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Vertical Movement from a Threaded Rod

  1. Mar 5, 2008 #1
    I want to move 30lbs vertically on a platform. To do this I will have a rotating threaded rod driven by a motor. How much torque, hp, rpm do I need? I'm guessing the pitch of the threads will be important. As far as speed goes, I only want the platform to move about an inch in 1-2 seconds.

    Please let me know if you can help me determine what equations might be necessary.
    Thanks
     
  2. jcsd
  3. Mar 6, 2008 #2

    Danger

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    Welcome to PF, Stephii.
    I'm afraid that I can't help you with any of the math stuff, but I do know that frictional factors matter almost as much as the thread pitch. Generally, the shallower the pitch, the more mechanical advantage you have. Things like nylon followers, though, can make a big difference as to how steep you can go before things bind up.
     
  4. Mar 6, 2008 #3

    Q_Goest

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    Hi Stephii,
    Gotta love the way Danger always puts a disclaimer in front of his posts. lol
    But what he's saying about friction and pitch is true. The torque-load relationship for a threaded fastener is:

    T = F * (P / (2*pi) + ut * rt / cos(B) + un * rn)
    Where:
    T = torque applied to fastener (in lb or mm-N)
    F = force created by fastener in direction of axis (lb or N)
    P = pitch of the threads (in or mm)
    ut = coefficient of friction between nut and bolt threads
    rt = effective contact radius of the threads (in or mm)
    B = the half-angle of the threads (30 degrees for UN or ISO threads)
    un = coefficient of friction between the face of the nut and the upper surface of the joint
    rn = effective radius of contact between the nut and joint surface (in or mm)

    So if you ignore the head of the nut since it isn't in contact, the last set of values, un and rn drop out.

    That's the torque needed to produce the force of 30 pounds you want, so that should also give you enough information to determine the power needed to move the platform an inch in 1 or 2 seconds.
     
  5. Mar 6, 2008 #4

    Danger

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    That's getting to be a bit of a cliche, ain't it? :biggrin:
    It's the only way that I can post anything without compromising the integrity of the site. Since I have no education, I have to make sure that everyone knows that my opinions are just that... my opinions. I love to help out when possible, but it would be potentially disastrous if someone were to take my word for something without hearing from real scientists first. It's critical that people like you confirm or debunk those opinions before action is taken. :smile:
     
  6. Mar 6, 2008 #5

    Q_Goest

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    I work with a lot of technicians who come up with lots of very practical ideas. Unfortunately, many engineers have a tendancy of wanting to stay in the ivory tower and design something at the end of a telephone or computer program. The end result is the typical, "Who the hell designed this!!!???" comments from the field.

    Everyone has good ideas and should be heard. A good engineer puts their pride in a box and burys it, then talks to people who have to live with their designs. That isn't to say an engineer should simply do whatever it is the technician comes up with, but a good engineer will evaluate ideas for technical merit and chose the best solution as opposed to simply doing it his/her way because they have the responsibility for the design. Don't worry, Danger, you are loved! :biggrin:
     
  7. Mar 6, 2008 #6
    Thanks guys. I'm an engineer-in-training so I appreciate the heartfelt comments about engineers. I will try to not be a behind-the-desk type of engineer. I did some research of my own and also found a possible way to calculate it. So, I will look at my options and choose the one I think will best suit my purpose.

    Thanks again for all your help.
     
  8. Mar 9, 2008 #7
    I may be a moron but if Torque is the produce of mass and velocity than we can find the velocity at the point by the cross product of the angular velocity and radius. In short Torque= mass(angular velocity of the radius X radius) or T= m*(w x r)
     
    Last edited: Mar 9, 2008
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