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 slowbug
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| posted on May 1st, 2003 at 06:24 PM |
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power verses torque
I've asked this question numerous times and have never got a clear answer.
OK I know how both are measured, but, what is the difference in drivability.
What gives a car acceleration off the line? is it power or torque or a combination of both?
What if you had a high horsepower engine with pathetic torque....and visa versa.
Cheers
Wayne
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 kombi_kid
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| posted on May 1st, 2003 at 06:42 PM |
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hey
torque gives it launch speed gives it top end power.
many highhorse power engines will have torque but torque is really givin by what g'box ratios you have etc!!!
cheers
rhys
it aint just cool its aircool'd
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hool
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| posted on May 1st, 2003 at 07:30 PM |
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i don't think i've ever read a simple enough explanation for the difference between power and torque...
horsepower helps your engine rev faster (not necessarily higher, just reach the redline faster). if you have two cars with the same weight and gearing
etc, only difference being horsepower, the more powerful car will obviously rev faster and therefore accelerate faster.
torque is usually described as 'low-down grunt', or pulling power. peak torque is always developed at lower revs than peak power.
take two cars with identical power, weight gearing etc, but one develops more torque. put them both on a flat stretch of road at the same speed and
there's no difference... once the flat road becomes a steep climb the less torquey engine will need to drop down a gear, where the torquier
engine will be able to maintain speed longer without changing down.
in terms of drivability it depends on your driving style. an engine thats designed for max power will usually sacrifice torque, and a torquey engine
will not rev as well.
if you want a car for racing then focus more on power. if you want something thats more pleasant in traffic then focus on torque...
the other day i got to drive a peugeot 405 turbo diesel wagon - i loved it! this engine redlines around 5,000rpm but the acceleration in traffic was
brilliant - it went really hard, but you don't have to drive it hard, like a high power engine. if you want to accelerate you put your foot down
and it goes, rather then having to change down to accelerate...
hope this ramble makes some sense!
julian
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70AutoStik
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| posted on May 1st, 2003 at 09:17 PM |
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A simple explanation would be a little too long for a post, but here's some points.
Torque is what accelerates the car, power is useful in that in can be geared down to provide torque. If you look at a stock VW engine's torque
curve, you will find it peaks at about 2000rpm, then gently drops off until about 4000. This is close to ideal for an economical street engine, as it
encourages you to shift to a higher gear. If you look at the torque curve for a racing engine, you will find it a fairly narrow hump at high revs.
This is what is wanted in a race car, as a close-ratio gearbox and low diff ratio can be used to provide maximum torque (and fuel economy doesn't
really matter 1-2 mpg can be acceptable on a racing car.)
I think aussiebug or dave have a link to a site that explains a bit more.
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OvalGlen
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| posted on May 1st, 2003 at 11:59 PM |
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hool....take two cars with identical power, weight gearing etc, but one develops more torque. put them both on a flat stretch of road at the same
speed and there's no difference.
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I would argue that the torquier car would pull away from the other if they were to accelerate.
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Setting up a car for Bigger horsepower , developed by extremely high reving can tend to make the car annoying to drive , as you feel you have to
thrash the thing to get it to accelerate decently.
Wear in the motor on such a car is considerable, even the gearbox suffers from having to row it through the gears all the time.
The greatest force within a motor is the centrifigal force on the piston (trying to escape the confines of the barrel/head), this increases
dramatically as revs are increased.
I prefer lower revs with more torque.
Regards,Glenn>
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hool
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| posted on May 2nd, 2003 at 12:45 AM |
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yep Glen, you're probably right - the other would need to drop back a gear to get the revs back up into the power band...
a friend of mine a few years ago bought a datsun 1600 already set up for club level racing like supersprints.. it had a very well prepared 2.4litre
with twin dells on it. the engine was a 2 litre block stroked to 2.4l, it wasn't a particularly revvy engine but it had masses of torque. he took
me for a drive around the 'burbs, at lights if he accelerated too hard the tyres would light up, but the engine was hardly ticking over - it was
sitting in the mid-range while the tyres were getting fried
i watched him at a supersprint at eastern creek, first time out in the car he was lapping at around 1'45". later on he worked out that the
car was doing 150mph (over 240k's) down the straight..
which is pretty good for a car set up more as a 'grunter' than a 'screamer'...
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Andy
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| posted on May 2nd, 2003 at 01:13 AM |
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Reading these enquiries always gives me a good laugh 
I shall refrain myself and only clarify one point for Slowbug. Torque is measured, Power is calculated. Power = Torque x RPM, not measured.
