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?

(more to follow)

On 3/20/2016 9:55 PM, Vir Campestris wrote:
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?

(more to follow)

Not enough information. Torque at what RPM?

On 20/03/2016 21:55, Vir Campestris wrote:
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?

(more to follow)

How long is a piece of string?

The answer to your question depends on a lot of factors. The gearing is
clearly very important. Unless the gearing is right, the engine may not
be at the speed at which it generates peak torque when the car speed is
10m/S.

Then you have to take account of transmission losses and tyre rolling
resistance.

I spent many hundreds of happy hours doing such calculations when I
worked for a motor manufacturer in the 1960's - mainly without the
benefit of computers!
--
Cheers,
Roger
____________
Please reply to Newsgroup. Whilst email address is valid, it is seldom
checked.

In article ,
Vir Campestris wrote:
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?

Insufficient data to calculate. By miles. Sorry, metres.

--
*War does not determine who is right - only who is left.

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

On Sun, 20 Mar 2016 21:55:48 +0000, Vir Campestris wrote:

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?

(more to follow)

Like an explanation of the strange units you appear to be using along
with at least one more parameter such as power output at the peak torque
speed from which to derive the engine rpm or else a simple statement of
the actual rpm at this speed.

Assuming nM means N-m and m/S means m/s, we only need to know the actual
engine rpm figure at which its peak torque was measured, along with an
assumed figure of gearbox efficiency (your question is phrased so as to
imply a motor vehicle of some sort or other), in order to calculate a
valid result.

For all I know, this might be exactly the 'answer' you were fishing
for. :-)


--
Johnny B Good

Gear ratios and friction etc.
Brian

--
----- -
This newsgroup posting comes to you directly from...
The Sofa of Brian Gaff...

Blind user, so no pictures please!
"newshound" wrote in message
news
On 3/20/2016 9:55 PM, Vir Campestris wrote:
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?

(more to follow)

Not enough information. Torque at what RPM?

On 21/03/2016 03:29, Johnny B Good wrote:
On Sun, 20 Mar 2016 21:55:48 +0000, Vir Campestris wrote:

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?

(more to follow)

Like an explanation of the strange units you appear to be using along
with at least one more parameter such as power output at the peak torque
speed from which to derive the engine rpm or else a simple statement of
the actual rpm at this speed.

Assuming nM means N-m and m/S means m/s, we only need to know the actual
engine rpm figure at which its peak torque was measured, along with an
assumed figure of gearbox efficiency (your question is phrased so as to
imply a motor vehicle of some sort or other), in order to calculate a
valid result.


I thought we needed a bit more information than that given I'm unclear
what eg "force available at the wheels" means if the car is in a 4-wheel
drift depositing large amounts of rubber[1]: the speed doesn't tell us
the direction of motion relative to the wheels

[1] Cenotaphs optional!


--
Robin
reply-to address is (intended to be) valid

On 21/03/16 10:24, Robin wrote:
On 21/03/2016 03:29, Johnny B Good wrote:
On Sun, 20 Mar 2016 21:55:48 +0000, Vir Campestris wrote:

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?

(more to follow)

Like an explanation of the strange units you appear to be using along
with at least one more parameter such as power output at the peak torque
speed from which to derive the engine rpm or else a simple statement of
the actual rpm at this speed.

Assuming nM means N-m and m/S means m/s, we only need to know the
actual
engine rpm figure at which its peak torque was measured, along with an
assumed figure of gearbox efficiency (your question is phrased so as to
imply a motor vehicle of some sort or other), in order to calculate a
valid result.


I thought we needed a bit more information than that given I'm unclear
what eg "force available at the wheels" means if the car is in a 4-wheel
drift depositing large amounts of rubber[1]: the speed doesn't tell us
the direction of motion relative to the wheels

[1] Cenotaphs optional!


All that the question betrays is ignorance on the part of its poser.

If the surface of the sun is 5700 degrees, how hot will it get on
midsummer's day in Wyoming?

--
"What do you think about Gay Marriage?"
"I don't."
"Don't what?"
"Think about Gay Marriage."

On 21/03/2016 12:13, The Natural Philosopher wrote:

If the surface of the sun is 5700 degrees, how hot will it get on
midsummer's day in Wyoming?


Wyoming is in merka, so you need to give the answer in Fahrenheit. The
temperature you gave for the sun's surface is in degrees kelvin, so
surely that's a fail already?

