In article ,
Roger Mills wrote:
I take it you don't remember the original thread some weeks ago? it was
obvious that Vir Campestris was attempting to prove he was now right about
that.

So are you telling me that everything you've written relates to
*another* thread rather than this one? I can only cope with one thread
at a time!

But it's exactly the same. Which part of an engine's output produces
maximum acceleration. Those with any knowledge of mechanics know it's at
maximum torque. Those who just read about engine specs think it's at
maximum power.

And then those who don't understand the subject at all bring in red
herrings like gearboxes.

--
*Drugs may lead to nowhere, but at least it's the scenic route *

Dave Plowman London SW
To e-mail, change noise into sound.

In article ,
Vir Campestris wrote:
On 05/04/2016 12:37, Roger Mills wrote:
The very first post which started this thread *actually* said the
following:

"Q1: If an engine is capable of a peak torque of 400nM, what is the
force available at the wheels at a speed of 10m/S?"

. . . which is impossible to answer without knowing anything about
the gearing - and which makes no mention whatsoever about acceleration.

The whole point was that it's impossible to say. Torque on its own is
useless - you need revs too, and that means it's power. Force times
speed.

No car engine produces torque without revs. Any more red herrings you'd
like to introduce?

--
*Why can't women put on mascara with their mouth closed?

Dave Plowman London SW
To e-mail, change noise into sound.

On 05/04/2016 21:40, michael adams wrote:
"Roger wrote in message
...
On 05/04/2016 15:35, michael adams wrote:


Surely maximising the amount of power transmitted to the wheels
to change the speed from say 1 mph to say 5mph would simply
result in wheelspin. As many boy racers at traffic lights know
to their cost, never mind many stars in reasonably prices cars*.
And the same would presumably also apply in wet or icy conditions*
at much higher speeds

Not that the same doesn't apply equally well to torque.

Well yes, of course. But I think we're assuming we have a dry road with plenty of
adhesion.

Indeed, but how would those two conditions prevent wheelspin between
say 1mph and 5mph ?


They wouldn't. The argument about whether you get max acceleration by
using max torque from the engine or max power with a lower gear only
applies when there is sufficient adhesion to use full throttle without
wheelspin. If there isn't, you get maximum traction when the wheel is on
the point of slipping but not slipping. To achieve this you need to
reduce the power output and/or apply the brakes to the driving wheels.
Many modern cars do this automatically by integrating engine control
with the (anti-lock) braking system.
--
Cheers,
Roger
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On 05/04/16 23:37, Roger Mills wrote:
They wouldn't. The argument about whether you get max acceleration by
using max torque from the engine or max power with a lower gear only
applies when there is sufficient adhesion to use full throttle without
wheelspin.

Strong smell of spin here, in an effort to divert attention away from
people who haven't got this far?


--
Ideas are more powerful than guns. We would not let our enemies have
guns, why should we let them have ideas?

Josef Stalin

On Sat, 02 Apr 2016 11:20:15 +0100, Dave Plowman (News) wrote:

In article ,
Johnny B Good wrote:
On Fri, 01 Apr 2016 17:16:40 +0100, Dave Plowman (News) wrote:

In article ,
Roger Mills wrote:
Indeed they are. The rate of acceleration is dependent on gearbox
*output* torque - not on engine torque.

And the two are directly related.

Why is it impossible to see this? A gearbox simply multiplies the
torque of the engine.

I think we can all accept that as "A Given". However, for a given road
speed where the selected gearbox ratio matches the peak engine power
revs, you will gain a modest increase in torque applied to the driven
wheels compared to a ratio selected to allow the engine to run at peak
torque revs.

Of course A gearbox simply multiplies the torque of the engine.
Regardless of the engine or road speed. But if it increases the torque
at peak BHP, it will also do the same at peak torque, so you'll have
more torque at the wheels at peak torque. Therefore better
acceleration...

But at a lower road speed.



If we assume the theoretical case of a perfect stepless gearbox which
can automatically adjust the ratio to hold the engine to its peak
torque rpm as the road speed increases, you'll find that the resulting
acceleration will be less than one configured to hold the engine to its
peak power output rpm (it will operate at a larger reduction ratio in
this case which increases the torque applied to the road wheels
compared to the ratio range used to keep the engine operating at its
maximum torque rpm).

The type of gearbox makes no difference to the principle.

In the case of a manual gearbox, we approximate this mode of operation
by choosing max power rpm as the next change up point rather than the
lower peak torque rpm point.

A decent box will then plonk you at or near the peak torque point in the
next higher gear - to give you the best acceleration.

I did have that in mind but I thought I'd leave that little factoid for
someone else to consider. It doesn't matter who else was going make that
observation but it does help that you're well aware of this effect. :-)

Choosing to use the engine's peak torque point does optimise something.
It's just not maximum possible acceleration. What it's most likely to
optimise is most *fuel economic* maximum acceleration to any given target
cruising speed since max torque revs approximates very closely to maximum
engine efficiency revs.

Normally, assuming maximum fuel efficient acceleration is not your prime
concern (and this is generally true for almost every fuel consumption
conscious driver on the planet), minimising journey fuel consumption is
achieved by 'short changing' so as to achieve the most economic gear
ratio (top gear) for the cruise part of the journey as soon as possible
without labouring the engine and without using higher revs than necessary.

Obviously, this depends on road conditions (starting off in an uphill
direction may require more engine revs and in a 30mph limited section of
road may force you to stay in a lower gear until you hit the level or a
downhill section) but despite this, the aim remains the same, don't
dawdle in an unnecessarily low ratio unless road conditions leave you no
choice.

Since typical urban driving conditions require far less acceleration
demands than the capabilities of even modestly specced motor cars, the
change up points are often well below the peak engine torque revs anyway.
It matters not that the engine is reving comfortably below its maximum
torque revs on part throttle when changing up, the importance point is
that you still get the best possible fuel economy in spite of the less
than ideal driving conditions when choosing the tallest gear ratio the
engine can comfortably handle.

