Quote:
Originally Posted by RyanCahoon
Imagine that you have a beam supported on both ends, and you apply a downward force in the middle.
There are reaction forces on both ends of the beam, and the sum of all three force must be 0. I see this as the same case here: the beam is the module, one end is supported by the wheel bearings (and transitively by the wheel on the carpet), the other end is supported by the the bearings on the drive axle. |
In this case, there is no reaction force on the left end of the beam as shown in my diagram; the left wheel is off of the carpet. The beam in this case would be the module bracket. The forces on it are gravity and those applied at each of the three axle bearings.
- When no torque is applied through the drive axle, the forces on the wheel bearings reduce to the weight of the wheels and wheel gears. Presumably the CoG of the module including the wheels and gears is somewhat offset from the drive axle, so the module will enter a pendular motion to place that CoG below the drive axle.
- When torque is applied through the drive axle, note that (assuming good bearings) it is not applied to the module frame, but to the wheel gears. This torque is then transmitted to the module frame until a wheel touches the carpet and encounters a normal force. This is the case I analyzed in the paper. Note that the module cannot apply any significant torque to the robot, except through the drive shaft (and particularly not through the bearings). Assuming that the applied torque is appropriate to moving the robot, the weight of the wheels and module are relatively insignificant. This means that the force B applied by the module to the wheel on the floor is balanced by a force -B applied by the module to the robot. This force provides all of the translational force from the module to the robot, and a portion of the weight bearing of the wheel (the remainder is through the reaction of the drive gear to providing force D).
Quote:
Originally Posted by RyanCahoon
I'm not sure if this is the correct analysis, but as a general point, I'm bothered that your derivation doesn't incorporate the distance between the drive axle and the wheel axle. If you increase this distance, then the torque load on the motor imposed by turning the wheel won't increase, but the torque load imposed by the normal force does. At some point, turning the wheel will be much easier for the motor than applying normal force, so the wheel will spin with very little normal force applied, causing the wheel to slip and the robot won't move.
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If the distance between the drive axle and the wheel axle is increased (and the wheel and wheel gear geometry remains constant), this means that the force D is being applied by the drive gear at the end of a longer moment arm from the drive axle, so of course it will require a greater torque from the drive shaft, though it will turn at a lower angular speed. For a given robot acceleration, the torque load required by the motor system will increase, though D remains the same.