On 21/03/16 18:23, GB wrote:
On 21/03/2016 12:13, The Natural Philosopher wrote:

If the surface of the sun is 5700 degrees, how hot will it get on
midsummer's day in Wyoming?


Wyoming is in merka, so you need to give the answer in Fahrenheit. The
temperature you gave for the sun's surface is in degrees kelvin, so
surely that's a fail already?


Almost certainly

--
New Socialism consists essentially in being seen to have your heart in
the right place whilst your head is in the clouds and your hand is in
someone else's pocket.

On 21/03/2016 12:13, The Natural Philosopher wrote:
On 21/03/16 10:24, Robin wrote:
On 21/03/2016 03:29, Johnny B Good wrote:
On Sun, 20 Mar 2016 21:55:48 +0000, Vir Campestris wrote:

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?

(more to follow)

Like an explanation of the strange units you appear to be using along
with at least one more parameter such as power output at the peak torque
speed from which to derive the engine rpm or else a simple statement of
the actual rpm at this speed.

Assuming nM means N-m and m/S means m/s, we only need to know the
actual
engine rpm figure at which its peak torque was measured, along with an
assumed figure of gearbox efficiency (your question is phrased so as to
imply a motor vehicle of some sort or other), in order to calculate a
valid result.


I thought we needed a bit more information than that given I'm unclear
what eg "force available at the wheels" means if the car is in a 4-wheel
drift depositing large amounts of rubber[1]: the speed doesn't tell us
the direction of motion relative to the wheels

[1] Cenotaphs optional!


All that the question betrays is ignorance on the part of its poser.

If the surface of the sun is 5700 degrees, how hot will it get on
midsummer's day in Wyoming?


Easy, 5700 degrees.
What was the real question?

On 21/03/2016 03:29, Johnny B Good wrote:
For all I know, this might be exactly the 'answer' you were fishing
for.:-)

It is indeed, and I'm sorry question two was so long coming. I've been busy.

If a car is generating 400nM of torque at 2000RPM, what is the power
output of the engine?

Andy

On 24/03/16 21:22, Vir Campestris wrote:
On 21/03/2016 03:29, Johnny B Good wrote:
For all I know, this might be exactly the 'answer' you were fishing
for.:-)

It is indeed, and I'm sorry question two was so long coming. I've been
busy.

If a car is generating 400nM of torque at 2000RPM, what is the power
output of the engine?

Andy
800000 nm per minute :-)



--
Future generations will wonder in bemused amazement that the early
twenty-first centurys developed world went into hysterical panic over a
globally average temperature increase of a few tenths of a degree, and,
on the basis of gross exaggerations of highly uncertain computer
projections combined into implausible chains of inference, proceeded to
contemplate a rollback of the industrial age.

Richard Lindzen

In article ,
Vir Campestris wrote:
If a car is generating 400nM of torque at 2000RPM, what is the power
output of the engine?

Have you not Googled to find a calculator to do just this?

The formula I was taught many many years ago is for imperial units.

Easy to remember. Two pies 'n' tea.

So:-
2 x pi x RPM x torque(lb.ft)
---------------------------- = BHP
33,000

Result is approx 112 bhp.

--
*You're never too old to learn something stupid.
Dave Plowman London SW
To e-mail, change noise into sound.

On 25/03/2016 11:14, Dave Plowman (News) wrote:
In articleJ6OdnTIlls8dx2nLnZ2dnUU78V2dnZ2d@brightvie w.co.uk,
Vir wrote:
If a car is generating 400nM of torque at 2000RPM, what is the power
output of the engine?

Have you not Googled to find a calculator to do just this?

The formula I was taught many many years ago is for imperial units.

Easy to remember. Two pies 'n' tea.

So:-
2 x pi x RPM x torque(lb.ft)
---------------------------- = BHP
33,000

Result is approx 112 bhp.


I used rpm x lb.ft / 5252 - which is more or less the same thing.

Some of the engines I was working on developed their max power at about
5250 rpm - meaning that the power was numerically equal to the torque at
that speed.
--
Cheers,
Roger
____________
Please reply to Newsgroup. Whilst email address is valid, it is seldom
checked.

In article ,
Roger Mills wrote:
On 25/03/2016 11:14, Dave Plowman (News) wrote:
In articleJ6OdnTIlls8dx2nLnZ2dnUU78V2dnZ2d@brightvie w.co.uk,
Vir wrote:
If a car is generating 400nM of torque at 2000RPM, what is the power
output of the engine?

