Quote:
Originally Posted by Ether
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The force is 10*sqrt(2) because the wheel is constrained from moving sideways by the counter-balancing sideways force from the wheel on the other side of the bot.
This 10*sqrt(2) force can be split into forward and sideways components of 10 pounds each.
If the command is forward on all 4 wheels and the robot is not moving, yes. We haven't yet discussed the case where the robot is moving forward.
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Aha, I was under the assumption that it's not possible to put a linear force on something via a wheel without the wheel moving forward, even by a miniscule amount ... my bad

. I'll ignore the fact that realistically this specific situation happens maybe one in 100 times on the field and indulge the discussion some more.
For a 100% stationary Mecanum wheel, the forces that move sideways put strain on the robot frame. Experimentally, one can see this by doing the aforementioned spinning with of a Mecanum wheel one one's hand while it touches the ground. In the experiment, it moves the arm of the person holding the wheel sideways as it rolls forward. On a 4-wheeled Mecanum robot, it simply puts strain on all of the intervening frame members creating
micro stress fractures. Over time and with enough reptition, the frame will fatigue and eventually give way assuming the wheels hold up.
As such, since the sideways force is absorbed by the frame it cannot contribute to the forward component force of the robot. Since the input torque applied to the roller is split into two component forces, the forward force must be less than the input force since the sideways forces is absorbed.