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-   -   4" Wooden Mecanum Wheel (http://www.chiefdelphi.com/forums/showthread.php?t=91726)

Ether 14-02-2011 18:31

Re: 4" Wooden Mecanum Wheel
 
Quote:

Originally Posted by TheOtherGuy (Post 1022968)
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?




Manoel 14-02-2011 18:48

Re: 4" Wooden Mecanum Wheel
 
Quote:

Originally Posted by squirrel (Post 1022828)
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

TheOtherGuy 14-02-2011 22:03

Re: 4" Wooden Mecanum Wheel
 
Quote:

Originally Posted by Ether (Post 1022974)
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.

MrForbes 14-02-2011 22:04

Re: 4" Wooden Mecanum Wheel
 
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.

Matt H. 14-02-2011 23:05

Re: 4" Wooden Mecanum Wheel
 
Quote:

Originally Posted by TheOtherGuy (Post 1022968)
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 14-02-2011 23:06

Re: 4" Wooden Mecanum Wheel
 
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 :cool:

Ether 14-02-2011 23:15

Re: 4" Wooden Mecanum Wheel
 
Quote:

Originally Posted by TheOtherGuy (Post 1023158)
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.





Ether 15-02-2011 07:59

Re: 4" Wooden Mecanum Wheel
 
3 Attachment(s)
Quote:

Originally Posted by Ether (Post 1023230)
Your parabola plot is correct...

I found the problem: the ellipse you plotted was not correct.

The ellipse you plotted was y=sqrt(4-x^2/2)-1.5 (see attachment 1).

The ellipse should be (sqrt(64-2x^2)-7)/2 (see equations #1 and #4 of attachment 2).

The ellipse in attachment 2 is plotted in attachment 3. It is a close (but not exact) fit for the parabola you plotted.

If your rollers are indeed contoured per the ellipse in attachment 1, then they are quite a bit off.


[edit]The good news is, this means a larger radius for your end fastener[/edit]




TheOtherGuy 15-02-2011 10:48

Re: 4" Wooden Mecanum Wheel
 
From eq.#1, it looks like you used the diameter of the mecanum wheel as the radius instead of the radius. The equation for the ellipse without translation should be y^2/4 + x^2/8 = 1. The second ellipse needs to be translated up 3 units in order to give the roller a diameter of 1 in the middle, so the second equation is (y-3)^2/4 + x^2/8 = 1. Solving for y, you should get 3/2, and plugging that back into the first equation gives roots of x as +/- sqrt(7/2).

Intersection of two ellipses

Ellipse shifted down 1.5

Roots of ellipse shifted down

I did a quick check on the wheel I CADed, and the rollers do indeed follow these ellipses (and have the correct side profile on the wheel itself).

Heh, just checked back on the parabola equation, seems I made a small error in setting the radius of the wheel R equal to 4" instead of 2"! That would do it. Here is the fixed parabolic equation with the ellipse above. Sorry about the confusion.

Ether 15-02-2011 12:27

Re: 4" Wooden Mecanum Wheel
 
Quote:

Originally Posted by TheOtherGuy (Post 1023465)
From eq.#1, it looks like you used the diameter of the mecanum wheel as the radius instead of the radius.

I used the radius that you gave me:

Quote:

Originally Posted by TheOtherGuy (Post 1023158)
Specifically, the length of the roller L is 2sqrt(7/2)" (from ellipse equation), R is 4", and r is .5"


Quote:

Heh, just checked back on the parabola equation, seems I made a small error in setting the radius of the wheel R equal to 4" instead of 2"! That would do it. Sorry about the confusion.
Not to worry. The difference between the ellipse and parabola is probably negligible for these size rollers anyway.

I did the analysis mostly to satisfy my own curiousity. However, for larger rollers requiring expensive tooling to be commercially produced, it might be worth using the parabola instead of the ellipse.



MrForbes 15-02-2011 14:05

Re: 4" Wooden Mecanum Wheel
 
it gives me a warm fuzzy feeling to see my son arguing about (I mean discussing) equations

joek 15-02-2011 14:56

Re: 4" Wooden Mecanum Wheel
 
Quote:

Originally Posted by squirrel (Post 1023600)
it gives me a warm fuzzy feeling to see my son arguing about (I mean discussing) equations

only in FRC....
wood=fragile, carve them from solid plexi

MrForbes 15-02-2011 14:59

Re: 4" Wooden Mecanum Wheel
 
We try not to solve problems by throwing technology at them...we like to think of ways to make stuff using supplies/equipment that we have already, or can find easily. Big chunks of plastic and the machines needed to machine them to complex shapes, are not stuff we have or can get locally.

joek 15-02-2011 16:54

Re: 4" Wooden Mecanum Wheel
 
Quote:

Originally Posted by squirrel (Post 1023632)
We try not to solve problems by throwing technology at them...we like to think of ways to make stuff using supplies/equipment that we have already, or can find easily. Big chunks of plastic and the machines needed to machine them to complex shapes, are not stuff we have or can get locally.

good point, then i suggest metal


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