Have you not Googled to find a calculator to do just this?

The formula I was taught many many years ago is for imperial units.

Easy to remember. Two pies 'n' tea.

So:-
2 x pi x RPM x torque(lb.ft)
---------------------------- = BHP
33,000

Result is approx 112 bhp.


I used rpm x lb.ft / 5252 - which is more or less the same thing.

Yes - you can use a constant. But the formula explains things to those who
are interested.

Some of the engines I was working on developed their max power at about
5250 rpm - meaning that the power was numerically equal to the torque at
that speed.

I'd say this is near impossible (or desirable) in practice.

--
*Why is the man who invests all your money called a broker?

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

On 25/03/2016 14:20, Dave Plowman (News) wrote:
In ,
Roger wrote:



I used rpm x lb.ft / 5252 - which is more or less the same thing.

Yes - you can use a constant. But the formula explains things to those who
are interested.

True. But 33000 is a constant, anyway. If you express the speed in
radians per second rather than RPM, you're just left with 550 at the
bottom - since 1 HP is defined as 550 ft-lb per second.

Some of the engines I was working on developed their max power at about
5250 rpm - meaning that the power was numerically equal to the torque at
that speed.

I'd say this is near impossible (or desirable) in practice.


Sorry, you've lost me. What's near impossible?

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

On 25/03/2016 11:14, Dave Plowman (News) wrote:
In article ,
Vir Campestris wrote:
If a car is generating 400nM of torque at 2000RPM, what is the power
output of the engine?

Have you not Googled to find a calculator to do just this?

The formula I was taught many many years ago is for imperial units.

Easy to remember. Two pies 'n' tea.

So:-
2 x pi x RPM x torque(lb.ft)
---------------------------- = BHP
33,000

Result is approx 112 bhp.


OK, now if it's generating 200BHP at 5000RPM, what's the torque? (I'd
have stuck with kW, about 84, with fewer conversions but that's not
important)

Andy

On 25/03/2016 21:16, Vir Campestris wrote:
On 25/03/2016 11:14, Dave Plowman (News) wrote:
In article ,
Vir Campestris wrote:
If a car is generating 400nM of torque at 2000RPM, what is the power
output of the engine?

Have you not Googled to find a calculator to do just this?

The formula I was taught many many years ago is for imperial units.

Easy to remember. Two pies 'n' tea.

So:-
2 x pi x RPM x torque(lb.ft)
---------------------------- = BHP
33,000

Result is approx 112 bhp.


OK, now if it's generating 200BHP at 5000RPM, what's the torque? (I'd
have stuck with kW, about 84, with fewer conversions but that's not
important)

Andy

About 210 lb-ft

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

In article ,
Vir Campestris wrote:
Easy to remember. Two pies 'n' tea.

So:-
2 x pi x RPM x torque(lb.ft)
---------------------------- = BHP
33,000

Result is approx 112 bhp.


OK, now if it's generating 200BHP at 5000RPM, what's the torque? (I'd
have stuck with kW, about 84, with fewer conversions but that's not
important)

Think you may need to get your maths books out and revise. ;-)

--
*I don't work here. I'm a consultant

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

On 25/03/2016 22:48, Roger Mills wrote:
On 25/03/2016 21:16, Vir Campestris wrote:
On 25/03/2016 11:14, Dave Plowman (News) wrote:
In article ,
Vir Campestris wrote:
If a car is generating 400nM of torque at 2000RPM, what is the power
output of the engine?

Have you not Googled to find a calculator to do just this?

The formula I was taught many many years ago is for imperial units.

Easy to remember. Two pies 'n' tea.

So:-
2 x pi x RPM x torque(lb.ft)
---------------------------- = BHP
33,000

Result is approx 112 bhp.


OK, now if it's generating 200BHP at 5000RPM, what's the torque? (I'd
have stuck with kW, about 84, with fewer conversions but that's not
important)

Andy

About 210 lb-ft

210lbft at 5000 doesn't compare well with 400nM at 2000RPM

Andy

On 26/03/2016 14:53, Dave Plowman (News) wrote:
In article ,
Vir Campestris wrote:
Easy to remember. Two pies 'n' tea.

So:-
2 x pi x RPM x torque(lb.ft)
---------------------------- = BHP
33,000

Result is approx 112 bhp.


