If now the instrument, perfectly levelled, is adjusted to have its centre wire on one of the marks, then when elevated to the star, the star will traverse the wire, and its exact position in the field at any moment can be measured by the micrometer wire. Alternate observations of the star and the terrestrial mark, combined with careful level readings and reversals of the instrument, will enable one, even with only one mark, to determine the direction of the meridian in the course of an hour with a probable error of less than a second. The second mark enables one to complete the station more rapidly and gives a check upon the work. As an instance, at Findlay Seat, in latitude 57 deg. 35', the resulting azimuths of the two marks were 177 deg. 45' 37".29 [+-] 0".20 and 182 deg. 17' 15".61 [+-] 0".13, while the angle between the two marks directly measured by a theodolite was found to be 4 deg. 31' 37".43 [+-] 0".23. [Illustration: FIG. 3.] We now come to the consideration of the determination of time with the transit instrument. Let fig. 3 represent the sphere stereographically projected on the plane of the horizon,--ns being the meridian, we the prime vertical, Z, P the zenith and the pole. Let p be the point in which the production of the axis of the instrument meets the celestial sphere, S the position of a star when observed on a wire whose distance from the collimation centre is c. Let a be the azimuthal deviation, namely, the angle wZp, b the level error so that Zp = 90 deg. - b. Let also the hour angle corresponding to p be 90 deg. - n, and the declination of the same = m, the star's declination being [delta], and the latitude [phi]. Then to find the hour angle ZPS = [tau] of the star when observed, in the triangles pPS, pPZ we have, since pPS = 90 + [tau] - n, -Sin c = sin m sin [delta] + cos m cos [delta] sin (n - [tau]), Sin m = sin b sin [phi] - cos b cos [phi] sin a, Cos m sin n = sin b cos [phi] + cos b sin [phi] sin a. And these equations solve the problem, however large be the errors of the instrument. Supposing, as usual, a, b, m, n to be small, we have at once [tau] = n + c sec [delta] + m tan [delta], which is the correction to the observed time of transit. Or, eliminating m and n by means of the second and third equations, and putting z for the zenith distance of the star, t for the observed time of transit, the corrected time is t + (a sin z + b cos z + c) / cos [delta]. Another very convenient form for stars near the zenith is [tau] = b sec [phi] + c sec [delta] + m (tan [delta] - tan [phi]).