# 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?