Pounds - metric

Outback bloke · Post by **Outback bloke** » Fri May 12, 2006 8:30 pm

How does 1027lbs of torque convert to metric?

AndrewT · Post by **AndrewT** » Fri May 12, 2006 8:36 pm

1 pounds = 4.07668492 × 10^16 newton meters
1 027 pounds = 4.18675541 × 10^19 newton meters

Outback bloke · Post by **Outback bloke** » Fri May 12, 2006 8:46 pm

http://www.unitconversion.org/unit_converter/force.html

1 pound-force = 0.45359237 kilogram-force is what I have found.

Ben · Post by **Ben** » Fri May 12, 2006 9:36 pm

Brett,

torque is the measurment of force applied at a distance measured from the application point.

1027 pounds of torque is not a real measurement, it should be over a length - usually feet for pounds and metres for kilograms.

1 pound = .454kg

1 pound at one foot distance from the fulcrum (lever point) is .454kg at one foot, but is .454/3.281 (feet per metre) = 0.138 kilograms at one metre

hence to translate 1027 lb-ft is equal to 141.7 kg-m

Kilogram is a unit of weight as a mass applies on earthsubject to gravity. All objects accelerate towards the centre of the earth at 9.81m per second. If you don't account for the acceleration of gravity as it applies to a mass on a body then you multiply the mass by the value of acceleration - hence 1kg is 9.81 newtons. Therefore, to convert 1 kg-m to n-m (nm) you just multiply it by 9.81 and so 1027 lb-ft = 141.7 kg-m = 1390 nm

Interesingly, if you put a 1 metre bar (on a nut say) and then hang your full body weight off it (I am 80kg so will work on that) I am applying 80kgm force to that nut - that is 785nm torque. The engine in my wagon can develop only 181nm torque maximum. In theory, a weight of 19kg hung off a 1 metre bar is sufficient to stop the engine dead. Unfortunately that doesn't take into account momentum, inertia, etc, etc and is actually why we have flywheels - that's just to keep the bloody thing running!!!

explain things better?

Outback bloke · Post by **Outback bloke** » Fri May 12, 2006 11:13 pm

I thought that torque is measured over distance myself until I was at the dyno today. I was told that "the car has 850 pounds of torque and that is not foot pounds". I was then told that due to the gearing of the big tyres that it would be about 1027 pounds on standard rubber.

That kind of confused me but I thought it must have been some different way that things are measured on dynos.

Ben · Post by **Ben** » Sat May 13, 2006 5:25 am

BYB-01 wrote:I was told that "the car has 850 pounds of torque and that is not foot pounds". I was then told that due to the gearing of the big tyres that it would be about 1027 pounds on standard rubber.

I don't know the torque relationship as it acts on a dyno and 850 ft-lb is quite possible, especially in lower gears (was it a 3rd gear run?)

Maybe he was talking foot-yards??? :P

Outback bloke · Post by **Outback bloke** » Sat May 13, 2006 10:31 am

4th gear. I think I will drop round there next week and ask what it is in Nm.

Wilbur · Post by **Wilbur** » Sat May 13, 2006 11:05 am

Torque is force by distance. So would the distance be a rotation of your tyres ?

I just rang him and he said they don't know the conversion as the dyno software does it. Ask him for the Kw print out and not the HP one he has obviously given you.

Ben · Post by **Ben** » Sat May 13, 2006 3:57 pm

The distance is the radius of the tyres, but that is just the torque applied to the drive shaft...

Ben · Post by **Ben** » Fri May 19, 2006 8:01 pm

BYB-01 wrote:4th gear. I think I will drop round there next week and ask what it is in Nm.

What did you find out?

Sti_Brumby · Post by **Sti_Brumby** » Sat May 20, 2006 8:32 am

http://www.yousendit.com/transfer.php?a ... FC2C1317F2

Little program that help converts