However, when attempting to outdrag another vehicle on a clear motorway
(there are very few of those about - the only one that comes to mind is
the M58 into Liverpool early to late evening on a Sunday), you change up
just at or after the maximum BHP revs which, as you observed, will put
you close to the maximum torque point, probably providing close to the
torque you had at the road wheels just prior to 'running out of steam' in
the previous gear.

Further acceleration will see the accelerating force from the driven
wheels continue to fall off as you increase road speed until you finally
reach a limiting speed in top gear where the thrust finally balances the
steady counterforce of drag and friction. Strangely, top gear isn't
designed to limit the engine to maximum torque revs at the maximum road
speed rating of the car if the manufacturer wants to offer the highest
maximum road speed possible from the engine.

However, the manufacturer may well offer an 'overdrive' ratio to
maximise motorway cruising economy at speeds between 70 and 80mph against
modest headwinds and gentle uphill slopes where the engine runs close to
maximum torque revs at 80mph but, aside from such considerations of fuel
economy, gearing is generally configured for drive-ability at a small
sacrifice in best possible fuel economy.

Since the product of thrust from the road wheels and road speed is an
expression of power (energy delivery rate) translated from the engine's
output shaft via the transmission system, you need to ideally arrange for
the engine to be delivering its maximum power output throughout the whole
range of road speeds if you wish to maximise the acceleration performance
of the vehicle either by using a perfect continuously variable ratio
automatic gearbox that holds the engine close to its maximum power output
revs or else by approximating this ideal with skilfully chosen gear
changes with a manual box.

If we observe the performance of such a vehicle with a perfect
continuous ratio box automatically holding the engine to its peak power
output at maximum throttle opening on a dynamometer rolling road you will
observe an inverse relationship between thrust generated at the wheels
and the speed of the rollers.

If we observe, for example 300 Lbs thrust at 40mph, it will drop to
150Lbs at 80 mph representing the constant steady state maximum engine
power translated via our intelligent automatic gearbox into speed/thrust
combinations with equal products of power delivered to the rolling road.

With a perfect gearbox, we can plot a continuous curve of falling thrust
with increase of road speed up to the car's transmission system / max
engine power output limit which we might well expect to correspond to the
car's maximum rated road speed due to drag and rolling friction forces
anticipated by the designers of the car.

If we reconfigure the gearbox to hold the engine at its peak torque rpm
during such a test run we will see a lower product of road speed and
thrust which corresponds to less acceleration and a slower top speed.

At the end of the day, it's all a matter of imparting kinetic energy
into the moving mass of the vehicle at the highest possible rate (maximum
engine power) to maximise the vehicle's acceleration courtesy of a
variable gear ratio box (manual or perfect continuously variable
automatic box) to provide as much effective thrust which declines with
the increasing speed being imparted to the vehicle. Since it's engine
power and not torque alone that's required to accelerate the vehicle,
shifting up at peak torque output of any piston ICE powered car does not
offer the best acceleration performance.

--
Johnny B Good

On Tue, 5 Apr 2016 21:32:01 +0100, Vir Campestris
wrote:

The whole point was that it's impossible to say. Torque on its own is
useless - you need revs too, and that means it's power. Force times speed.

Hm.

A standing car would never accelerate...


Thomas Prufer

In article ,
The Natural Philosopher wrote:
On 05/04/16 23:37, Roger Mills wrote:
They wouldn't. The argument about whether you get max acceleration by
using max torque from the engine or max power with a lower gear only
applies when there is sufficient adhesion to use full throttle without
wheelspin.

Strong smell of spin here, in an effort to divert attention away from
people who haven't got this far?

That's what this thread has all been about. Red herrings galore.

If there's a danger of wheelspin, simply do the test in a higher gear.

The result will be the same. Maximum acceleration at maximum torque, not
maximum BHP.

--
*Make it idiot-proof and someone will make a better idiot.

Dave Plowman London SW
To e-mail, change noise into sound.

On 06/04/2016 04:08, Johnny B Good wrote:


However, when attempting to outdrag another vehicle on a clear motorway
(there are very few of those about - the only one that comes to mind is
the M58 into Liverpool early to late evening on a Sunday), you change up
just at or after the maximum BHP revs which, as you observed, will put
you close to the maximum torque point, probably providing close to the
torque you had at the road wheels just prior to 'running out of steam' in
the previous gear.

Indeed.

Further acceleration will see the accelerating force from the driven
wheels continue to fall off as you increase road speed until you finally
reach a limiting speed in top gear where the thrust finally balances the
steady counterforce of drag and friction. Strangely, top gear isn't
designed to limit the engine to maximum torque revs at the maximum road
speed rating of the car if the manufacturer wants to offer the highest
maximum road speed possible from the engine.

However, the manufacturer may well offer an 'overdrive' ratio to
maximise motorway cruising economy at speeds between 70 and 80mph against
modest headwinds and gentle uphill slopes where the engine runs close to
maximum torque revs at 80mph but, aside from such considerations of fuel
economy, gearing is generally configured for drive-ability at a small
sacrifice in best possible fuel economy.

Yes. Maximum speed occurs when the thrust vs speed curve intersects the
drag vs speed curve. This is not necessarily in the highest available
gear if that gear is an "overdrive".

Since the product of thrust from the road wheels and road speed is an
expression of power (energy delivery rate) translated from the engine's
output shaft via the transmission system, you need to ideally arrange for
the engine to be delivering its maximum power output throughout the whole
range of road speeds if you wish to maximise the acceleration performance
of the vehicle either by using a perfect continuously variable ratio
automatic gearbox that holds the engine close to its maximum power output
revs or else by approximating this ideal with skilfully chosen gear
changes with a manual box.

If we observe the performance of such a vehicle with a perfect
continuous ratio box automatically holding the engine to its peak power
output at maximum throttle opening on a dynamometer rolling road you will
observe an inverse relationship between thrust generated at the wheels
and the speed of the rollers.

If we observe, for example 300 Lbs thrust at 40mph, it will drop to
150Lbs at 80 mph representing the constant steady state maximum engine
power translated via our intelligent automatic gearbox into speed/thrust
combinations with equal products of power delivered to the rolling road.

