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How to calculate the torque to pick the motor for climbing hall robot

  • Darknight2
  • March 11, 2024 at 6:03 PM
  • Thread is Unresolved
  • Darknight2
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    • March 11, 2024 at 6:03 PM
    • #1

    Hi everyone. I have a big project to build a mobile robot that can climb and clear the hall ship container. However, I have a problem with calculating the torque for the motor. As you can see in this picture, Fr=545N is the thrust of the pressure nozzle. Fms1 is the friction force of the wheel. FA=2000N is a magnetic attraction. H is the distance between the hall ship and the vehicle's center of gravity. It is a 3-wheel mobile robot with 2 driving wheels and 1 multi-directional wheel ( caster wheel). So I don't know how to calculate the torque to pick the motor. Can anyone give me an idea to solve that? Thanks very much. Sorry for my bad English.r/MobileRobots -  picture of dynamics robot

  • Lemster68 March 11, 2024 at 6:16 PM

    Approved the thread.
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    • March 11, 2024 at 9:33 PM
    • #2

    Is this a homework question? Certainly looks like one.

    Check out the Fanuc position converter I wrote here! Now open source!

    Check out my example Fanuc Ethernet/IP Explicit Messaging program here!

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    • March 11, 2024 at 11:23 PM
    • #3

    if you are climbing, you need to overcome couple of things:

    1. gravity

    2. friction

    3. air resistance etc.

    the air resistance is not going to be the problem due to (assumed) low velocity.

    friction if a function of contact force and friction coefficient.

    the gravity is by far the largest component (if this are done correctly).

    this can be done differently with varying degrees of complexity. the quickest and simplest way is to use energy.... height is a factor in potential energy. speed is a factor in kinetic energy. mass is a factor in both of them. energies add up. power is a rate at which energy changes. so knowing how fast the climb need to be along with how much energy is needed you can calculate power.

    from there you can get torque. you can use mechanical advantage (gearbox, pulleys, wheel size...) to get more torque. tradeoff is the speed.

    1) read pinned topic: READ FIRST...

    2) if you have an issue with robot, post question in the correct forum section... do NOT contact me directly

    3) read 1 and 2

  • Darknight2
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    • March 12, 2024 at 6:28 AM
    • #4
    Quote from panic mode

    if you are climbing, you need to overcome couple of things:

    1. gravity

    2. friction

    3. air resistance etc.

    the air resistance is not going to be the problem due to (assumed) low velocity.

    friction if a function of contact force and friction coefficient.

    the gravity is by far the largest component (if this are done correctly).

    this can be done differently with varying degrees of complexity. the quickest and simplest way is to use energy.... height is a factor in potential energy. speed is a factor in kinetic energy. mass is a factor in both of them. energies add up. power is a rate at which energy changes. so knowing how fast the climb need to be along with how much energy is needed you can calculate power.

    from there you can get torque. you can use mechanical advantage (gearbox, pulleys, wheel size...) to get more torque. tradeoff is the speed.

    Display More

    thanks u very much

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Thread Tag Cloud

  • abb
  • Backup
  • calibration
  • Communication
  • CRX
  • DCS
  • dx100
  • dx200
  • error
  • Ethernet
  • Ethernet IP
  • external axis
  • Fanuc
  • help
  • hmi
  • I/O
  • irc5
  • IRVIsion
  • karel
  • kawasaki
  • KRC2
  • KRC4
  • KRC 4
  • krc5
  • KRL
  • KUKA
  • motoman
  • Offset
  • PLC
  • PROFINET
  • Program
  • Programming
  • RAPID
  • roboguide
  • robot
  • robotstudio
  • RSI
  • safety
  • Siemens
  • simulation
  • SPEED
  • staubli
  • tcp
  • TCP/IP
  • teach pendant
  • vision
  • Welding
  • workvisual
  • yaskawa
  • YRC1000

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