My personal preference is always loads of stump pulling torque, but for you, as has been mentioned it all depends on your preference and driving
style.
Andy
:thumb
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slowbug
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| posted on May 2nd, 2003 at 01:23 AM |
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Ok, now i'm starting to get a grasp of this.
having to change down a gear or not...thats clinched it for me....i think
cheers
Wayne
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Baja Wes
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| posted on May 2nd, 2003 at 09:31 AM |
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Andy is correct, autostik is on the right track, and the other posts are confusing.
HP ~ torque x RPM.
Torque is what accelerates you. It translates directly to the force you wheels push your car forward with. Your gear ratios will multiply the engine
torque up or down, and effectively change the force your car is being pushed forward with. Higher gears (4th eg = 0.82:1) are not multiplying the
torque by the same ratio as lower gears (1st eg = 3.86:1), and therefore is pushing the car forward with proportionally less force.
So by using very low gearing you could get incredible amounts of torque at the wheels, and push the car forward very very hard. There is one prob,
that low gearing will make the engines torque occur at very slow car speeds. So all that torque ain't much use if the engine revs out at 5kph.
Obviously there needed to be a way of determining what force the torque could push the car with, at equivalent car speeds. Somehow the speed at which
that force was pushing needed to be included in the equation.
Therefore we have POWER. Power is essentially a measure of FORCE x VELOCITY. Which for some rotating can be rearranged to give torque x RPM. This
means power is not effected by gear ratios.
So now you can directly compared engines with different torques and different gearbox ratios. eg
Take a car (1) with 50HP @ 4000rpm (with say 4000rpm redline), using the power eqn you will find that engine will have 66 ft.lbs of torque.
Take another car (2) with 50HP @ 8000rpm (with say 8000rpm redline), using the power eqn you will find that engine will have 33 ft.lbs of torque.
So they have the same HP, but different torque. Which one is fastest?
If you have the same gear ratio's in both cars, car 1 will accelerate much faster but then reach it's RPM limit. car 2 will accelerate half
as fast, but will reach it's RPM limit at twice the speed of car 1. Obviously not a fair comparison.
So gear car 2 twice as low as car 1, so they both will reach their RPM limits at the same speed. So that since car 2 is geared twice as low, it's
engine torque would be doubled at the wheels compared to car 1. So now they have the same torque at the wheels, and will do the same speed at their
RPM limit. Obviously they will now have identical acceleration and speed.
So simply, torque doesn't really mean much for 1/4 mile times. If a car with high HP is geared correctly, it will always beat a car (of the same
weight) with lower HP but higher torque (obviously at lower RPM).
However if the high torque and high HP cars are on the same gearing, the torquey car will out accelerate the high HP car in the same gears.
From there it gets complicated. When people say the engine is torquey, or another engine is all revs no torque, they are refering to the entire power
curve, not just peak figures. And it is the entire curve that will effect everyday driving.
I'll use my engine options for the baja as an example. My two main options were the WRX engine (2.0lt turbo), and the KLZE (V6 2.5lt NA). Both
have almost identical peak torque figures, but the WRX makes it peak torque at slightly higher RPM, so it makes slightly more peak HP.
What does it mean to acceleration? If you select either motor in my acceleration calc program http://offroadvw.net/tech/calc.htm and set them to the same gearbox with high stall speeds, they will give about the same 0-100
and 1/4 mile acceleration.
However if you set the stall speed (take off RPM in first gear) to something low like 2000rpm, the KLZE is not effected because it has a really wide
torque band. The WRX engine however is significantly slower when driven like this, because at 2000rpm it is well below it's torque band and will
be slow until the turbo kicks in at higher RPM. Basically with the V6 you can flattened it in any gear at any RPM and it will go. If the flatten the
WRX at low rpm not much will happen until the boost comes on, so it you want to go you'd have to drop it back a gear to get it in it's peak
torque area.
SO you would describe the V6 as a torquey motor (due to it's wide torque band) and the WRX engine as a revvy HP motor (due to it's lack of
low RPM torque) even though they both have almost identical peak torque and peak HP figures.
[Edited on 1-5-2003 by Baja Wes]
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slowbug
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| posted on May 2nd, 2003 at 09:45 AM |
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:thumb thanks for that detailed explanation Wes....thats excellent.
I am going to print that out......so if anyone ever asks me, i can show them
Cheers
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Baja Wes
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| posted on May 2nd, 2003 at 10:20 AM |
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I should probably post it on my website as a technical article because it is a common question.
