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Image from page 45 of "Astronomy for high schools and colleges" (1881) | by Internet Archive Book Images
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Image from page 45 of "Astronomy for high schools and colleges" (1881)

Identifier: astronomyforhigh00newc

Title: Astronomy for high schools and colleges

Year: 1881 (1880s)

Authors: Newcomb, Simon, 1835-1909 Holden, Edward Singleton, 1846-1914, joint author

Subjects: Astronomy

Publisher: New York, H. Holt and Company

Contributing Library: The Library of Congress

Digitizing Sponsor: The Library of Congress


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Text Appearing Before Image:

^ a sweep of a pencil in amoment mark out the semicircle of our meridian uponthe heavens by a line from the north j^ole through thezenith and south horizon, to the south pole. At the endof an hour this semicircle would have apparently moved15° toward the east by the diurnal motion. Then im-agine that we again mark our meridian on the celestialsphere. The two semicircles will meet at each pole andbe widest apart at the equator. Continuing the processfor twenty-four hours, we should have twenty-four semi-circles, all diverging from one pole and meeting at the 36 ASTRONOMY. otlier, as shown in Fig. 11. The circles thus formed arecalled hour circles. Hence the definition : Hour circles on the celestial sphere are circles passingthrough the two poles and therefore cutting the equatorat right angles. The hour circle of any particular star is the hour circlepassing through that star. In Figure 11a let the outlinerepresent the celestial sphere ; Z being the zenith and P


Text Appearing After Image:

Fig. 11<».—circles of the sphere. the north pole. Let A be the position of a star.Then PAB is, by definition, part of the hour circle ofthe star A. The angle ZPA is then called the hourangle of the star. Hence the definition ; The hour angle of a star is the angle which its hourcircle makes with the meridian of the place. The hour angle of a star is, therefore, continually chang-ing in consequence of the diurnal motion. The declination of a star is its distance from the equa-tor north or south. Thus, in the figure, CWDE isthe celestial equator and the arc ^ ^ is the declination of RIGHT ASCENSION AND DECLINATION. 27 the star A. By tlie previous definition, P A h tlie polardistance of tlie star. Because P B is 90°, it follows thatthe sum of the polar distance and declination of a starmake 90°. Therefore, if we put ^, the polar distance of the star,and d its declination, we shall have : ^ + (J = 90°. d = 90° — p. The declination and hour angle of a star are two co-ordinates


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Taken circa 1881