Turning a Tank Drivetrain

Guide

  • Drive left and right sides different speeds to turn
  • Drive left and right sides at opposite speeds to "turn on the spot"
  • Location of the "spot" is dependant on wheel material and the Centre of Gravity
  • Since the wheels are not facing the direction the robot is trying to turn there will be some scrub
    • Scrub is the amount of friction resisting the turning motion
  • This scrub is useful when being defended or defending
  • To ease analysis, we resolve the friction force into x (horizontal) and y (vertical) components
  • Since wheels are able to rotate, the friction force perpendicular to the axis of rotation of the wheels is 0
  • Assuming pairs of wheels (one left and one right wheel)
    • Have the same traction material
    • Are inline (distance from centre of robot is equal)
  • The forces can be simplified
  • We get a torque diagram
  • The pivot point of the torque arm is not fixed, it will be located where the torques balance
    • τ1 = τ2
    • FF,1 · d1 = FF,2 · d2
    • d1/d2 = FF,2 / FF,1
    • d1/d2 = μ2 · FN,2/ μ>1 · FN,1
    • d1 + d2 = L
  • d1/d2 = μ2 · FN,2
    • What does this mean?
      • If μ2 > μ1 then d2 < d1
        • The drivetrain will turn on a "spot" closer to the higher traction wheels
      • If FN,2 > FN,1 then d2 < d1
        • The drivetrain will turn on a "spot" closer to the wheels that support more weight
    • How much friction force do the wheels need to over come in order to turn?
      • τdriven = τ1 + τs
      • τdriven = FF,1 · d1 + FF,2 · d2
    • Recall
      • d1/d2 = FF,2/FF,1 or d1 = FF,2/FF,1 · d2
    • Combine
      • τdriven = FF,1 · FF,2/FF,1 + FF,2 · d2
      • τdriven = 2 · FF,2 · d2
    • Also
      • τdriven = 2 · FF,1 · d1
      • τdriven = 2 · FF,2 · d2
    • What does this mean?
      • The smaller you make FF,1 or FF,2, the less force is needed to overcome friction
      • FF = μ · FN
      • Therefore must μ1, μ2, FN,1, or FN,2 to ease turning

Friction Force

  • Often it requires too much power to turn 4 wheel drivetrains due to friction force
    • τdriven = 2 · μi · FN,i · di
  • How can this be over come?
    • Reduce FN,1 or FN,2
      • Means shifting CoG to one end
      • May not be able to shift far enough to be effective
      • Increase likeihood of tipping
    • Reduce μ1 or μ2
      • Means using low traction material on one set of wheels
      • Lose traction for pushing
    • Reduce d1 and d2 (reduce L)
      • Wide drivetrains are difficult to navigate through narrow spaces
      • May need length for robot functions
    • What about combining points 1 and 3?