Return to main pageBecause of the way the Saros-Inex Panorama (and from it the Saros series numbers) is defined - by fitting existing data into a regular but not entirely logical grid - there is no simple way of determining which Saros series a given eclipse belongs to. However, because the Panorama is a grid, with a regular time progression across the rows and down the columns (one Inex and one Saros respectively), the time between any two eclipses can be expressed as a simple sum or difference of Inex and Saros periods. This relationship is expressed by the formula T = (a x I) + (b x S), where T is the time between the eclipses in some convenient unit, I is the Inex interval, S is the Saros interval and a & b are whole numbers (either positive or negative). The best unit to choose for T is the number of lunations (new Moon to new Moon periods) as then I and S become simply 358 and 223. By the definition of the Panorama, the number "a" will be the number of Saros series between the two eclipses. One can thus move from the known to the unknown i.e. if one knows the Saros series number of a particular eclipse it is possible to deduce that of another eclipse as long as the time between them is also known.
There are two ways of using the formula. Firstly, one can derive the difference in series number for some common eclipse intervals and use that to move from eclipse to eclipse. For example, the most common interval between successive eclipses, 6 lunar months, is 5 Inexes minus 8 Saroses [ as 5 x 358 - 8 x 223 = 6 ]. Such eclipses will thus be placed on the Panorama five columns apart and so will have Saros series numbers that differ by 5. This is the reason that successive eclipses do not have successive Saros numbers. The situation for eclipses separated by 5 and 1 lunar months is not so simple, unfortunately - they have Saros numbers differing by -33 and +38 respectively [ -33 x 358 + 53 x 223 = 5 and 38 x 358 - 61 x 223 = 1 ].
One can also use the formula with arbitrary intervals by solving it mathematically i.e. for a given T find a and b such that the formula holds and both a and b are whole numbers. This is not easy using "paper and pencil" but very simple using a computer program. Unfortunately, because you have one formula and two variables, the answer will not be unique. For example, for T=11896 one solution is a=27 and b=10 [27 x 358 + 10 x 223 = 11896] and thus two eclipses this number of lunations apart will have Saros series numbers differing by 27. However, a=250, b=-348 is also a solution, as the reader should quickly be able to verify. The eclipses in question could thus have series numbers differing by 250. In general, if a=X generates a solution, a=X+223, a=X+446 etc. will also generate solutions, and in fact X-223, X-446 ... will do so as well. This problem requires a little more information to solve.
The eclipses in question are 11896 lunations apart, and as this is only 961.82yrs a difference of 250 series would mean 3.85yrs per series. Given that we know that corresponding eclipses in adjacent series are one Inex (29yrs) apart, this is clearly impossible. The required answer must therefore be a difference of 27 series, yielding a much more plausible value of 35.62yrs per series. In fact, the T value given corresponds to the interval between the first eclipses of series 33 and 60. [It is, of course, entirely possible for the series numbers of two eclipses to differ by 250, it's just that the T value would have to be very much larger in this case.]