4" Wooden Mecanum Wheel

[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

[MrForbes]

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]

Originally Posted by squirrel

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