Delambre English

 METHODS to find the Sextiles of the French Calendar.

By C. DELAMBRE.

       THE way to find the leap years in the Julian and Gregorian calendars is extremely simple, because the intercalations are placed there at regular intervals.  It is not the same in the French calendar, the years of which are regulated by the true equinox.

       The difficulty of determining the moment of the equinox, and consequently the first day of the year, a long time in advance, is the strongest objection that has been made against the new calendar.  It has been said that it would be almost impossible to adjust our astronomical tables to the new form of the year, calculated so far over years of an average and invariable length.

       I will not conceal this inconvenience, against which I protested with the majority of astronomers;  my intention is only to examine if, to determine the moment of an equinox, there is not a simpler means than the direct calculation made on the tables of the sun; for if we succeeded in fixing a large number of equinoxes in advance, there would remain no difficulty in constructing the tables of epochs, and we could keep the astronomical tables in the form they had until now.

       If the years were all of equal length;  one could, by a very simple calculation, find the beginning of each, starting from a given equinox.

       The year varies in length, due to the planetary equations of the solar theory, the change in eccentricity, and the movement of the apogee, which means that after one revolution, the sun not having  quite the same mean anomaly as at the beginning, the equation of the center is no longer quite the same;  which, for the present, and for many years to come, will increase the longitude of the sun, cause it to arrive more quickly at the equinox, and diminish the length of the year.

       Let us examine the effect of these various causes.

      According to my solar tables, which are used for the Knowledge of Time, the inequalities of the sun can go to nearly 50" in six equations. It is almost impossible that all these equations are at the same time at the maximum and of the same sign.  Let us assume, however, it to have the limit of uncertainty.

        The sun takes a little more than 20 minutes to travel 50". The planetary equations can therefore advance or delay an equinox by 20'. Most often the equations will be easier and will partially compensate each other; but it is certain that we cannot answer, within 20', of a calculated equinox, without taking these equations into account.

       The central equation decreases every year by 0.188” sin z, z being the mean anomaly.

       The hourly movement is 147.8472" - 4.9644" cos z.  The excess of the mean year over the true one, is therefore in this respect,         0.188" sin z 3600"            =

                   147.8472" - 449644 cos z

 4.5776872" sin z + 0.076855" sin 2z.

       The movement of the apogee is 62.146" per year. Thus let A be the mean anomaly at the epoch of the French era, i the number of years elapsed, we will have z = (A – i•62.146”) = (82° 31’ 40” – i•62.146”).

       Substituting this value of z, expanding and reducing, we will have for the quantity by which any year is shorter than the mean 4.55863” – i•0.0002241” due to the change of eccentricity.

       The movement of the apogee being 62.146” per year, at the end of any year the average anomaly is smaller than at the beginning by this same quantity; the equation of the center is lower than the variation  due to 62" decrease in the anomaly;  the year is shortened by this variation converted into time.  Now the equation of the center is - 1° 55’ 26.352” sin z + 72.679” sin 2z – etc., whose differential is – 1° 55’ 26” cos  z d z + 72.7” cos 2z d (2z), put for z its value above, expand, and reduce in time, we will have

                     7.6816" + i•0.015148"; 

Let's combine this quantity with that which is produced by the change of eccentricity, we will have

                     12.2402” + i•0.014924”. 

        Let a = 12.2402”, and b = 0.014924;  the decrease will be a + i b.  Thus we will have

 for the first year   a + b

 for the second..... a + 2 b

 for the third.....       a + 3 b.

Thus the decrease of the first year is a + b, the sum of the decreases of the first two 2a + 3b, the sum of the first three 3a + 6b, and the sum for a number i of years i•a + 1/2 i(i + 1) b, and the annual decrease by a medium between a number i of years a + 1/2 b (i + 1) = a + 1/2 b + 1/2 b•i =  12.24766” + i•0.007462;  thus the annual decrease, by a medium between the first 100 years, will be.  ..

                                                                          12.9939”

                                                                                                   0.7462".

     Between the first 200                            13.7401

                                                                                                   0.7462".