With a perfect gearbox, we can plot a continuous curve of falling thrust
with increase of road speed up to the car's transmission system / max
engine power output limit which we might well expect to correspond to the
car's maximum rated road speed due to drag and rolling friction forces
anticipated by the designers of the car.

If we reconfigure the gearbox to hold the engine at its peak torque rpm
during such a test run we will see a lower product of road speed and
thrust which corresponds to less acceleration and a slower top speed.


Indeed. But my head is black and blue from being hit against a brick
wall trying to get Mr Plowman to understand that!
--
Cheers,
Roger
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"Tim Streater" wrote in message
.. .
In article , michael adams
wrote:

"Tim Streater" wrote in message
...
In article , michael
adams wrote:

"Tim Streater" wrote in message
et...
In article , michael
adams wrote:

"Roger Mills" wrote in message
...

And the fact remains that - at any given speed - the highest
instantaneous
rate of change in speed (acceleration) is obtained by maximising the
amount of power transmitted to the wheels. Simple laws of Physics!

Surely maximising the amount of power transmitted to the wheels
to change the speed from say 1 mph to say 5mph would simply
result in wheelspin.

Doesn't necessarily follow. Studded tires come to mind.

But depending on weight distribultion, and assuming the studded tires
dont simply carve a big hole in the road surface, all that that might achieve
might be to send the front of the car into the air, if not flip it over
completely.

Front wheel drive.

And overal studded tires would be a specal case dependent on an
optimum road surface, and optimum weigh distribution and would only
be applicable up to a certain speed.

All cases are special cases and depend on all the factors you mention.

What was being proposed, was a general principle.

If all cases were special cases, then there would be little point in
formulating general principles at all, would there ?

It might spin or it might not. Depending on all the special case
variables. So a statement of the bleeding obvious, really.

Which relates to my statement:

"If all cases were special cases, then there would be little point in
formulating general principles at all, would there ?"

( A response to your above claim that "all cases are special cases" )

and to which I assume you were responding, how exactly ?


michael adams

....

In article , Thomas Prufer
writes
On Tue, 5 Apr 2016 21:32:01 +0100, Vir Campestris
wrote:

The whole point was that it's impossible to say. Torque on its own is
useless - you need revs too, and that means it's power. Force times speed.

Hm.

A standing car would never accelerate...


Thomas Prufer
Let's bring clutches into the argument just for a laugh.
--
bert

And the fact remains that - at any given speed - the highest
instantaneous rate of change in speed (acceleration) is obtained by
maximising the amount of power transmitted to the wheels. Simple laws of
Physics!


It seems that it is agreed that Max Speed is only attained at Max Power.
At any speed less than Max Speed, Max Power is unavailable.
WOT (Wide Open Throttle) is available at any speed, but does not equal Max Power, but, at, say, 3000 RPM, WOT would supply Overall Max Torque. Remember, WOT will give maximum effort at EVERY engine speed.
So, the above quotation should be re-written to read "... maximising the amount of power, available at that engine speed by opening the thottle fully, transmitted to the wheels."

On 06/04/2016 21:02, wrote:

And the fact remains that - at any given speed - the highest
instantaneous rate of change in speed (acceleration) is obtained by
maximising the amount of power transmitted to the wheels. Simple laws of
Physics!


It seems that it is agreed that Max Speed is only attained at Max Power.
At any speed less than Max Speed, Max Power is unavailable.

No true. Max speed is usually defined as the maximum speed which can be
obtained on a level road on a windless day - and occurs when the max
available driving forces match the resistive forces from aerodynamic
drag, etc.

But you may need max power to climb a steep hill at a lower road speed.
In this case, the driving force is also being resisted by the component
of the vehicle mass acting down the slope.

And you can use maximum power - albeit transiently - whilst accelerating
through the gears. In this case, the driving force is rested by the
inertia of the vehicle - F = M x A etc.

WOT (Wide Open Throttle) is available at any speed, but does not equal Max Power, but, at, say, 3000 RPM, WOT would supply Overall Max Torque. Remember, WOT will give maximum effort at EVERY engine speed.
So, the above quotation should be re-written to read "... maximising the amount of power, available at that engine speed by opening the thottle fully, transmitted to the wheels."

Wide open throttle at any engine speed will give the same torque at the
crankshaft as you could measure on a dynamometer at the same steady
speed. But it may not all get out to the outside world in a dynamic
situation because some of the torque will be absorbed accelerating the
engine itself.
--
Cheers,
Roger
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On Wed, 06 Apr 2016 11:01:35 +0100, Roger Mills wrote:

On 06/04/2016 04:08, Johnny B Good wrote:

====snip====


If we reconfigure the gearbox to hold the engine at its peak torque
rpm
during such a test run we will see a lower product of road speed and
thrust which corresponds to less acceleration and a slower top speed.


Indeed. But my head is black and blue from being hit against a brick
wall trying to get Mr Plowman to understand that!

Well, if Dave wishes to continue arguing otherwise, I don't see any
point in further correspondence on the matter.

The long and the short of it is that accelerating a mass is a matter of
raising its kinetic energy and the quickest way to achieve that is the
use of a range of gear ratios such that the engine operates as close as
possible to its peak BHP rpms during the change up sequence required to
accelerate the vehicle as quickly as possible.

Peak torque revs with any practical piston ICE are unlikely to produce
more than 60 or 70 or so percent of its maximum power output so that's
most definitely the wrong choice of gear change point as far as
maximizing a car's acceleration performance is concerned.

It's a pity I couldn't have distilled it to just those two paragraphs
sooner rather than later. :-(

--
Johnny B Good

On 07/04/2016 02:39, Johnny B Good wrote:
On Wed, 06 Apr 2016 11:01:35 +0100, Roger Mills wrote:

On 06/04/2016 04:08, Johnny B Good wrote:

====snip====


If we reconfigure the gearbox to hold the engine at its peak torque
rpm
during such a test run we will see a lower product of road speed and
thrust which corresponds to less acceleration and a slower top speed.


