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Re: pic: A mechanum wheel that shouldn't bump around at all.
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Changing the angles will change the vectors of the wheels. A smaller angle (assuming the maximum angle, 90 degrees, is just a roller mounted straight lengthwise) should theoretically give more strafing motion. Vice versa for a larger angle and forward/back motion. |
Re: pic: A mechanum wheel that shouldn't bump around at all.
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Justin, if you are interested let me know and I'll start another thread. ~ |
Re: pic: A mechanum wheel that shouldn't bump around at all.
no, if the rollers are at 45 degrees, then the strafe speed is half of what the forward reverse speed is, because going forward, the wheels act like normal wheels, however when they are strafing, they are 2 vectors at 45 degree angles, so only half of the magnitude of the vector is in the direction of the strafe.
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Re: pic: A mechanum wheel that shouldn't bump around at all.
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This is what the vectors of a mecanum wheel look like. http://lib.store.yahoo.net/lib/yhst-...lSpecSheet.pdf |
Re: pic: A mechanum wheel that shouldn't bump around at all.
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Re: pic: A mechanum wheel that shouldn't bump around at all.
hmmmm, i know that it is less speedy/ powerful strafing than side to side though.
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Re: pic: A mechanum wheel that shouldn't bump around at all.
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Re: pic: A mechanum wheel that shouldn't bump around at all.
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Ether, I posted the part I think led Justin "astray"; namely the logical conclusion that if X< than Y, Z is more effective one way and X > Y, Z is more effective another way, therefore if X=Y, Z is optimized for both directions, and that he simply misstated what he meant. It's up to him to correct me if I'm wrong on that. (Also note that I'm in college, and I don't know/don't remember how to do a matrix transformation. I wouldn't be surprised if that reference confused a number of people. Some explanation may be in order, either in here or in another thread, as to a) what they are, b) how to do them, and c) why they apply here.) |
Re: pic: A mechanum wheel that shouldn't bump around at all.
A request was made for a more detailed explanation of my earlier post concerning kinematic analysis of a mecanum wheeled-robot, and the derivation of the inverse and forward transformation matrices.
I decided to start a new thread for this discussion in order not to divert the focus away from this thread's OP's intent. http://www.chiefdelphi.com/forums/sh...51&postcount=1 The analysis answers many of the questions posted here about the effect of roller angle and whether fwd/rev vs strafe should be the same speed etc. ~ |
Re: pic: A mechanum wheel that shouldn't bump around at all.
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Re: pic: A mechanum wheel that shouldn't bump around at all.
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Re: pic: A mechanum wheel that shouldn't bump around at all.
When running front-back, all 4 wheels are contributing "full power" in one direction. When running side-side, all 4 are contributing "full power" in another direction, but two of them have to work against the other two to do it. This will amplify any effects of friction and other inefficiencies that were masked by running all together.
The problem with theory is that it often leaves out reality. This is why engineers use safety factors and try to account for reality wherever they can. Theoretically, there is no friction and no inefficiency, and inefficiency (where it does exist) is uniform any way you look at it. In reality? Not only is there friction and inefficiency, but it's hard to give them a nice, easy number. |
Re: pic: A mechanum wheel that shouldn't bump around at all.
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The issue making the two directions act differently in the real world is the friction of the rollers. It helps in the forward direction, and hurts in the sideways direction. If you use sensors to maintain the same wheel speed, your front-back speed will be the same as your left-right speed. With non-frictionless rollers, though, you have to apply more power to maintain that speed when traveling sideways. |
Re: pic: A mechanum wheel that shouldn't bump around at all.
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In the backwards/forwards motion, the wheels are still acting against each other. Simple vector analysis tells us that perpendicular components of each force vector cancel each other out. Vector analysis also tells us that the same exact thing happens in the strafing motion, however, this time, the backwards/forwards vectors are the ones that cancel each other out. The two motions have the exact same oppositions, just in different directions. The only thing that I have been trying to prove is that each component of the force vectors on the wheels are theoretically exactly equal in magnitude. I have only been trying to prove that beliefs like this: Quote:
Believe me, I understand the fine line between reality theory/inefficiencies. However, I'm just trying to say that the inefficiencies aren't caused by things you described. The only factor at work is the friction of the rollers. If you tighten each roller so that it no longer rolls, you will essentially have four "normal" traction, but you will no longer be able to strafe. This is because the omnidirectionality of mecanum wheels is dependent on the rollers "slipping". However, because there will never be a frictionless roller, the strafing motion will never be equal in force to the forward/backwards motion. |
Re: pic: A mechanum wheel that shouldn't bump around at all.
In the real world of aerospace engineering, systems engineers try to model friction and inefficiency as closely as possible when they are building dynamic models to allow them to build accurate procurement specs for suppliers. So theory does include friction and inefficiency.
Roller bearing friction plays a large and asymmetric role for mecanum wheels. In the fore/aft direction, roller bearing friction is a good thing. It moves the reaction force of the floor closer to the plane of the wheel, so it takes less reaction force for a given forward force. The result is better traction. For a frictionless roller bearing, the reaction force is aligned with the roller axis. As roller bearing friction increases from zero to locked roller, the reaction force moves from being aligned with the roller axis to lying in the plane of the wheel. When the reaction force lies in the plane of the wheel, that's essentially the same as a standard wheel. So, the reason a mecanum wheel doesn't have as much pushing force (in the fore/aft direction) as a standard wheel is not because there's less forward force available, but because the reaction force is larger than it would be for a standard wheel, and therefore the mecanum wheel starts to slip before a standard wheel would. In the sideways direction, roller bearing friction moves the reaction force toward the plane of the wheel, which reduces the force vector component in the sideways direction. To compensate, the motor must output more torque. This increases the reaction force, which reduces available traction. So in the sideways direction, roller bearing friction is a bad thing. It causes increased motor power consumption (for a given desired force) and it reduces available traction. ~ |
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