     Between the first 300.                           14.4863

                                                                                                   0.7462".

     Between the first 400.                           15.2325

                                                                                                   0.7462".

     Between the first 500.                           15.9787

 These quantities form an arithmetic progression whose difference is 50 b = 0.7462”.

       It will be more than enough to determine the equinoxes for four hundred years, so we will assume that the year decreases by 15.2” or in decimals of a day.                        0.0001759.

     The average year is.  ..... 365.2422222.

     The excess of the year over 365 days

 so will be.                                         0.2420463.

     The quadruple is.                     0.9681852.

     The quintuple.                            1.2102315.

       If the excess of the year over 365 days were 0.25 instead of 0.2420463, it would produce one day every four years, and the fourth year would invariably be sextile.  But, because of the missing 0.0079537, the sextile will sometimes not arrive until the fifth year.

        We only want to know the sextiles, so it is useless to calculate all the equinoxes;  it will suffice for us to have them every 4 or 5, according to the circumstances.  I will give two methods to achieve this.

       I first calculate the equinox of the 1st year on my tables, neglecting the planetary equations, since we are forced to neglect them for the following ones. I think the first of vendémiaire arrived,  at.. .  .  .  .... 0.3846181 true time.  I count the first equinox in true time, in order to have the others also in true time; because the equation of time at the equinox depends only on the eccentricity of the sun and the mean anomaly, and the amount by which it varies in four centuries is imperceptible, otherwise there was no difficulty in bringing this additional element in the previous determinations.

To the first equinox add the

 annual delay, or.  .. .  .  .  .  .  .  .  .  ......                     0.2420463.

       We will have the equinox of the year 2... 0.6266644.

       Add a second time the same

 quantity, we will have for the year 3 ..             0.8687107.

       By adding it a third time,

 we will have for year 4...........                                1.1107570.

       Here we see a whole day, we will give it to year 3 which therefore will be sextile.

       Now that we know a sextile, we will deduce the following ones, without going through the intermediate equinoxes.

       We found above that the lag for four years is 0.96818.2, and for five years it is 1.2102315;  but to avoid the subtraction of a day at each addition, I will write for four years 9.9681852, and for five years 0.2102315, and I will omit the tens as in the logarithmic calculation.

       Here is now the calculation of the sextiles:

       At the equinox of the first sextile, I add the delay for four years;  I have the equinox of the following sextile, that is to say, of the year 7: to this last I add the same quantity, I have the equinox of the year 11, third sextile;  the same operation gives me that of the year 15, fourth sextile.  The hour of this equinox is ................                                   0.7732663

 less therefore than.....                                      0.7897685

 which is the arithmetic complement of

 delay for five years, or.  .  .  .  .  .  .  .  .             0.2102315.

       So I see that I can add the delay for five years, without having a whole day, and that the sextile will be delayed to the year 20, or there will be an interval of five years.

       After year 20, I start adding the lag of the equinox again for 4 years, and mark the sextiles from every 4 until I get to a number below 0.7897685, which takes  place in the year 48;  then I add the delay for five years, and the next sextile falls on the year 53;  after which I start again, and so on until the year 400.

       Although this calculation is extremely simple, it is good to check it at certain intervals;  or at least finishing.  Thus, to verify the year 172 at the equinox of the year 3 or.  .. .  .  .  ........             0.8687107

 I add 100 times the annual delay or.  .    24.2046300

                    60 times this same delay or.  .  14.5227780

                       9 times this same delay or.    .  2.1784167

                  169...・・・・・・.                         41.7745354

 the whole number 41 shows that the year 172 is the 41st sextile after the year 3, that is to say, the 42nd of the French era;  and the fraction 0.7745354 is the equinox of the year 172, this is what the preceding calculation had given, which, thereby, is verified.

       The number 169 = 172 - 3, is found by subtracting the starting year from the one you want to verify.

       I checked in this way the year 324, starting from 172;  and the year 400, starting from 334: we can check everything starting from year 1.

Departure year.    1.       0.3846181

     For......           300 ..   72.6138900

     For.                    90.      21.7841670

     For.  ...                  9.        2.1784167

    Are.  .  .  .  .  .  .400 ....96.9610918

       We will therefore have 97 sextiles in these first 4 centuries.

