4" Wooden Mecanum Wheel

[Ether]

Originally Posted by TheOtherGuy

The third link you provided gave a ghastly equation (well, a few relatively nice equations with lots of substitution ) on the last page for estimating the parabola, yet when I plot it, the length is off by quite a bit (should go to sqrt(7/2))...

I don't follow you. What are you plotting and what are you comparing it to?

[TheOtherGuy]

Originally Posted by Ether

I don't follow you. What are you plotting and what are you comparing it to?

Sorry, should have elaborated. The blue line is the ellipse I'm currently using for the roller contour, and the purple is the equation found on the last page of the third link. Specifically, the length of the roller L is 2sqrt(7/2)" (from ellipse equation), R is 4", and r is .5". The equations are then:
D = R – r = 4-.5 = 3.5
F = (sqrt(2*3.5^2+(1/2*sqrt(7/2))^2))
G = (sqrt(4*3.5^2+(1/2*sqrt(7/2))^2))
T = (4sqrt(2)/ sqrt(2*3.5^2+(1/2*sqrt(7/2))^2))
A = 32*(2*r-G*(T-1)) / (L^2*(T+1)^2)

But when I plot it, the roots aren't +/- sqrt(7/2). Oh well, maybe I'm just no good at copy-pasting.

[Ether]

Originally Posted by TheOtherGuy

Sorry, should have elaborated. The blue line is the ellipse I'm currently using for the roller contour, and the purple is the equation found on the last page of the third link. Specifically, the length of the roller L is 2sqrt(7/2)" (from ellipse equation), R is 4", and r is .5". The equations are then:
D = R – r = 4-.5 = 3.5
F = (sqrt(2*3.5^2+(1/2*sqrt(7/2))^2))
G = (sqrt(4*3.5^2+(1/2*sqrt(7/2))^2))
T = (4sqrt(2)/ sqrt(2*3.5^2+(1/2*sqrt(7/2))^2))
A = 32*(2*r-G*(T-1)) / (L^2*(T+1)^2)

But when I plot it, the roots aren't +/- sqrt(7/2). Oh well, maybe I'm just no good at copy-pasting.

Your parabola plot is correct.

The equation simplifies to y= 0.5-0.066683598*x^2

The roots are not supposed to be +/-sqrt(7/2). Why do you think they should be?

If you want the radius to go to zero, you need a longer roller.

[Matt H.]

Originally Posted by TheOtherGuy

Hey Matt! How's MIT?
Mr. Forbes mentioned those as a first solution, too. It may be worth looking into. I just had a crazy idea that may or may not work, but if it's possible to drill and tap into the end of a #10 threaded rod, maybe I can screw on a small washer onto the ends of the rod.

MIT is challenging, but also fun.

Working from http://www.engineersedge.com/screw_threads_chart.htm, tapping the threaded rods seems as though it will present its own set of difficulties.

Your 10-24 threaded rod has a 0.138" minor diameter (total diameter minus the depth of the threads). This means you will need to use 4-48 or small fasteners to have any material between the inner screw and the outer threaded rod.
Tapping #4 machine screw holes in steel by hand is a pain and repeating the process 48 (or is it 96?) times would likely result in several broken taps/drill bits.

Also, I believe you get into issues with loading on the screws and threaded rod. Speaking qualitatively #4 screws are at risk of breaking during shock loading.

[MrForbes]

Kevin and I talked on the phone about it, we came up with a few other solutions...like using round, internally threaded, 1/4" od spacers. And a few other ideas.

Good to hear you're keeping challenged, that was always your problem

[Manoel]

Originally Posted by squirrel

heh...I used the compound rest. I suggested that the error would be pretty low if we made the half roller as two cones. Kevin figured the angles at 6 and 16 degrees, I set the compound rest to those angles to turn it. then we sanded the lump in the middle to radius it.

Squirrel,

This is how we turn the rollers for our mecanum wheels: http://www.youtube.com/watch?v=1-Zvib5nZVw&hd=1

[MrForbes]

Thank you for posting a link to the video! I thought about using a coarse abrasive if we used rubber rollers, but that would work well with wood too. It makes a mess, but it's an easy way to make the parts.