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#1
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Hi. I've been planning a 4x8 cnc router for quite some time now. It just dawned on me that I may not have enough space to have a horizontal table in my shop. Has anyone tried an upright design? I would appreciate if someone would point out some of the difficulties of a vertical design. Would save lots of space, and I may be able to convert the back side of the table into a panel saw. I'm using a 2" ballscrew with a .2 pitch. Wondering how much torque needed to drive a 75lb gantry vertically with this screw. Last edited by jimluu; 12-05-2007 at 03:21 PM. |
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#2
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| Don't have any numbers for you, but just spinning a 2" screw requires quite a bit of torque. And your gantry will probably weigh quite a bit more than 75lbs. All the vertical routers I've seen have the gantry moving side to side, so you don't have to lift the gantry, just the Z axis.
__________________ Gerry Mach3 2010 Screenset http://home.comcast.net/~cncwoodworker/2010.html (Note: The opinions expressed in this post are my own and are not necessarily those of CNCzone and its management) |
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#3
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| you may look at this : .... ;-) http://www.cnczone.com/forums/showthread.php?t=6840 its a wall mounted and off course DIY ... |
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#4
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| Thanks for the response! I've thought about a horizontal design, but if I can make a vertical design work, I'll save half the space. I'm planning to use a NEMA 34 motor with around 1200oz of torque or more. Do you think this is still not enough? I have not bought the motor yet and can always buy bigger ones. I'm wondering what happens if there is no power to the motor...will the gantry slide down? |
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#5
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| Its an interesting concept though. Have you though about turning it on its side? That might save some space. Of course you would have to worry about the Z-axis assembly back driving lol. You also have to think about how your will hold the piece you are working on. Gravity is not longer on your side. Keep us informed! Regards,
__________________ Brian WWW.CNCRouterSource.com |
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#6
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| If anyone out there knows how to calculate torque requirements...please help. My gantry will likely weight 75lbs, but lets use 100lbs to err on the safe side. My ballscrew is 2" in diameter and has a pitch of .2. I'll be mainly cutting plywood and MDF. I'm not familiar with any types of cnc physics and have a hard time interpreting some of the formulas posted else where in the forum. I would appreciate any help to see if it's at all feasible to go forward with this vertical design. |
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#7
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| Jim, I would be glad to help. Below is the basic process most people use for claculating how much torque is required or how much linear force you can get from torque. To find how much linear force you will get from your lead screw and motor assembly, use this formula. 1. Convert ball screw/lead screw turns per inch to pulley radius: Example: ½-10 2 start = 5 turns per inch convert this 1 inch/5 turns per inch = 1/5= .2 = pitch So pitch = .200 Then divide pitch by 2pi or 6.28 Example: 0.200 / (2pi) = 0.200 / 6.283 = 0.0318" (this is the equliv. radius pulley) 2. Convert motor torque into linear force: divide torque of your motor by the puller radius example: Torque = 200 Oz.-In. 200oz-in / .0318 in = 6289.3 oz. (linear force) 6289.3 / 16 = 393 lb (linear force) 3. Multiply result by the efficiency: Ball screws have efficiency of .75-0.95 (I usually use .8) Acme Lead screws have efficiency of about .45-.60 (I use .5 for rolled ACME screws and .4 for basic ACME threads with anti-backlash nuts) 393 * 0.6 (efficiency for ball screw) = 235.8 lb linear force. You can see that in this example that at 200 oz in. you would have 235.8 pounds of linear force. Remember though, the rated motor torque for stepper motors is at holding torque. You would need to know about how much torque you would have at a certain RPM that would give you the desired feed rate. Again, this is a basic calculation but it works well most of the time. The correct way is to break everything into inertial forces. The above calculations does not account for acceleration of the machine. You case would be interesting as you would need more torque in on direction and very little in the other. I hope this helps. I am making an excel spread sheet that will do a lot of this for you. When I’m done I’ll send it your way. Regards
__________________ Brian WWW.CNCRouterSource.com |
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#8
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| Brian, Thank you for your help. I would never have figured all that out by myself. Seems like the 1200oz motor that I have in mind should work ok. I'm going to hire out a local company for the frame then I'll post the results here. |
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#9
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| Jim, I was just about to post and tell you that for startup you roughly need double the torque, and that would be even more true for your applications. However, with that much torque, you shouldn't have a problem lol. Thats a lot. I look forward to hearing an update. Keep us informed. Regards,
__________________ Brian WWW.CNCRouterSource.com |
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#11
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| For the mounting bearings, be sure and use a proper thrust bearing on each end in conjunction with a radial bearing. This will take the thrust off the radial bearings and the motor. If you do this, your end mount bearings should be fine. If you need a diagram on how to use a thrust bearing in conjunction with radial bearings, just let me know. Regards,
__________________ Brian WWW.CNCRouterSource.com |
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