ORTHOTENY – A LOST CAUSE: Part 2

By Dr. Donald E. Menzel

Dr. Menzel is Director of Harvard College Observatory and Professor of Astrophysics at Harvard University.  In Part 1 of his article, which appeared in the May/June, 1965, number of this REVIEW, he asked Aimé Michel several questions concerning his straight line theory and the plotting of the renowned BAVIC line.  Michel’s reply appeared in the same issue.  Dr. Menzel’s statistical formulas, which are presented in this Part, will be examined by a prominent mathematician in the next issue. 

[The mathematician was André Avez.  His article was circulated privately but not received by FSR (see the editor’s apology, September/October 1965, p. 14).]

THE general confusion that exists concerning the statistical basis of Orthoteny, the tendency reported by Michel of Flying Saucer sightings to align themselves along straight lines, indicates the need for a more rigorous derivation of the formulas involved.  I originally used Mebane’s formulas and figures to minimise possible criticism for changing the rules.  However, since his expressions do require correction for the dependence of the corridors upon the observations, I shall here derive the exact formulas for the study. 

[Menzel’s reply to criticism by Seevior & Huxley, students at Cambridge University, is omitted here.]

Upon a map of France draw some arbitrary corridor, consisting of two parallel lines.  Let this corridor occupy the fraction ƒ of the total area of France.  Now consider the random dispersal of n points upon this map.  The chances that the first point will fall into the corridor is ƒ.  The chance that the first two points will fall in the corridor is ƒ².  The chance that the first m points will fall in the corridor is ƒ^m.  Similarly, the chance that the remaining n−m points fall in the area outside the corridor is (1−ƒ)^n−m.  Hence the “a priori probability” of getting exactly m points into the corridor in some specified order is the product, ƒ^m(1−ƒ)^n−m. 

In our statistical problem, only the ultimate distribution of the points, not their order of arrival in the corridor, concerns us.  The number of ways of distributing n particles on a map so that m of them falls into the corridor and n−m outside the corridor is equal to the number of combinations of n things taken m at a time, or

( n ) = n! m m!(n−m)!

(1)

where the symbol “n!” denotes 1×2×3 … n, or n factorial.  Thus the probability of getting exactly m points in the corridor is the product

P(n,m)= ( n ) ƒm(1−ƒ)n−m m

(2)

[Remark about Seevior & Huxley’s criticism is omitted here.]

It is well known that this probability is a term in the binomial expansion of [(1−ƒ)+ƒ]ⁿ.  Hence

∑ n P(n,m)=[(1−ƒ)+ƒ]n=1n=1 m=0

(3)

In other words the sum of the probabilities over all conceivable distributions is 1. 

If we repeat the experiment W times, either with the same or with different corridors, the probable number of m-point lines becomes:

N(m)=WP(n,m)

(4)

If the corridors do not have the same weight,

N(m)= ∑P(n,m)   w

(5)

where ∑ denotes the summation over the W corridors. 

[Equation (5) is given above as printed in FSR.  I am not sure what was intended, or what Menzel means by the “weight” of a corridor.  However, the equation does not seem to be used below.]

In the foregoing review of basic statistics, we have assumed that the corridor was specified in advance.  How does the problem change when the observations themselves determine the corridor?  We use any pair of corridor points to fix the corridor.  These points do not have to fall in the corridor by chance.  We merely put them arbitrarily into the corridor and then randomly distribute the remaining n−2 points.  The correct formula for the probability of getting exactly m points into the corridor, when two of them are put in directly, is:

P(n,m)= ( n−2 ) ƒm−2(1−ƒ)n−m m−2

(6)

These probabilities are terms in the binomial expansion of [(1−ƒ)+ƒ]ⁿ⁻² and sum to

∑ n P(n,m)=1 m=2

(7)

as before.  The first term, P(n,2)=(1−ƒ)ⁿ⁻², is equal to the probability of finding only 2-point lines on the map.  Since all lines have at least two points, the factor (1−ƒ)ⁿ⁻² is simply the probability that no extra points fall into the corridor. 

As before, the expected number of m-point lines is equal to the probability of finding an m-point line times the number of corridors.  For n independent points, the number of corridors is equal to the number of pairs of points, or

W= n(n−1) = ( n ) 2 2

(8)

However, this situation would arise only if no 3-point or higher-point lines existed. 

