Info

5,57; 0

0;

0,48

0;29,29

5;58,26

In an anomalistic month of 27d 13;18,36h the Moon, initially at apogee, returns to it and travels 6 physical signs; the intermediary tabulated values of V(t - t0, 2n) represent the increment of the true lunar longitude at the beginning of each of the 27 days listed. This increment is not counted from Aries 0°; rather, it is an increment over the true longitude at the time when the mean anomaly was 0°. The true argument of latitude, ro(t - t0, 2n), is related to the increment in true longitude by means of the expression:

ro(t - to, 2n) = A'(t - to, 2n) - An, where AN is the longitude of the lunar node which, in Alfonsine astronomy, moves at the rate of -0;3,10,38°/d. Indeed, this is the quantity that can be deduced from the table where, for any ro and A', the following relation holds:

ro(t - t0, 2n) = A'(t - t0, 2n) - (t - t0) ■ 0;3,10,38.

By inspection of the table we see that the tabulated lunar equation was derived from the lunar longitude according to the expression:

q(t - t0, 2n) = [A'(t - t0, 2n) - A'(t - t0, 2n + 10°)]/10, and this is why all entries for the lunar equation in the double argument table are multiples of 6. Thus, for example, q(1,0) = [A'(1, 0) - A'(1, 10)]/10 = (11;52° - 11;43°)/10 = 0;9°/10 = 0;0,54°.

Similarly, the tabulated hourly velocity results from the following expression:

v(t - t0, 2n) = [A'(t - t0 + 1d, 2n) - A'(t - t0, 2n)]/24.

Thus, for example, v(1, 0) = [A'(2, 0) - A'(1, 0)]/24 = (23;40° - 11;52°)/24 = 11;48°/24 = 0;29,30°.

Both these quantities are used for interpolation purposes: q(t - t0, 2n) is for interpolation between consecutive days in an anomalistic month to take care of the hours and minutes at the time under consideration, and v(t - t0, 2n) is for interpolation between consecutive sets of entries for double elongation (at intervals of 10°) at the time under consideration. The combined effect of these two factors is to be added both to the increment in lunar longitude and to the argument for lunar latitude derived from the table.

Chapter 11 provides a worked example for the use of the lunar tables for 1447, July 18, at 12;30h. Although not stated explicitly, the general method to compute the true position of the Moon (true longitude and true argument of latitude) by means of Bianchini's tables consists, first, in determining in the tables for mean motions the mean lunar anomaly, the double elongation, the mean lunar longitude, and the mean argument of lunar latitude. The key move for Bianchini is to determine the double elongation at the time when the mean anomaly was most recently equal to 0°. To facilitate this computation Bianchini set up these special tables for determining the time when the mean lunar anomaly is 0°. Then one enters into the double argument table with time (mean anomaly) and the double elongation at the time when the mean anomaly was most recently equal to 0° and looks for the true lunar longitude and the true argument of lunar latitude closest to the given mean anomaly and mean double elongation. Next, one interpolates in the double argument table to take into account the excesses of time (anomaly) and double elongation. The interpolation is greatly facilitated by the tabulated quantities. In the worked example, we are first told to find in Tables 12, 13, and 16 the entries for 1440y, 6y, and June for the four quantities: time, double elongation, lunar longitude, and argument of lunar longitude. We are also told to add 18d 12;30h to time. We note again that Bianchini uses complete days. After adding the corresponding radices for the Incarnation found in Table 15, the resulting values are:

time = 89d 16;23h = 26d 3;1h (1440y) + 14d 4;33h (6y) + 15d 16;9h

(June) + 18d 12;30h (date) + 15d 4;10h (Incarn.) double elongation = 117;16° = 214;58° (1440y) + 153;53° (6y) + 70;56°

(June) + 37;29h (Incarn.) mean longitude = 1,35;20° = 4,32;19° (1440y) + 4,2;31° (6y) + 18;25°

(June) + 4,42;5h (Incarn.) mean argument of latitude = 1,20;55° = 42;46° (1440y) + 5,57;47° (6y) + 27;10° (June) + 13;12h (Incarn.).