Indeed. But my head is black and blue from being hit against a brick
wall trying to get Mr Plowman to understand that!

Well, if Dave wishes to continue arguing otherwise, I don't see any
point in further correspondence on the matter.

The long and the short of it is that accelerating a mass is a matter of
raising its kinetic energy and the quickest way to achieve that is the
use of a range of gear ratios such that the engine operates as close as
possible to its peak BHP rpms during the change up sequence required to
accelerate the vehicle as quickly as possible.

Peak torque revs with any practical piston ICE are unlikely to produce
more than 60 or 70 or so percent of its maximum power output so that's
most definitely the wrong choice of gear change point as far as
maximizing a car's acceleration performance is concerned.

It's a pity I couldn't have distilled it to just those two paragraphs
sooner rather than later. :-(


What you say is absolutely right, but I'm afraid that you're wasting
your breath as far as Mr P is concerned. He believes that torque is GOD
- period!

I wonder how he would view the electrical analogy - where torque equates
to current and speed to voltage. Would he consider that 10 amps at 12
volts is just as useful as 10 amps at 240 volts?
--
Cheers,
Roger
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"Roger Mills" wrote in message
...

What you say is absolutely right, but I'm afraid that you're wasting your breath as far
as Mr P is concerned. He believes that torque is GOD - period!

quote

"Torque, moment, or moment of force (see the terminology below) is the tendency
of a force to rotate an object about an axis,[1]

https://en.wikipedia.org/wiki/Torque

quote

Given that torque is the power transmitted via the drive axle which causes the
wheels to rotate, its difficult to see what's wrong with his claim, quite honestly.

Except maybe in cars fitted with sails.


michael adams

....

On 07/04/2016 10:19, michael adams wrote:

Given that torque is the power transmitted via the drive axle which causes the
wheels to rotate, its difficult to see what's wrong with his claim, quite honestly.

Energy of moving object = mass x velocity squared.

If you want it to go faster you have to input energy.

If the energy source has higher output at high revs you have more to put in.

In a car this quite often means changing gear and possible slipping the
clutch and is therefore a dynamic system and not a static one like
having an engine running at one speed at maximum torque.

See its easy to argue.

"dennis@home" wrote in message
eb.com...
On 07/04/2016 10:19, michael adams wrote:

Given that torque is the power transmitted via the drive axle which causes the
wheels to rotate, its difficult to see what's wrong with his claim, quite honestly.

Energy of moving object = mass x velocity squared.

If you want it to go faster you have to input energy.

Yes.


If the energy source has higher output at high revs you have more to put in.


Yes.

In a car this quite often means changing gear and possible slipping the clutch and is
therefore a dynamic system and not a static one like having an engine running at one
speed at maximum torque.

See its easy to argue.

Only if you choose to deliberately conflate engine torque, which is
presumably measured at the flywheel, with driven axle torque which
actually turns the wheels.

I'm not concerned about the engine.

For the wheels to be turning or accelerating fastest -
in the absence of the car being powered by external
factors such as the wind propelling the body, they
need to be subject to the maximum available torque.

Quite how it got there, is an entirely different question.


michael adams

....

On 07/04/2016 12:38, michael adams wrote:


I'm not concerned about the engine.

That's the problem - Mr Plowman *is*.

For the wheels to be turning or accelerating fastest -
in the absence of the car being powered by external
factors such as the wind propelling the body, they
need to be subject to the maximum available torque.


Indeed. And you obtain that by running the engine at max power and using
a suitable gear ratio. But Mr Plowman doesn't accept that. He says that
you have to run the *engine* at max *torque*. So he would end up using a
lower engine speed - producing less *power* - with higher gearing which
provides *less* torque multiplication - resulting in less torque at the
wheels. Mad!
--
Cheers,
Roger
____________
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checked.

"Roger Mills" wrote in message
...
On 07/04/2016 12:38, michael adams wrote:


I'm not concerned about the engine.

That's the problem - Mr Plowman *is*.

For the wheels to be turning or accelerating fastest -
in the absence of the car being powered by external
factors such as the wind propelling the body, they
need to be subject to the maximum available torque.


Indeed. And you obtain that by running the engine at max power and using a suitable
gear ratio. But Mr Plowman doesn't accept that. He says that you have to run the
*engine* at max *torque*. So he would end up using a lower engine speed - producing
less *power* - with higher gearing which provides *less* torque multiplication -
resulting in less torque at the wheels. Mad!

Sorry, but maybe I'm missing something here.

Basically the engine is pistons on con rods going up and down
which is converted to rotary motion by cams on the crankshaft
which is attached to a flywheeel.

And as I understand it anyway, its from the flywheel that engine
torque, is measured.

And so my question is this.

If the engine is running at maximum power, and thus the crankshaft
is turning as fast as possible, then how is it possible for less than
maximum torque to be delivered to the flywheel ?


michael adams

....

On Mon, 04 Apr 2016 23:09:22 +0100, Roger Mills wrote:

On 04/04/2016 21:42, Vir Campestris wrote:
On 04/04/2016 18:59, Roger Mills wrote:

Of course not. It largely follows the torque curve (minus the drag!)
in each gear, and then there's a step change (downwards!) when you
change to the next gear.

I thought the idea was to run so far past peak power that when you
changed up you were at the same power output the other side of the
peak?
So no steps?

Andy

That would be the ideal, and may be achievable if you have a lot of
closely spaced gears. Otherwise you're likely to reach valve bounce or,
in the case of a diesel, hit the governor before reaching the point
where the next gear up will allow the engine to produce the same amount
of power.

Wow! it seems I missed a few dozen posts due to excessive indentation in
the thread list. I mention this to explain my absence from this part of
the discussion.

What you mention in the paragraph above is all true but the argument by
Dave over max torque versus max power rpm change up points for maximum
acceleration didn't need these extra real world complications which *do*
further aid the fact that best acceleration is achieved by changing to
the next higher gear at or just above the peak power revs.


In all the calculations I did when I was doing this for a living -
albeit a few decades ago - the next gear up always dropped you further
down the power curve. When doing a practical road test, you could
counter that to some extent by not letting the engine revs drop during
the gear-change so that you got an inertial 'kick' when you let the
clutch in.

