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Torque calculation(4)

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. 

If I understand correctly, your "torque multiplier" will multiply the input torque by 1. Unless the multiplier is somehow hidden inside that "bulge" in the main shaft, that is. Three gears of the same size will not multiply torque at all. handleman, CSWP (The new, easy test) 

The arrows in that drawing point to the little shaft as being the torque multiplier. I don't see how unless there's something clever inside it. My worry is that this is a "Proprietary and Confidential" drawing published on a public web site.  Steve 

Hello handleman,
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 cutout 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. 

You math is wrong. You need a 1:5 ratio, not a 5" radius. Multiplying 2400 ftlbs by 5" gives you nonsense, ftlbsin. Ted 

minerk (Mechanical) 
4 Nov 08 8:29 
The unit as shown has a total reduction of 125:1, entirely due to the torque multiplier. As previously pointed out, gears of the same diameter do not change torque. To get the required input torque divide the desired output torque by the total gear ratio (125). 

I thought that length from the central axis of the drive shaft, to the 1st driving point, must have some bearing on the output?
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. 

No. If all three of your gears are the same, they do not multiply the torque at all. Carry your units through each calculation and you will see that you come up with nonsense. If this is not your field, what is, as you list yourself as Mechanical? handleman, CSWP (The new, easy test) 

rb1957 (Aerospace) 
4 Nov 08 9:21 
"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." ... what's wrong with 12000/125 = 96 ?
"... = 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" 

Torque=Force X radius of the Gear
For example,
You apply 2400 ftlb of torque
Force = 2400 ftlb X 12"/5"= 5760 lb
You need 12000 lbft
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 ftlb to 96 ftlb


re: handleman,
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. 

From print :"Torque multiplier is used to drive shaft". It is asking what input torque is needed to a 1:125 multiplier on the input shaft to apply 12000 ft lbs to the threaded faces.
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.


minerk (Mechanical) 
4 Nov 08 12:27 
Torque is Force (lb) x distance (ft). Where you're getting confused is not understanding that the torque in the shaft/sprocket is the same everywhere in the shaft/sprocket. The tangential force is different depending on the perpendicular distance to the axis of rotation. Here's how the calculation actually should look doing it the way you're trying:
12000 ftlb x 12 in/ft = 144000 inlb. Using a gear with a 5" radius, the tangential FORCE at the rim of the gear is 144000 inlb / 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 lnlb.
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 ftlbs / 125 = 96 ft lbs input. 

You need to apply 96 ftlbs at the input to get 12000 ftlbs at the output, to answer your question. See minerk's conclusion above. Ted 

But where does the 1:125 come from? There is nothing on the drawing that suggests any sort of torque multiplier.
I find it interesting that there are three 5inch radius gears, and that 5 x 5 x 5 = 125. 

minerk (Mechanical) 
4 Nov 08 13:10 
Taking another look at the drawing I see a notation that the input shaft is actually DRIVEN by a torque multiplier. 

rb1957 (Aerospace) 
4 Nov 08 13:10 
yeah, but you need a 5:1 ratio at each gear ... maybe they're double gear, with a 5" rad gear meshing with a 1" rad gear 

Is this similar to the torque multiplier? If so then this finally makes sense. http://www.geartronics.com/x4page2.htmIf it's a 125:1 ratio, you need to input 96 ftlbs 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) 

I have to eat a little crow as a 125:1 torque multiplier is readily available from Norbar. 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#standardIf 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. 

grumpyblade, are you trying to create a torque multiplier? Or are you arranging gears to drive a prepackaged torque multiplier? Ted 

Seems to me that everyone is overthinking the problem. The answer to the specific question asked on the drawing is "You need to apply 12000 ftlb at that point to achieve 12000 ftlb at the output." Let's not forget that this is a question given to the OP by his management. TTFN
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I would like to thank all contributors to this thread, for their assistance, I hpoe I didn't get anyone too confused, as unclesyd says torque multipliers are readily available, the one we have is quite an impressive piece of kit.