       We will see in the following page the additions up to the year 400, numbers which correspond to 4 years and 5 years alternately;  because on the second line we see the number 9.968182 which gives 4 years, and on the sixth line the number 0.2102315 which gives 5 years;  we have only marked 4 and 5 twice, since it is easy to see by the repetition of the same numbers, and by the sums which result from them, which are the places where there are 4 and those where there are 5.

       It would be easy to extend this table for the following centuries, but this will suffice to serve as an example.

                                       TABLE 326–327

       We see, by going through the previous table, that the delay of the sextile returns after periods of 29 and 33 years;  never less than 29 or more than 33;  except before the year 20, because the French era begins in the middle of one of these periods.  It is easy to feel the reason.

       If we reduce to a continued fraction the average delay of the equinox 0.2422222, we will have for successive approximations, the fractions 1/4, 7/29, 8/33, 31/128, 39/161, and finally 109/450 exact value.

       The first fraction assumes the delay of just 6 hours;  which is far too much: however, it causes sextiles to return more often than every 4 years.

       The second 7/29, gives 7 sextiles in 29 years;  it assumes 5ʰ 47' 35" delay, and that's too little.

       The third 8/33 gives 8 sextiles in 33 years;  it assumes 5ʰ 49' 6" delay, and that's too much.

       It is therefore not enough that the sextile delays once, in 33 years, and it is too much once in 29;  this is why delays sometimes occur after 29 and sometimes after 33 years.

      The average delay of the equinox is approximately 5ʰ 48' 48"; but the average delay of the first four centuries is, according to the preceding calculations, lower by 15", or 5ʰ 48' 33", which holds almost the middle between 5ʰ 47' 35", and 5ʰ 49' 6"

       It suffices to know the years in which the sextile experiences this delay;  because from there we always have six or seven sextiles spaced every 4 years;  thus knowing that two successive delays take place in the year 20 and in the year 53, we conclude that the intermediate sextiles are 24, 28, 32, 36, 40, 44 and 48;  knowing that the following delay falls on 82, we will have for intermediate sextiles, 57, 61, 65, 69, 73, 77.

       It's a matter of sorting out under what circumstances the period should be 29 or 33 years.

       29 annual delays of the equinox, or

                          29 x 0.2420463 = 7.0193427

 or rejecting the whole days..…. 0.0193427 this is what must be added for 29 years, which gives 7 sextiles;  and as the sum must not make a whole day, it is necessary, so that the period is 29 years, that the equinox of the departure is less than 0.9806573, supplementing the delay for 29 years;  this is what takes place in 53, 115, 210, 272 and 367;  thus these years will begin periods of 29 years.

       The 33 annual delays or 33 x 0.2420463 = 7.9875279, or discarding the integers 0.98752791;  this is what to add for 33 years, if the starting equinox is 0.9806573 or above.

       So here is the rule of these periods.  If the equinox of a delayed sextile is below 0.9806573, the period will be 29 years, otherwise it will be 33.

       This consideration greatly shortens the calculations.

                                                         EQUINOXES.  delayed sextiles.

   The equinox of the year 10, first

 lagged sextile, is....                  0.9834978..           20

   It is over the limit;

 so I add for .  .  33 Years.  .  .  9.9875279

   This one is below.                   0.9710257.              53

 etc….. 

                  TABLE 329–330

       Knowing by this calculation all the delayed sextiles, we can easily deduce the intermediaries, which follow each other every 4 years.

                           TABLE 331

All these equinoxes were determined by neglecting the small equations of the sun, and we said that these equations could accelerate or retard the equinox by 20' = 0.0138888 etc.

