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torque req'd to climb over the bump

torque req'd to climb over the bump

torque req'd to climb over the bump

(OP)
Hi,

I need to design a cart that carry 200 kips of load and the cart is only four wheels; a 2-wheeled Driver is used with one 2-wheeled Idler. The concrete floor is not perfect and the cart needs to climb over the rail (channel cut in the concrete, 2" wide, 1" deep). I would like to know how i can calculate torque require to go over the rail. The wheels are solid rubber.

RE: torque req'd to climb over the bump

Depends on the wheel diameter.  A 2" diameter wheel will never be able to get out of the channel no matter how much torque you throw at it.  A 2' diameter wheel will hardly even notice the bump.

RE: torque req'd to climb over the bump

(OP)
The wheel will be 16" diameter. thanks for the help. Right now the existing cart with 0.5 hp had a hard time to go over the rail.

RE: torque req'd to climb over the bump

You need a bit more than 50 inch-kip at each wheel that is stuck in the rail, assuming the 200 kip load is evenly distributed on the cart.  Of course, you'll have to account for losses in your gearbox, etc.  

For a static situation (cart has no momentum) the required torque at the wheel is the same no matter what the diameter is.  However, the amount of work required to roll out of the rail is a function of the wheel diameter.  The smaller the wheel the further down into the rail it falls, the more energy (work) required to get it out.

RE: torque req'd to climb over the bump

(OP)
Hi handleman

Can you show me how you get the 50 kips-in to get out the rail and also the equation need to calculate work to get out from the rail? Thanks.

Dennis

RE: torque req'd to climb over the bump

200 kips = 100 tons

the work done is 400 kip.in (raising the cart up over the channel), i guess you're rasing only 1/2 the cart at a time, so lets go with 200 kip.in.

picture the 16" dia wheel up against the 2" step, draw in the contact radius.  as the wheel climbs over the step this contact radius moves to the vertical; it moves through an angle of acos(14/16) = 0.5rad.

the work done by torque is T*theta ... 100 = T*0.5,
torque is 200 kip.in,
on an 8" radius, the contact force would be 25 kip.

any help ?

RE: torque req'd to climb over the bump

rb1957,
I believe we're visualizing this differently.  Here's a sketch of how I interpret what's going on.  Dennis, please correct if wrong.



RE: torque req'd to climb over the bump

(OP)
thx handleman

it is much more clear with the picture. Now the Force is known so i can figure out the torque needed depends on the size of the wheel. Thanks.

Dennis

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