OK, now if it's generating 200BHP at 5000RPM, what's the torque? (I'd
have stuck with kW, about 84, with fewer conversions but that's not
important)

Think you may need to get your maths books out and revise. ;-)

Well, if it's too hard for you...

400nM at 2000RPM = 84kW (to the nearest integer) ~= 112HP.

Reverse the transformation, and from 200HP you get 210lb-ft, or 284nM. A
little over half. Right?

Now, if you have a gearbox attached to this engine, and you set it to
reduce the output shaft to 1000RPM, what is the resultant torque at
those two engine speeds?

Andy

In article ,
Vir Campestris wrote:
On 26/03/2016 14:53, Dave Plowman (News) wrote:
In article ,
Vir Campestris wrote:
Easy to remember. Two pies 'n' tea.

So:-
2 x pi x RPM x torque(lb.ft)
---------------------------- = BHP
33,000

Result is approx 112 bhp.


OK, now if it's generating 200BHP at 5000RPM, what's the torque? (I'd
have stuck with kW, about 84, with fewer conversions but that's not
important)

Think you may need to get your maths books out and revise. ;-)

Well, if it's too hard for you...

I'm not the one asking. And I've given you the means to calculate BHP or
torque at a given RPM. Which is why I suggested you revise your maths.

400nM at 2000RPM = 84kW (to the nearest integer) ~= 112HP.

Be better if you stuck to the same units.

Reverse the transformation, and from 200HP you get 210lb-ft, or 284nM. A
little over half. Right?

You are the one doing the calcs.

Now, if you have a gearbox attached to this engine, and you set it to
reduce the output shaft to 1000RPM, what is the resultant torque at
those two engine speeds?

What is the point of all this? A gearbox multiplies the torque in
relationship to the ratio of the gears, less any friction.

Andy

--
*Rehab is for quitters

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

On 28/03/2016 00:29, Dave Plowman (News) wrote:
In article ,
Vir Campestris wrote:
On 26/03/2016 14:53, Dave Plowman (News) wrote:
In article ,
Vir Campestris wrote:
Easy to remember. Two pies 'n' tea.

So:-
2 x pi x RPM x torque(lb.ft)
---------------------------- = BHP
33,000

Result is approx 112 bhp.


OK, now if it's generating 200BHP at 5000RPM, what's the torque? (I'd
have stuck with kW, about 84, with fewer conversions but that's not
important)

Think you may need to get your maths books out and revise. ;-)

Well, if it's too hard for you...

I'm not the one asking. And I've given you the means to calculate BHP or
torque at a given RPM. Which is why I suggested you revise your maths.

400nM at 2000RPM = 84kW (to the nearest integer) ~= 112HP.

Be better if you stuck to the same units.

I didn't choose to insert imperial. You did.

Reverse the transformation, and from 200HP you get 210lb-ft, or 284nM. A
little over half. Right?

You are the one doing the calcs.

Now, if you have a gearbox attached to this engine, and you set it to
reduce the output shaft to 1000RPM, what is the resultant torque at
those two engine speeds?

What is the point of all this? A gearbox multiplies the torque in
relationship to the ratio of the gears, less any friction.


The point is, Dave, that you are unable to apply these calculations to
real world situations when they disagree with your prejudices. I'd hoped
to be able to talk you through it.

112BHP / 400nM at 2000 RPM geared down to 1000 RPM gives a little under
(because of friction) 800nM at the output.

200 BHP / 284nM at 5000 RPM geared down to 1000 RPM gives a little under
1420nM at the output.

This clearly shows you get more force by choosing peak power in a low
gear over peak torque in a higher one.

But, hey, if you want to believe otherwise when you've got the
mathematical skills to prove to yourself that your long held prejudice
is wrong - I'm not going to argue with you any more.

Andy

On 29/03/2016 21:04, Vir Campestris wrote:


This clearly shows you get more force by choosing peak power in a low
gear over peak torque in a higher one.


That's a no-brainer, anyway. Just as you can express engine power in
terms of engine speed (times) engine torque (divided by) a suitable
constant, you can also express propulsive power as road speed (times)
thrust at the contact patch (divided by) a different constant - but they
should equate to the same thing apart from transmission losses. So when
the engine is developing its peak power, you'll get peak thrust at the
wheels.

Having said that, it's not particularly useful unless you've got an
infinitely variable transmission because fixed gearing will only allow
max power to be applied at the road at a few discrete speeds - not over
the whole speed range.
--
Cheers,
Roger
____________
Please reply to Newsgroup. Whilst email address is valid, it is seldom
checked.

