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How Do mecanum wheels handle the bumps?
Someone mentioned that mecanum wheels don't have good friction and that might be a problem when going over the bumps in breakaway. I wondering if we will have any problems if there is a similar obstacle in the years to come. i know mecanum wheels offer great maneuverability but i don't know if its worth the loss of ability to handle bumps and similar obstacles (assuming there is enough loss to cause concern). please list any disadvantages of mecanum wheels. thanks :)
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Actually, mecanum wheels, at least for my team, have no trouble going over the bumps. The main reason people think they have trouble seems to be that the driver must get the robot lined up fairly straight, but 'fairly straight' appears to be about 10 degrees margin of error in each direction. Our robot can simply drive up to the bumps, and drive over, no problemo.
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thanks a lot. i will definatly look into mechanum wheels now :)
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mecanum wheels are absolutely wonderful for maneuverability, as for going over the bump, they don't really have a problem as long as you're going forward and not diagonal or sideways. The only other thing is that mecanums work best when all four wheels are touching the ground at all times, so an active suspension is fairly important
hope that helps :) |
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When developing a design for this years robot our team questioned mecanum wheels' ability to push or pin and avoid being pushed or pinned.
In the end we decided to implement a crab drive. While slightly more complicated we felt that it would give us a pushing advantage. My question is whether or not these asumptions were accurate and if so to what extent? What other advantages or dissadvantages do mecanums have over some other forms of drive trains? |
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mecanums make it very simple to avoid pinning by scooting off in whatever direction you choose, it is also surprising how well you can push/pin, but only in the forward direction. It's easy to be pushed when pushed from the side, but from front or back you can just strafe away. the biggest pro (to me at least) between mechs and crab is that mechs are much simple mechanically, than crab, and are usually lighter as well. over other drive trains, mechs have the maneuverability. They can go any direction they please, and turning is not difficult, like it can be for tank or skid steer.
active suspension -- basically it is some kind of force that keeps the wheels touching the ground. if there's a dip under only one wheel, the suspension pushes that wheel further down while the rest of the wheels a pulled a bit higher. Think of the shocks on a bicycle, it's a similar concept. For our team, the way we implement suspension is by using springs captured between a point on the wheel module and the frame. If you'd like, i can dig up a picture of the robot for you |
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Mechanums can push and go over the bump very well. You do have to watch you weight distribuition though. Most videos show a stripped down robot that is not at the full weight plus battery and bumpers. A full wieght robot that is well ballanced will go right over but one that is out of ballance will struggle.
Our video http://www.youtube.com/watch?v=7_rtqg838Nc |
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An "active" suspension would be a pretty neat trick for a FIRST robot. http://en.wikipedia.org/wiki/Active_suspension ~ |
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Our mechanum wheels worked wonders over the bump. We lowered our center of gravity as far as possible though, to avoid tipping, and to help maneuverability over the bump.
We had no problems. [: |
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Mecanums generally try to avoid defense rather than overpowering it. This doesn't work particularly well in my opinion. Teams using mecanum should maximize their use in portions of the game with limited defensive interaction. For example, team 230 straddles the bump using mecanums in order to hang. |
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we're in team 2669 using mecanum wheels (2010 edition).
When our weight was almost the max (1-2 pounds less) we had a little problem when passing the bumpers, we needed some momentum and then it passed them nicely. Right now, after we lost more than 20 pounds, we can stand 1mm next to the bump and then it passes it smoothly without any problems |
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All seriousness aside, where did you get the 71% number? ~ |
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We've got mecanums and have no problem going over the bump. We've found that if the robot is mis-aligned with the bump simply driving into it the mecanum wheels will slide around to let everything line up, then we climb over the bump without a problem.
