On average, how often (once
every N years) does a blue moon
(2nd full moon in the same month)
fall on New Year’s Eve in Los
Angeles?
This appeared in BENT for Fall, 2019 (the publication of Tau Beta Pi).
Their answer (19 years), in the next issue, referenced the fact that phases of the moon generally follow an inexact period of 19 years, so that 19 years after a full moon another will occur on the same date. But this is not exact; when one is missed the interval is longer, throwing off the mean. But 19 years is the mode of the times between occurrences, not the mean.
Ordinarily we take "average" to be the same as "mean", not "mode".
My answer is that full moon on January 31 will occur on average every 29.5 years, just as it would for any other date of the year. This is the necessary and sufficient condition for it to be a blue moon, as another full moon will have taken place early in January.
The 19-year Metonic cycle does cause enough regularity as to interfere with the easy match to 29.5 days as some dates do repeat for a long series of cycles, but then miss getting them when the cycle veers of to other dates. The long run is indeed longer than we wish to calculate them.
Using formulae from Jean Meeus's "Astronomical Formulae", I first calculated new moons on January 31 (thereby being blue also) using the mean position of the moon relative to the sun, i.e., not considering the irregularities in the motion).
Using mean moon and sun:
1599
1618 19
1637 19
1675 38
1694 19
1713 19
1732 19
1770 38
1789 19
1819 30
1838 19
1895 57
1914 19
1933 19
1952 19
1971 19
1990 19
2009 19
2047 38
2066 19
2085 19
2104 19
2161 57
2180 19
2210 30
2229 19
2267 38
2286 19
2305 19
2324 19
2362 38
2381 19
2400 19
2419 19
2438 19
2457 19
2495 38
2552 57
2601 49
2639 38
2658 19
2677 19
2753 76
2772 19
2810 38
2829 19
2848 19
2867 19
2886 19
2924 38
2935 11
3011 76
3030 19
3049 19
ans =
27.3584905660377 Average distance in years.
There are many 19's, but the average is 27.4.
Using major irregularities in lunar and solar motion in the celestial sphere:
1599
1618 19
1637 19
1656 19
1675 19
1694 19
1713 19
1732 19
1751 19
1789 38
1819 30
1865 46
1884 19
1895 11
1933 38
1971 38
1990 19
2009 19
2028 19
2047 19
2066 19
2104 38
2115 11
2180 65
2267 87
2286 19
2305 19
2324 19
2343 19
2362 19
2381 19
2400 19
2438 38
2495 57
2506 11
2552 46
2590 38
2601 11
2620 19
2639 19
2658 19
2772 114
2810 38
2829 19
2886 57
2924 38
2962 38
2981 19
3030 49
ans =
29.8125 avg.
%BENT Fall 2019
clearvars
prev=0; diff=[];
for k= -13*410.5:13000
T=k/1236.85;
j=(2451550.09766+29.530588861*k+.000154*T^2-1/3); % -1/3 for L.A.
M=0; Mp=0;
M=2.5534+29.1053567*k;
Mp=201.5643+385.81693528*k+.0107582*T^2;
j=j-.40614*sind(Mp)+.17302*sind(M);
d=datetime(j,'convertfrom','juliandate');
dv=datevec(d);
dv=dv(1:3);
if isequal(dv(2:3),[12,31])
if prev>0
diff(end+1)=dv(1)-prev;
end
disp([dv(1) dv(1)-prev])
prev=dv(1);
end
end
mean(diff)
To verify all are "blue"
%BENT Fall 2019
clearvars
prev=0; diff=[];
for k= -13*410.5:13000
T=k/1236.85;
j=(2451550.09766+29.530588861*k+.000154*T^2-1/3); % -1/3 for L.A.
M=0; Mp=0;
M=2.5534+29.1053567*k;
Mp=201.5643+385.81693528*k+.0107582*T^2;
j=j-.40614*sind(Mp)+.17302*sind(M);
d=datetime(j,'convertfrom','juliandate');
dv=datevec(d);
dv=dv(1:3);
if isequal(dv(2:3),[12,31])
if prev>0
diff(end+1)=dv(1)-prev;
end
disp([dv(1) dv(1)-prev])
T=(k-1)/1236.85;
j=(2451550.09766+29.530588861*(k-1)+.000154*T^2-1/3); % -1/3 for L.A.
M=0; Mp=0;
% M=2.5534+29.1053567*(k-1);
% Mp=201.5643+385.81693528*(k-1)+.0107582*T^2;
j=j-.40614*sind(Mp)+.17302*sind(M);
d=datetime(j,'convertfrom','juliandate');
dv=datevec(d);
dv=dv(1:3);
disp(dv(2:3)) % one lunation earlier
prev=dv(1);
end
end
mean(diff)
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Posted by Charlie
on 2024-11-23 23:11:18 |
solution
