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breadth and thickness must be large enough to keep the shadows of the gnomons from ever falling quite out of the sides of the hollows, even when the sun's declination is at the greatest. Lastly, draw the equinoctial dial in the middle, all the hours of which are equidistant from each other; and the dial will be finished.

66. As the sun goes round, the broad end of the shadow of the stile acbd will show the hours in the quadrant Ac, from the sun-rise till VI in the morning; the shadow from the end M will show the hours on the side Lq from V to IX in the inorning; the shadow of the stile efgh in the quadrant Dg in the long days, will show the hours from sun-rise till VI in the morning; and the shadow of the end N will show the morning hours, on the side Or, from III to VII. Just as the shadow of the northern stile abtd goes off the quadrant Ac, the shadow of the southern stile iklm begins to fall within the quadrant Fl, at VI in the morning; and shows the time, in that quadrant, from VI to XII at noon; and from noon till VI in the evening, in the quadrant m E. And the shadow of the end O shows the time from X1 in the forenoon till III in the afternoon, on the side r N; as the shadow of the end P shows the time from IX in the morning till I o'clock in the afternoon on the side Qs.

67 At noon, when the shadow of the eastern stile efgh goes off the quadrant h C, in which it showed the time from VI in the morning till noon, as it did in the quadrant g D, from sunrise till VI in the morning, the shadow of the western stile nopq begins to enter the quadrant Hp; and shows the hours thereon from XII at noon till VI in the evening and after that till sun-set, in the quadrant qG; and the end Q casts a shadow on the side Ps, from V in the evening till IX at night, if the sun be not set before that time. The shadow of the end I shows the time on the side Kp from III till VII in the afternoon; and the shadow of the stile abcd shows the time from VI in the evening till the sun sets. The shadow of the upright central wire, that supports the globe at top, shows the time of the day, in the middle or equinoctial dial, all the summer half-year, when the sun is on the north side of the equator.

DIALLING BY SPHERICAL TRIGONOMETRY.

68. The construction of sun-dials on all planes whatever, may be included in one general rule; sufficiently intelligible, if that of a horizontal dial for any given latitude be well understood. For there is no plane, however obliquely situated with respect to any given place, but what is parallel to the horizon of some other place; and, therefore, if we can find that other place, by a problem, on the terrestrial globe, or by a trigonometrical calculation, and construct a horizontal dial for it; that dial applied to the place where it is to serve will be a true dial for that place. Thus, an erect direct south dial in 51° 30' N. lat. would be a horizontal dial on the same meridian, 90° southward of 51° 30′ N. lat.: which falls in with 38° 30′ S. lat. But if the upright plane declines from facing the south at the given

place, it would still be a horizontal plane 90° from that place, but for a different longitude, which would alter the reckoning of the hours accordingly.

69. To calculate the angles which the hour lines of a horizontal dial make with the meridian or twelve o'clock line, see fig. 3. Let NES W represent the horizon of any place, PSN the meridian, and P the N. pole of the sphere: let KPH be any hour circle, for example, the circle which makes with the meridian an angle of 15°, then the arch of the horizon intercepted between N, the north, and PH the hour circle, in the plane of which the sun is at X1 or I o'clock, measures the angle contained by the substile of the dial, and the hour lines corresponding to these hours. In the spherical triangle P N H, right angled at N, there are given the side PN, which is the elevation of the pole above the horizon, and the angle NPH which is contained by the meridian and hour circle, to find NH the arch of the horizon opposite that angle. By spherical trigonometry, radius is to the sine of PN as the tangent of NPH to the tangent of NH the side required. Hence we have this practical rule. To find the angle which any hour line of a horizontal dial makes with the meridian, or which is the same, to find the angle which the hour lines on any dial make with the substile.-To the logarithmic sine of the latitude of the place for which the dial is made, add the logarithmic tangent of the sun's distance from the meridian, for the hour required, the sum, 1-10, is the logarithmic tangent of the angle required.

70. Example.-To find the angles which the hour lines of XI or I make with the meridian of a horizontal dial for the latitude of London, which is 51° 30'.

To logarithmic sine of 51° 30′
Add logarithmic tangent of 15°

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9.89354 9.42805

9-32159

Sum, rejecting 10, is which is the tangent of 11° 51′ nearly. In like manner it will be found, that the hour lines of X and II make each with the meridian an angle of 24° 18', &c. And by computing in this manner, with the sine of the latitude, and the tangents of 30°, 45°, 60°, and 75°, for the hours of II, III, IV, and V in the afternoon; or of X, IX, VIII, and VII in the forenoon; you will find their angular distances from XII to be 24° 18′, 38° 3′, 53° 35', and 71° 6'; which are all that there is occasion to reckon. And these distances may be set off from XII by a line of chords; or rather, by taking 1000 from a scale of equal parts, and setting that extent as a radius from C to XII, fig. 4, and then, taking 209 of the same parts, which are the natural tangents of 11° 50′, and setting them from XII to XI and I, on the line H O, which is perpendicular to C XII: and so for the rest of the hour lines, which in the table of natural tangents, against the above distances, are 452, 782, 1355, and 2920, of such equal parts from XII, as the radius C XII contains 1000. And, lastly, set off 1257, the natural tangent of 51° 30', for the angle of the

stile's height, which is equal to the latitude of vertical plane at Z is parallel, we proceed to the

the place.