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Think about a free body diagram for physics. You may have 9 N of force in one direction and 2 N in the opposite direction, so you are letting off 11 N of force, but you only have a net of 7 N in the forward direction. Mechanum wheels seem nice in theory, but if I'm not mistaken, a team has never won nationals with a mechanum drive. If they meet tough defense they can easily be out pushed. They're good for qual matches with low defense, but I'll be interested to see how they fair in elims in Atlanta. |
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The derivation is left as an exercise to the reader. Think of the angle of the rollers. |
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But that's beside the point. The question on the table is whether or not "only 71% of the output torque is transmitted in the direction of travel" when the bot is going straight forward (or backward). Quote:
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http://www.chiefdelphi.com/forums/sh...ad.php?t=84187 As for mechanums, there is always a question as to the loss of power vs manueverability issue. You have to trade it carefully based on how you think the game will play out. There is no straightforward answer. You need to decide what the strategy your team wants to execute is and whether or not that system fits into your strategy. This is why in the first part of build season should set criteria your robot needs before you build anything figure out how to play the game to win and develop a strategy before you build your robot and these decisions become easy now that you can buy mechanum and swerve as COTS. |
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Thanks Joe. I know that 71% is half the square root of 2 :-) The question is, does this number have anything to do with the claim that "Only ~71% of the output torque is transmitted in the direction of travel", when the bot is being commanded straight forward or backward. I say no. I say all the torque being applied to the wheel gets reacted by the carpet. I am challenging what appears to be the conventional wisdom on this topic. Think about it. ~ |
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My team uses the 6" version of AndyMark Mecanum wheels (the ones with the lower coeff of friction), and we have no problems going over the bump. The only problem to date (that I'm aware of) is when our mentor was driving the drive train during the build season, and he drove the robot full speed into and over the bump. We lost one roller; considering the way he was driving it I'd say it did 3x better than expected. |
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If mecanum drive trains lack pushing power, why not have the opportunity to lock the rollers, effectively turning mecanum into 4 wheel tank drive?
While I was away in Thailand during build season, 190 did just that. |
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Ty -- it would be possible to pull that off, but would takea lot of research and testing (off-season project anyone?) where as, in my opinion, mecanums are still good for defense, but not the pushing kind. They're much better for the manueverable, get-in-the-way kind
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190's an exception to the rule. That's not exactly an easy to design feature. |
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All the forward torque applied to the wheel is transmitted to the carpet. The torque is not reduced by cos(45). Quote:
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Mecanums have the same math / efficiency setups as normal "omni" drives, with the same vector math. You'll pretty quickly figure out where 71% comes from. |
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Edit: "Disclaimer" I was in no way involved with 190's robot this year so, essentially, this is just me analyzing another cool robot I saw at a regional. |
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Ether, the problem with your logic is that Mecanum wheel rollers are not effectively locked when pushing against a hard object. Properly designed rollers still 'roll' at the off angle due to a good bushing that allows the roller to free-roll anyway. The underlying issue isn't that Mecanums can't push; even 70% torque of 4 CIMs is enough to push an object to get out of most situations. It's even good enough to play defense in most situations since defense can usually be done just by turning another robot or getting in the way of another robot.
Conventional wisdom says Mecanums can't push; this is due to software control and a driver's lack of understanding of how a Mecanum design works holistically, OR that the drive train was designed with speed in mind and no high-speed drive train can push worth a durn. We've found that with 4 CIMs a good balance is 10-11 fps before losses. Under the covers, when on a diagonal, software tells 2 motors to drive the diagonal while the other two motors sit idle. This allows the diagonal strafe. Yet when pushing while on this diagonal, the bot effectively has 70% * 1/2 (due to 2 motors out of 4) = 35% of the original torque of the 4 CIM setup. Ergo, Mecanum drive trains with true holonomic control only ever apply 35% to 70% of 4CIMs to any direction. Thus 'Mecanums can't push' should be turned into 'Mecanum drivers don't know how to drive' or 'the whole holonomic idea of Mecanums is terrible for competition robots'. Teams who try to do a full field-centric Mecanum holonomic drive can probably anecdotally attest to this. This year 1885 did Mecanum, yet we only did standard tank/skid steer with the ability to strafe 90 degree from forward -- nowhere in between. Rarely did we get into a situation where we needed to push, yet we also didn't have a problem since it wasn't setup for true holonomics. |
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The rollers do spin noticeably when you're driving straight forward. This is where the obvious loss of pushing power manifests itself. If you go through the vector analysis of how much torque is pushing in which directions for an ideal rectangular-layout mecanum or omniwheel drivebase, you find that ~70% goes to movement in the forward direction, and ~70% goes to trying to squash or stretch the frame in the sideways direction. All the torque is available in the diagonal direction. The torque in the forward direction is only cos(45) of the maximum. |