        Suppose the equations delay the equinox by 20', it will be necessary to add 0.014 to the time of the equinox found by our operation.  If this addition can be done without the sum being an integer, the sextile is safe.  So in the years 20, 53, 115, 777, 210, 272, 334, 367 and 429, there is no doubt these years will be sextiles.  But in 82, 144, 239, 301, 396, 458, 491, if the equinox is delayed by 20' by the perturbations, the time of the equinox will pass 24 hours, there will be one more day to give to the years  81, 143, 238, 300, 395, 457 and 490, which will be sextiles, and then 82, 144, etc.  will be common.  This doubt can only be removed by calculating the perturbations;  it takes place for all the equinoxes which exceed 0.986.  I marked with the sign — all those sextiles which could be carried over to the year preceding them:

       Suppose now that the perturbations advance the equinox, it will be necessary to subtract 0.014 from the hour of the calculation;  but the remainder must not be below 0.7579537, otherwise the sextile would be rejected to the following year.  There will therefore be some doubt on the sextile, when the equinox does not pass 0.772, consequently on those of the years 48, 110, 205, 267 and 362, which I have marked with the sign +, because the sextile could be  delayed until the following year, that is, 49, 111, etc.

       It will usually suffice to know the sign of the perturbations to remove this doubt.

       If the sum of the perturbations is additive, it will increase the longitude of the sun, will advance the equinox;  the doubt will vanish in the years marked -, but it will increase for the years marked +, and then it will need be necessary to complete the calculation.

       If they are negative, they will dispel the doubt in the years marked +, and increase it in the others.  In general, if they are of a sign contrary to that which marks the year, they will dissipate the doubt;  if they are of the same sign, they will make the whole calculation necessary.

       I have made these calculations, and I have not found in the first four centuries any sextile to be displaced, by the effect of these perturbations;  but the equinoxes of the years 144, 301, and 362 will arrive so near midnight, that it will perhaps never be possible to say, even after observation, on what day they will have arrived.  This is the true defect of the French calendar.  It will be necessary to stick to the result of the calculation, and this result is still subordinated to the degree of precision which the tables can have.

       When one calculates the equinoxes of all the sextiles, it is easy, by what precedes, to distinguish those which are doubtful;  when we only calculate the sextiles, delayed by the periods of 29 and 33 years, we only see the doubtful sextiles, in —;  but it is easy to conclude from them those which are doubtful in +, because the latter always precede by five years a delayed sextile, so that the doubt in general bears only on the year in which the delay must take place.

      Doubt in + takes place when the equinox does not pass.  .  .0.772.

       The movement for five years, is.....                                                   0.210.

      Thus, when an equinox does not exceed                                       0.982, the sextile preceding it by five years is doubtful.  This is what happens for (53 - 5) = 48, (115 - 5) = 110;  (210 - 5) = 205, (272 - 5) = 267, (367 - 5) = 362, (429 - 5) = 424. In this case we calculate for the year which precedes by 4 years only instead of  5, i.e. for 49, 111, 206, 268, 363 and 425;  then, if the rigorously calculated equinoxes fall before 0.242, the preceding years, that is to say 48, 110, etc. will be sextile.  If they fall in the last quarter, it will be the year for which we calculate that will be sextile.

       To directly find the equinox of any year, without going through the intermediaries, or in general to find the true entry of the sun in any sign, we can use the following formula, which can easily be put into tables.

       When the mean sun enters any sign, the whole equation of the center is far from the true sun entering this same sign.  The time interval between the entrance of the true sun and the mean sun, at the same sign, is equal to the equation of the center calculated for the entrance of the mean sun, and converted into time.

     The true hourly motion = 147.8472" - 4.9644" cos z + 01042" cos 2z - 0.002" cos 3z.

     The movement for 1" of time = 0.04106867" - 0.00137933" cos z + 0.00002894" cos 2z - 0.00000055" cos 3z.

     The interval sought will therefore be …..[ ]

 dividing, expanding and reducing, I found 168730.4" sin z + 1064.4" sin 2z - 15.8" sin 3z + 0.98" sin 4z - i•4.57778" sin z.  This quantity added to the mean time of the entry of the mean sun in any sign or in any degree, will give the mean time of the entry of the true sun in the same sign or the same degree.

       At the equinox and at the solstice, the time equation has only one part = -1/15 (center equation) = +461"8 sin. z - 4"8 sin.  2 z + 0"7 sin. 3 z - i. 0”0125 sin z.

