Pounds/foot (and other nonsense) MEW 226

The confusion arises because there is no easy way to distinguish that with torque the force is acting at right angles (or tangential if you prefer) to the distance measured, while with work the force is acting through the distance measured. So with torque, the multiplication should be a vector multiplication, not a scalar multiplication.

If that is too confusing, think of it like this...if you pull on your spanner lengthways, you will not undo the nut. You have to pull sideways.

On the other hand, if you pull sideways on a (miniature) train carriage, nothing will happen, but if you pull along the track it will move and you will have done work.

So whether we use pounds-foot or foot-pounds for torque, (or SI units) we really need a symbol in there that means that the force is acting at right angles to the moment arm. Otherwise it will always be ambiguous.

It gets even worse if you are looking at radio control servos, where they habitually leave out the distance and just talk about a servo being so many kg. (Yes, they use kg, not Newtons....)

On the motor thing, note that Jasons rule of thumb is fine for DC motors but does not apply to Induction motors.

John

I just read the bit that Capstan wrote "force in a radial direction aka torque" which leaves me starting to wonder if centrifugal force might be sneaking a look in as well perhaps?

Mark

Hey Les, I just read your post as well. I hold my hands up, you are right - it would have to be a square metre or even a randomly shaped one but it would defiantly need some area. It would however not have much pressure compared to most things in daily life. Now, if you could make a balloon big enough to trap lots of hot air you would have plenty of pressure if you got collecting in the right places....

Mark

John,

Perhaps we could call it Theta? 

Mark

PS. or even cos theta 

Edited By Mark C on 03/03/2015 00:34:32

Earlier in the thread, Harold, I was with you all the way but not this.

Power = work done per second

Work = force x distance moved through (in the direction of the force incidentally)

By convention (at least amongst Engineers), lb(f).ft is used when referring to a torque, ft.lb(f) when referring to work (or energy) and ft.lb(f)/sec when referring to power.

Torque and work have the same dimensions, yes, but they are not the same thing (not even close) so it is useful to express the units differently to avoid confusion .... I think we agree on this.

(with the "f" written as a subscript which I can't figure how to do in this editor).

It's pretty much implied by the use of lb(f) as the force unit rather than poundals.

> The calculations show the required power ..at what rpm is not considered. ..

Oh yes it was,! How do you think I calculated speed? I estimated the no-load speed at 3000 rpm and used 2000rpm and 60% efficiency for the calculations, knowing I needed a margin for going uphill.

Neil

I saw sin theta suggested, which as theta is 90 degrees equals 1...

So next time you calculate torque, don't forget to multiply by 1 to get the right answer

It wasn't there Jason, I just took one corner of the spreadsheet as the whole thing is a mess of gear ratios and centre distances and scale factors etc.

The most surprising thing was how close the actual performance was (subjectively) to the calculations, it was certainly in the right ball park.

Neil

.

Noting that one Newton [approximately 0.225 pounds force] is exerted by a small apple on planet Earth

... But Greengrocers understand the gravity of the situation

MichaelG.

Should there be an apostrophe in greengrocers ? To get even more confusion try reading a chinese manual ,for say a generator or tractor. Loads of unknown (to me ) units. And what did happen to the drill?

.

NO ... But I accept that they might expect one.

MichaelG.

Neil,

You didn't really mean to type "I saw sin theta suggested, which as theta is 90 degrees equals 1..." that did you.

You meant Cos really?

Mark

I bet nobody's taken wind speed and coastal drift into consideration!

Let alone the conservation of angular momentum.

JA

> You meant Cos really?

No, it refers to the angle between the force and the lever. If the force was pulling or pushing the lever it would be an angle of 0 or 180 degrees and therefore no torque and 0 in magnitude.

I imagine you are thinking of the angle the force makes with a tangent of the circle, so we are at cross purposes, or at least 90 degrees out of phase with each other.

Neil

Nope, I reckon if you are applying torque by definition it is a tangential force (it always acts perpendicular to the lever arm) but you might not pull at exactly a right angle so some of the effort gets lost. You then need to think of the force acting on a point on the lever arm and then you need some idea of direction but the direction is relative to the lever arm and hence rotated 90 degrees in a sense. But if it is not torque, then it is a force acting on a body and you then need to know the direction of the force which would be cos theta.

I suppose it depends where on the curve you start!

Mark

PS, being in the northern hemisphere, do we need to increase or reduce the torque load on threaded components depending on whether the torque wrench is turning clockwise or anticlockwise? And does that mean Australian machines fall apart easier 'cos the bolts aren't tight enough?

Edited By Mark C on 03/03/2015 19:41:20

That has more to do with them being Australian....

And you reckon you have trouble with sparkplugs in your IC engines oiling up, up there. Should try it down here where the plug is at the bottom of the cylinder.