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Thread: Need help with torque to force conversion

  1. #1

    Need help with torque to force conversion

    Scenario:
    Electric motor, 30 ftlb @ 300 rpm
    6000 lb load (13200 N) fixed to "this" machine
    Motor drives a shaft that drives tracks (like a tank)
    Will need to go up an incline (0 - 10°)
    Weight of machine is not a factor
    Assume frictionless system

    I can't figure out how to determine if my motor can drive that load on a flat surface, let alone an incline.
    I can design a gearbox to get the power needed but I need help to see if I can run it on a flat surface.
    Do I have enough info to see if this motor will move the machine?

  2. #2
    Yes, except for the one other thing that was probably the purpose of the assignment.

  3. #3
    This isn't an assignment. This is part of a prototype design I'm working on.
    The fact that I've been out of school for 6 years and haven't really used these type of equations since is why I need help now.
    Last edited by aakrusen; 07-06-2012 at 12:34 PM.

  4. #4
    Administrator Kelly Bramble's Avatar
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    Quote Originally Posted by aakrusen View Post
    Scenario:
    Electric motor, 30 ftlb @ 300 rpm
    6000 lb load (13200 N) fixed to "this" machine
    Motor drives a shaft that drives tracks (like a tank)
    Will need to go up an incline (0 - 10°)
    Weight of machine is not a factor
    Assume frictionless system

    I can't figure out how to determine if my motor can drive that load on a flat surface, let alone an incline.
    I can design a gearbox to get the power needed but I need help to see if I can run it on a flat surface.
    Do I have enough info to see if this motor will move the machine?
    Allmost sounds like a conveyor system...

    Check out..

    http://www.engineersedge.com/motors/...w_cylinder.htm

    http://www.engineersedge.com/motors/...e_equation.htm

  5. #5

  6. #6
    Quote Originally Posted by aakrusen View Post
    This isn't an assignment. This is part of a prototype design I'm working on.
    The fact that I've been out of school for 6 years and haven't really used these type of equations since is why I need help now.
    Woops, my mistake for jumping to assumptions. The "assume frictionless system" seemed too familiar :P

  7. #7
    Quote Originally Posted by mbanerj View Post
    Woops, my mistake for jumping to assumptions. The "assume frictionless system" seemed too familiar :P
    No biggy, I would have come to sites like this when I was in school if I knew about them. But then again, when I was in school I was doing this stuff everyday and I would have been able to do the easy stuff like this without a lot of help.
    The "frictionless" part was an attempt to stay with the jargon/lingo so I don't sound like such a noob. Too late.
    To be honest, I feel a little silly asking these questions because I should know this stuff, but it's like I said, I've been out of the game for so long I need a little reminding to setup the equations, FBD, etc.

  8. #8
    Kelly,
    Thanks for the links, there's a lot of good info in them, it might have what I'm looking for. At the very least I think it will help me set up me equations.

  9. #9
    So here's what I'm getting from the Engineer's Edge website:
    T=(WK˛ * ΔN)/308t
    WK˛=W(V/2*PI*N)˛
    W=6000 lbs
    V=44 feet/min
    t=5 seconds
    N=300
    ΔN=300

    I come up with .64 lb-ft of torque to move my 6000 object from 0 to 44 feet/min. Does anyone else get that? Does it seem low? It feels like the equation is not asking for all the variables.
    When I put in that I want to take 1 second to get up to speed it says I only need 3.18 lb-ft.
    Why do I not need to enter in the radius of my "conveyor belt"?
    Also, my RPMs will not be constant through the acceleration, right?
    I do know that without gearing and friction, I have 30lb-ft available at stall (zero RPMS) with my choosen motor, but I'm trying to find out if it's enough to now this load. Then I can add friction to determine gearing and continue down that path.

  10. #10
    Senior Engineer
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    this is what I would do:

    The gear that is connected to the motor is pulling the entire load tangentially as it rotates the track wheel. Therefore, the force that the the track wheel must exert on FLAT ground should be theoretically equal to the load*Coefficient of Friction for the conveyor/track system (Include a safety factor/be generous with your Coefficient). On an Incline you must add on the gravitational force it must also overcome. Hopefully you remember how to set up a free body diagram and solve for you incline force (m*g*sin10...I believe). Once you have your total force, multiply it by the moment arm which will be your track wheel radius and BOOM! you have your operating torque... Then you can play around with your gearbox ratios/HP/RPMs.

    ** You cannot say frictionless ** look around at tables of conveyor belts and then just beef up your coefficient.
    Last edited by wderrick; 07-09-2012 at 03:56 PM. Reason: Coefficient of Friction

  11. #11
    Administrator Kelly Bramble's Avatar
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    Quote Originally Posted by aakrusen View Post
    So here's what I'm getting from the Engineer's Edge website:
    T=(WK˛ * ΔN)/308t
    WK˛=W(V/2*PI*N)˛
    W=6000 lbs
    V=44 feet/min
    t=5 seconds
    N=300
    ΔN=300

    I come up with .64 lb-ft of torque to move my 6000 object from 0 to 44 feet/min. Does anyone else get that? Does it seem low? It feels like the equation is not asking for all the variables.
    When I put in that I want to take 1 second to get up to speed it says I only need 3.18 lb-ft.
    Why do I not need to enter in the radius of my "conveyor belt"?
    Also, my RPMs will not be constant through the acceleration, right?
    I do know that without gearing and friction, I have 30lb-ft available at stall (zero RPMS) with my choosen motor, but I'm trying to find out if it's enough to now this load. Then I can add friction to determine gearing and continue down that path.
    OK, I just looked at the equations and webpage in question and here's what is going on..

    http://www.engineersedge.com/motors/...e_equation.htm

    The equation application was miss-identified. What it actually calculates is the tangential torque at the OD of the a pulley driving a conveyor. I did update the web page for clarity.

    First, you need to draw/post a conveyor system you intend to evaluate. The conveyor system will have many components, such as:

    Conveyor belt of known length and weight
    Drive pulleys
    Idler Pulleys
    Weight of materials
    Any incline angles defined
    Any gearing and torque reduction
    Other components requiring torque and moment of inertia evaluations

    After you know the system then you can estimate (calculate) the torque and force requirements for each of the system components less applicable friction. Then you will add up your cumulative start and running torque/horse power requirements for the required minimum system motor(s).

    Conclusively, you will also add a FOS in your operating motors requirements.

  12. #12
    Administrator Kelly Bramble's Avatar
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  13. #13
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    Hi I also need help for a similar application.
    motor torque is 0.005 nm at 7000 rpm.
    i need to convert this inot a thrust.
    the thrust force direction is at 90 deg. to motor shaft.
    what is the force that i can get with this kind of arrangement and if need we can place a gearhead to enhance the force.

    can any one help me in this

  14. #14
    Senior Engineer
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    Quote Originally Posted by venkat.k View Post
    Hi I also need help for a similar application.
    motor torque is 0.005 nm at 7000 rpm.
    i need to convert this inot a thrust.
    the thrust force direction is at 90 deg. to motor shaft.
    what is the force that i can get with this kind of arrangement and if need we can place a gearhead to enhance the force.

    can any one help me in this
    Your force will be dependent on the moment arm (radius) of your driven gear.

  15. #15
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    Ummm, not quite so. As the arm is rotating the angular moment will be reducing and thus the thrust will drop.

    Depending on what you are trying to achieve, a simple wheel and arm may not do what you want if constant thrust (force) is important for the entire stroke.

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