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Robot design Torque required?

  1. Jul 8, 2011 #1
    I am building a robot with a total mass of 835g, using two servo motors conected to the tires made of rubber which have a 3.33cm radius, and i wanted to know how to find out the total torque required to move the robot taking friction into account, the only things i know are:

    Torque=Force*radius

    Friction Force =Coefficient of Rolling ressistance*Normal force
     
  2. jcsd
  3. Jul 9, 2011 #2

    jack action

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    Mixing both of your equations:

    Torque/radius = Coefficient of Rolling resistance * Normal force

    Or:

    Torque = radius * Coefficient of Rolling resistance * Normal force

    And:

    Normal force = Robot mass * g

    This will be the torque required at constant speed (acceleration = 0). When you will accelerate, the torque needed will be:

    Torque = radius * (Coefficient of Rolling resistance * Robot mass * g + Robot mass * acceleration)

    Or:

    Torque = radius * Robot mass * (Coefficient of Rolling resistance * g + acceleration)

    What is more important is the power required by your motors, which is:

    Power = Robot velocity * Robot mass * (Coefficient of Rolling resistance * g + acceleration)
     
  4. Jul 10, 2011 #3
    ok thanks but im still a little confused
    when you said

    Power = Robot velocity * Robot mass * (Coefficient of Rolling resistance * g + acceleration)

    can we assume a random reasonable velocity and an reasonable acceleration?
     
    Last edited: Jul 10, 2011
  5. Jul 10, 2011 #4

    jack action

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    Well, when you will be «cruising», your velocity is constant, meaning the acceleration is zero. If you put the maximum velocity you wish to achieve, the acceleration is also zero and the equation will give you the minimum power you need from your motor.

    You can check http://hpwizard.com/car-performance.html" [Broken] for more detailed info. It is for cars and it also includes aerodynamic forces, but the theory applies to your robot as well (find the theory explained at the bottom of the page). The graph labeled «MAXIMUM ACCELERATION & SPEED» is basically the equation I gave you for power where the acceleration available drops as velocity increases for a given power.

    For example, using the simplified version of the calculator, with a 2 W motor (or motors), a mass of 835 g and a coefficient of friction of 0.8, your robot should be able to reach a top speed of 13-14 km/h and have enough acceleration to reach 10 km/h in about 2-3 s (depending how you connect your motor to the wheels).

    The power you need depends only on the performance you want to achieve (top speed and time taken to reach your cruising speed).
     
    Last edited by a moderator: May 5, 2017
  6. Feb 10, 2012 #5
    What if a drag force is known. How should power and torque formulas be modefied?
     
  7. Feb 10, 2012 #6

    jack action

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    You add the drag force (FD) this way:

    Torque = radius * (Robot mass * (Coefficient of Rolling resistance * g + acceleration) + FD)

    Power = Robot velocity * (Robot mass * (Coefficient of Rolling resistance * g + acceleration) + FD)

    Where FD = 0.5 * air density * drag coefficient * Robot frontal area * (Robot velocity)²
     
  8. Feb 12, 2012 #7
    You need much more information to be able to determine the required torque at the actuated joints of a robot. Typically, an inverse dynamic model is used to relate the dynamics of an input force
    \begin{equation}
    f_i(t)
    \end{equation}
    to the end effector acceleration
    \begin{equation}
    a_p(t)
    \end{equation}
     
  9. Feb 13, 2012 #8
    Thank you, Jack!
     
  10. Feb 13, 2012 #9
  11. Feb 13, 2012 #10
    Hey, Fex32!

    Thank you for MIT's links. I'll look them through. But looks like that complicated theory is more about multi-joint assemblies, like robot arms and manipulators.
    There are a couple of drives and wheels to move my bot and no plans to land it on Mars. I believe simplified method will do. It's also implemented in Drive Motor Sizing Tool http://www.robotshop.com/dc-motor-selection.html
     
  12. Feb 13, 2012 #11
    Ohhhh, It's a mobile robot. lol.
    I thought it was a robot manipulator.
    Then I agree with the above.

    Cheers,
     
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