Force calculation on handle turning a shaft

[coreytroy]

I have a manual brake winch with a 320mm handle that simply turns a shaft to operate the winch, either up or down.
The specifications state that 19kg of hand force is required to operate the winch(assume it refers to max load applied).

If i was to remove the handle and use an impact wrench instead, what sort of forces would be required to turn the nut/shaft to the same force as the 19kg handle ?

Is there a simple calulcated formula i can use for finding the result ?

 

[MintJulep]

No, there is not. Because kg is not a unit of force.

However, if we guess that "we know what you mean", then yes, there is.

 

[coreytroy]

If i was to locate the handlle into the horizontal position, and place a 19kg weight on the handle, that would not equate to a force of 19kg ?

 

[KENAT]

Mint's point is that kg is a unit of mass.

The standard metric unit of force is the Newton - though I've seen other people use kg force analogous to lbf (and don't get me started on slugs).

Assuming you're on earth with a crude approximation of gravity as 10 m/s^2 then you probably have about 190N of hand force.

You need to worry about equivalent torque/moment.

http://en.wikipedia.org/wiki/Torque

Use force times distance (I suggest you convert the distance to meters) to get the torque you need to match. Plus if your 'handle' is actually a wheel you need to use the radius not diameter.

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[mikekilroy]

Once you get past the purest symantics discussion, your answer is

T=F*D
so in your units T = 19kg * (320mm/2) = 3,040Kg-mm

In case your impact wrench is not calibrated in Kg-mm units, simply convert with (1kg-force=2.2 pound-force) & 1" distance =25.4mm distance

So your impact wrench will need to put out 263 #-in torque.

The internet is full of folks who call weight kg force units, including lots of conversion tools. plenty, such as

http://wiki.answers.com/Q/If_you_weigh_120_pounds_how_much_do_you_weigh_in_killograms

, even spell it wrong but get the idea across just fine.

 

[coreytroy]

Thanks for the reply.
Still trying to get my head aropund this.

The wrench say's it is good for 625 ft/lbs.
So would that be:

(3.04 x 2.2)=6.69lb per 25.4x12=304.8ft :: 2,039.12lb/ft, making the wrench less that 1.3 of the required power ?

 

[KENAT]

Well Mike, one might hope an 'Engineering Professional' could have worked it out from what I gave them. However, presuming your assumptions are correct then I'm sure the OP will be grateful enough to give you a star.

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[dvd]

That 320mm probably represents a handle swing-radius and not a diameter, so dividing by 2 is possibly an error. Torque = r x F.

 

[coreytroy]

That 320mm probably represents a handle swing-radius and not a diameter, so dividing by 2 is possibly an error. Torque = r x F.

Correct.

 

[DRWeig]

Kenat, I second that motion.

Coreytroy, please read mikekilroy's post again carefully, one line at a time.

Best to you,

Goober Dave

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[coreytroy]

Thank you.

 

[dvd]

Here is a nice conversion program. Probably would help you. Always write out your units.

http://joshmadison.com/convert-for-windows/

 

[mikekilroy]

Coreytroy, I made the assumption that your 320mm was a diameter, not a radius, and from further posts here it appears I may be wrong; of course if your plan is to REMOVE the "320mm" handle and use an impact wrench, then YOU could use a ruler and measure the distance from center of shaft to where that 19kg would be hung yourself........no need for you you to assume anything.....

If your impact wrench says it is good for a max output torque of 625 ft/lbs, then I respectfully suggest it is probably only good for 90-100 lbs-ft torque. My reason for THIS assumption is that if it indeed reads ft/lbs then it is probably a low cost chinese import since torque units are not lbs/ft but rather lbs * ft as you saw from the Torque equation = Force * distance, and I have NEVER seen one of those cheap chinese rated 625 good for more than 100 #-ft!

So if you use the assumed real output of your impact wrench of 100#-ft, that 100 #-ft x (12inches/ft) you see you still have 1200 #-in torque available so you can easily rip that shaft off the winch and destroy it very fast. This may not be such a good idea at all....

 

[DRWeig]

Mike,

If you get your impact wrench at Harbor Freight, it will give the full rated torque for about one hour before it begins its precipitous drop to about what you said.

Their Earthquake 1/2" is rated 700 ft-lb. When it drops off to 100, it's still good for rotating your tires and it stays there a long time. That is, it stays there a long time unless you let your teenage son use it to low-ride his 300Z. He'll kill it on those suspension parts.

Torque-limiting extensions seem to be fairly consistent for that job. Always double-check with a clicker or bending-beam.


Best to you,

Goober Dave

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[MintJulep]

Mike,

So suddenly correct units become important?

 

[mikekilroy]

as a clue to solve a puzzle.....