In article ,
Vir Campestris wrote:
On 28/03/2016 00:29, Dave Plowman (News) wrote:
In article ,
Vir Campestris wrote:
On 26/03/2016 14:53, Dave Plowman (News) wrote:
In article ,
Vir Campestris wrote:
Easy to remember. Two pies 'n' tea.

So:-
2 x pi x RPM x torque(lb.ft)
---------------------------- = BHP
33,000

Result is approx 112 bhp.


OK, now if it's generating 200BHP at 5000RPM, what's the torque? (I'd
have stuck with kW, about 84, with fewer conversions but that's not
important)

Think you may need to get your maths books out and revise. ;-)

Well, if it's too hard for you...

I'm not the one asking. And I've given you the means to calculate BHP or
torque at a given RPM. Which is why I suggested you revise your maths.

400nM at 2000RPM = 84kW (to the nearest integer) ~= 112HP.

Be better if you stuck to the same units.

I didn't choose to insert imperial. You did.

I gave you the way to work things out using the data I know. A little
Googling will probably give you one using other units.

Kw means nothing to me as regards a car engine's output. So I'll stick to
the units I know.

Reverse the transformation, and from 200HP you get 210lb-ft, or 284nM. A
little over half. Right?

You are the one doing the calcs.

Now, if you have a gearbox attached to this engine, and you set it to
reduce the output shaft to 1000RPM, what is the resultant torque at
those two engine speeds?

What is the point of all this? A gearbox multiplies the torque in
relationship to the ratio of the gears, less any friction.


The point is, Dave, that you are unable to apply these calculations to
real world situations when they disagree with your prejudices. I'd hoped
to be able to talk you through it.

I don't need to be talked through it.

112BHP / 400nM at 2000 RPM geared down to 1000 RPM gives a little under
(because of friction) 800nM at the output.

200 BHP / 284nM at 5000 RPM geared down to 1000 RPM gives a little under
1420nM at the output.

Doesn't come of much of a surprise here that 5:1 gearing gives more torque
at the wheels than 2:1, if the engines are similar.


This clearly shows you get more force by choosing peak power in a low
gear over peak torque in a higher one.

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.

But, hey, if you want to believe otherwise when you've got the
mathematical skills to prove to yourself that your long held prejudice
is wrong - I'm not going to argue with you any more.

Think you need to revisit basic mechanics.

--
*Don't squat with your spurs on *

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

In article ,
Roger Mills wrote:
This clearly shows you get more force by choosing peak power in a low
gear over peak torque in a higher one.


That's a no-brainer, anyway. Just as you can express engine power in
terms of engine speed (times) engine torque (divided by) a suitable
constant, you can also express propulsive power as road speed (times)
thrust at the contact patch (divided by) a different constant - but they
should equate to the same thing apart from transmission losses. So when
the engine is developing its peak power, you'll get peak thrust at the
wheels.

Are you at it too? ;-)

Peak torque at the wheels happens when the engine is at peak torque - not
peak BHP, unless the two coincide. Regardless of which gear you choose.

--
*Why were the Indians here first? They had reservations.*

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

On Thursday, 24 March 2016 21:22:11 UTC, Vir Campestris wrote:
On 21/03/2016 03:29, Johnny B Good wrote:
For all I know, this might be exactly the 'answer' you were fishing
for.:-)

It is indeed, and I'm sorry question two was so long coming. I've been busy.

If a car is generating 400nM of torque at 2000RPM, what is the power
output of the engine?

Andy

The general formula for mechanical power. (in imperial units) is:-

HP= 2Ï€ Nt divided by 33,000

HP = horsepower.
N = rpm
t = torque in lb ft.

746 watts in a horsepower.

On Tue, 29 Mar 2016 22:21:52 +0100, Roger Mills wrote:

Having said that, it's not particularly useful unless you've got an
infinitely variable transmission because fixed gearing will only allow
max power to be applied at the road at a few discrete speeds - not over
the whole speed range.

Well...

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

(Not that DAFs were known for neck-snapping acceleration, or anything other than
going backwards real fast...)


Thomas Prufer

In article ,
Thomas Prufer wrote:
On Tue, 29 Mar 2016 22:21:52 +0100, Roger Mills wrote:

Having said that, it's not particularly useful unless you've got an
infinitely variable transmission because fixed gearing will only allow
max power to be applied at the road at a few discrete speeds - not over
the whole speed range.

