The physics of cars - a question sequence.

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)

Reply to
Vir Campestris
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Not enough information. Torque at what RPM?

Reply to
newshound

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!

Reply to
Roger Mills

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

Reply to
Dave Plowman (News)

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. :-)

Reply to
Johnny B Good

Gear ratios and friction etc. Brian

Reply to
Brian Gaff

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!
Reply to
Robin

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?

Reply to
The Natural Philosopher

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?

Reply to
GB

Almost certainly:)

Reply to
The Natural Philosopher

Easy, 5700 degrees. What was the real question?

Reply to
dennis

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

Reply to
Vir Campestris

800000 nm per minute :-)
Reply to
The Natural Philosopher

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.

Reply to
Dave Plowman (News)

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.
Reply to
Roger Mills

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

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

Reply to
Dave Plowman (News)

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.

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

Reply to
Roger Mills

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

Reply to
Vir Campestris

About 210 lb-ft

Reply to
Roger Mills

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

Reply to
Dave Plowman (News)

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