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hreadth and thickness must be large enough to place, it would still be a horizontal plane 90° keep he shadows of the gnomons from ever falling from that place, but for a different longitude, quite out of the sides of the hollows, even when which would alter the reckoning of the hours the sun's declination is at the greatest. Lastly, accordingly. draw the equinoctial dial in the middle, all the 69. To calculate the angles which the hour hours of which are equidistant from each other; lines of a horizontal dial make with the meridian and the dial will be finished.

or twelve o'clock line, see fig. 3. Let NESW 66. As the sun goes round, the broad end of represent the horizon of any place, PSN the the shadow of the stile acbd will show the meridian, and P the N. pole of the sphere : hours in the quadrant Ac, from the sun-rise till let KPH be any hour circle, for example, the VI in the morning; the shadow from the end M circle which makes with the meridian an angle will show the hours on the side Lq from V to of 15°, then the arch of the horizon interIX in the inorning; the shadow of the stile cepted between N, the north, and PH the efgh in the quadrant Dg in the long days, will hour circle, in the plane of which the sun is at show the hours from sun-rise till VI in the XI or I o'clock, measures the angle contained by morning; and the shadow of the end N will the substile of the dial, and the hour lines corshow the morning hours, on the side Or, from responding to these hours. In the spherical III 10 VII. Just as the shadow of the northern triangle PNH, right angled at N, there are given stile abtd goes off the quadrant Ac, the shadow the side PN, which is the elevation of the pole of the southern stile i klm begins to fall within above the horizon, and the angle N PH which is the quadrant Fl, at VI in the morning; and contained hy the meridian and hour circle, to shows the time, in that quadrant, from VI to find NH the arch of the horizon opposite that XII at noon; and from noon till VI in the angle. By spherical trigonometry, radius is to evening, in the quadrant m E. And the shadow the sine of PN as the tangent of NPH to the of the end O shows the time from X1 in the tangent of N H the side required. Hence we forenoon till III in the afternoon, on the sider N; have this practical rule. To find the angle which as the shadow of the end P shows the time from any hour line of a horizontal dial makes with the IX in the morning till I o'clock in the afternoon meridian, or which is the same, to find the angle on the side Q s.

which the hour lines on any dial make with the 67 At noon, when the shadow of the eastern substile.—To the logarithmic sine of the latitude stile e f g h goes off the quadrant h C, in which it of the place for which the dial is made, add the showed the time from VI the morning till logarithmic tangent of the sun's distance from noon, as it did in the quadrant g, from sun- the meridian, for the hour required, the sum, risc till VI in the morning, the shadow of the 1-10, is the logarithmic tangent of the angle western stile nopq begins to enter the quadrant required. Hp; and shows the hours thereon from XII at 70. Erample.To find the angles which the noon till VI in the evening : and after that till hour lines of XI or I make with the meridian of sun-set, in the quadrant 9G; and the end Q casts a horizontal dial for the latitude of London, which a shadow on the side Ps, from V in the evening is 51° 30'. till IX at night, if the sun be not set before that time. The shadow of the end I shows the time To logarithmic sine of 51° 30' 9.89354 on the side Kp from III till VII in the after Add logarithmic tangent of 15° 9:42805 noon; and the shadow of the stile abcd shows the time from VI in the evening till the sun sets. Sum, rejecting 10, is

9.3215 The shadow of the upright central wire, that supports the globe at top, shows the time of the day, which is the tangent of 11° 51' nearly. In like in the middle or equinoctial dial, all the summer manner it will be found, that the hour lines of X half-year, when the sun is on the north side of and II make each with the meridian an angle of the equator.

24° 18', &c. And by computing in this manner,

with the sine of the latitude, and the tangents of DiallinG BY SPHERICAL TRIGONOMETRY.