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My reply to your post also assumes we are talking about mecanum wheels with rollers which are free to spin. Whether or not they actually spin in a given scenario is a different question. For example, consider the following thought experiment: There are two identical robots, RobotA and RobotB, which are absolutely identical in every detail except that RobotA has mecanum wheels and RobotB has standard wheels. The tread on RobotB's standard wheels is the same material as the rollers on RobotA's mecanum wheels. The standard wheels have the same effective diameter as the mecanum wheels. RobotA and RobotB are each facing a brick wall, and there is a load cell on the front of each robot to measure how hard it is pushing on the wall. Each robot is given a slowly increasing identical forward command. I claim the following happens: - Up to a certain point, both robots push with exactly the same force. There is no 71% factor involved. - If the motors are sufficiently powerful, a point will be reached where the wheels start to slip. RobotA, with the mecanum wheels, will reach this point before RobotB, with the standard wheels. ~ |
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(The original LabVIEW omni/mecanum drive vi took that into account and reduced the maximum power in the forward/sideways directions to 70% of what it could be. That gave 50% of full motor torque in every direction. It was a good idea for making consistently-controllable omnidirectional movement easy, but it was a terrible idea for being able to call on the most performance when necessary.) |
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It is the same logic as the situation where a box sits on an inclined plane. Gravity pull the box straight down, yet only a component of the gravitational force creates the force normal to the ramp that keeps the box on the ramp. the other component is parallel to the ramp. Same logic, different situation. Interestingly enough, the speed of a Mecanum drive train is the same as standard skid steer, and perhaps that's the confusion? |
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If the robot is pushing against a brick wall, and the wheels are not turning, then the rollers will not be spinning. Everything looks static. Try it. Quote:
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Refer to RobotA mentioned in the thought experiment in the post at the link shown above. If the motor is providing a torque of 40 in-lbf to a wheel, and the wheel is not slipping, then the carpet MUST be reacting with a force of 10 pounds (for a 8" diameter wheel) in the plane of the wheel, acting against the wheel trying to push it forward. The forward component is 10 pounds (the exact same forward component you would get with a standard wheel). It cannot be otherwise. And yes, with the mecanum wheel there are additional vector forces at play. But they do not subtract from the 10 pounds. I encourage those of you who are using the "vector" argument to think a bit more about how the vectors are created. ~ |
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Hi Alan, Please take a look at my post at the above link and see if I can convince you otherwise. ~ |
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This is a real-world effect, and it does explain the pushing force of a stalled mecanum-wheeled robot in the forward direction. ~ |
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thanks to everyone who is replying. i'm learning a lot about drivetrains. unfortunately im wishing i paid more attention in trigonometry -_-.... but please continue. this is good stuff :)
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Consider a wheel with rollers free to spin on the same axis as the main axle. It doesn't matter how grippy the material is for the rollers, these wheels can't apply a force no matter how hard they drive or how fast they spin. Consider a wheel with rollers free to spin on an axis perpendicular to the main axle (traditional trick/omni wheels). It doesn't matter how hard they drive, they'll have the same coefficient of friction as wheels without the roller but made with the same material. Mecanum wheels are in the middle - they spin on a 45. If you take a mecanum wheel and spin it, you'll find that it pushes you on a 45. Its hard to describe the feel until you actually do it. No matter which way the wheels spin, they're fighting each other. |
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This year as a mid fielder, I think we play very aggressive defense and can push people around with our mecanums. Also defense isn't always about having the ability to just push people, there is more to it. -Clinton- |
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Try your thought experiment with a single mecanum wheel. Better yet, try a real experiment with a single mecanum wheel, applying a spinning torque to it with your hand, and trying to hold it in place at the same time. You'll definitely find as much force pointing sideways as pointing forward. |
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anyone can respond to this |
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Refer to the scenario described in this post... http://www.chiefdelphi.com/forums/sh...8&postcount=42 ... and consider the following: In the case of 45 degree rollers, the 40 in-lbf of "forward" torque applied to each wheel creates a forward force on each wheel of 10 pounds (due to the reaction force of the carpet) AND a sideways force of 10 pounds (also due to the reaction force of the carpet). These two vectors add (via vector addition) to a force along the axis of the roller of 10*sqrt(2). THAT is how the vector forces work. Consider your second example above with wheels with rollers free to spin on an axis perpendicular to the main axle. The 40 in-lbf of torque applied to each wheel will create a forward force on each wheel of 10 pounds (due to the reaction force of the carpet) and NO sideways force. Consider your first example above with wheels with rollers free to spin on the same axis as the main axle. If you try to apply torque to a wheel, all it will do is spin. Consider a new example with rollers aligned not at a 45 angle, but a 60 degree angle (relative to axis of the wheel). The 40 in-lbf of "forward" torque applied to each wheel creates a forward force on each wheel of 10 pounds (due to the reaction force of the carpet) AND a sideways force of 10/sqrt(3) pounds (also due to the reaction force of the carpet). These two vectors add (via vector addition) to a force along the axis of the roller of 20/sqrt(3). THAT is how the vector forces work. The forward force on each wheel in all cases (except the degenerate case where the roller axis is parallel to the wheel axis) is 10 pounds. 100% of the applied torque shows up as the corresponding forward force. ~ |
