Grill crank calculation 1

[PEU]

Hi, I just finished a meat grill and it heavier than I expected, the grill is moved up/down by a crank.

I would like to know how to calculate the length of the crank so the force needed to move is not to much, for example I need my wife to be able to turn it.

Here is a schematic, the axle diameter is 15mm(5/8"):

The CAD drawing is here:

http://peu.net/temp/parrillaguido.jpg

Thanks a lot in advance!

Pablo

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I use SW2007, no need to upgrade yet

 

[Timelord]

Iqnoring friction:

37.5 lb/2 = 18.75 lbs tension each cable
18.75 x 5/8 = 11.72 in-lbs torque required
11.72 / X = lbs force on handle
i.e. If handle is 8 inches long the force to turn it is:
11.72/8 = 1.46 lbs

Timelord

 

[hydtools]

Do not divide by 2. The total load of 37.5 lbs. must be lifted. Both cables pull on the shaft.

Ted

 

[hydtools]

Divide the diameter of the 5/8 shaft by 2 and use the radius for the weight lever arm.

Ted

 

[Timelord]

Since I had to divide by two for the radius and multiply for the two cables, I left out the canceling twos. I didn't want to complicate things for the OP. I stand by my answer.

Timelord

 

[FeX32]

1.46 lbs. I think my girlfriend could lift that.





Fe

 

[PEU]

Thanks!
I came to the same conclusion as Timelord but via the empirical route, welded a 20 inches bar to the shaft and began to crank it, it became comfortable around 10 inches.

Pablo

---
I use SW2007, no need to upgrade yet

 

[FeX32]

Have you thought about putting a safety on that rod. I think if you have it rolled all the way up and then it is accidentally let go the lever will have enough energy to break your arm.

Fe

 

[dvd]

Good luck with that cable wound around a 5/8" diameter rod. And in a heated space as well. You might want to look into what the minimum radius is for your particular diameter of cable. Also might want to consider how to attach to the shaft.