Allan Pratt <[EMAIL PROTECTED]> writes: > According to a source I read, Hipparchus, a 2nd C BC astronomer > calculated the length of the year to within six minutes of accuracy. > Considering that at best he had a sundial and a water clock, how did he > do this?
I hope a historian will answer this, but I am willing to speculate. H's minutes were surely defined not with respect to a cesium clock but as a fraction of a day. The year is defined by the seasons, i.e., the declination of the sun. The declination is most sensitive to the date around the equinoxes. Since the equinox is one of the most fundamental and easily observed astronomical events, it is plausible that the equinox had been determined and recorded, at least to the nearest day, for hundreds if not thousands of years before Hipparchus. If he had available an uninterrupted calendar and a record of an equinox 240 years earlier, then, by counting the number of days between that and a contemporaneous observation of an equinox and dividing by the number of years, he could calculate the length of the year to a the claimed accuracy: (1 dy) / (240 yrs X 365 dys/yr) = 1/87,600 = (6 min / 1 yr) / (60 min/hr X 24 hrs/dy X 365 dys/yr). Alternatively, if he knew what he was about, he could by careful naked-eye observation determine the time of the equinox to within a fraction of a day. If his observations had an accuracy of 0.1 day, then he would only need observations 24 years apart, easily within a professional lifetime even in those days. The observation must not necessarily be of the equinox. One could use solar eclipses in a similar way, or simply the date in spring on which the sun first becomes visible in a notch between two mountains. Note that you don't even need a sundial or a water clock for any of this! --Art
