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

Not enough information. Torque at what RPM?
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Gear ratios and friction etc. Brian
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On 20/03/2016 21:55, Vir Campestris wrote:

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!
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Insufficient data to calculate. By miles. Sorry, metres.
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On Sun, 20 Mar 2016 21:55:48 +0000, Vir Campestris wrote:

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. :-)
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On 21/03/2016 03:29, Johnny B Good wrote:

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!
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On 21/03/16 10:24, Robin wrote:

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

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?
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On 21/03/16 18:23, GB wrote:

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

Easy, 5700 degrees. What was the real question?
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On 21/03/2016 03:29, Johnny B Good wrote:

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
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On 24/03/16 21:22, Vir Campestris wrote:

800000 nm per minute :-)
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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.
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On 25/03/2016 11:14, Dave Plowman (News) wrote:

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

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?
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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.
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On 26/03/2016 15:12, Dave Plowman (News) wrote:

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 http://www.gbclassiccars.co.uk/rover_p6_3500.html 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?
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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.
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