New sw method: "extreme scale"

[DEMOREP1 at aol.com]

Mr. Eppley wrote in part:
The interesting feature about this new method is that it's not a
ranked ballot method.  It's a rating method.

Define E to be the number of eligible voters.  
Define C to be the number of candidates.

Define S to be 2EC rounded up to the next power of 10.  This may
be a huge number.  For example, with C = 3 candidates and E = 100
voters, 2EC is 600, so S = 1000.

1. Each voter rates each candidate on a scale ranging from 0 to S.
2. The score of each candidate is the sum of the ratings assigned it 
by the voters, normalized by dividing by S.  The winner is the 
candidate with the highest score.
--------
D- A possible U.S. election for U.S. President--  6 candidates (C),  101
million voters (E), 2EC= 1,212 million= 1.212 billion, so S= 10 billion.
Good luck to many voters to avoid making errors in the number of millions or
billions of ratings that they give to each candidate.
---------
Mr. Eppley wrote more:

The advantage of the huge scale is that even though smart voters will
vote near the extremes (0 or S) for most or all candidates due to
strategy concerns, it's possible for the voter to rate a true
favorite higher than a lesser evil without fear of electing the
greater evil, if the voter knows s/he is part of a smart majority 
which prefer the lesser evil more than the greater evil.

Example:  3 candidates A, B, and C.  99 voters.  S = 1000

   49: A > B
   50: B > A

Suppose some of the 50 also prefer C more than B.  
They can vote:  
   A =    0
   B =  999
   C = 1000

As long as all 50 of the majority rate B at least 999, B's score will
be at least 49.950.  If they also rate A=0, then even if all 49 A
supporters vote A = 1000, A's score will be at most 49.000.
----
D- 
In plain Condorcet, some of the 50 B voters could vote C ahead of B and A.

Example--
49 AB
40 BA
10 CBA
B wins in plain Condorcet. The 10 CBA voters expressed their opinion.

There is also the factor of *if the voter knows s/he is part of a smart
majority*. A slight error in estimating the size of the smart majority might
cause a major strategic error.
Example--
 the initial estimate of 
   49: A > B
   50: B > A
could easily be the reverse
50: A > B
49: B > A
due to polling errors.

For B to win he/she must average just above 980 of a maximum of 1,000 among
the 50 nominal B voters. (980 x 50 = 49,000, maximum of A)

Example-
49 A=1000
48 B= 1000
1   C= 1000
     B=   500
1   C= 1000
     B=   499
A= 49,000, B= 48,999, C= 2,000

A wins. B's average is 979.88.

With even more candidates, would B get his/her above 980 average per voter ?

Mr. Eppley's proposal seems to be a variant of Borda's method.

I point out again that nice examples can be thought up in any method but that
 not so nice Arrow strategic voting possibilities must be brought up.

Mr. Eppley's proposal again brings out the fact that even second choices
might be very weak relatively on the +100 percent for to -100 percent against
scale (and the tendancy to vote the + or - extremes).