       Add this series to the previous one, and we have 169192"2 sin. z＋1059"6 sin.  2z ー15"1 sin. 3 z + 0”92 sin. 4 z - i. 4"59 sin.  z, or, in day decimals, 1.95824"3 sin. z + 1226"4 sin.  2zー18'2 sin.  3z + 1”134 sin.  4z-i.  5”284 sin.  z.

      This quantity added to the mean time of the mean equinox or solstice, will give the true time of the true equinox or solstice.  For the other 8 signs, it would still be necessary to add the second part of the time equation, that is to say, -1/16 (reduction at the equator).

       This formula therefore provides us with a new means of determining an equinox and any sextile.  For this purpose I built four tables, of which here is the explanation and the use:

       Table I is that of epochs.  It gives for the year 1 the mean time of the entry of the mean sun in each sign;  this time is counted from midnight, and expressed in decimals.

       Next to it is the mean anomaly of the sun for the corresponding day, expressed in decimal degrees and counted from perigee, instead of in the above formula the z's were counted from apogee. To count them from perigee, it suffices to change all the signs.

       The last column gives the second part of the equation of time.  The first part is fused into Table III.

       Table II gives what must be added to the epochs, if one wishes to have, for any year, the entry of the mean sun at the same sign with the corresponding mean anomaly.

       Table III, built on the above formula, (changing the signs) gives the amount by which the entry of the true sun at the same sign precedes or follows the entry of the mean sun.  It gives separately the effect of the change of eccentricity for 100 years.

       Example I. We ask for the true equinox of year 1, or entry at the sign ♎️.

                                                  Epoch.  Mean Anom. 

         Table I gives              - 1.550   289.601

         Table III.  .  .  .             + 1.936

         True Equinox.  .  .  .  .  0.386;  [= 9:15:50 in the morning old style]

which shows that the true equinox followed the beginning of the year by 0.386ᵈ, or that it arrived on 1st vendémiaire at 3ʰ 86’.  [new.  style 9:15:50 in the morning old style of Jan 22, 1792]

Example II.  We ask for the entry of the Sun at the sign ♒️ for year 4.

         Table I…..                                            120.197    022.928

         Equation of time.  .                     -               7

         Year 4...                                          +     0.727    399.942

         Table III.  .  .  .  .  ...                         -     0.681    022.870

       Entry at the sign of the ♒️ . . . .  120.236. The 120 days make four full months;  the fraction 0.236 belongs to the first day of the fifth month, or to the first pluviôse.

       Example III.  We ask for the autumnal equinox for the year 30.

             Table I……..                             - 1.550     289.601

             Table II for 30 years.  .         7.024     399.444 

              Table III.....                            + 1.933     289.045

              Variation for 30 years     -           2

              Equinox of ♎️ year 30……. 7.405.

The 7 days indicate that there were seven sextiles before the year 30, and that the equinox arrived on the 1st vendémiaire at 4ʰ 5';  finally the equinox arriving in the second quarter of the day, shows that the year 30 is the second year after the sextile which took place in 28.

        Example V. We ask for the equinox of ♎️, year 49.

           Table I..                                     - 1.550     289.601

           Table II.  49 years.                 11.627     399.079

           Table III.                                    + 1.931     288.680

           Variation.  for 49 years.  -             2

            Equinox ♎️, year 49.          12.006.

There were therefore 12 sextiles before 49. The equinox falls in the first quarter of the day, so 49 follows a sextile.  48 will therefore be a sextile, and the following sextile will be 53 and not 52, because at the equinox of the year 49 we can add 0.968, movement for 4 years, without having a whole day.

     Example VI.  We ask for the equinox of the year 82.

          Table I..............              -   1.550     289.601

          Table II.  82 years.  .. .   19.620     398.446

          Table III.  .  .  .  .  ....        + 11.928     288.047

          Variation.  82 years.   -            4

          Equinox ♎️, year 82.... 19.994.

So there will be 19 sextiles before 82, and 82 will be the 20th sextile, because the equinox is very close to midnight.

