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#41
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By 'real' i meant professionally made with either low friction non-bearing rollers or bearings for the rollers. Also, professionally produced omniwheels are more likely to withstand a beating.
And, i believe it is possible to get 100% motor torque using four wheels. look at it this way, ._ {_} . where the '{' and '_' are the wheels. say you power the {} wheels in the same direction. you get full power/torque out of them. Say you power the '_' wheels in the same dircetion, same deal. Now lets say we geared the motors so that when only one set of wheels are powered, you get 6 feet/second and whatever push that translates to. If you power BOTH sets you get the robot traveling in a diagonal, at the new speed 6rt(2) feet per second, and a new power that is sgrt(2) times the power of an individual set of wheels. so, you get 2x the power of one set of wheels, and one set of wheels gives you the same power from he motors that a tank drive would, so with twices the motors ou get twice the power of a four-motor tank drive, which gives you 100% power except for gearing and wheel inefficiencies. I don't see where your 70% is coming from . . perhaps I can get a link to an explanation . . . ? at the 45' diagonals you get 100% power out of the motors. The only way it isn't as powerful as a tankdrive at same geared to he same speed is that you are loosing power to make the rollers turn, which is a near-negligable loss. I don't understand how it could be different . . I accept that you only get full power at the 45' diagonals, but that is still another axis of freedom on which full power is available over a tank-drive. edit: your idea about using a compass/angular acceleration sensor: We have been planning that since last season. It will be in place using one or two solid-state gyros. Error will add up, but we may build a control that allows for the secondary driver to manually adjust the robot's percieved heading so the error can be corrected as the match progresses. edit: I looked over your post a second time, and it is true that at the diagonals you get sgrt(2) times the power of each individual set of wheels, but I think you failed to consider that you ALSO get sqrt(2) times the speed, so twice the power. You gotta look at both. And you need a driver who is aware of all of this. Last edited by Frank(Aflak) : 30-09-2003 at 10:25. |
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