# gear ratios

• Jan 4th 2010, 12:44 AM
rcanen
gear ratios
I was hoping somebody could help me. I'm trying to get a cordless drill (18volt) to turn a 1500 lbs round hay bale. I have a 2 inch gear (gear a) diameter sprocket which the cordless drill attaches to. I learned that the drill produces about 480 inch/lbs of torque. A roller chain connects gear a to my large diameter sprocket (gear b). Gear b is attached to the round bale end. The bale is six feet in diameter and would weigh between 1000 and 1500 lbs. I have thought about using a 32 inch diameter sprocket for gear b which i believe gives me a 16 to 1 gear ratio. Is this set up sufficiently enough to produce enough (force) to turn this bale on the ground? I also would like it to have enough torque to turn this round bale up a short ramp onto a trailer (10-20 degree slope). What diameter for gear b would you suggest? I'am not concerned about speed but if i get the correct sprocket i'll get the best of both worlds.
Thank you........
• Jan 8th 2010, 02:43 PM
Kiwi_Dave
Usually as mathematicians when we see a problem like this we assume some things are ideal until the problem is simple enough that we can solve it.

For example in this problem we might assume that the bale is perfectly cylindrical, that there is no friction that the ground is perfectly flat etc. The trouble with this problem is that if make all of those simplifications we find that you can use any gear ratio you like and the bale will move (speed may vary), yet we know that in practice this is not true.

Probably the one simplification that we really cannot make is that the bale is circular. Instead it will have a flat side on the ground and this means that as it starts to turn it will tend to rise going over the corner between the flat "side" and the circular circumference. It will also deform so that the flat side moves around the circumference and this will consume energy.

Now I think this is a practical problem that you actually want to build, so I suggest you need to find a way of measuring (in the paddock!) the torque required to turn the bale. Then you can easily find the required gearing ratio.

With a strong arm somehow attached to your male and a spring balance (perhaps fisherman's scale?) you should be able to find a way to calculate the torque.

You will probably find that the torque required to start the bale turning will be considerably bigger than the torque required to keep it turning. And of course you may find that the cordless drill simply is not powerful enough.
• Jan 16th 2010, 04:38 PM
rcanen
gear ratio part 2
Thank you Kiwi Dave for posting your reply to my question. I did what you suggested however my spring scale maxed out at 220 pounds of force. It was close to rolling but never did. Can we make an assumption that 400 pounds would do it then design a gear ratio for that assumption. When I convert 480 inch/lbs to foot/lbs I get 40. If I use four gears, a two inch (gear a) to a 10" (gear b). Gear B is welded to 2inch (gear c) which is chained to a 18" (gear D). This sequence gives me a 225 to 1 ratio doesn't it? Would a 225 to 1 gear ratio be enough to amplify the orginal force of 40 foot/lbs?

To answer your previous question the bales are basically cyclindrical. When they are first made they are very round. After they sit around they do flatten out. I would like to build this so I can roll them up a short ramp onto a trailer. I also learned that a 24volt drill produces 50 ft/lbs of torque. Thank you Kiwi Dave..........

rcanen
• Jan 17th 2010, 02:58 AM
Kiwi_Dave
The first gear pair has a ratio of 2:10 and the second pair is 2:18 so you have a ratio of 5*9=45.

Your drill develops 480 inch-pounds or 40 foot-pounds of torque so you can expect up to 40*45=1800 foot-pounds of torque to be applied to your bale.

You did not say how long your test lever arm was. But so long as
400 pound * lever arm length (in feet) is less than 1800 then you should be OK.