We are then told to subtract from the time shown above the entry corresponding to 3 anomalistic months in Table 14, and to add to the other quantities shown above the corresponding entries for 3 anomalistic months. Note that, as a result of the organization Bianchini gave to his tables, one has to subtract from the time the entry for 3 anomalistic months. The reason is that in 3 anomalistic months the anomaly advances by 3 ■ 13;3,53,57° ■ (27d 13;18,36h) = 1080;0° = 0°; hence, the motion in anomaly in 3 anomalistic months can be disregarded. But in the time corresponding to 3 anomalistic months the increments in the other quantities are not 0° and so their values have to be added. The resulting values are:

time = 7d 0;27h = 89d 16;23h - 82d 15;56h (3 anom. months) double elongation = 332;44° = 117;16° + 215;28° (3 anom. months) mean longitude = 1,44;33° = 1,35;20° + 9;13° (3 anom. months) mean argument of latitude = 1,34;30° = 1,20;55° + 13;35° (3 anom. months).

As we said above, let t be the time in question and t0 the time when the mean anomaly was most recently 0°; in order to use Table 17 we need to find t - to, at which time the mean anomaly was the same as it was at t. We then find the double elongation at t - t0, for it is needed as the second argument in the table. In this case, t - t0 = 7d 0;27h, and the double elongation, 2n = 332;44°. Moreover, the mean longitude at t - t0 is X0 = 1,44;33°, and the mean argument of latitude is ro = 1,34;30°. Since the arguments in Table 17 are given in integer days and increments of 10° of double elongations, we will have to interpolate.

Thus, one enters Table 17 with t - t0 = 7d and 2n = 330°. The corresponding entries in the double argument table are:

X'(7, 330) = 1,25;0° q(7, 330) = 0;0,48° v(7, 330) = 0;32,40°/h ro(7, 330) = 1,25;22°.

Next we are told to take the excess of time over 7d (0;27h) and the excess of double elongation over 330° (2;44°) and multiply them by v(7, 330) and q(7, 330), respectively. The result is

0;12,19° = 0;27h ■ 0;32,40°/h - 2;44° ■ 0;0,48° (0;12,18°, if correctly computed).

This amount has to be added both to X'(7, 330) and ro(7, 330), to obtain X'(t - t0, 2n) = X.'(7d 0;27h, 332;44°) = 1,25;12,19° (= 1,25;0° +0;12,19°) and ro '(t - t), 2n) = ro'(7d 0;27h, 332;44°) = 1,25;34,19° (= 1,25;22° + 0;12,19°), respectively. Finally, the true lunar longitude (on the 9th sphere) at the time sought is:

X = X' (t - t0, 2n) + X, where X = 1,44;33°, the mean longitude found before, plays the role of an intermediary radix. Similarly, the true argument of lunar latitude, ro, is:

3,0;4° = 1,25;34,19° + 1,34;30°, that is, ro = ro'(t - t0, 2n) + ro, where ro = 1,34;30°, the mean argument of lunar latitude found before, plays the role of an intermediary radix.

Using the standard Alfonsine Tables, we have recomputed the entry for t - t0 = 7d and 2n = 330° in this table, X'(7, 330) = 1,25;0° (= 85;0°). After 190 days, the increment in double elongation is 170;40,18° (= 7d ■ 2 ■ 12;11,27°/d) and the increment in anomaly is 91;27,18° (= 7d ■ 13;3,54°/d). Thus, the double elongation, 2n, is 170;40° + 330° = 140;40°. The equation of center, c3(2n), corresponding to 140;40°, is +11;5°; and the value for the minutes of proportion, c4(2n), is 52. To recompute this entry where t - t0 (7d in our case) is the time elapsed since the instant when the mean anomaly was 0°, we need to use the true anomaly, a = 102;32° (= 91;27° + 11;5°). The equation of argument, c6(a), corresponding to 102;32°, is -4;53°, and the correction to this equation, c5(a), is listed as -2;39°. According to the rules for computing a lunar position with the standard Alfonsine Tables, we have to compute:

X'(t - t0, 2n) = AX + c6(a) + c4(2n) ■ c5(a), where ci refers to a value computed with interpolation from the entries in column i in the standard Alfonsine Tables. In Fig. 2, the observer is at O, the center of the epicycle at C, the mean epicyclic apogee is at Ae (directed towards D'), the true epicyclic apogee is at Ae, and the Moon is at L. There are two equations: q1 is approximated by c6(a), and q2 is approximated by c4(2n) ■ c5(a).

Now, after 7 days, the increment in longitude, AX, is 7d ■ 13;10,35° = 92;14,5°. Thus, the total increment in longitude, X', is 85;3°, for:

X'(7, 330) = 92;14° - 4;53° - 0;52 ■ 2;39° = 92;14° - 4;53° - 2;18° = 85;3°, in close agreement with the given entry (85;0°). All our other recomputa-tions with the standard Alfonsine Tables also yield very good agreement, often exactly to the minute.

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