30°, 45°, 60°, and 75°, for the hours of II, III,

IV,' and 'V in the afternoon; or of X, IX, VIII, 68. The construction of sun-dials on all planes and VII in the forenoon ; you will find their anwhatever, may be included in one general rule; gular distances from XII to be 24° 18', 38° 3', sufficiently intelligible, if that of a horizontal 53° 35', and 71° 6'; which are all that there is dial for any given latitude be well understood. occasion to reckon. And these distances may For there is no plane, however obliquely situated be set off from XII by a line of chords ; or with respect to any given place, but what is pa- rather, by taking 1000 from a scale of equal parts, rallel to the horizon of some other place; and, and setting that extent as a radius from C to therefore, if we can find that other place, by á XII, fig. 4, and then, taking 209 of the same problem, on the terrestrial globe, or by a trigo- parts, which are the natural tangents of 11° 50', nometrical calculation, and construct a horizontal and setting them from XII to XI and I, on the dial for it; that dial applied to the place where line H 0, which is perpendicular to C XII: and it is to serve will be a true dial for that place.- so for the rest of the hour lines, which in the Thus, an erect direct south dial in 51° 30' N. lat. table of natural tangents, against the above diskould be a horizontal dial on the same meridian, tances, are 452, 782, 1355, and 2920, of such 90° southward of 51° 30' N. lat.: which falls in equal parts from XII, as the radius C XII conwith 38° 30' S. lat. But if the upright plane tains 1000. And, lastly, set off 1257, the nadeclines from facing the south at the given tural tangent of 51° 30', for the angle of the

stile's height, which is equal to the latitude of vertical plane at 2 is parallel, we proceed to the the place.

construction of a horizontal dial for the place

D, whose latitude is 30° 14' south; but anticiDECLINING DIALS.

pating the time at D by 2 h. 51 m., neglecting the 71. Let us suppose that an upright plane at half minute in practice, because D is so far London declines 36° westward from facing the westward in longitude from the meridian of south, and that it is required to find a place on

London ; and this will be a true vertical dial at the globe to the horizon of which the said plane London, declining westward 36o. is parallel; and also the difference of longitude 76. Assume any right line CSL, fig. 4, for between Loudon and that place.

the substile of the dial, and make the angle 72. Let NES W be the horizon of London, KCP equal to the latitude of the place, viz. fig. 5, whose zenith is Z, and P the N. Polé 30° 14', to the horizon of which the plane of of the sphere; and let Zh be the position of a the dial is parallel; then CRP will be the axis vertical plane at Z, declining westward from S of the stile, or edge that casts the shadow on the (the south) by an angle of 36°; on which plane hours of the day, in the dial. This done, draw an erect dial for London at Z is to be described. the contingent line E Q, cutting the substilar line Make the semi-diameter 2 D perpendicular to Zh, at right angles in K; and from K make KR and it will cut the horizon in D, 36° west of the perpendicular to the axis CRP. Then KG = south S. Then a plane, in the tangent HD, touch- KR being made radius, that is, equal to the ing the sphere in D, will be parallel to the plane chord of 60°, or tangent of 45° on a good sector, Zh; and the axis of the sphere will be equally take 42° 52' (the difference of longitude of the inclined to both these places. Let WQ E be the places 2 and D) from the tangents, and having equinoctial, whose elevation above the horizon of set it from K to M, draw CM for the hour line Z (London) is 38° 30': and PRD be the me- of XII. Take KN, equal to the tangent of an ridian of the place D, cutting the equinoctial angle less by 15° than KM; that is, the tangent in R. Then it is evident, that the arc RD is of 27° 52': and through the point N draw CN the latitude of the place D, where the plane for the hour line of I. The tangent of 12° 52 Zh would be horizontal, and the arc R Q is (which is 15° less than 27° 52'), set off the same the difference of longitude of the planes Zh way, will give a point between K and N, through and DH.

