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Dear All,

There is no rational for using the centre of gravity as a meeting point for the
group, or at least none has been brought to light so far in the course of this
discussion. It seems to me that there is a broad consensus amongst members of
this list that a reasonable goal is to minimize travel distance in one way or
another. But the centre of gravity does not minimize travel distances in any
way! The beach-ball-with-coins paradigm is therefore misguided and so is any s/w
implementation thereof.

It might be a good idea to define the problem before we attempt a solution.
Somebody suggested to minimize the total travel distance. ("If this were a
company trying to hold a meeting..."). Clearly, this solution would be identical
to minimizing the average distance, since the total distance travelled equals
(by definition!) the average distance that each individual travels multiplied by
the number of participants. While this strategy would be the optimum for the
group as a whole, it might be unfair to some individual. To see this, consider
two participants: Any meeting point on the great circle connecting them would
minimize the total travel distance. If indeed these were two employees of a
company to be sent to a meeting, it would not matter where on that great circle
they meet, the company would always pay the same cost in travel time and fare.
But if both participants shall be burdened equally, only the mid point would be
"fair". The mid point would be the solution to a least square sum algorithm, and
such an algorithm would also deliver a "fair" solution for more than 2
participants, although in general each participant would no longer be burdened
equally. Obviously, for more than 3 participants there is no way of burdening
each participant equally.

The question is thus (and note that this is not a mathematical but an ethical
question): Are we looking for a minimum solution, and if so, shall we minimize
average or total travel distance, or what else?

For disbelievers who still think the centre of gravity might do something for
us, here is an example that proves that the c.o.g. is not minimal w.r.t. the
above. This new example does not suffer from the "flaw" of my previous ones in
an other message that dealt with degenerate cases:

Let A be located at 0N/0E, B at 30N/90E, C at 30S/90E. Convince yourself that
the centre of gravity G is located at 0N/60E by using one of the two posted
computer programs. Now find the distances from G to A, B and C as 60 deg,
41.40962 deg and 41.40962 deg respectively. Hence the average is 47.60641 and
the square root of the sum of the squares is 83.84219.

Now assume M 1 deg to the east of G at 0N/61E. The distances become 61 deg,
40.76072 deg and 40.76072 deg deg. The average travel distance is reduced to
47.50714. The solution is better for the group as a whole. Total distance
travelled by all is nearly 2 deg less than with the c.o.g. solution.

Then assume N 1 deg to the west of G at 0N/59E. The distances become 59 deg,
42.06984 deg and 42.06984 deg. The square root of the sum of the squares becomes
83.78987 and is smaller than for G. The solution is fairer to all, because A,
who travels the farthest, saves 1 deg, while B and C both have to travel a
little over half a deg farther, compared with the c.o.g. solution.

This shows that the centre of gravity does not yield a minimal solution in
either sense.

Herbert Prinz (from 1368950/-4603950/4182550 ECEF)

   

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