Let N°(m) be the number of m-point lines observed in some special example.  It is easy to show that the effective number of corridors resulting from n points, is

W= ∑ n N°(m)= ( n ) − ∑ n (m−2)(m+1) N°(m) m=2 2 m=3 2

(9)

This is the actual number of corridors resulting from any given distribution of lines.  The expected number of m-point lines, therefore, is

N(m)=P(n,m)W

(10)

If the observed distribution is random, we should expect N(m) to equal N°(m).  Since the quantity W is constant for such distribution, the ratio

N(m) = P(n,m) = ( n−2 ) ( ƒ ) m−2 N(2) P(n,2) m−2 1−ƒ

(11)

is independent of W.  For the same ratio, Mebane’s formula gives

N(m) = ( n−2 ) ƒm−2 /( m ) N(2) m−2 2

(12)

which contains the extra neons [sic, extraneous?] and sometimes appreciable factor, ( m 2 ) in the denominator.  Specifically, his formula will predict too few lines of large m. 

[Statistical discussion of the width of a corridor is omitted here.  Concludes that “the corridor should contain … 68 per cent of the observations that define it.” I do not understand this.]

The length of the corridor will vary, of course, from only a few miles up to the longest possible line crossing France.  The length, l, of the average line will depend on the shape of the border.  Approximating it with a square, one easily determines that the average line bounded by a square is almost exactly equal to the length of one of its sides.  France, with an area of 213,000 square miles, is equivalent to a square 460 miles on a side.  The fraction of this area contained in a rectangle 2D miles wide by 460 miles long is

ƒ=2Dl/l2=2D/l=D/230

(17)

The value of ƒ, therefore, depends on the width we assign to the corridor. 

Michel suggests a corridor width of 5 miles, which gives

ƒ=1/92

However, in re-defining it he obtains a value of

ƒ=1440/65000=1/45

Mebane, defining as straight two lines that meet at an angle of 178½° or greater, suggests

ƒ=1/80

Nevertheless, I point out that a triangle with such a flat angle would fit neatly inside a corridor of

ƒ=1/150

I pointed out in my earlier article that Mebane did not properly define his “corridor”.  His discussion requires that ƒ decrease with m, whereas I think that ƒ might show some slight increase.  Three lines meeting at the same angle and bent in the same direction would require a corridor of double width.  And the lines of many points tend to be longer than those of fewer points. 

[Remarks about Seevior & Huxley’s criticism are omitted here.]

As a matter of fact, the only way I can see of finding ƒ is by experiment, as Mebane himself has suggested.  A least-squares solution would yield ƒ as a by-product and clearly ƒ will vary from corridor to corridor.  An inspection of Michel’s straight lines indicates that he has been extremely careless in his drawing of the lines.  Sometimes he permits large discrepancies.  Often he omits many lines fully as good as those he has drawn.  On map 7, for October 2, 1954, I have, for example, more than doubled the number of lines given by Michel. 

With this fact in mind I re-investigated Michel’s famous map 10, for October 7.  He gave, as statistics, 19 3-point lines, 3 4-point lines, and 1 7-point line.  Michel omitted far more lines than he drew.  My final recount gives forty-two 3-point lines, six 4-point lines and one 7-point line. 

[Further criticism of Michel is omitted here.]

Now what do the statistics say?  By equation (9), with n=27, W=208.  The values of N(3), N(4), etc., calculated from (10) are very sensitive to the unknown corridor factor, ƒ.  However, let us arbitrarily adopt Mebane’s value of ƒ=1/80.  Then we find, in turn, the values in the following table:

	2	3	4	5	6	7	Total
Observed	158	42	6	1	0	1	208
Theoretical	152	48	7	0.7	0.05	.0025	208

The only remarkable fact about this table is the unexpectedly close agreement between the theoretical and observed numbers of lines.  Certainly there is no evidence that the alignments could have arisen other than by chance.  The 7-point line is the only possible exception.  Examination of the map shows that 3 points of this line are unusually close.  Also, the line is extremely long, one of the longest that can be drawn in France.  Clearly, this special corridor deserves a larger value of ƒ than most of the others.  The low figure for the theoretical probability is not significant. 

Re-study of all of Michel’s cases leads me to the same conclusion.  The vast networks, the stars and spiderwebs to which he attaches such significance, completely disappear.  Not only are the alignments he draws due to chance, but he fails for some reason to draw in dozens of other lines, equally good. 

[Menzel’s reply to criticism by Michel is omitted here.]

I submit that Orthoteny has failed and must be thrown out.