## Torque calculation

## Torque calculation

(OP)

I am having a problem with clarifying the following question. I need to know what torque to apply at the shaft to acheive the required torque at the faces indicated, see attached sketch.

The torque multiplier is an enclosed arrangement of gears, with an input socket (to take a torque wrench), and output sprocket,(which fits the drive shaft), designed so you can set your torque wrench to 125th of the final output.

I don't know whether I am right but if I apply 19.2 ft/lb x 125 = 2400 x 5" (radius of gear) = 12000 ft/lb.

The torque multiplier is an enclosed arrangement of gears, with an input socket (to take a torque wrench), and output sprocket,(which fits the drive shaft), designed so you can set your torque wrench to 125th of the final output.

I don't know whether I am right but if I apply 19.2 ft/lb x 125 = 2400 x 5" (radius of gear) = 12000 ft/lb.

## RE: Torque calculation

-handleman, CSWP (The new, easy test)

## RE: Torque calculation

My worry is that this is a "Proprietary and Confidential" drawing published on a public web site.

- Steve

## RE: Torque calculation

No, the torque multiplier is an enclosed arrangemant of gears that multiplies the input 125 times, a standard torque wrench fits into one side and a sprocket fits into the drive shaft @ the square cut-out in sketch.

I thought that the radius of the 1st gear on the drive shaft(in this case 5") would increase the amount of torque applied.

## RE: Torque calculation

Multiplying 2400 ft-lbs by 5" gives you nonsense, ft-lbs-in.

Ted

## RE: Torque calculation

## RE: Torque calculation

My calculation, as per comment by hydtool, was based on 12000ft/Lb x 12(into inches) = 144000in/Lb divided by 125(torque multiplier) = 1152 divide by 5 (length from axis) = 230.4 divided by 12 (back to feet) = 19.2ft/Lb.

This is not really my field, I am just guessing, trying to look at it logically, rather than with any real expertise.

## RE: Torque calculation

-handleman, CSWP (The new, easy test)

## RE: Torque calculation

"... = 1152 divide by 5 (length from axis) = 230.4" ... isn't this calc determining the force on an arm of 5" to produce the torque (of 1152in.lbs) ?

and being picky, "ft/lb" looks like "ft per lbs" rather than "ft times lbs" ... i'd suggest either "ft.lbs" or "ft*lbs"

## RE: Torque calculation

For example,

You apply 2400 ft-lb of torque

Force = 2400 ft-lb X 12"/5"= 5760 lb

You need 12000 lb-ft

The first Gear is 5" in radius

the second Gear you need 8"

The third Gear you need 15.625"

Or you can increase the torque from 19.2 ft-lb to 96 ft-lb

## RE: Torque calculation

My field is CAD/CAM, I was asked to produce drawings and CNC programs for the parts, for a machine to torque up one componemt on to another, I was also asked if I coiuld find out what torque would need to be applied to the torque wrench, hence the question,

I joined this forum for that reason, and as the question was a mechanical one, I joined a mechanical forum, of course.

## RE: Torque calculation

Wouldn't the gears cancel out any multiplication factor which would allow a 96 ft lbs input to the torque multiplier to apply 12000 ft lbs to the faces as stated above.

One thing that bothers me is where did they get a 1:125 multiplier. The reaction arm would be awful.

## RE: Torque calculation

12000 ft-lb x 12 in/ft = 144000 in-lb. Using a gear with a 5" radius, the tangential FORCE at the rim of the gear is 144000 in-lb / 5" = 28800 lb.

This tangential force is what drives the next gear in the train. So the torque in the next shaft, using another gear with 5" rad. is 28800 lb x 5" = 144000 ln-lb.

Do that again for the next gear and you'll find the same result. The only way to change the torque in the shafts is to change the diameter of the gears. Since we already knew all this without going through the calculation we suggested the shortcut of dividing your output torque by your total gear ratio (125). 12000 ft-lbs / 125 = 96 ft lbs input.

## RE: Torque calculation

Ted

## RE: Torque calculation

I find it interesting that there are three 5-inch radius gears, and that 5 x 5 x 5 = 125.

## RE: Torque calculation

## RE: Torque calculation

## RE: Torque calculation

http://www.geartronics.com/x4page2.htm

If it's a 125:1 ratio, you need to input 96 ft-lbs of torque. And make sure that the "anchored" or "braced" part of the torque multiplier has a very secure and strong seat somewhere on the machine.

-handleman, CSWP (The new, easy test)

## RE: Torque calculation

What is interesting is the size of the input sockets as small as 1/2". I also noted that they have an anti windup feature on the bigger multipliers. I happened to be around a 1:10 multiplier when it let go.

I was close about one thing as it has a nice size reaction arm.

http://www.ephtools.com/ht35000.html#standard

If you ever use any type of torque multiplier make sure the reaction arm is reacting against something substantial enough to restrain it. In another incident involving a multiplier I was on an investigating committee where two idiots tried to manually hold an arm reacting against 1600 lbs ft. The result was one broken leg on the guy that was turning the wrench.

## RE: Torque calculation

Ted

## RE: Torque calculation

Let's not forget that this is a question given to the OP by his management.

TTFN

FAQ731-376: Eng-Tips.com Forum Policies

## RE: Torque calculation