Mathematics: Compiled from the Best Authors, and Intended to be the Text-book of the Course of Private Lectures on These Sciences in the University at Cambridge, Volume 2W. Hilliard, 1808 - Mathematics |
From inside the book
Results 1-5 of 21
Page 463
... right ascension and hour circles , are secondaries to the equator or equinoe- tial . 16. Circles of celestial longitude are secondaries to the ecliptic . 17. The zenith and nadir are the poles of the rational horizon ; the former being ...
... right ascension and hour circles , are secondaries to the equator or equinoe- tial . 16. Circles of celestial longitude are secondaries to the ecliptic . 17. The zenith and nadir are the poles of the rational horizon ; the former being ...
Page 464
... ascension is the sum or difference of the right ascension and ascentional difference . Or , it is the arc of the equinoctial , contained between the vernal equinoctial point and the point , that rises with the star . And the oblique ...
... ascension is the sum or difference of the right ascension and ascentional difference . Or , it is the arc of the equinoctial , contained between the vernal equinoctial point and the point , that rises with the star . And the oblique ...
Page 465
... right ascension , and declination . EXAMPLE . Given the sun's longitude 1 ° 11 ° 1 ' 44 " , and the obliquity of the ecliptic = 23 ° 28 ' ; required the sun's right ascension and declination . VOL . II . LII PROJECTION . Draw the ...
... right ascension , and declination . EXAMPLE . Given the sun's longitude 1 ° 11 ° 1 ' 44 " , and the obliquity of the ecliptic = 23 ° 28 ' ; required the sun's right ascension and declination . VOL . II . LII PROJECTION . Draw the ...
Page 466
... right ascension YB , Rx cos . BYO = cot . VO x tang . B. Cot . YO : R 41 ° 1 ' 44 " ar . co . 9'9396053 90 10 ° :: Cos . BYO 23 28 9'9625076 : Tang . B 38 35 49 9 * 9021129 Hence the right ascension = 38 ° 35 ′ 49 466 MATHEMATICS .
... right ascension YB , Rx cos . BYO = cot . VO x tang . B. Cot . YO : R 41 ° 1 ' 44 " ar . co . 9'9396053 90 10 ° :: Cos . BYO 23 28 9'9625076 : Tang . B 38 35 49 9 * 9021129 Hence the right ascension = 38 ° 35 ′ 49 466 MATHEMATICS .
Page 467
... right ascension = 38 ° 35 ′ 49 ′′ , which , reduced to time , 2h . 34 ′ 23 ′′ 16 " " . * PROBLEM II . Given the obliquity of the ecliptic , and the sun's declination ; to find the sun's longitude and right ascension . EXAMPLE . The ...
... right ascension = 38 ° 35 ′ 49 ′′ , which , reduced to time , 2h . 34 ′ 23 ′′ 16 " " . * PROBLEM II . Given the obliquity of the ecliptic , and the sun's declination ; to find the sun's longitude and right ascension . EXAMPLE . The ...
Other editions - View all
Common terms and phrases
abscisses altitude axis azimuth base Ca² cask centre complement cone conjugate cosine course curve DE³ declination departure describe dial diameter diff difference of latitude difference of longitude distance divide draw the parallel drawn ecliptic ellipse equal equinoctial EXAMPLES feet figure find the rest frustum height Hence horizon hour angle hour lines hyperbola hypotenuse inches intersection LATITUDE SAILING length measure Mercator's meridional difference middle latitude miles multiply NOTE oblique circle opposite ordinates parabola parallel of latitude parallel sailing perpendicular plane sailing pole prime vertical primitive Prob PROBLEM projection Prop proportional Q. E. D. COR quadrant radius rectangle Required the content rhumb right ascension right circle right line rule secant segment Side AC sine sphere spheric triangle spindle square star station Stereographic Projection stile sun's tance tang tangent THEOREM vertical
Popular passages
Page 3 - A sphere is a solid bounded by a curved surface, every point of which is equally distant from a point within called the center.
Page 147 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Page 8 - Take the length of the keel within board (so much as she treads on the ground) and the breadth within board by the midship beam, from plank to plank, and half the breadth for the depth, then multiply the length by the breadth, and that product by the depth, and divide the whole by 94; the quotient will give the true contents of the tonnage.
Page 59 - ... small statue, the head of which is 97 feet from the summit of the higher, and 86 feet from the top of the lower column, the base of which measures just 16 feet to the centre of the figure's base. Required the distance between the tops of the two columns ? Ans.
Page 61 - A gentleman has a garden 100 feet long, and 80 feet broad ; and a gravel walk is to be made of an equal width half round it ; what must the breadth of the walk be to take up just half the ground? Ans. 25-968 feet.
Page 63 - If a heavy sphere, whose diameter is 4 inches, be let fall into a conical glass, full of water, whose diameter is 5, and altitude 6 inches ; it is required to determine how much water will run over ? AHS.
Page 62 - Ans. the upper part 13'867. the middle part 3 '605. the lower part 2-528. QUES J. 48. A gentleman has a bowling green, 300 feet long, and 200 feet broad, which he would raise 1 foot higher, by means of the earth to be dug out of a ditch that goes round it: to what depth must the ditch be dug, supposing its breadth to be every where 8 feet ? Ans.
Page 21 - ... 07958 in using the circumferences ; then taking one-third of the product, to multiply by the length, for the content. Ex. 1. To find the number of solid feet in a piece of timber, whose bases are squares, each side of the greater end being 15 inches, and each side of the less end 6 inches ; also, the length or the perpendicular altitude 24 feet.
Page 187 - AC 2AC nearly ; that is, the difference between the true and apparent level is equal to the square of the distance between the places, divided by the diameter of the earth ; and consequently it is always proportional to the square of the distance.
Page 29 - ... -5236, for the content. RULE II. To 3 times the square of the radius of the segment's base, add the square of its height ; then multiply the sum by the height, and the product by -5236, for the content.