[EM] Clock Methods

[Forest Simmons]

> Date: Mon, 14 Feb 2005 01:46:38 +0000
> From: Gervase Lam <gervase.lam at group.force9.co.uk>

<snip>

>
>>>                 A
>>>                 .
>>>       A>C>B .       .[4] A>B>C
>>>
>>>   Not(B) .              . Not(C)
>>>
>>> C>A>B [2].                . B>A>C
>>>
>>>        C .              . B
>>>
>>>       C>B>A .       .[3] B>C>A
>>>                 .
>>>
>>>               Not(A)
>
> [C>A>B was written at the 7 o'clock position.  I have now corrected it to
> C>B>A.]
>
> I don't think the centre of mass is on B.
>
> The total of all the votes is 2 + 3 + 4 = 9.
>
> Draw a vertical and horizontal lines so that they are perpendicular to
> each other and cross each other at the centre of the circle.  With respect
> to these lines, the co-ordinates of C>A>B is obviously (-1, 0).
>
> The A>B>C and B>C>A voters are respectively 60 degrees above and below the
> horizontal line.  Therefore, their x co-ordinates are respectively cos +60
> and cos -60 (i.e. 1/2 and 1/2).  Also, their y co-ordinates are
> respectively sin +60 and sin -60 (i.e. sqrt(3)/2 and -sqrt(3)/2).
>
> The x co-ordinates of each of the rankings is weighted by the number of
> votes cast for each ranking.  This is then used to work out the weighted
> mean x co-ordinate of the clock or spinning-top.
>
> ((2 * -1) + (4 * 1/2) + (3 * 1/2)) / 9 = 3/18
>
> The weighted mean y co-ordinate of the spinning-top is worked out in the
> same way except that the y co-ordinates of the rankings are used instead
> of the x co-ordinates.
>
> ((2 * 0) + (4 * sqrt(3)/2) + (3 * -sqrt(3)/2)) / 9 = sqrt(3)/18
>
> Therefore, the co-ordinates of the centre of mass is (3/18, sqrt(3)/18)).
>
> Now to work out the angle of the line from the centre of the circle to the
> centre of mass:
>
> Tan [Angle] = (sqrt(3)/18) / (3/18) = sqrt(3)/3 = 1/sqrt(3).
>
> Therefore, the angle of the line is 30 degrees above the horizontal line.
> This means the line points at Not(C).
>
> This is not nice in this case because the voters are equally split between
> candidates A and B.  There is nothing between them.


Here's another way to arrive at this same result:

The respective Borda counts of A, B, and C are 10, 10, and 7.

Think of A, B, and C as being vectors, and find the resultant of

      10A+10B+7C

Which can also be expressed as 3(A+B)+7(A+B+C)
which equals 3(A+B) since A+B+C=0 by symmetry.

Of course A+B is the same as not(C), so the resultant vector points in the 
direction of not(C), as you also found.

If we interpret not(C) as a toss up between A and B, then this result 
makes sense.

The problem with Borda (hence this center of gravity approach) is that it 
is clone dependent.

But sprucing up removes the clone dependence and reduces public elections 
to the three candidate case, so this method is still worthy of 
consideration.

Forest