not an engineer but i think this is right....please confirm.
th88,
Since you are desperate enough to post the same post twice, despite being given the complete information the first time, I'll give Kelly a rest and have a go at it. Do you know the difference between force and torque? Do you know how force creates torque? if so, you can calculate the torque on the 20" wheel from the 1500# load. Can you do that? If you can, then just apply the ratio to determine the torque on the small gear.
The 1500# load is on the cable drums 20 inch dia.Originally Posted by jboggs
To lift load, an greater than 1500# must be exerted on the opposite side of the cable drum.
If the 2 inch diameter pinion drive dear is 10:1 ratio to the cable drum, a force slightly greater than 150# will lift the load
Since the pinion lever arm radius is 1 inch, a torque slight greater than 150 in/lb will move the load.
Right????
The 1500# load is on the cable drums 20 inch dia.
To lift load, an greater than 1500# must be exerted on the opposite side of the cable drum.
If the 2 inch diameter pinion drive dear is 10:1 ratio to the cable drum, a force slightly greater than 150# will lift the load
Since the pinion lever arm radius is 1 inch, a torque slight greater than 150 in/lb will move the load.
Right????
Not quite....
The gear ratio is equal to 20/2 or 10:1. Output divided by input - the small gear is driving the large gear to lift the 1500 lb load..
Gear ratio = diameter out/diameter in
10 = diameter out/diameter in
and
Torque out = lbs x (radius in)
Torque out = 1500 lbs x 10 in = 15,000 lb-in
Again:
(torque out)/(torque in) = 10
Using algebra
Therefore:
15,000 lb-in/10 = torque in
1,500 lb-in = torque in (small gear)
Last edited by Kelly_Bramble; 11-15-2015 at 11:55 AM.
Tell me and I forget. Teach me and I remember. Involve me and I learn.
Don't forget that the torque and loading is actually applied at the gear pitch diameter.
Tell me and I forget. Teach me and I remember. Involve me and I learn.
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