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Unit II — Book II: Parallel Circles, Polar Arcs & Astronomical Inequalities¶

Original English Translation from Edmond Halley’s Oxford Edition (1758, pp. 48–79)

From Intrinsic Triangles to Celestial Rising-Time Arcs¶

Where Book I establishes the pure intrinsic geometry of a single spherical triangle, Book II (Propositions II.1–II.12, Halley pp. 48–79) applies those triangle inequalities to families of great circles cutting across an oblique great circle (such as the ecliptic) and a system of parallel circles (diurnal parallels to the celestial equator). As Menelaus explicitly announces on p. 75 before Proposition II.10, his aim is to replace the multi-case three-dimensional chord arguments of Theodosius’s Sphaerica Book III (ed. Hunt, 1707) with universal, strictly intrinsic spherical triangle proofs—together with their converses.

Selected Propositions & Programmatic Passages of Book II (pp. 48–79)¶

Proposition II.1 — Problem: Oblique Spherical Angle Construction (p. 48)¶

Enunciation (Halley 1758, p. 48)

From a point taken on the side of an obtuse-angled spherical triangle opposite to the obtuse angle, to draw an arc of a great circle which, with one of the arcs containing the obtuse angle, shall comprehend an angle equal to that same obtuse angle.

[§2.1 · Construction & Opening Demonstration, Halley p. 48] Let \(\triangle \mathrm{AB\Gamma}\) be an obtuse-angled spherical triangle as described, such that \(\widehat{\mathrm{AB}}\) and \(\widehat{\mathrm{B\Gamma}}\) are each less than a semicircle, with the obtuse angle at \(\mathrm{A}\) (\(\angle \mathrm{BA\Gamma}\)); and let a point \(\mathrm{\Delta}\) be taken on arc \(\mathrm{B\Gamma}\). It is required to draw from point \(\mathrm{\Delta}\) to arc \(\mathrm{A\Gamma}\) an arc of a great circle comprehending with \(\mathrm{A\Gamma}\) an angle equal to the obtuse angle \(\angle \mathrm{BA\Gamma}\).

Let \(\mathrm{M}\) be the pole of arc \(\mathrm{A\Gamma}\), and through points \(\mathrm{M}, \mathrm{\Delta}\) let the great-circle arc \(\mathrm{M\Delta I}\) be drawn (meeting \(\mathrm{A\Gamma}\) orthogonally at \(\mathrm{I}\)); produce arc \(\mathrm{A\Gamma}\) to \(\mathrm{E}\) so that \(\widehat{\mathrm{AE}}\) and \(\widehat{\mathrm{\Gamma I}}\) are equal, and draw the great-circle arc \(\mathrm{ME}\). Because \(\widehat{\mathrm{\Gamma B}} + \widehat{\mathrm{BA}}\) taken together are less than a semicircle, Menelaus applies Book I Proposition I.13 (and subsequent cevian separation lemmas in Props. II.2–II.9) to locate the unique transversal arc through \(\mathrm{\Delta}\) reproducing angle \(\angle \mathrm{BA\Gamma}\).

Menelaus’s Programmatic Critique of Theodosius (p. 75)¶

[§2.2 · Halley 1758, p. 75] Immediately after completing the monotonic arc-difference inequalities for spherical triangles (Props. II.4–II.9), Menelaus pauses to state the methodological purpose of Book II:

“Having therefore premised these demonstrations, we have subjoined what follows in order to verify the propositions contained in the book of Theodosius On the Sphere, together with their converses—yet in a manner more universal and such as compels assent, both for establishing the first principles and for demonstrating the things written previously, some of which indeed scarcely appear sufficiently firm.”

—Menelaus of Alexandria, Sphaerica Book II (Halley 1758, p. 75)

Proposition II.10 — Universal Theorem Superseding Theodosius III.5, III.6 & III.9 (pp. 75–76)¶

Enunciation (Halley 1758, p. 75)

If on a sphere there be a great circle touched by one of the parallel circles, and on that great circle two equal arcs be taken between the point of contact and the greatest of the parallel circles; and through the endpoints of those arcs there be described both parallel circles and great circles issuing from the pole: then those parallel circles shall cut off unequal arcs from the great circles drawn from the pole, of which the arc nearer to the greatest of the parallels shall be greater than the more remote. Moreover, that portion of the greatest of the parallels cut off by the great circles drawn from the pole which lies nearer to the common intersection of the oblique circle and the greatest parallel shall be less than the portion more remote from the same.