Well...

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

(Not that DAFs were known for neck-snapping acceleration, or anything
other than going backwards real fast...)

Infinitely variable is a misnomer. It has a limit to the ratios available.
But is variable between the lowest and highest.

--
*Everyone has a photographic memory. Some just don't have film*

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

On 3/27/2016 9:48 PM, Vir Campestris wrote:
On 25/03/2016 22:48, Roger Mills wrote:
On 25/03/2016 21:16, Vir Campestris wrote:
On 25/03/2016 11:14, Dave Plowman (News) wrote:
In article ,
Vir Campestris wrote:
If a car is generating 400nM of torque at 2000RPM, what is the power
output of the engine?

Have you not Googled to find a calculator to do just this?

The formula I was taught many many years ago is for imperial units.

Easy to remember. Two pies 'n' tea.

So:-
2 x pi x RPM x torque(lb.ft)
---------------------------- = BHP
33,000

Result is approx 112 bhp.


OK, now if it's generating 200BHP at 5000RPM, what's the torque? (I'd
have stuck with kW, about 84, with fewer conversions but that's not
important)

Andy

About 210 lb-ft

210lbft at 5000 doesn't compare well with 400nM at 2000RPM

Andy

Why not? As one of the earlier posters pointed out, peak torque is
normally at relatively low revs.

On 3/30/2016 9:05 AM, Thomas Prufer wrote:
On Tue, 29 Mar 2016 22:21:52 +0100, Roger Mills wrote:

Having said that, it's not particularly useful unless you've got an
infinitely variable transmission because fixed gearing will only allow
max power to be applied at the road at a few discrete speeds - not over
the whole speed range.

Well...

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

(Not that DAFs were known for neck-snapping acceleration, or anything other than
going backwards real fast...)


Thomas Prufer

Actually the Jazz hybrid with CVT (with less than 100 BHP including the
electric motor) is pretty nippy off the line if you floor it. Regularly
sees off 5 series BMWs showing off at the lights :-)

On 3/30/2016 12:30 AM, Dave Plowman (News) wrote:
In article ,
Vir Campestris wrote:
On 28/03/2016 00:29, Dave Plowman (News) wrote:
In article ,
Vir Campestris wrote:
On 26/03/2016 14:53, Dave Plowman (News) wrote:
In article ,
Vir Campestris wrote:
Easy to remember. Two pies 'n' tea.

So:-
2 x pi x RPM x torque(lb.ft)
---------------------------- = BHP
33,000

Result is approx 112 bhp.


OK, now if it's generating 200BHP at 5000RPM, what's the torque? (I'd
have stuck with kW, about 84, with fewer conversions but that's not
important)

Think you may need to get your maths books out and revise. ;-)

Well, if it's too hard for you...

I'm not the one asking. And I've given you the means to calculate BHP or
torque at a given RPM. Which is why I suggested you revise your maths.

400nM at 2000RPM = 84kW (to the nearest integer) ~= 112HP.

Be better if you stuck to the same units.

I didn't choose to insert imperial. You did.

I gave you the way to work things out using the data I know. A little
Googling will probably give you one using other units.

Kw means nothing to me as regards a car engine's output. So I'll stick to
the units I know.

Reverse the transformation, and from 200HP you get 210lb-ft, or 284nM. A
little over half. Right?

You are the one doing the calcs.

Now, if you have a gearbox attached to this engine, and you set it to
reduce the output shaft to 1000RPM, what is the resultant torque at
those two engine speeds?

What is the point of all this? A gearbox multiplies the torque in
relationship to the ratio of the gears, less any friction.


The point is, Dave, that you are unable to apply these calculations to
real world situations when they disagree with your prejudices. I'd hoped
to be able to talk you through it.

I don't need to be talked through it.

112BHP / 400nM at 2000 RPM geared down to 1000 RPM gives a little under
(because of friction) 800nM at the output.

200 BHP / 284nM at 5000 RPM geared down to 1000 RPM gives a little under
1420nM at the output.

Doesn't come of much of a surprise here that 5:1 gearing gives more torque
at the wheels than 2:1, if the engines are similar.


This clearly shows you get more force by choosing peak power in a low
gear over peak torque in a higher one.

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.

