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Hand Cranking for a Assembly fixture rotation 3

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Gopalakrishnan C

Mechanical
Jul 17, 2019
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Hi,

I need a hand cranking methods to rotate by 1600NM. I was searching in Boston gears but unable to find out.
The fixture rotation rpm is 0.5.
Thanks!
 
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I think you want a speed reduction gear-train … 80Nm at 10rpm would drive 1600Nm at 0.5rpm

another day in paradise, or is paradise one day closer ?
 
Yeah - the HP continuous output of a pair of human's legs for someone in good shape is about 0.1Hp. Hand cranking will work for about 10-30 seconds or so for anyone not on a sailing team.

They make gear stages so just add them together if the input force is too high, but each adds friction and eventually that goes south. Lowering the input force means increasing the input rate, so you get a different set of metabolic problems.
 
If the stab is turning 1/2 RPM then things have totally gone south. And selecting from Boston Gear on an aircraft application is the wrong place to begin.
 
Wait - you have a random design which seems impractical based on power availability from any normal person and you would like me to go and find documentation for a random some mechanical thing? To say what about it, exactly?
 
1600 newton-meters (torque) at 0.5 RPM.

When you see torque and RPM, you should instinctively calculate the associated power.

Wolfram Alpha says that's 0.112 hp.

If it's continuous, then you'll need one of these (offered as a visual aid to assist with clarifying thoughts):

5687252.jpg


If it's just to adjust a fixture once in a while, then either back away from the 0.5 RPM requirement, or add an electric motor.
 
Untitled_glgqwc.jpg

Sorry for the in-proper information!
Actually i need to rotate a heavy loaded fixture overhanging on an axis. And this one needs to be rotated by hand crank method.
I calculated the maximum torque 1600 NM.
And it is not a continuous rotation.
I searched and found the Manual crank (Haacon Type 242.3).
 
I think you're misleading us with "output rpm = 0.5", when you add "it is not a continuous rotation" (which is implied by stating an output rpm).

I think you should say you have to rotate the mass through 180deg? in 15sec?

Sketches still very simplistic. If you think you've found a handle to work, why ask us ?

another day in paradise, or is paradise one day closer ?
 
Ideally, a fixture would be designed so that the axis of rotation is in-line with the system's center of mass. Then the item could be easily rotated with a fingertip.

If that's impossible here, then it's a lifting exercise. The rotation is essentially incidental. So one should be searching for a winch assembly. Or a jack.

...And redundant straps in case someone bumps into the rachet pawl.
 
@SAITAETGrad
This is not for a School project!
I am designing a tool to rotate and position the heavy equipment
I am finding difficulties to calculate the torque for eccentric load of 680kg.
The axis of rotation is horizontal and the distance between Center of mass and Center of rotation is .25m

I am using τ=Iα
Here is the confusion, when i calculate Alpha:
I am using manual crank and there will not be any continues rotation. I just want to rotate and position the equipment. In this case when i change the rotational speed(n) my output torque value is drastically varying.

If any one can give me the clarity that how to calculate the Speed of rotation for manual crank? Or we simply take 60RPM for Torque calculation?

Thanks!
 
A manual crank back at the axis of "rotation" is not the correct approach.

By the time you have the gearing ratio to lift 680kg, upon release it'll spin at "a million" RPM killing someone.

 
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