Astronomical Images : Problem 2: Measuring a comet's parallax
Johannes Regiomontanus
Astronomical Images
<p style='text-align: justify;'>Johannes Regiomontanus died in 1475, leaving behind a printing press, instruments and a library containing printed books and manuscripts. Most of the library was bought by Bernhard Walther (1430-1504), the merchant-astronomer at Nuremberg and then ended up in the possession of Willibald Pirckheimer (1470-1530), the patrician friend of Albrecht Duerer. Pirckheimer sold on several of Regiomontanus's works to Johannes Schoener (1477-1547), who taught mathematics at the gymnasium in Nuremberg. Regiomontanus's work on comets, which was listed in his own printing advertisement, was first edited and published by Schoener in 1531 as <i>Sixteen Problems on the Magnitude, Longitude and True Position of Comets</i>. It was printed again, with several other works of Regiomontanus in 1544. The second problem offers a general solution to finding the parallax of a comet, when its apparent positions are observed on the same side of the meridian, from the same observation position over a known interval of time. The advantage of this solution is that it does not require the true positions of the comet (as measured from the centre of the Earth) to be known. It is assumed that the readers will be familiar with spherical triangles: once three elements (out of three angles and three arcs) are known, the remaining three may be derived. In this edition, the diagram is faulty: HO (great arc circle) has been drawn as HK (not used in the solution) and LN (great arc circle) is missing. The diagram is set up as follows: Circle ABCD: meridian Z: celestial north pole O: apparent place of the comet at first observation G: true place of the comet at first observation M: apparent place of the comet at second observation L: true place of the comet at second observation. N: defined as angle GHL = angle OHN (i.e. where O would be, given the same interval of time lapsed between G and L) (a) L and G are on a semidiurnal circle, thus HL = HG (b) QS is the semidiurnal circle from O, thus HN = HO (c) The following are known: ZO: altitude of the comet at first observation (d) Angle BZK: azimuth of the comet at first observation (e) ZM: altitude of the comet at second observation (f) Angle BZR: azimuth of the comet at second observation (g) Angle GHL: time lapsed between the two observations (h) ZH: 90° - latitude (i) Solution to finding LM and GO This solution involves showing that triangle LHN is equal to triangle GHO (I), and solving the triangles ZOH (II), ZHN (III), MZN (IV), and LMN (V). I. Triangle LHN = (want three lines) triangle GHO LH = GH (from b) HN = HO (from c) Angle LHN = angle GHO Because: angle LHG = angle NHO (from a) Angle LHG ' angle NHG = angle NHO ' angle NHG Thus, angle GOH = angle LNH (1) LN = GO (1') II. Triangle ZOH The following are known: ZO (from d), ZH (from i), angle BZK (from e) Thus derive: HO (2) Angle ZHO (3) Angle ZOH (3') III. Triangle ZHN The following are known: ZH (from i), HN (from c and 2). Angle ZHN = angle ZHO (from 3) ' angle NHO (from a) Thus, derive: Angle ZNH (4) Angle NZH (5) ZN (6) IV. Triangle MZN The following are known: ZM (from f), ZN (from 6) Angle MZN = angle BZP ' angle BZR (from g) = {angle BZH (180°) ' angle NZH (from 5)} ' angle BZR Thus, derive: Angle ZNM (7) Angle ZMN (8) MN (9) V. Triangle LMN The following are known: MN (from 9) and angle LMN (from 8) Angle LNM = angle MNH ' angle LNH Angle MNH = angle MNZ (from 7) + angle ZNH (from 4) Angle LNH = angle GOH (from 1) = angle ZOH (from 3') Thus, derive LM and LN, but LN = GO (from 1'), So LM and GO have been derived.</p>