speed).

Edited by Kelpiecross, 22 April 2013 - 07:23.

Just take a run at it.

The rate at which things are raised against the force of gravity is the actual definition of horsepower. So the power needed by a car going up a hill is directly related to its speed and thus how fast it is raising its own weight and its load as it climbs the hill. Presumably at 65mph the rate of vertical movement is greater than the power the 1800 develops at that speed in top gear. Presumably again the power developed at 50mph by the 1800 matches the power needed to raise its total weight vertically at that speed (plus the lesser aero drag and rolling resistance at that speed).

Thanks k.c. that is good food for thought. Tried to drop my brain into a lower gear to consider the question but got valve bounce.

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.

The rate at which things are raised against the force of gravity is the actual definition of horsepower. So the power needed by a car going up a hill is directly related to its speed and thus how fast it is raising its own weight and its load as it climbs the hill. Presumably at 65mph the rate of vertical movement is greater than the power the 1800 develops at that speed in top gear. Presumably again the power developed at 50mph by the 1800 matches the power needed to raise its total weight vertically at that speed (plus the lesser aero drag and rolling resistance at that speed).

Johnny, you're almost on the money here... Your intuition is absolutely right, and as I've mentioned before, this is what causes all the confusion! In a given gear, you will get maximum acceleration at peak torque. However, at a given road speed, you will get maximum acceleration from maximizing engine power. It is that simple.

Yes, when you hit a gradient, the car will slow down until the power from the engine balances the force x velocity of the vehicle (force coming from the weight of the vehicle on an incline and any aero/tyre resistance). And yes, you will maintain a higher speed if you change down - because you are increasing this power. You say its because you are shortening the ratio, and therefore scaling up the torque, but this will always be maximised at peak engine power. So if at some road speed, you had a choice of any ratio you like (CVT), you would always put it at peak power, period. Therefore, when developing a race engine for outright speed, companies are always tuning for power.

The post manolis made on page 1 is the clearest possible way of showing this - torque at the wheels vs. carspeed. The maximum will always coincide with peak power on the engine curve.
http://forums.autosp...w...t&p=6216883

The post manolis made on page 1 is the clearest possible way of showing this - torque at the wheels vs. carspeed. The maximum will always coincide with peak power on the engine curve.
http://forums.autosp...w...t&p=6216883

.....or the max available torque "at" the RPM of peak power on the torque curve, which is around 6,300 RPM!

.....or the max available torque "at" the RPM of peak power on the torque curve, which is around 6,300 RPM!

At any given Road Speed,yes you will get maximum acceleration at power peak rpm,but the reason this is true
is the practical consideration that if you're at 6300 rpm ,say,you must be in a lower gear than if you were at peak torque
of say 5000rpm. And that lower gear will multiply the lesser torque figure at 6300 so that it exceeds the torque at 5000rpm
in the "other gear".

However instantaneous acceleration is not as useful as an extended surge of acceleration for as long as possible--thats why
diesels without a cvt are so damned annoying with their narrow torque band.

At any given Road Speed,yes you will get maximum acceleration at power peak rpm,but the reason this is true
is the practical consideration that if you're at 6300 rpm ,say,you must be in a lower gear than if you were at peak torque
of say 5000rpm. And that lower gear will multiply the lesser torque figure at 6300 so that it exceeds the torque at 5000rpm
in the "other gear".

However instantaneous acceleration is not as useful as an extended surge of acceleration for as long as possible--thats why
diesels without a cvt are so damned annoying with their narrow torque band.

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!