      Example VII.  Equinox of the year 110.

        Table I. ...                                 -   1.550      289.601

        Table II,                 100 yrs.    24.222      398.081

                                            10 yrs.       2.180      399.827

        Table III.                                  +   1.926      287.509

                                                           -             6

        Equinox ♎️, year 110...         26.772.

There were 26 sextiles before 110. The year 110 is sextile, and the year 115 will be likewise, because we can add the movement for 5 years or 0.210, without having a whole day.

      These tables will therefore make it possible to find, by a very simple calculation,

       1. The equinox of any year, and generally the entry of the sun into any sign;

       2. How many sextiles there will have been before the year for which we are calculating;  the number of sextiles which will have preceded will always be equal to the number of whole days found by the calculation;

       3. If the proposed year is sextile, because then the equinox will fall between 0.758 and 1.0;

       If the proposed year follows a sextile, because then the equinox will fall between 0.0 and 0.242;

       If the proposed year is placed two years after the sextile, because then the equinox will fall between 0.242 and 0.484;

       If the proposed year comes three years after a sextile, because the equinox will be between 0.484 and 0.726;

       If the proposed year comes four years after a sextile, because the equinox will be between 0.726 and 0.968;

       If the proposed year comes five years after a sextile, then the equinox will be between 0.968 and 1.0.

       In the latter case, we will be sure of at least six sextiles after the proposed year, which will be the sextile itself;  the next six sextiles will follow each other every four years.

       Finally, these tables will give, for an indefinite time, the sextiles of the French calendar, saving the errors of the elements of the sun and the perturbations which have been neglected.

       The use of these tables is safer than the approximate formula used above to determine the sextiles.  For the rest, the formula and the tables agree as well as one might expect.  To prove it, let us calculate by the tables the equinox of the year 524, the furthest from those which we have calculated by the formula.

         Table I........                              -         1.550      289.601

         Table II.  500 years.                   121.111        390.406

                              24 years.                       5.571       399.559

         Table III.                                    +       1.866       279.566

         Variation for 524 years.    -              26

         Equinox of 524..                        126.972

         The formula had given.               0.9748

                         ————————————

TABLES to find, in any year, the true time of the entry of the Sun into the twelve signs of the Zodiac, and particularly the sextiles of the French Calendar.

                  TABLE I, or Table of Epochs.

                  TABLE II.  Movements for the years.

       The movement for 100 years is marked in two ways in this table.  The first is only used to calculate the entry of the sun into the signs for the year 100 only.  The second way will be used for all the following years, that is to say, when the movement for these years is taken in several parts.  The reason is, that the movement of the first hundred years has been diminished from the movement for one year;  for example, the movement for 35 years, in the tables, is really the movement of 34;  so do others.

       But the movement for centuries, at the end of the table, has not been diminished.  This decrease in the first hundred years was made to simplify the calculation.

                            ————————————

                                  TABLE III.

Difference between the mean time of the entry of the mean Sun at the signs of the Zodiac, and the true time of the entry of the true Sun at the same signs.

       By means of the preceding rules, we can therefore calculate in advance, for a fairly long time, the sextiles of the French calendar, and keep our astronomical tables in their accustomed form.  However, it would have been desirable if the mean year had been retained and the intercalations fixed invariably.  

      I have given three plans for this.

      The first was to keep the Gregorian intercalation, which gives a sextile every 4 years for ordinary years, and every 400 years for centennial years;  but as this rule assumes the year be too long, I proposed to suppress the sextile of the year 3600 and its multiples.

      The second was to place a sextile every four years in ordinary years.  In centennial years the sextiles would have returned after 400 and 500 alternately, and would have fallen on the years 400, 900, 1300, 1800, etc. , so that to recognize a centennial sextile, it would be necessary to divide the number of centuries by 9.  If the remainder of the division was 0 or 4, the year would be sextile.  If the remainder were another number, the year would be common.

      The third plan was to omit the centennial sextile in the year 4000 and in all its multiples, but the error would have been one day in 400000 years; to correct it, it would have been necessary to reduce the year 400000 and its multiples to 364 days.