If you don't mind abusing the clutch, that does provide additional
acceleration impetus. It was a trick I often used with the works van when
accelerating up to speed after hopping onto a motorway section. :-)


That reminds me of one factor which we had to include in performance
calculations which hasn't been discussed here - namely the effect of the
engine's own moment of inertia. In a low gear, quite of lot of engine
torque is used up accelerating the engine itself - reducing the amount
available for accelerating the vehicle. The "effective mass" of the
engine, which was different for each gear, had to be added to the
vehicle mass when calculating acceleration.

Yet another consideration. Most noticed when driving a naturally
aspirated diesel powered 3 1/2 tonner when starting off in bottom gear.
You soon learned to live with the almost total absence of perceivable
acceleration you'd normally experience with a lightweight petrol engined
van.

Again, I excluded this extra consideration from my arguing the case for
max power revs. Indeed, the perfect automatic continuous ratio gearbox
holding the engine to its max power revs rather neatly eliminates this
specific issue. :-)

There are many real world effects that complicate the issue being
discussed here. One of them being Dave's assertion that the DAF's CVT
system was optimised to hold the engine to it's max torque rpm under full
throttle acceleration or hill climbing conditions. If true (and it might
well be), this optimises for best fuel efficiency / acceleration ratio
because, almost without exception, peak torque revs in a piston ICE
coincides with maximum efficiency. This won't maximise acceleration
performance but it does reduce strain on the transmission components
somewhat which would be a major consideration in such a choice.

In short, to boil it down to its essence, accelerating a body from rest
to a different speed imparts kinetic *energy* to that body. The key thing
in this being the requirement to impart *energy* into said body. The
greater the acceleration you wish to impart, the greater the rate of
energy delivery that needs to be imparted. Energy delivery rates are
expressed as 'power' Kilowatts or horsepower. The higher the power level,
the greater the acceleration hence the choice to keep your prime mover
(the engine) operating at or close to its maximum power output revs by
use of a variable ratio transmission system.

--
Johnny B Good

On 07/04/16 15:25, michael adams wrote:
"Roger Mills" wrote in message
...
On 07/04/2016 12:38, michael adams wrote:


I'm not concerned about the engine.

That's the problem - Mr Plowman *is*.

For the wheels to be turning or accelerating fastest -
in the absence of the car being powered by external
factors such as the wind propelling the body, they
need to be subject to the maximum available torque.


Indeed. And you obtain that by running the engine at max power and using a suitable
gear ratio. But Mr Plowman doesn't accept that. He says that you have to run the
*engine* at max *torque*. So he would end up using a lower engine speed - producing
less *power* - with higher gearing which provides *less* torque multiplication -
resulting in less torque at the wheels. Mad!

Sorry, but maybe I'm missing something here.

Basically the engine is pistons on con rods going up and down
which is converted to rotary motion by cams on the crankshaft
which is attached to a flywheeel.

And as I understand it anyway, its from the flywheel that engine
torque, is measured.

And so my question is this.

If the engine is running at maximum power, and thus the crankshaft
is turning as fast as possible, then how is it possible for less than
maximum torque to be delivered to the flywheel ?

Torque != angular momentum.

You could have it spinning at 6k RPM but the slightest load (in terms of
a braking torque opposing the motion) slows it down.

You could have the same engine running at 3k RPM and it takes a much
much larger braking torque to slow the engine.

"michael adams" wrote in message
o.uk...

"Roger Mills" wrote in message
...
On 07/04/2016 12:38, michael adams wrote:


I'm not concerned about the engine.

That's the problem - Mr Plowman *is*.

For the wheels to be turning or accelerating fastest -
in the absence of the car being powered by external
factors such as the wind propelling the body, they
need to be subject to the maximum available torque.


Indeed. And you obtain that by running the engine at max power and using a suitable
gear ratio. But Mr Plowman doesn't accept that. He says that you have to run the
*engine* at max *torque*. So he would end up using a lower engine speed - producing
less *power* - with higher gearing which provides *less* torque multiplication -
resulting in less torque at the wheels. Mad!

Sorry, but maybe I'm missing something here.

Basically the engine is pistons on con rods going up and down
which is converted to rotary motion by cams on the crankshaft
which is attached to a flywheeel.

And as I understand it anyway, its from the flywheel that engine
torque, is measured.

And so my question is this.

If the engine is running at maximum power, and thus the crankshaft
is turning as fast as possible, then how is it possible for less than
maximum torque to be delivered to the flywheel ?

Thanks anyway, I've answered my own question.

" Typically, the torque peak will occur at a substantially lower RPM
than the power peak."

At least, somebody else has.


michael adams

....

On Mon, 04 Apr 2016 14:57:57 +0100, Dave Plowman (News) wrote:

In article ,
Roger Mills wrote:
On 04/04/2016 11:07, Dave Plowman (News) wrote:


Adding in drag etc is simply attempting to move the goalposts yet
again. Drag depends on speed. And that is irrelevant to the
discussion.


If you are trying to maximise acceleration, drag is highly relevant.

From 0-10 mph? Doubt you could measure it. From 150-160 mph it could be
very significant. Depending on the design of the vehicle. Which you
haven't specified.

If you simply want to maximise thrust at the wheels, it isn't.

No such things as thrust at the wheels. It's a term used for jet
engines.

That's because it's the most effective way to measure the performance of
a jet engine. However, the use of 'thrust measurement' isn't exclusively
reserved for jet engines. You can measure or calculate thrust imparted by
the driven wheels of a road vehicle readily enough and is a valid way to
measure performance. It can easily be measured on a rolling road
dynamometer for example, which when multiplied by the rollers'
circumferential speed, calculates the effective output power being
employed to accelerate the vehicle under test after the losses in the
transmission system have taken their toll.


But the force which actually accelerates the car is the *difference*
between the thrust at the wheels and the aerodynamic drag.

You're making this up as you go along...

