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Mecanum Wheels - How do they work?
My team and I are looking at using Mecanum wheels for this year's competition as a means of strafing. My few questions are:
Any and all diagrams/pictures/screenshots of LabVIEW are welcome! |
Re: Mecanum Wheels - How do they work?
Mecanum wheels have rollers positioned at 45 degrees from the axis of rotation. This transfers the force from the rotation of the wheel by the motor 45 degrees from the axis of rotation. When you power each wheel independently and vary the speed and direction the wheels rotate, you can achieve omnidirectional motion. So, to answer you first question, you need 4 wheels (2 left, 2 right) to drive forward, backward, sideways, and anything in between. The rollers should be oriented such that they form an X when you look from above. If you search "mecanum" in the white papers section, you will find more in depth explanations and example code.
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Re: Mecanum Wheels - How do they work?
What is the "White Papers Section," and with that answer arose another question - do each of the four wheels need to have their own motor/gearbox config.?
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Re: Mecanum Wheels - How do they work?
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There is a good whitepaper here on Chief Delphi describing some of the physics concepts behind the wheel and clearing away some of the misconceptions about the wheels. It was written based off some of Ether's posts but I don't remember who it's by (can someone post a link to this whitepaper if you know which one I'm referring to?). As for programming, there should be a "holonomic drive" VI in the WPI library VIs section that does the math for mecanum automatically. You have one joystick axis for forward and backward movement, one for strafing, and one for rotation (much like arcade). |
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Also to sketch a rough text-based diagram: Where "///" represents the 45* rollers of the Mec Wheels. And "|" "-" represent the Robot's Frame. |---|------------|---| |\\\|...............|///| |....|...............|....| |....|...............|....| |....|...............|....| |///|...............|\\\| |---|------------|---| |
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Re: Mecanum Wheels - How do they work?
You'll also want some form of suspension or flex in your frame to keep each wheel in contact with the ground (at a minimum), you'll do best with an equal weight distribution on your wheels, and the wheels should be in a square.
Best of luck. |
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FIRST may try their best to keep the carpet flat, but every year there seems to be bumps and places to get hung up on. The kinematics aren't easy to resolve when only 3 of the 4 wheels touch the ground. |
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Holonomic drive has two options, Polar and Cartesian. In Polar you specify a direction, magnitude and rotation. In Cartesian you specify x, y, and rotation. Personally, I like Cartesian better than Polar because it is more similar to a traditional arcade drive. Try both and see which you like better. The code I've attached shows how our team used mecanum in 2011. The left joystick controls forward-back and strafing. The right joystick does turning. |
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Refer to attached sketch First, the kinematics (ideal): Let P be the perimeter of the rectangle formed by the centers of the 4 wheels: P = 2*(Lwb+Ltw); Let r be the wheel radius For a given vehicle rotation speed omega_v (radians/sec) of a vehicle rotating in-place about the center of the aforementioned rectangle (i.e. Vx=0 and Vy=0), the wheel rotational speed omega_w (rad/sec) will be given by omega_w = (1/r)*K*omega_v .... (see page 7 of my mec kinematics paper) ... where K = (Lwb+Ltw)/2 = P/4; solving for omega_v and substituting for K: omega_v = (1/K)*r*omega_w = (4/P)*r*omega_w; So you can see that the vehicle rotation speed for a given wheel rotational speed, is the same for all rectangles with the same perimeter. Now, the forces and torques (ideal): wheel torque: tau; carpet force component in plane of wheel and floor: Ff = tau/r; total carpet force in direction of mec roller axis: Fr = Ff*sqrt(2) = (tau/r)*sqrt(2); Let the ratio f be defined as: f = Ltw/Lwb; theta = atan(f); .... (see sketch) alpha = pi/4 - theta; carpet force component in direction of vehicle rotation: Fv = Fr*cos(alpha) = (tau/r)*sqrt(2)*cos(alpha); distance from center of rectangle to center of wheel: D = (1/2)*sqrt(Lwb^2+Ltw^2); P = 2*(Ltw+Lwb) = 2*Lwb*(f+1) => Lwb = (1/2)*(P/(1+f)) & Ltw = f*(1/2)*(P/(1+f)); so D becomes: D = (1/2)*sqrt(((1/2)*(P/(1+f)))^2+(f*(1/2)*(P/(1+f)))^2); D = P*sqrt(1+f^2)/(4*(1+f)); torque about center of rectangle: Tv = Fv*D; Tv = ((tau*sqrt(2))/r)*cos(alpha)*(P*sqrt(1+f^2)/(4*(1+f))); Tv = ((tau*sqrt(2))/r)*cos(pi/4 - theta)*(P*sqrt(1+f^2)/(4*(1+f))); Tv = ((tau*sqrt(2))/r)*cos(pi/4 - (atan(f)))*(P*sqrt(1+f^2)/(4*(1+f))); using: cos(pi/4-atan(f)) = (1+f)/(sqrt(2)*sqrt(1+f^2)); ... Tv simplifies to: Tv = ((tau*sqrt(2))/r)*((1+f)/(sqrt(2)*sqrt(1+f^2)))*(P*sqrt(1+f^2)/(4*(1+f))); Tv = (P*tau)/(4*r); Now that I've done all that trig and algebra, here's a much quicker way: So the torque on the vehicle is the same for all rectangles with the same perimeter. Now consider the non-ideal case with roller free play, roller friction, and carpet compression and stretching. This will be an intuitive explanation. In the case where Ltw >> Lwb (very wide configuration), the fore/aft component of the wheel translational motion is much larger than the strafing component. In the case where Lwb >> Ltw (very long narrow configuration), the strafing component of the wheel translational motion is much larger than the fore/aft component. Since a non-ideal mec wheel is considerably less efficient in the strafing direction, you'll see more losses when trying to rotate when Lwb >> Ltw. Bottom line: For the small deviations from square typically seen with FRC mec bots, and for properly functioning mec wheels on a frame that's sufficiently flexible to maintain traction on all 4 wheels, the bot should turn fine - and certainly better than a skid-steer with the same dimensions. |
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This is great stuff. |
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