4" Wooden Mecanum Wheel

[MrForbes]

So I need to get the parabola attachment for my 1946 South Bend 9" lathe?

[Ether]

Originally Posted by squirrel

So I need to get the parabola attachment for my 1946 South Bend 9" lathe?

You can get it from the same place you got the ellipse attachment :-)

[MrForbes]

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.

[Ether]

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.

You could do a similar thing for the parabola.

Or if you had time to kill you could cut a series of steps.

Or not :-)

Anyway, I thought the math might be of interest.

[Fe_Will]

Have you considered a T nut driven into the end of each roller? You may need to modify the flange diameter but that can be accomplished with a grinder if needed.

[Matt H.]

They make products for this situation!

Undersized Machine Screw Hex Nuts are made slightly smaller than regular hex nuts:
http://www.mcmaster.com/#undersized-...x-nuts/=b192e8

Using the 10-24 undersized nut will reduce the corner to corner diameter by ~0.1 in.

If that's not enough you can use Allen nuts which would reduce the corner to corner diameter by ~0.12. However they are pricey ($0.90/nut).

http://www.mcmaster.com/#allen-nuts/=b18xlg
http://en.wikipedia.org/wiki/Internal_wrenching_nut

[TheOtherGuy]

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.

Looking back at it, the dowel comes slightly smaller than an inch at the largest, and our estimations cut it down on the inside by another fraction of an inch, which may have made the end of the roller near the nut too small a diameter. Then again, I haven't wrapped it yet... maybe I'll go buy some tape and try that out.

The whole parabola thing is very interesting, thanks for bringing that up! The difference is hardly noticeable in CAD, but there is a slight bulge near the ends of the rollers, despite the rollers meshing perfectly. 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))... http://www.wolframalpha.com/input/?i=plot+y%3dsqrt(4-(x^2)/2)-1.5,+y%3d.5-(32(2*.5-(sqrt(4*3.5^2%2b(1/2*sqrt(7/2))^2))(+(4sqrt(2)/+sqrt(2*3.5^2%2b(1/2*sqrt(7/2))^2))-1))/(14*((4sqrt(2)/+sqrt(2*3.5^2%2b(1/2*sqrt(7/2))^2))%2b1)^2))x^2,+x%3d-2+to+2&incParTime=true

T-nuts are a possibility, although that would require changing the entire setup of the rollers so the axle is live. I'm not sure if that's a good idea with 3/16" plywood, but press fitting a small piece of the aluminum rod into the hole might give it enough strength.

Originally Posted by Matt H.

They make products for this situation!

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.

[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.

[Ether]

Originally Posted by Ether

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]

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]

Originally Posted by TheOtherGuy

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:

Originally Posted by TheOtherGuy

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

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]

it gives me a warm fuzzy feeling to see my son arguing about (I mean discussing) equations

[joek]

Originally Posted by squirrel

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