For the purpose of this particular argument, the real world
consideration of rolling resistance and aerodynamic drag on a car's
performance is an unnecessary complication even though these factors are
essential in determining the required engine power to drive the car to a
specified maximum speed.


That's why every car has a maximum speed - when it reaches the point
where the drag equals the maximum thrust available so there's nothing
left to accelerate it any more.

And you may care to reflect on why it is that the maximum speed
invariably coincides with the engine's maximum *power* output in an
appropriate gear - not at its max torque point.

None of which has any bearing on the point.

True enough when all that's being discussed is the optimum change up
point as you accelerate the car to some predetermined speed in the
shortest time possible, assuming the target speed is below the drag
limited speed.

When this assumption is met, the best acceleration performance,
regardless of fuel economy, will always occur when the engine is allowed
to operate at or close to its maximum power output revs since
accelerating a mass involves raising its kinetic *energy* at as high a
rate as possible.

Since energy delivery rate is simply a matter of *power level* usually
expressed in KW or BHP, it follows that the best acceleration, regardless
of other complications, can only be achieved by operating the prime mover
(car engine) at its maximum power revs courtesy of a variable ratio
transmission system.

In practice, no piston ICE reaches its limiting maximum power output at
the same maximum torque rpm speed unless deliberately governed to prevent
it going faster than the maximum torque rpm speed (which is a distinct
possibility with the diesel engines used in 38 tonners). Even in this
case, all that achieves is the coincidence that the optimum change up
point matches both max torque and max bhp rpms.

--
Johnny B Good

On Mon, 04 Apr 2016 21:43:22 +0100, Vir Campestris wrote:

On 04/04/2016 19:04, Roger Mills wrote:
Sorry - if you don't understand that, there's really no point in
discussing this any further.

I already said that!


As did I elsewhere in this thread. However, it's an exercise that's
rather akin to (and almost as distasteful as) picking at a scab. Somehow
or other, we just can't resist 'having one more go'. :-(

--
Johnny B Good

On Tue, 05 Apr 2016 10:52:36 +0100, Dave Plowman (News) wrote:

Just to see if I was missing something that others found so obvious, I
asked this question on a forum designated to engine building and tuning,
etc.

'At what point on an engines output do you get maximum acceleration?'

And the answers were near unanimous. Peak torque.

I'll take the views of those guys over the bar room mechanics here any
day.

Ah! I see your problem now. If, as you've seemingly posed the question,
we're discussing, "When does the maximum instantaneous acceleration over
the operating rpm range of the prime mover occur in any *one* gear ratio
at a time?", then the answer to *that* question *will indeed* be at
maximum torque rpms.

However, since you used this little factoid to claim that the fastest
acceleration of a car can therefore only be achieved by making each
successive up-change at peak torque revs rather than at peak power revs,
you've immediately posed an entirely different more complex question
which now involves best gear selection choice versus engine revs to
maximise the acceleration of the whole vehicle.

I'm sure that, if you care to pose this completely different question to
the engine building and tuning forum, you'll get a different response,
one that matches what most in this NG have been trying to tell you,
seemingly to no avail over the past 8 days ever since you posted these
statements:

"And you'd get even more force at peak torque in that gear..."


You get the best acceleration with the maximum torque *at the wheels*. And
in any given gear, this will be when the engine develops maximum torque."

Whilst, on careful examination, both statements *are* strictly true, the
implied maximum acceleration requiring the engine to be held to its max
torque rpm as you progress through the necessary gear changes is patently
*untrue* since accelerating a mass involves raising its kinetic energy
which means its acceleration depends on the how swiftly you can increase
its kinetic energy which in turn is a function of energy rate or power.

I've been re-reading your subsequent posts and you seem to be purposely
avoiding this particular aspect of your implied question over best use of
gear changes versus max torque or BHP rev points to accelerate a car to
speed. Indeed, I spotted a statement where you seem to be trying to limit
the discussion to the engine alone sans the complications introduced by
the pesky transmission system.

Whether you were merely trying to introduce a confusingly phrased
assertion to stir things up or were confused yourself still isn't
entirely clear but I do have my suspicions that you were trying to 'Stir
up a Lively Debate' between mostly just yourself and the rest of the
group. If my suspicion is correct, may I be the first to congratulate you
"On a Job Well Done". :-)

--
Johnny B Good

On Thu, 07 Apr 2016 12:38:58 +0100, michael adams wrote:

"dennis@home" wrote in message
eb.com...
On 07/04/2016 10:19, michael adams wrote:

Given that torque is the power transmitted via the drive axle which
causes the wheels to rotate, its difficult to see what's wrong with
his claim, quite honestly.

Energy of moving object = mass x velocity squared.

If you want it to go faster you have to input energy.

Yes.


If the energy source has higher output at high revs you have more to
put in.


Yes.

In a car this quite often means changing gear and possible slipping the
clutch and is therefore a dynamic system and not a static one like
having an engine running at one speed at maximum torque.

See its easy to argue.

Only if you choose to deliberately conflate engine torque, which is
presumably measured at the flywheel, with driven axle torque which
actually turns the wheels.

I'm not concerned about the engine.

Yes but Dave was. If you carefully examine his statements in the post he
made 8 days back which I've quoted he

"And you'd get even more force at peak torque in that gear...


You get the best acceleration with the maximum torque *at the wheels*. And
in any given gear, this will be when the engine develops maximum torque."

You'll observe that, strictly speaking, they're actually true! :-)

He can't help it if others want to misinterpret the second statement as
implying best acceleration of a car being achieved by choosing to
maintain the engine speed at or close to peak torque revs rather than
peak power revs by choosing gear ratios to maintain this state. :-)

I suspect Dave was merely trying to 'pep up' the debate, relying on our
human propensity to read into such bare statements their own
interpretation of the 'question' thus posed. Even I failed to observe the
strictness applied to those statements (Damn! Where *is* a "Sheldon
Cooper" when you need one?)


For the wheels to be turning or accelerating fastest -
in the absence of the car being powered by external factors such as the
wind propelling the body, they need to be subject to the maximum
available torque.

More accurately, that last statement would be better paraphrased as,

"they need to be subject to the maximum torque available at that road
speed."


