WHY THE "ROULETTE WHEEL" APPROXIMATIONS WORK FOR THE EARTH AND MOON
by
Mary Anne Weaver
Copyright (c) 1999 by Delphi Technologies
THE MOON'S CHANGING ALIGNMENTS
Because the Moon's motion in space, its pole changes with respect to the positions of stars over a span of two decades. This in turn changes the star motion calculations slightly for each year in question. Further, it complicates matters by creating different "recurring alignments" during different years. For example, some years, Sirius will be at 0 degrees at one Apollo landing site and, at the same time, 33 degrees at the other site. During other years this alignment will not occur.
However, this is a predictable and cyclic occurrence, meaning that the aforementioned Sirius alignments reoccur on a regular and predictable basis. This is a cycle that occurs one full time in the decades I'm interested in, though -- how can I ignore it over the course of analyzing the 50's through the 70's (the 82 launches I chose vary between 1958 and 1978, approx.)?
I don't ignore it, but it is not very significant because when I subtract out the recurring alignment odds, the difference in the resulting probability values are only about 0.01 for all the recurring alignments taken together, for the Moon. Additionally, I can simply take into account in my close approximation, that the subtractions only occur a specific percent of the time. This is because I am "averaging out" the times when these simultaneous alignments happen, with the times when they do not.
DIFFERING ROTATION SPEEDS BETWEEN THE EARTH AND MOON
Planets do differ from roulette wheels in a few important ways, one of which is they rotate at different rates. The moon takes a full 27.322 days to rotate once, while the Earth only takes 24 hours! So how is this taken into account?
Recall how the probabilities are defined:
Time at angle of interest, Earth location
P_earth = -----------------------------------------------
Total Rotation Time of Earth
Notice that we have "time, Earth" divided by "rotation, Earth." So, even though the Earth rotates faster than the Moon, this is already taken into account because the star moves much faster through the degree of interest, and the rotation time is also that much faster. The two are directly proportional, so when you divide the transition time of the star through the angle, by the planetary rotation time, the differences in rotation speeds factor out.
Are there any cases where I have to be careful about computing probabilities? If I was only interested in finding the probability of an event on one particular Earth day -- say, August 11, 1999 -- then I could not make the above assumption! That’s because on the Moon, one day of Earth time is a only 1/27th of its own cycle, or equivalent to less than one Earth hour! So, if I only care about one specific Earth day, then I have to compute what the probability is in the following manner:
Sum_Moon = Sum of times, of all stars that will transit any angles of
interest
Sum_Moon
P(moon) = ----------------------------
Total Time (24 hours)
This may or may not equal the result I get when I take into account how much time stars spend at angles of interest with respect to the total rotation time of the Moon (i.e. 27.322 days), because the Moon only has a fraction of its cycle to go through. During that fraction, I may not have any alignments occur at all, or I might have more per unit time than I would in a complete cycle.
However, I’ve set this model up to work for long periods of time, not just one day or one month. If there is truly a pattern in the 19.5 and 33.0 degree stellar alignments, only this kind of approach will reveal it. If this paper is to effectively determine whether a pattern exists or not, then the probability model needs to be accurate over long periods, i.e. in the case of the space program, decades, and the pattern needs to continue throughout those years (if it is a pattern).
Suppose I was just interested in one lunar month of time (one of the Moon's rotations), occurring in the month of July, 1969. Then I'd write the following equation,
Sum_cycle_Jul_1969 = Sum, time it takes for all stars at all lunar
sites to pass through angles of interest in
a lunar cycle taking place in July, 1969
Sum_cycle_Jul_1969
P(moon) = ---------------------------------------------
One Lunar Rotation Cycle for July 1969
But then, suppose I wanted to determine the probability for the entire year 1969. Then,
Sum_lunar_month_1 = Sum, time it takes for all stars at all lunar
sites to pass through angles of interest in
first lunar cycle of 1969 (starting January 1)
.
.
.
Sum_lunar_month_13 = Sum, time it takes for all stars at all lunar
sites to pass through angles of interest in
last lunar cycle of 1969 (in Nov/Dec)
But, there aren't exactly 13 Moon cycles every year. There are 13.36 cycles, approximately. So, to include this "tail end", I'd have to also add:
Sum_lunar_leftover = Sum, time it takes for all stars at all lunar
sites to pass through angles of interest in
last few days of 1969, left over from unfinished
cycle.
Now, we have:
Sum_1969 = Sum_lunar_month_1
+ Sum_lunar_month_2
.
.
.
+ Sum_lunar_month_13
+ Sum_lunar_leftover
So,
Sum_1969
P(Moon) = ---------------------------------
Total Time interval (365 days)
However, one cycle is much like another in that 365 day period. The amount of time a star stays at a specific angle does not change appreciably during that time, during each lunar cycle of a year. Therefore, let us dispense with the "leftover" portion of this, and rather, because each cycle is "the same" in one year, write the equation thus:
Sum, time it takes for all stars at all lunar sites
to pass through angles of interest in one cycle
P_Moon = ------------------------------------------------------------
Time, One Lunar rotation cycle
Because I am looking at, "what is the probability that an event will occur at any moment on any day within the decades 1950 through the 1990s", I can use the above equation for the Moon, without having to worry about what day it is (though I do have to be concerned about what year it is, due to the "recurring alignments" problem).
I can use this equation for the Moon because it adequately expresses the probability that, at any given day of Earth time during an Earth year, a stellar alignment will be occurring on the Moon. That's because if I pick any Earth day during an Earth year at random, I don’t know what "phase" of the Moon's cycle I'm in. It could be during any part of the Moon’s rotation cycle. Therefore, during a series of Earth years, the probability is P(moon), that a lunar alignment will be occurring at that Earth day and time.
If the Moon were to rotate so slowly that, in a number of Earth years (in this case, again, decades), the Moon didn't rotate a full cycle or the Moon only rotated a few times, then I could not make this assumption. But the Moon completes 13.36 cycles every Earth year, and since I'm interested in the probability an alignment occurring on any Earth day during the decades 1950 through 1970's, this motion averages out. It doesn't matter that it happens slower or faster than the Earth, because it happens often enough during the time period that I'm looking at, that I can use the above equations.