PLEASE HELP! Lead Screw Ratios

[Ridgebotics]

Pretty close to 140 pounds as fast as possible. We're going to be using 2 cims

[lonnie lascelle]

With a trade off between speed and torque you will have to pick a starting point on your desired end result of time of actuation, needed torque, distance, force and then work backward from there. The number of starts is not enough information, You will have to determine how many turns per inch you have/ or will need, to get the desired speed and mechanical advantage. This works in conjunction with the mechanical advantage/reduction in the gear box etc. Without knowing what your mechanical application is it is hard to make any suggestions with out more deails and specifics on the application, there are many factors. Provide more details and a sketch or something and people may be able to help you more. If you have more specific questions post them too.

[Jibri Wright]

I honestly think a lead screw will be too slow. The manufacturer said running the lead screw we had with 60 pounds of wait any faster than they advised would cause the thread to bend. We didn't want to take the risk.

[NOV8R]

We've got a ACME 1/2 inch lead screw with a 1/4 inch lift per revolution hooked up to a mini CIM/ Banebot P-80 12:1 . In addition the output of the transmission is hooked to the lead screw via a 24 tooth to 12 tooth chain/sprocket drive. The whole gear reduction is 6:1. The setup works great and can lift 34 inches in just under 20 sec. We tried a 775 motor (AM 0915 gearbox) and it was way too slow and almost wouldn't lift at all.

[Jibri Wright]

Originally Posted by NOV8R

We've got a ACME 1/2 inch lead screw with a 1/4 inch lift per revolution hooked up to a mini CIM/ Banebot P-80 12:1 . In addition the output of the transmission is hooked to the lead screw via a 24 tooth to 12 tooth chain/sprocket drive. The whole gear reduction is 6:1. The setup works great and can lift 34 inches in just under 20 sec. We tried a 775 motor (AM 0915 gearbox) and it was way too slow and almost wouldn't lift at all.

We used 5/8 inch lead screw but it should still work. Whats the gear reduction of the gearbox did you use with the Bane Bots motor? Also i don't know if a mini-CIM with a 6:1 gear reduction will work if the Bane Bots motor with a large gear reduction didn't work. Then again, we didn't test the mini-CIMs for a lead screw, so I could be wrong.

[JesseK]

It's possible on a rudimentary level to figure out the torque requirements of a lead screw by figuring out the angle of the threads as if projected on a flat plane. I figured this out as we went along in 2011, so I don't know if it's the best way to do it:

e.g. There are 2" of travel in 1 turn, PI*0.625" of circumference rotation in 1 turn, thus Theta = Tan(2/(PI*0.625)) should give the angle of thread incline in radians. Sin(Theta) will then correlate to the amount of force due to gravity the motor must overcome just to move the load. You'll also want to add in a buffer for the losses within the threaded nut, thrust bearings and the gearbox. If I remember correctly, I did it as a ratio [sin(Theta)/(sin(Theta)+cos(Theta)) * weight of load] was the force at the tip of the sprocket on the threaded rod, and it seemed to be close to what we experimented with under heavy load.

Note that the '2" of travel in 1 turn' is the rating after the # of starts is taken into account.

As far as how fast an acme rod can spin under load -- if the ends have thrust bearings and the radial bearings are also mounted properly, then there's no real reason to not go right up to the rod's rated limits (we went beyond it in 2011, yet that was due to a very light load that year). One end's radial bearing should have a little play (ours was in a block mount that used a 1/4" bolts in a oversized 3/8" hole) so that when it gets to the end it doesn't seize/bend the rod.

[Jibri Wright]

Originally Posted by JesseK

It's possible on a rudimentary level to figure out the torque requirements of a lead screw by figuring out the angle of the threads as if projected on a flat plane. I figured this out as we went along in 2011, so I don't know if it's the best way to do it:

e.g. There are 2" of travel in 1 turn, PI*0.625" of circumference rotation in 1 turn, thus Theta = Tan(2/(PI*0.625)) should give the angle of thread incline in radians. Sin(Theta) will then correlate to the amount of force due to gravity the motor must overcome just to move the load. You'll also want to add in a buffer for the losses within the threaded nut, thrust bearings and the gearbox. If I remember correctly, I did it as a ratio [sin(Theta)/(sin(Theta)+cos(Theta)) * weight of load] was the force at the tip of the sprocket on the threaded rod, and it seemed to be close to what we experimented with under heavy load.

Note that the '2" of travel in 1 turn' is the rating after the # of starts is taken into account.

As far as how fast an acme rod can spin under load -- if the ends have thrust bearings and the radial bearings are also mounted properly, then there's no real reason to not go right up to the rod's rated limits (we went beyond it in 2011, yet that was due to a very light load that year). One end's radial bearing should have a little play (ours was in a block mount that used a 1/4" bolts in a oversized 3/8" hole) so that when it gets to the end it doesn't seize/bend the rod.

We never tested if we can go faster than the suggested speed with a load of 120 pounds, mainly because we saw seizing in the threaded rod and were scared to go any faster. Because of this, I can't give you a definitive of if going any faster will cause problems or not. Use the calculations above to see how fast you need to go and test it. If it doesn't cause that much of a problem, I would bring a couple of replicas of the entire system with me to competition just in case. Then there's that pesky money issue...

[kevin.li.rit]

Originally Posted by JesseK

It's possible on a rudimentary level to figure out the torque requirements of a lead screw by figuring out the angle of the threads as if projected on a flat plane. I figured this out as we went along in 2011, so I don't know if it's the best way to do it:

e.g. There are 2" of travel in 1 turn, PI*0.625" of circumference rotation in 1 turn, thus Theta = Tan(2/(PI*0.625)) should give the angle of thread incline in radians. Sin(Theta) will then correlate to the amount of force due to gravity the motor must overcome just to move the load. You'll also want to add in a buffer for the losses within the threaded nut, thrust bearings and the gearbox. If I remember correctly, I did it as a ratio [sin(Theta)/(sin(Theta)+cos(Theta)) * weight of load] was the force at the tip of the sprocket on the threaded rod, and it seemed to be close to what we experimented with under heavy load.

Note that the '2" of travel in 1 turn' is the rating after the # of starts is taken into account.

This is assuming that thread is square. The other thread angle(the incline directly out of the axis of the screw) is not flat on a standard acme screw. This adds to the amount of torque you will need and complicates the equations a little more.

But back to the OP, a quick and dirty calculation you can always do to figure out how fast your motor can lift your load(force) is take your motors' theoretical output power (Watts) at its max current draw for FIRST (eg 40A) and use the below fiziks equation. Of course this will only work on your 100% efficient system.

Power=Force*Distance/(Time)

[Dad1279]

You can calculate by using TPI (turns per inch). For example, 10 TPI screw(1/2"-10 Acme screw) would be equal to a winch with a pulley circumference of 1" and a gear reduction of 10:1. (but with more friction/less efficiency than a winch)

[kevin.li.rit]

Originally Posted by Ridgebotics

Pretty close to 140 pounds as fast as possible. We're going to be using 2 cims

As fast as possible is hard to calculate without knowing very specifically what type and size of screw you are using since the screw itself is providing a reduction.