Glad to help
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aussiebug
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| posted on May 2nd, 2003 at 04:37 PM |
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Good post Wes.
Except that the true conversion is
Power (hp) = Torque (ft-lb) x RPM / 5252
(you left off the /5252 bit).
I like to look at it like this.
Torque is the twisting action on the part you are measuring, and for our cars, the relevant part is the rear axle.
The harder you can twist that axle, the faster the car will accelerate.
But all engines have a preferred rpm range, and so we need a gearbox to match the engine's requirements to the car which will be moving from zero
to xxxkmh.
The early buses were pushed along quite happily (well - sort of) by a 36hp engine. Now a heavy bus needs a lot of torque at the axles to accelerate
it, and the only way to get enough torque was to have very low gearing - the engine is revving it's head off for very little road speed. The
early buses all have reduction hubs to accomplish this, so no matter what gear you were in, the engine would be spinning at about 1.2 times the
equivalent engine in a lighter bug.
Higher rpm means (in general) more hp, so the gearing translated the high rpm (high horsepower) into useful torque at the wheels.
But as Wes says, that kind of gearing means the engine will run out of puff at lowish speeds - in fact the early buses were hard pressed to find
90kmh.
Of to think of it in just beetle terms - lets say you have the same 1600 engine in two cars of the same weight, but one has the 4.375 final drive
(1200/1300 gearbox) and one has the 4.125 final drive (1500 gearbox). Everything else is the same.
The first car will accelerate faster than the second car, because the low gearing will allow more rpm (hp) at lower speeds - converting it to more
useful torque at the wheels; but the 1st car will have a lower top speed than the second car, because the engine will reach it's maximum rpm at a
lower speed.
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Baja Wes
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| posted on May 2nd, 2003 at 04:46 PM |
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| Quote: | Originally
posted by aussiebugExcept that the true conversion is
Power (hp) = Torque (ft-lb) x RPM / 5252
(you left off the /5252 bit).
|
Actually rob you will notice I used a proportional symbol ~ (well as close to the symbol as I could find on my keyboard) meaning there were other
factors in the equation. If you use imperial units you use the silly 5252 imperial conversion.
I normally use SI units, power in watts, torque in Nm, and motor speed in radians/second. In this form of the equation there is no factors, it is
simply watts = N.m x radians/sec.
You'll notice my examples, 50HP, 4000rpm, 66ft.lbs are correct. :thumb
[Edited on 2-5-2003 by Baja Wes]
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1972super
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| posted on May 2nd, 2003 at 04:55 PM |
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While on the subject of power, how many kilowatts did the original Super bug 60 hp 1600cc engine give out? |
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Bizarre
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| posted on May 2nd, 2003 at 05:20 PM |
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If you have a look under the Dyno Day posts for Sydney and Melbourne you will see a 1600 in fresh condition puts out about 40 ponies AT THE WHEELS.
So about 30 real life kilowatts
Futue te ipsum!!!
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1972super
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| posted on May 2nd, 2003 at 05:26 PM |
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Not bad, 30k/w's for a 800kg car. I'm yet to drive my Beetle and can't wait to feel how it drives. |
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jboy82
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| posted on May 3rd, 2003 at 01:54 PM |
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Hi,
Just as a very average guide(not quoting please).
Power "lost" from the engine to wheels usually is about 1/3 of the engine actual power(from the flywheel).
I have observed this while looking at many dyno figures.
This just the average and it varies with different cars especially 4WD like Subies.
So 40hp at the wheels is probably around 60hp "true" engine power. Just a guesstimate,
I'm not asking for this:vader> in the company of vw masters.
Josh
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The_Bronze.
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| posted on May 4th, 2003 at 02:37 AM |
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Interesting reading - even though I have had this explained to me before I enjoy rereading it to clarify.
A point of interest - My baja feels torquey when full loaded with tools, kids, camping gear and sand. Took it for a spin today with none of those
things including the loss of linings, seats (Excluding the drivers) and it felt really racie - even for a baja. It hook into the corners, spun the
tyres on grass and dug in on acceleration - so weigh has a lot to do with it all.
[size=4]Bronze.[/size]
Rathmines, Lake Macquarie, NSW.
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Che Castro
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| posted on May 4th, 2003 at 01:11 PM |
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hehe good post wes... u set em straight
A 4wd drivetrain will lose about 1/3 a transaxle (where the engine and box are on the same side) will lose around 1/5.
my 1600sp got 39hp at the wheels @ ~4900 rpm or so.
the torque curve of the engine is also an indicator of how full the cylinders are getting as it is a measurement of force at an instant
Jon
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