Quite how it got there, is an entirely different question.

Quite! For most of us here, including myself, this is a matter dealt
with by optimal use of the gearbox to operate the engine as close to its
maximum power output revs rather than the, as implied by Dave's
statements, maximum engine torque revs.

--
Johnny B Good

On 06/04/16 21:50, Roger Mills wrote:
t seems that it is agreed that Max Speed is only attained at Max Power.
At any speed less than Max Speed, Max Power is unavailable.

No true. Max speed is usually defined as the maximum speed which can be
obtained on a level road on a windless day - and occurs when the max
available driving forces match the resistive forces from aerodynamic
drag, etc.

Wrong. If you want max speed that will correspond to max power with
exactly the right gear ratio.

If the gear ratio is not optimal, it wont, but that's not what was said


--
No Apple devices were knowingly used in the preparation of this post.

On 07/04/2016 18:59, The Natural Philosopher wrote:
On 06/04/16 21:50, Roger Mills wrote:
t seems that it is agreed that Max Speed is only attained at Max Power.
At any speed less than Max Speed, Max Power is unavailable.

No true. Max speed is usually defined as the maximum speed which can be
obtained on a level road on a windless day - and occurs when the max
available driving forces match the resistive forces from aerodynamic
drag, etc.

Wrong. If you want max speed that will correspond to max power with
exactly the right gear ratio.

If the gear ratio is not optimal, it wont, but that's not what was said



What was said was that you couldn't use max power at any speed lower
than max speed. That is wrong - and I explained why, but you
conveniently snipped the relevant bit of my post!
--
Cheers,
Roger
____________
Please reply to Newsgroup. Whilst email address is valid, it is seldom
checked.

On 07/04/2016 15:25, michael adams wrote:


And so my question is this.

If the engine is running at maximum power, and thus the crankshaft
is turning as fast as possible, then how is it possible for less than
maximum torque to be delivered to the flywheel ?


Because the torque which the engine produces at wide open throttle
varies with speed. If you plot torque against engine speed, you get a
curve which is convex upwards. Peak torque typically occurs at 3000 -
3500 rpm.

If you then calculate power at each speed (speed x torque) and plot
that, you get a different curve. That rises steadily with speed until it
reaches its peak - maybe at 5000 rpm, and then starts to drop off again.

So, at peak torque, the engine is producing more torque than it is at
peak power. Because power is the product of speed and torque, it
continues to rise even after the torque has peaked.

So, to return to your question, you can get more torque - but less power
- out of an engine by running it a speed lower than its max power speed.

Or, to put it another way, peak torque and peak power don't occur at the
same speed as each other.

Hope that helps!
--
Cheers,
Roger
____________
Please reply to Newsgroup. Whilst email address is valid, it is seldom
checked.

On 07/04/2016 15:26, Johnny B Good wrote:


In short, to boil it down to its essence, accelerating a body from rest
to a different speed imparts kinetic *energy* to that body. The key thing
in this being the requirement to impart *energy* into said body. The
greater the acceleration you wish to impart, the greater the rate of
energy delivery that needs to be imparted. Energy delivery rates are
expressed as 'power' Kilowatts or horsepower. The higher the power level,
the greater the acceleration hence the choice to keep your prime mover
(the engine) operating at or close to its maximum power output revs by
use of a variable ratio transmission system.


Absolutely. What puzzles me is why that isn't patently obvious to anyone
who claims to know anything about cars and engineering!
--
Cheers,
Roger
____________
Please reply to Newsgroup. Whilst email address is valid, it is seldom
checked.

"Roger Mills" wrote in message
...
On 07/04/2016 15:25, michael adams wrote:


And so my question is this.

If the engine is running at maximum power, and thus the crankshaft
is turning as fast as possible, then how is it possible for less than
maximum torque to be delivered to the flywheel ?


Because the torque which the engine produces at wide open throttle varies with speed.
If you plot torque against engine speed, you get a curve which is convex upwards. Peak
torque typically occurs at 3000 - 3500 rpm.

If you then calculate power at each speed (speed x torque) and plot that, you get a
different curve. That rises steadily with speed until it reaches its peak - maybe at
5000 rpm, and then starts to drop off again.

So, at peak torque, the engine is producing more torque than it is at peak power.
Because power is the product of speed and torque, it continues to rise even after the
torque has peaked.

So, to return to your question, you can get more torque - but less power - out of an
engine by running it a speed lower than its max power speed.

Or, to put it another way, peak torque and peak power don't occur at the same speed as
each other.

Hope that helps!

Thanks. I found a website with the curves etc.

" Typically, the torque peak will occur at a substantially lower RPM
than the power peak."

My problem really is that the above statement isn't intuitively
obvious. To me, at least.

I suspect understanding these things requires the ability
to visualise the interplay of at least two different
cocepts at the same time; that once that's been achieved
then it all fits into place; but for people who've never
bothered with this, beyond a certain age, without 3 D models
and intensive tuition it may be a bit of a lost cause.

Especially when there's so much else out there to get
confused about.

Which people who have been familiar with these concepts
since their youth may find a bit difficult to understand.


michael adams

....

On 06/04/2016 19:23, bert wrote:
Let's bring clutches into the argument just for a laugh.

No, let's not.

You can get better acceleration off the line by slipping the clutch. But
you'll waste fuel and wreck your clutch.

Andy.

On Thursday, 7 April 2016 02:39:38 UTC+1, Johnny B Good wrote:
On Wed, 06 Apr 2016 11:01:35 +0100, Roger Mills wrote:

On 06/04/2016 04:08, Johnny B Good wrote:

====snip====


If we reconfigure the gearbox to hold the engine at its peak torque
rpm
during such a test run we will see a lower product of road speed and
thrust which corresponds to less acceleration and a slower top speed.


Indeed. But my head is black and blue from being hit against a brick
wall trying to get Mr Plowman to understand that!

Well, if Dave wishes to continue arguing otherwise, I don't see any
point in further correspondence on the matter.