But, hey, if you want to believe otherwise when you've got the
mathematical skills to prove to yourself that your long held prejudice
is wrong - I'm not going to argue with you any more.

Think you need to revisit basic mechanics.

+1

On 30/03/16 00:30, Dave Plowman (News) wrote:
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.

In the one sentence is encapsulated Lefty's are inevitably stupid
politics

'in any given gear' is irrelevant, when you can choose your gear.


--
€œit should be clear by now to everyone that activist environmentalism
(or environmental activism) is becoming a general ideology about humans,
about their freedom, about the relationship between the individual and
the state, and about the manipulation of people under the guise of a
'noble' idea. It is not an honest pursuit of 'sustainable development,'
a matter of elementary environmental protection, or a search for
rational mechanisms designed to achieve a healthy environment. Yet
things do occur that make you shake your head and remind yourself that
you live neither in Joseph Stalins Communist era, nor in the Orwellian
utopia of 1984.€�

Vaclav Klaus

On 30/03/16 00:37, Dave Plowman (News) wrote:

Peak torque at the wheels happens when the engine is at peak torque - not
peak BHP, unless the two coincide. Regardless of which gear you choose.

No, it doesn't, because we have a car at speed and a gear box.

No wonder you vote Labia.

Peak torque at the wheels is when the gear ratio is lowest.

If you ignore car speed.

Which is 'as true' as your original fatuous statement


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

Josef Stalin

In article ,
The Natural Philosopher wrote:
On 30/03/16 00:30, Dave Plowman (News) wrote:
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.

In the one sentence is encapsulated Lefty's are inevitably stupid
politics

I'm sure you think that a very pithy comment. Now try it again in English.

'in any given gear' is irrelevant, when you can choose your gear.

Thanks from proving once more you can't follow a discussion.

Let me see if I can put it in terms even you can understand.

Regardless of the gear chosen, the peak acceleration *in that gear* will
occur when the engine is developing peak torque.

Please feel free to say what you don't understand now.

--
*Don't worry; it only seems kinky the first time.*

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

In article ,
The Natural Philosopher wrote:
Peak torque at the wheels happens when the engine is at peak torque -
not peak BHP, unless the two coincide. Regardless of which gear you
choose.

No, it doesn't, because we have a car at speed and a gear box.

Ok. You've now confirmed you are an idiot.

--
*Since light travels faster than sound, some people appear bright until you hear them speak.

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

On Wed, 30 Mar 2016 11:20:47 +0100, "Dave Plowman (News)"
wrote:

Infinitely variable is a misnomer. It has a limit to the ratios available.

Marketing-speak would say it's got non-discrete ratios, hence they are
continuous, hence infinitely many ratios, hence infinitely variable.:-)

But, yeah, it's bounded.

(And the Wiki entry on the Variomatic efficiency may be a poor translation from
another language? It's oddly phrased: "Due the fact the engine runs most of the
time in its best fuel economic speed, the fuel consumption of this car was at
accepted level...)

But is variable between the lowest and highest.

Plus the two extra-special modes, outside this range! Only available for a short
period!

(These are "goes nowhere with the engine running" and "goes somewhere even with
the engine off" -- if pushed, or rolling downhill.)

I'll concede that the extra-special modes are a bit spurious. (But one is very
fuel-efficient!)


Thomas Prufer

In article ,
Thomas Prufer wrote:
Infinitely variable is a misnomer. It has a limit to the ratios
available.

Marketing-speak would say it's got non-discrete ratios, hence they are
continuous, hence infinitely many ratios, hence infinitely variable.:-)

But it doesn't have an infinite number of ratios. That's my point. It's
just the usual adspeak to take in those who don't understand mechanics.
Which seems to include quite a few on here. ;-)

--
*I must always remember that I'm unique, just like everyone else. *

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

On 30/03/2016 00:30, Dave Plowman (News) wrote:



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.


That is true of *that* gear at *that* road speed.

HOWEVER, the engine won't be developing its maximum power when running
at its max torque speed. So - at that roadspeed - you could actually get
*more* thrust and *more* acceleration by choosing a different gear which
allowed the engine to run at at its max power speed. The engine would be
producing less torque than in the first case, but this would be more
than offset by extra torque multiplication from the lower gearing.

Tell me, in your boy racer days, how did you decide when to change gear
in order to achieve the best possible 0-100 time?
--
Cheers,
Roger
____________
Please reply to Newsgroup. Whilst email address is valid, it is seldom
checked.