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http://www.chiefdelphi.com/forums/sh...5&postcount=50 ~ |
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190 designed a mecanum wheel with double internal yokes (this means that the end pieces of the rollers are cantilevered, and the middle piece is supported on both ends) that is driven like any other mecanum wheel via sprocket on one side. On the other side of the wheel, however, is a friction plate with a face profile that matches the side of the rollers closest the the wheel's axis of rotation. When the friction disk is pressed into the rollers via pneumatic cylinder, the disk prevents the rollers from spinning while allowing the wheel to rotate. When they want to return to mecanum drive, they disengage the friction plate. I will now attempt to illustrate this on my computer... EDIT: Here's the pic. ![]() |
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It seems from your explanation that you might be able to put a carpet-treaded wheel underneath the mecanum wheel at a 45 degree angle and get greater torque at its axle than you put into the original mecanum wheel. Then use that torque to drive another mecanum wheel driving another carpet wheel and keep multiplying forces ad infinitum. It's apparent that something is wrong with your model. |
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Or I could just use the circumference of a small standard wheel to drive the circumference of a larger standard wheel. This would create greater torque on the larger wheel than was input to the smaller wheel. [quote] Then use that torque to drive another mecanum wheel driving another carpet wheel and keep multiplying forces ad infinitum. /quote] Yes, you could do that. Just like I could use the "load" end of one first-class lever to drive the "effort" end of a another second-class lever, ad infinitum. http://www.sciencebyjones.com/first_...rs_drawing.gif There is no "conservation of torque" law in physics either. Quote:
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Power = Torque * Rotational Speed, and is Energy per Unit Time -- in the example of pushing against a wall, the mechanical input power into the wheel is constant. The rollers translate that power 45 degrees from the angle at which it's applied to the wheel (due to the direction of the traction and tangental free spin). The power at 45 degrees cannot be greater than the input power, lest the Law CoE be broken. Thus, to figure out how much tractive force is applied axialy to the roller (given that the roller free spins tagental to its axle), vector components are necessary. Assuming that when the wheel spins a minute amount to push forward on the force sensor, the roller spins an equal amount, and Assuming that there are two wheels oriented such that their roller axles are tangental to each other (and 45 degrees from the axis wheel rotation) The forces generated into the robot are absorbed by the frame. The forces generated forward of the robot are absorbed by the sensor. The two forces, when summed via vector analysis, much equal the sum forces at the 45 degree angle of the respective rollers. Otherwise, the law of CoE is broken since this is in effect, a Power problem. |
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I believe Ether is right. I think you can see the "extra" force as coming from the sideways forces holding the wheels together (these forces must be there, or the wheels would fly apart). This force also gets reduced to the component along the roller, which then has a forward component. The bottom line is, if nothing is slipping sideways, the full torque is being converted to a forward force. It's kind of reminiscent of tacking in a sailboat.
If you're going at 45 degrees, two wheels slip completely so you only get power from half the wheels, so a swerve drive is probably better for a general angle. |
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I encourage you to mull over the vector analysis that I presented and see if it doesn't make sense to you. ~ |
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I'm trying to understand the vectors as you are trying to describe them, and I'm failing utterly. Can you explain how they work when a mecanum drivebase is traveling at 45 degrees? I wouldn't expect more than 100% of the force to be available in that case either. |
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In the case of the 45 degree rollers mentioned in my prior post... http://www.chiefdelphi.com/forums/sh...5&postcount=50 ... the force created by the 40 in-lbf drive torque on the wheel creates a force of 10*sqrt(2) along the roller axis. 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. Quote:
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If you look at a mechanum bot going at 45 degrees, you'll see two wheels on opposite corners don't turn at all, they just slip. A still robot trying to push at 45 degrees would have its force reduced to 1/sqrt(2) compared to the forward direction. |
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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. |
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The frame does more more than absorb a force: It transmits the force. All these forces balance or you'll get sideways acceleration. So, a wheel has more than one input force: From the motor transmitted as a torque around the axle, and from the other wheel transmitted through the frame. Add up all these inputs and you'll get Ether's result.