The long and the short of it is that accelerating a mass is a matter of
raising its kinetic energy and the quickest way to achieve that is the
use of a range of gear ratios such that the engine operates as close as
possible to its peak BHP rpms during the change up sequence required to
accelerate the vehicle as quickly as possible.

Peak torque revs with any practical piston ICE are unlikely to produce
more than 60 or 70 or so percent of its maximum power output so that's
most definitely the wrong choice of gear change point as far as
maximizing a car's acceleration performance is concerned.

It's a pity I couldn't have distilled it to just those two paragraphs
sooner rather than later. :-(

--
Johnny B Good

It seems that you're saying that becase Power is obtained at the TOP of the power curve and Peak Torque much lower on the power curve, that a car will accelerate more rapidly between 90 and 100 mph than it will between 40 and 50 MPH.

On 07/04/2016 10:19, michael adams wrote:
Given that torque is the power transmitted via the drive axle which causes the
wheels to rotate, its difficult to see what's wrong with his claim, quite honestly.

Torque is force, not power. I suspect he doesn't know the difference.

Andy

On 07/04/2016 21:03, michael adams wrote:
" Typically, the torque peak will occur at a substantially lower RPM
than the power peak."

My problem really is that the above statement isn't intuitively
obvious. To me, at least.

Power can be defined as force times speed.

For a rotating engine that's torque times revs.

If the torque is the same you will have more power at higher revs - in
fact, if you doubled the revs you'd get double the power.

Highest power comes when the engine is struggling to get enough air
through, and the torque has fallen off so much that the increase in revs
doesn't give you more power.

Andy

On 07/04/2016 22:24, wrote:


It seems that you're saying that becase Power is obtained at the TOP of the power curve and Peak Torque much lower on the power curve, that a car will accelerate more rapidly between 90 and 100 mph than it will between 40 and 50 MPH.

No-one is saying anything remotely like that!

For a start, you're not going to be using the same gear at 40 mph as at
90 mph - you select the best gear for the speed you're at, to enable the
engine to develop something approaching max power at all road speeds.

Since power = force x speed, the same amount of power at a high road
speed gives you less force to accelerate the car than is available at
low speed. *And* the aerodynamic drag is 4 times as high at 80 as it is
at 40. Putting that all together, a car accelerates much *less* rapidly
as high speed.

When it reaches it's maximum speed - the point at which drag equals the
available tractive effort - it doesn't accelerate at all!
--
Cheers,
Roger
____________
Please reply to Newsgroup. Whilst email address is valid, it is seldom
checked.

In article ,
michael adams wrote:

"Roger Mills" wrote in message
...

What you say is absolutely right, but I'm afraid that you're wasting
your breath as far as Mr P is concerned. He believes that torque is
GOD - period!

quote

"Torque, moment, or moment of force (see the terminology below) is the
tendency of a force to rotate an object about an axis,[1]

https://en.wikipedia.org/wiki/Torque

quote

Given that torque is the power transmitted via the drive axle which
causes the wheels to rotate, its difficult to see what's wrong with his
claim, quite honestly.

Hasn't stopped many trying. And failing.

Except maybe in cars fitted with sails.

Would that be 'thrust'? ;-)

--
*I finally got my head together, now my body is falling apart.

Dave Plowman London SW
To e-mail, change noise into sound.

In article ,
Roger Mills wrote:
On 07/04/2016 12:38, michael adams wrote:


I'm not concerned about the engine.

That's the problem - Mr Plowman *is*.

For the wheels to be turning or accelerating fastest -
in the absence of the car being powered by external
factors such as the wind propelling the body, they
need to be subject to the maximum available torque.


Indeed. And you obtain that by running the engine at max power and using
a suitable gear ratio. But Mr Plowman doesn't accept that. He says that
you have to run the *engine* at max *torque*. So he would end up using a
lower engine speed - producing less *power* - with higher gearing which
provides *less* torque multiplication - resulting in less torque at the
wheels. Mad!

How many many times do I have to say I'm only commenting on the best
acceleration *in any one gear* - so entirely due to engine output.

Perhaps there's something missing from the way I've made that point.
Anyone who's ever driven a car knows that you'll get better acceleration
by using a lower gear. It's so obvious it didn't ever need stating.

However, if making the point I am, saying to use a different gear is
simply nonsense.

And I'm utterly amazed so few understand the relationship between torque
and power.

--
*I love cats...they taste just like chicken.

Dave Plowman London SW
To e-mail, change noise into sound.

In article ,
michael adams wrote:
If the engine is running at maximum power, and thus the crankshaft
is turning as fast as possible, then how is it possible for less than
maximum torque to be delivered to the flywheel ?

I'm beginning to lose the will to live.

Power is a function of torque and engine speed. It therefore goes without
saying that maximum power will always be delivered at higher revs than
maximum torque in practice. An *ideal* engine would produce constant
torque throughout its rev range, so BHP would increase linearly with
engine speed. But no engine is ideal, therefore it will have both a torque
peak and BHP peak.

--
*When you get a bladder infection urine trouble.*

Dave Plowman London SW
To e-mail, change noise into sound.

In article ,
Johnny B Good wrote:
On Tue, 05 Apr 2016 10:52:36 +0100, Dave Plowman (News) wrote:

Just to see if I was missing something that others found so obvious, I
asked this question on a forum designated to engine building and tuning,
etc.

'At what point on an engines output do you get maximum acceleration?'

And the answers were near unanimous. Peak torque.

I'll take the views of those guys over the bar room mechanics here any
day.

Ah! I see your problem now. If, as you've seemingly posed the question,
we're discussing, "When does the maximum instantaneous acceleration over
the operating rpm range of the prime mover occur in any *one* gear ratio
at a time?", then the answer to *that* question *will indeed* be at
maximum torque rpms.

Have said this countless times. But didn't stop others saying this was
wrong - or introducing countless red herrings and moving of goal posts.

It's what started the original thread and caused Vir to start this one.

But it seems he's not alone in having little understanding of basic
mechanics.

--
*If you ate pasta and anti-pasta, would you still be hungry?

Dave Plowman London SW
To e-mail, change noise into sound.