DECLINING DIALS.

71. Let us suppose that an upright plane at London declines 36° westward from facing the south, and that it is required to find a place on the globe to the horizon of which the said plane is parallel; and also the difference of longitude between Loudon and that place.

72. Let NESW be the horizon of London, fig. 5, whose zenith is Z, and P the N. Pole of the sphere; and let Zh be the position of a vertical plane at Z, declining westward from S (the south) by an angle of 36°; on which plane an erect dial for London at Z is to be described. Make the semi-diameter ZD perpendicular to Zh, and it will cut the horizon in D, 36° west of the south S. Then a plane, in the tangent HD, touching the sphere in D, will be parallel to the plane Zh; and the axis of the sphere will be equally inclined to both these places. Let WQE be the equinoctial, whose elevation above the horizon of Z (London) is 38° 30′ and PRD be the meridian of the place D, cutting the equinoctial in R. Then it is evident, that the arc RD is the latitude of the place D, where the plane Zh would be horizontal, and the arc RQ is the difference of longitude of the planes Zh

and DH.

73. In the spherical triangle WDR, the arc WD is given, for it is the complement of the plane's declination from S the south; which complement is 54°, viz. 90°-36°: the angle at R, in which the meridian of the place D cuts the equator, is a right angle; and the angle RWD measures the elevation of the equinoctial above the horizon of Z, namely 38° 30'. Say, therefore, as radius is to the co-sine of the plane's declination from the south, so is the co-sine of the latitude of Z to the sine of RD the latitude of D: which is of a different denomination from the latitude of Z, because Z and D are on different sides of the equator.

As radius

To co-sine 36° 0′ — RQ.
So co-sine 51° 30′ QZ.

10.00000

9.90796
9.79415

To sine 30° 14′ = DR 9-70211 the latitude of D, whose horizon is parallel to the vertical plane Zh at Z.

construction of a horizontal dial for the place D, whose latitude is 30° 14′ south; but anticipating the time at D by 2 h. 51 m., neglecting the half minute in practice, because D is so far westward in longitude from the meridian of London; and this will be a true vertical dial at London, declining westward 36°.

76. Assume any right line CSL, fig. 4, for the substile of the dial, and make the angle KCP equal to the latitude of the place, viz. 30° 14', to the horizon of which the plane of the dial is parallel; then CRP will be the axis of the stile, or edge that casts the shadow on the hours of the day, in the dial. This done, draw the contingent line E Q, cutting the substilar line at right angles in K; and from K make KR perpendicular to the axis CRP. Then KG = KR being made radius, that is, equal to the chord of 60°, or tangent of 45° on a good sector, take 42° 52′ (the difference of longitude of the places Z and D) from the tangents, and having set it from K to M, draw CM for the hour line of XII. Take K N, equal to the tangent of an angle less by 15° than K M; that is, the tangent of 27° 52': and through the point N draw CN for the hour line of I. The tangent of 12° 52′ (which is 15° less than 27° 52′), set off the same way, will give a point between K and N, through which the hour line of II is to be drawn. The tangent of 2° 8', the difference between 45° and 50° 42′ placed on the other side of C L, will determine the point through which the hour-line of III is to be drawn; to which 2° 8', if the tangent of 15° be added, it will make 17° 8'; and this set off from K towards Q, on the line EQ, will give the point for the hour line of IV; and so of the rest. The forenoon hours line are drawn the same way, by the continual addition of the tangents 15°, 30°, 45°, &c., to 42° 52′ = the tangent K M for the hours of XI, X, IX, &c., as far as necessary; that is, until there be five hours on each side of the substile. The sixth hour, accounted from that hour or part of the hour on which the substile falls, will be always in a line perpendicular to the substile, and drawn through the centre C.

77. In all erect dials, CM, the hour line of for which the dial is to serve; for that line is the XII is perpendicular to the horizon of the place intersection of a vertical plane with the plane of 74. To find RQ the difference of longitude of pendicular to the plane of the horizon; and any the meridian of the place, both which are perthe places D and Z; say, as radius is to the co-line HO, or ho, perpendicular to CM, will be a sine of RWD 38° 30', the height of the equi- horizontal line on the plane of the dial, along noctial at Z, so is the co-tangent of DW 360 which line the hours may be numbered; and the plane's declination, to the co-tangent of RQ CM being set perpendicular to the horizon, the the difference of longitudes. Thus, dial will have its true position. To the logarithmic sine of 51° 30' Add the logarithmic tangent of 54° 0'

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9.89354

10.13874

10.03228

is the nearest tangent of 47° 8' WR; which is the co-tangent of 42° 52′ RQ, the difference of longitude sought. Which difference, being reduced to time, is 2 h. 514 m.

