The physics of cars - a question sequence.

It is, but part of the original definition of horsepower. 33000 ft.lb/min

I'm sticking with what I know and love as regards car engines. RPM, BHP and ft.lb. ;-)

I'd say it would be an odd torque curve to give that result. Falling off pretty steeply at peak BHP. But I'm only really guessing.

Reply to
Dave Plowman (News)
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Well, it was certainly true of the re-engineered Buick 3.5 litre V8 engine which Rover produced in the late 1960's.

This reference

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has it as 161 BHP at 5200 RPM, but my recollection is 160 at 5250.

Torque peaked much earlier - and was just over 200 lb.ft at around 3000 RPM - but power (the product of speed and torque) continued to rise - peaking at about 5250 rpm.

Note: At no time did I say that max torque coincided with max power! Did you think I did?

Reply to
Roger Mills

The BHP figure quoted in those days was gross. I'm not sure what that figure would be in today's money, but a lot less. The last carb engines in the SD1 - with much better heads and exhaust system - produced a quoted

155 BHP. Albeit on a lower CR.

Not after reading it carefully. ;-) It's just that I'd never come across curves giving the same torque figure and BHP figure. But I'd not claim to have seen all.

Reply to
Dave Plowman (News)

It only applies at that one speed - and only when using 'proper' units. Because BHP = RPM x lb.ft /5252 the power of *all* engines is numerically the same as the torque at 5252 RPM regardless of where they peak. [It's just that I remembered that the Rover V8 produced 160 lb.ft at 5250 RPM, and that was its peak power speed].

Reply to
Roger Mills

Ah. Right. I'm with you now. Sorry it took so long. ;-)

Reply to
Dave Plowman (News)

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

Andy

Reply to
Vir Campestris

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

Reply to
Vir Campestris

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.

Be better if you stuck to the same units.

You are the one doing the calcs.

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

Reply to
Dave Plowman (News)

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

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

Reply to
Vir Campestris

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.

Reply to
Roger Mills

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.

I don't need to be talked through it.

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.

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.

Think you need to revisit basic mechanics.

Reply to
Dave Plowman (News)

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.

Reply to
Dave Plowman (News)

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

HP= 2? Nt divided by 33,000

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

746 watts in a horsepower.
Reply to
harry

Well...

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

Thomas Prufer

Reply to
Thomas Prufer

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

Reply to
Dave Plowman (News)

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

Reply to
newshound

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 :-)

Reply to
newshound

+1
Reply to
newshound

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

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

Reply to
The Natural Philosopher

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

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.

Reply to
Dave Plowman (News)

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