which the hour line of II is to be drawn. The 73. In the spherical triangle WDR, the arc tangent of 2° 8', the difference between 45o and W D is given, for it is the complement of the 50° 42' placed on the other side of C L, will plane's declination from S the south ; which determine the point through which the hour-line complement is 54°, viz. 90°—36°: the angle at of III is to be drawn; to which 2° 8', if the tanR, in which the meridian of the place D cuts the gent of 15° be added, it will make 17° 8'; and equator, is a right angle; and the angle RWD this set off from K towards Q, on the line EQ, measures the elevation of the equinoctial above will give the point for the hour line of IV; and the horizon of Z, namely 38° 30'. Say, there- so of the rest. The forenoon hours line are fore, as radius is to the co-sine of the plane's de- drawn the same way, by the continual addition clination from the south, so is the co-sine of the of the tangents 15°, 30°, 45°, &c., to 42° 52' = latitude of Z to the sine of RD the latitude of the tangent KM for the hours of XI, X, IX, &c., D: which is of a different denomination from the as far as necessary; that is, until there be five latitude of Z, because Z and D are on different hours on each side of the substile. The sixth sides of the equator.

hour, accounted from that hour or part of the As radius

hour on which the substile falls, will be always

10.00000 To co-sine 3600=RQ: 9.90796

in a line perpendicular to the substile, and So co-sine 51° 30' = QZ , 9.79415

drawn through the centre C.

77. In all erect dials, CM, the hour line of To sine 30° 14' = DR 9.70211 = the latitude for which the dial is to serve; for that line is the

XII is perpendicular to the horizon of the place of D, whose horizon is parallel to the vertical intersection of a vertical plane with the plane or plane Zh at Z.

74. To find R Q the difference of longitude of the meridian of the place, both which are perthe places D and Z; say, as radius is to theąco- fine HO, or ho, perpendicular to CM, will be a

pendicular to the plane of the horizon; and any 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 36° which line the hours may be numbered ; and the plane's declination, to the co-tangent of R Q 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' 9.89354

78. If the plane of the dial had declined by . Add the logarithmic tangent of

an equal angle towards the east, its description 54° O


would have differed only in this, that the hour

line of XII would have fallen on the other side Their sum rejecting 10 . 10.03228

of the substile CL, and the line HO would is the nearest tangent of 47° 8' = WR; which have a subcontrary position to what it has in this is the co-tangent of 42° 52' = RQ, the difference figure. of longitude sought. Which difference, being 79. And these two dials, with the upper reduced to time, is 2 h. 51, m.

points of their stiles turned toward the N. 75. And thus having found the latitude and Pole, will serve for other two planes parallel to longitude of the place D, to whose horizon the them; the one declining from the N. towards


the E., and the other from the N. toward the tion, and H for the sine of the horary distance W., by the same quantity of angle. The like from VI. Then the relation of H to A will have holds true of all dials in general, whatever be three varieties. their declination and obliquity of their planes to 84. When the declination is towards the elethe horizon.

vated pole, and the hour of the day is between 80. If the plane of the dial not only declines, XII and VI; it is A=LD + Hld, and H = but also reclines, or inclines. Suppose its de- A - LD, clination from fronting the south S be equal to Id the arc S D, fig. 6, on the horizon; and its re

85. When the hour is after VI, it is A=LD clination be equal to the arc D d of the vertical

LD – A circle DZ: then it is plain, that if the quadrant - Hld, and H=

id of altitude 2d D on the globe cuts the point D

85.* When the declination is toward the dein the horizon, and the reclination is counted upon the quadrant from D to d; the intersection pressed pole, we have A = Hld - LD, and of the hour-circle PRd, with the equinoctial H=


Id WQ E, will determine Rd, the latitude of the place d, whose horizon is parallel to the given