75. And thus having found the latitude and longitude of the place D, to whose horizon the

78. If the plane of the dial had declined by an equal angle towards the east, its description would have differed only in this, that the hourline of XII would have fallen on the other side of the substile CL, and the line HO would have a subcontrary position to what it has in this figure.

79. And these two dials, with the upper points of their stiles turned toward the N. Pole, will serve for other two planes parallel to them; the one declining from the N. towards

the E., and the other from the N. toward the W., by the same quantity of angle. The like holds true of all dials in general, whatever be their declination and obliquity of their planes to the horizon.

80. If the plane of the dial not only declines, but also reclines, or inclines. Suppose its declination from fronting the south S be equal to the arc SD, fig. 6, on the horizon; and its reclination be equal to the arc D d of the vertical circle DZ: then it is plain, that if the quadrant of altitude Zd D on the globe cuts the point D in the horizon, and the reclination is counted upon the quadrant from D to d; the intersection of the hour-circle PRd, with the equinoctial W Q E, will determine Rd, the latitude of the place d, whose horizon is parallel to the given place Zh at Z; and R Q will be the difference in longitude of the places at d and Z. Trigonometrically thus:-Let a great circle pass through the three points, W, d, E; and in the triangle W Dd, right angled at D, the sides W D and Dd are given; and thence the angle DW d is found, and so is the hypothenuse W d. Again, the difference, or the sum, of D W d and DW R, the elevation of the equinoctial above the horizon of Z, gives the angle d WR; and the hypothenuse of the triangle W Rd was just now found; whence the sides Rd and WR are found, the former being the latitude of the place d, and the latter the complement of RQ, the difference of longitude sought. Thus, if the latitude of the place Z be 52° 30′ N. the declination SD of the plane Zh (which would be horizontal at d) be 36°, and the reclination be 15°, or equal to the arc Dd; the south latitude of the place d, that is, the arc Rd, will be 15° 9'; and RQ, the difference of the longitude, 36° 2'. From these data, therefore, let the dial, fig. 7, be described, as in the former example.

81. There are several things requisite in the practice of dialling; the chief of which shall be given in the form of arithmetical rules, simple and easy to those who have learned the elements of trigonometry. For in practical arts of this kind, arithmetic should be used as far as it can go; and scales never trusted to, except in the final construction, where they are absolutely necessary in laying down the calculated hour distances on the plane of the dial,

82. The latitude of the place, the sun's declination, and his hour distance from the meridian, being given, to find, first, his altitude, second, his azimuth. Let d, fig. 6, be the sun's place, dR, his declination; and in the triangle PZd, Pd the sum, or the difference of d R, and the quadrant PR, being given by the supposition, as also the complement of the latitude PZ, and the angle dP Z, which measures the horary distance of d from the meridian; we shall (by spheric trigonometry) find the base Zd, which is the sun's distance from the zenith, or the complement of his altitude. And, as sine Zd: sine Pd :: sine d PZ: dZ P, or of its supplement DZS, the azimuthal distance from the south.

83. Or the practical rule may be as follows: Write A for the sine of the sun's altitude, L and for the sine and co-sine of the latitude, D and d for the sine and co-sine of the sun's declina

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86. These theorems will be found useful and

expeditious enough for solving those problems, in geography and dialling, which depend on the relation of the sun's altitude to the hour of the day.

87. Example I. Suppose the latitude of the place to be 51° 30′ north: the time five hours distant from XII, that is, an hour after VI in the morning, or before VII in the evening; and the sun's declination 2° north. Required the sun's altitude?

Then to log. L-log. sin. 51° 30'-1'89354 add log. D=log. sin. 20° 0' -1.53405

Their sum -1.42759 gives LD logarithm of 0.267664, in the natural sines. And, to log. H=log. sin. 15° 0'-1.41300 log. log. sin. 38° 0'-1.79414 add { log. d=log. sin. 70° 0′-1-97300

Their sum -1.18014 gives Hld logarithm of 0-151408, in the natural sines. And these two numbers (0.267664 and 0.151408) make 0·419072A; which, in the table, is the nearest natural sine of 25° 47', the sun's altitude sought.

88. In these calculations the radius is considered as unity, and not 10-00000, by which, instead of the index 9, we have -1, which only makes the work a little easier.

89. The saine hour distance being assumed on the other side of VI, then L D-Hld is 0.116256, the sine of 60°40′ 30"; which is the sun's altitude at V in the morning, or VII in the evening, when his N. declination is 20°. But when the declination is 20° S. (or towards the depressed pole) the difference Hld-LD becomes negative; and thereby shows, that an hour before VI in the morning, or past VI in the evening, the sun's centre is 6° 40' 30" below the horizon.

90. Example II. From the same data to find the sun's azimuth. If H, L, and D, are given, then from H having found the altitude and its complement Zd: and the arc Pd (the distance from the pole) being given; say, As the co-sine of the altitude is to the sine of the distance from the pole, so is the sine of the hour distance from the meridian to the sine of the azimuth distance from the meridian. Let the latitude be 51° 30' N., the declination 15° 9′ S., and the time 2 h. 24 m. in the afternoon, when the sun begins to illuminate a vertical wall, and it is required to find the position of the wall. Then, by the fore

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