"Perpetual" motion - based on "perpetuum motion" video

"Perpetuum Motion" laser cut design

11-JAN-2016

Sadly, I've let my Techshop membership expire, but in a flurry of last-minute building before it went away, I got intrigued by "perpetual motion" machines that are depicted on youtube, and got in my head that I wanted to build one.

Ready for some good ol' geometry and trig?

The video

The starting point is to see the video at YouTube.  Search on "Perpetuum Mobile HD".

About 2m35s into it, you'll see a drawing attributed to Edward Somerset, and resulting machine that someone built.  I thought to make the whole thing, laser cut, but before doing that, I'd have to figure out the geometry and trig around it.

The same drawing is attributed as "Jacob Leupold’s Overbalanced Wheel" at http://jamesminshall.com/sustainability/energy-generation/perpetual-motion-generators/, though I personally find that page a bit overreaching in its claims.

I also started down the path of making one more like the "six orange segments cut out of a circle" in the first image of that page, which was easier to draw and cut, but building the "dumbbell" weights that slide along it proved to be too challenging for the scale I'd used.

The calculations

At the heart of the overbalanced wheel were a couple of variables up to anyone to choose.

The initial drawing looks sort of like this (drawn in Corel Draw 5).  This is a mirror image of what many online images/videos depict.  Laid out this way, it would rotate counter-clockwise.  I choose this representation since it makes the angle calculation visualization a bit less taxing.

Wheel geometry, 12 segments, 30-degree theta, R=5, D=8

 In this form, I see there being three degrees of freedom.  First is the R, the radius of the inner circle.  Next is what I call D, the length of the longer walls that rach the outside wall.  Finally, there is the number of segments -- in this case 12 -- and that determines theta, the angle of rotation from one R to the next (and therefore also about the origin, from one D to the next).

Definitions of R and D

Enlarged, we have this:

Labeling of theta, R, D

Calculating the outer circle radius

I want to compute the radius from the origin to the outside edge.  I chose A to be the angle (180-theta)/2, and D is the hypotenuse of that triangle.  The height from the horizontal to the first edge point (what I call d0) is H, and H = D*sin(A).

I then split R into two parts: XDL and XDR, and XDR is H*sin(A).  Therefore
A = (180-theta)/2
H = D*sin(A)
XDR = D*cos(A)
XDL = R-XDR
d0 = (XDL, H)
( AKA,d0x = XDL, d0y = H)
 Q = sqrt(XDL*XDL + H*H)

Labeling of A, H, XDL, XDR, Q, and d0

Where D meets the next D

The next intersection that's interesting is the one where the second R intersects our first D.  Thought of another way, that's where one D meets the next D.  I call that point (x1,y1) on the drawing below.  That can be found a number of ways.  The simple solution is
x1 = R*cos(theta)
y1 = R*sin(theta)
Then, based on similar triangles, we can find the ratio of distances to see where it lands, percentage-wise, along the length of D.  So, Base is to D as y1 is to H
Base/D = y1/H
Base = D*y1/H

"Base" also is the base length of the isosceles triangle formed by theta, so it can be computed that way, too.

Labels for Base, RestD, (x1,y1), qhyp

Calculating the distance between outer edge points

The last point that's interesting is (d1x, d1y).  This I chose to do using a normal rotation matrix, since all the edge points rotate by theta, just as all the other points on the wheel.

The simple rotation matrix approach says
x' = x*cos(theta) - y*sin(theta)
y' = x*sin(theta) + y*cos(theta)
(Ref. https://en.wikipedia.org/wiki/Rotation_matrix)

Thus
d1x = d0x*cos(theta) - d0y*sin(theta)
d1y = d0y*sin(theta) + d0y*cos(theta)

And the outer edge wall length (for some reason I called this "qhyp") is
qhyp = sqrt((d1x-d0x)^2 + (d1y-d0y)^2)

In retrospect, now that I draw it, I realize that the computation of qhyp could be much easier, because it's the base of a larger isosceles triange, still with theta as the top angle, and common leg lengths of Q.  Half the triangle base is Q*cos(A), and therefore qhyp = 2*Q*cos(A).

Or, for those of you who like the more obscure functions, qhyp == 2Q / sec(A).

Having qhyp lets me make each of the 1..nSegments outer walls.  The original video shows an outer wall more like the floor of a crude bridge, where individual narrow planks are placed at the edges of a circle of radius Q.  For my design, instead, I'm intending to cut a slightly larger circle, and have tab holes that hold the nSegments outer walls in place.

Knowing all the computations, I can create a program that generates all the lines and points and angles, and have it draw a laser-cuttable pattern for me, just to show where all the walls would go.

The intent is to create a disc with tab holes to cut out, so each wall could be put in at the right points.  Then, the walls of some given wallHeight could be constructed with knowledge of D (the inner edge walls), and qhyp (the outer edge walls).  The tabs and tab hole lengths would be some fraction of each wall length, but the tab hole width would have to be variable, based on the thickness of the measured material being cut.

And yeah, I already did write the program to do this, and have some notes on SVG generation and laser cutting experimentation to share, but I'll cover those topics in a later post.