- INDIA^261
formula
(f3438 - N/3438^2 - sin^2 è
2
We can therefore well understand why Aryabhata did not refine his table fur-
ther. Aryabhata's list of sine differences is the following:
225, 224, 222, 219, 215, 210, 205, 199, 191, 183, 174,
164, 154, 143, 131, 119, 106, 93, 79, 65, 51, 37, 22, 7.
A comparison with a computer-generated table for the same differences reveals
that Aryabhata's table is accurate except that his sixth entry should be 211 and the
eighth should be 198. But surely an error of less than half of 1% is not a practical
matter, and Aryabhata definitely took the practical approach. He explained that
his table of sine differences was computed by a recursive procedure, which can be
described in our terms as follows (Clark, 1930, p. 29). Starting with d\ = 225,
, _ , di + \-dn
«n + l — «ç j ,
"1
where each term is rounded to the nearest integer after being calculated from this
formula.
Aryabhata applied the sine function to determine the elevation of the Sun at
a given hour of the day. The procedure is illustrated in Fig. 14 for an observer
located at Ï in the northern hemisphere on a day in spring or summer. This
figure shows a portion of the celestial sphere. The arc RETSWV is the portion
of the great circle in which the observer's horizontal plane intersects the sphere.
The Sun will rise for this observer at the point R and set at the point V. The
arc is slightly larger than a semicircle, since we are assuming a day in spring or
summer. The chord RV runs from east to west. The Sun will move along the
small circle RHV at a uniform rate, and the plane of this circle is parallel to the
equatorial circle EMW. (At the equinox, the "day-circle" RV coincides with the
equatorial circle EW.) Aryabhata gave the correct formula for finding the radius
of this day-circle in terms of the elevation of the Sun above the celestial equator
and the radius of the celestial sphere. That radius is the sine of the co-declination
of the Sun. Although Aryabhata had the concept of co-latitude, which served him
in places where we would use the cosine function, for some reason he did not use
the analogous concept of co-declination. As a result, he had to subtract the square
of the sine of the declination from the square of the radius of the celestial sphere,
then take the square root.
The point Æ is the observer's zenith, Ì is the point on the celestial equator
that is due south to the observer, and S is the point due south on the horizon,
so that the arc ZM is the observer's latitude, and the two arcs Æ Í and MS are
both equal to the observer's co-latitude. The point Ç is the location of the Sun
at a given time, MF and HG are the projections of Ì and Ç respectively on
the horizontal plane, and HK is the projection of Ç on the chord RV. Finally,
the great-circle arc HT, which runs through Z, is the elevation of the Sun. The
problem is to determine its sine HG in terms of lengths that can be measured.
Because their sides are parallel lines, the triangles MOF and Ç KG are similar,
so that MO : HK = MF : HG. Hence we get
HK • MF
HG=~MO—-