[§2.3 · Demonstration & Halley’s Scholion, pp. 75–76] Let \(\mathrm{\Theta MB}\) be a great circle of the sphere (e.g., the ecliptic), touched by one of the parallel circles \(\mathrm{A\Delta E}\) (e.g., the tropic), and let the greatest of the parallel circles (the equator) be \(\mathrm{BT\Xi PX}\). On \(\mathrm{\Theta B}\) let two equal ecliptic arcs \(\widehat{\mathrm{\Theta K}} = \widehat{\mathrm{\Lambda M}}\) be taken; and through the pole \(\mathrm{\Gamma}\) and the points \(\mathrm{\Theta}, \mathrm{K}, \mathrm{\Lambda}, \mathrm{M}\) draw the meridian great-circle arcs \(\mathrm{\Gamma\Theta X}\), \(\mathrm{\Gamma KP}\), \(\mathrm{\Gamma\Lambda\Xi}\), \(\mathrm{\Gamma MT}\), together with the parallel circles \(\mathrm{K\Sigma}\), \(\mathrm{\Lambda O}\), \(\mathrm{M\Pi}\). Menelaus claims that the polar declination difference \(\widehat{\mathrm{\Pi O}} > \widehat{\mathrm{\Sigma\Theta}}\), and the equatorial right-ascension difference \(\widehat{\mathrm{XP}} > \widehat{\mathrm{\Xi T}}\) (so that equal ecliptic arcs near the solstice take longer to rise in right ascension than equal arcs near the equinox, while changing less rapidly in declination!).

Proof: Because \(\triangle \mathrm{M\Gamma\Theta}\) is a spherical triangle whose leg \(\widehat{\mathrm{M\Gamma}}\) is greater than leg \(\widehat{\mathrm{\Gamma\Theta}}\), and each leg is less than a quadrant, and equal arcs \(\widehat{\mathrm{\Theta K}} = \widehat{\mathrm{\Lambda M}}\) are taken on base \(\mathrm{\Theta M}\), it follows of necessity—by what was proved in Book I Prop. 33 and Book II Prop. 7—that angle \(\angle \mathrm{\Theta\Gamma K}\) is greater than angle \(\angle \mathrm{\Lambda\Gamma M}\), and consequently equatorial arc \(\widehat{\mathrm{XP}}\) is greater than \(\widehat{\mathrm{\Xi T}}\); and furthermore that \(\widehat{\mathrm{M\Gamma}} + \widehat{\mathrm{\Gamma\Theta}} > \widehat{\mathrm{\Lambda\Gamma}} + \widehat{\mathrm{\Gamma K}}\), whence \(\widehat{\mathrm{\Pi O}} > \widehat{\mathrm{\Sigma\Theta}}\).

Halley’s Note (p. 76): “N.B. Observe that this single proposition comprises the 5th, 6th, and 9th propositions of Book III of Theodosius’s Sphaerica.”

Proposition II.11 & II.12 — Symmetrical Equatorial & Horizon Arcs (pp. 76–79)¶

Proposition II.11 (p. 76): If two great circles on a sphere intersect one another, and on one of them two equal arcs be taken equidistant from their point of intersection, and through the endpoints of these arcs and through the poles of either great circle there pass great circles, those circles shall cut off equal arcs from the other of the first great circles.
Proposition II.12 (pp. 78–79): Extends Proposition II.10 to oblique horizon/parallel circle systems (rising times on the oblique sphere, ἀναφοραί) where the tangent parallel circle is larger than the tropical circle tangent to the ecliptic, together with its converse (“Et ex praecedentibus manifestum est quomodo consequatur conversa hujus”), bringing Book II to a close and preparing the transition from qualitative inequalities (Book II) to exact quantitative chord/sine ratios via the Menelaus Transversal Theorem in Book III.