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This is just bad physics. The math is plain wrong as I'm interpreting it nor does it explain how a frame 'transmits' the force tangent to how the force is applied. This isn't fluid dynamics, where the pressure builds up and escapes through the path of least resistance.
If this can be proven through experimentation and then explained with clear and concise diagram, then maybe I'd consider it as valid the next time I design a robot to sit still and push a wall. |
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Maybe I can explain it this way?
Put two CIMs into a single ToughBox. Run the CIMs in opposite directions. Do the forces balance? Yes. Is there net motion? No. Does that mean the motors won't self-immolate in a binge of stall current thirst? I think we both know the answer here. Many teams have inadvertently fashioned some very efficient OVENS out of the kit-supplied motors... it's not hard to do. Just because there is no net force, doesn't mean the energy isn't going somewhere other than motion. Now construct a gearbox where the two motors are fighting each other at 45 degrees, but work in the same direction at 45 degrees at the same time. In one component the motors face no resistance, and all is fine and dandy. In the other component, you have the same situation as above where the motors are essentially fighting each other, and neutralizing the work done entirely to 0. Does that shed some light into where that remaining ~30% goes? My team is running mecanums this year, and anecdotally that 30% is noticeable - I wouldn't try and push anyone with our robot. But the benefits... they are intriguing :D. |
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With our robot, the mechanum wheel is direct-driven by a toughbox. The motor creates torque around the axle. The rollers will convert that to a force with forward and sideways components. Our wheel is solidly attached to the axle, so the sideways force will be transmitted down the axis of the axle to the toughbox, through the toughbox mounting to a C-channel, etc. If we're not accelerating sideways, the axle is pushing back equally, with the force coming ultimately from the wheel on the other side, and by the same token, the rollers give this a forward component. For every action there is an equal and opposite reaction... A sailboat tacking into the wind is another interesting example. The wind blows on the sail, applying a force, ignoring drag, perpendicular to the sail surface. The sailboat can then accelerate in the direction of the keel. These two projections allow the wind to blow the boat upwind. If anyone's at the Minnesota regional later this week, we can draw some diagrams. |
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Ether, I re-read your first post. It looks like we may be talking past each other. Seems the key is *when* the rollers start slipping, which we've yet to determine. However, the fact remains that there are no mecanum wheels made out of traction tread. A swerve/crab drive robot with traction wheels, geared properly will beat out currently available mecanum wheels every time. Also, mecanum wheels in motion seem to have a built in inefficiency, not all the motor's power can be applied in one direction at once. |
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I was hoping to have a discussion of the forces/torques produced by mecanum wheels in motion after first obtaining some sort of general consensus concerning static analysis of mecanum wheels, but obtaining such consensus has proven quite a bit more problematic than anticipated. Theoretical arguments and gedanken experiments have not been persuasive. When the season is over I hope to get my hands on some mecanum wheels and gather some additional test data. I would be delighted however if someone else beats me to it. ~ |
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While the system does seem complex at first glance, it's really nothing more than a friction clutch system which is functionally identical to the dog clutches in most shifting transmissions used on many robots. |
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I've been looking into this locking mecanum design as seen here and i would like to know if anyone has found a way to make this or something similar work (in reference to the lockable rollers not the rotatable rollers). I would like to apply something like that our robot next year and I'm starting research early. anyway if anyone can help me understand this system your help would be greatly appreciated. :)
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To resurrect the discussion here. As in the examples suggested, one standard wheels would have 10 lb. of torque. Doesn't it follow that 45 degree mechanums would apply 10 lb. of force at that 45 degree angle?
You've agreed that the mechanum has seperate and equal sideways and forward forces, and the disagreement is what the total force is, as well as each seperate force (which would each follow from the other). Can we also agree that the total force is a seperate force*(sqrt)2? I'd also like to resuggest the idea of 60 degree mechanums. It was suggested they would still have 10 lb. forward force, with a lesser sideways force. What if you mounted 30 degree mechanums perpendicularly? With the rollers at the same angle, shouldn't forces be the same? The constant 10 lb. forward force model wouldn't work if I'm correct. I've taken only basic physics (I do understand vector math), so feel free to correct me if I'm way off here. |
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