86. These theorems will be found useful and place Z h at Z; and RQ will be the difference expeditious enough for solving those problems, in longitude of the places at d and Z. Trigono- in geography and dialling, which depend on the metrically thus :- Let a great circle pass through relation of the sun's altitude to the hour of the the three points, W, d, £; and in the triangle day. W Dd, right angled at D, the sides W D and 87. Example 1. Suppose the latitude of the Dd are given; and thence the angle D w d is place to be 51° 30' north : the time five hours found, and so is the hypothenuse W d. Again, distant from XII, that is, an hour after VI the difference, or the sum, of D w d and DWR, in the morning, or before VII in the evening; the elevation of the equinoctial above the horizon and the sun's declination 2° north. Required of Z, gives the angle d WR; and the hypothe- the sun's altitude ? puse of the triangle W Rd was just now found; Then to log. L=log.sin. 51° 30°-1'89354 whence the sides Rd and W R are found, the add log. D=log. sin. 20° 0' -1.53405 former being the latitude of the place d, and the latter the complement of RQ, the difference of

Their sum -1.42759 gives longitude sought. Thus, if the latitude of the place LD=logarithm of 0·267664, in the natural sines. Z be 52° 30° N. the declination SD of the plane And, to log. H=log. sin. 15° 0-1.41300 Zh (which would be horizontal at d) be 36°, and

add { log. l =log. sin. 38° 0ʻ–1.79414 the reclination be 15°, or equal to the arc Dd; I log. d=log. sin. 70° C -1.97300 the south latitude of the place d, that is, the arc Rd, will be 15° 9'; and 'RQ, the difference of

Their sum -1.18014 gives the longitude, 36° 2. From these data, there- Hld= logarithm of 0.151408, in the natural fore, let the dial, fig. 7, be described, as in the sines. And these two numbers (0·267664 and former example.

0·151408) make 0.419072=A; which, in the 81. There are several things requisite in the table, is the nearest natural sine of 25° 47', the practice of dialling; the chief of which shall sun's altitude sought. be given in the form of arithmetical rules, simple 88. In these calculations the radius is conand easy to those who have learned the elements sidered as unity, and not 10.00000, by which, of trigonometry. For in practical arts of this instead of the index 9, we have -1, which only kind, arithmetic should be used as far as it can makes the work a little easier. go; and scales never trusted to, except in the 89. The saine hour distance being assumed on final construction, where they are absolutely ne- the other side of VI, then LD- Hld is 0·116256, cessary in laying down the calculated hour dis- the side of 60°40' 30"; which is the sun's altitude tances on the plane of the dial,

at V in the morning, or VII in the evening, 82. The latitude of the place, the sun's decli- when his N. declination is 20o. But when the nation, and his hour distance from the meridian, declination is 20° S. (or towards the depressed being given, to find, first, his altitude, second, pole) the difference Hild-LD becomes negahis azimuth. Let d, fig. 6, be the sun's place, tive; and thereby shows, that an hour before VI dR, his declination; and in the triangle P Zd, in the morning, or past VI in the evening, the Pd the sum, or the difference of dr, and the sun's centre is 6° 40' 30" below the horizori. quadrant PR, being given by the supposition, 90. Erample II. From the same data to find as also the complement of the latitude PZ, and the sun's azimuth. If H, L, and D, are given, the angle d P Z, which measures the horary dis- then from H having found the altitude and its tance of d from the meridian; we shall (by complement Zd: and the arc Pd (the distance spheric trigonometry) find the base Zd, which is from the pole) being given; say, As the co-sine the sun's distance from the zenith, or the comple- of the altitude is to the sine of the distance from ment of his altitude. And, as sine Zd: sine Pd the pole, so is the sine of the hour distance from :: sine d PZ:dZP, or of its supplement DZS, the meridian to the sine of the azimuth distance the azimuthal distance from the south.

from the meridian. Let the latitude be 51° 30' 83. Or the practical rule may be as follows: N., the declination 15° 9' S., and the time 2 h. Write A for the sine of the sun's altitude, L and 24 m. in the afternoon, when the sun begins to l for the sine and co-sine of the latitude, D and illuminate a vertical wall, and it is required to de for the sine and co-sine of the sun's declina find the position of the wall. Then, by the fore

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