144

DIALING.

same ; and the angle of the triangle KEF should be made equal |]thereto. The hour-fines may he found by projection thus, on thecentre Z, Plate LVU. fig, 25, describe a circle as ENW i> to re-present the horizon of the given place,suppose London ; draw N. S.and E. W. at right angles to it, find the pole P be the stereogra-phic projection of the sphere, see Projection of the Sphere.then project the meridians 1 P u, 2 P b, 3 P c, Sec. making theangles N P a, N P />, N P c, &c. equal to lo°, 30°, 43°, &c.Lay a ruler from the points in which these hour-circles cut thehorizon, and draw the lines 1 l , 2 It., 3 III., 4 IV., &c and theywill be the true hour-lines of the dial. Set off the latitude from Sto A, in this case 51° 32', and draw Z A for the stile, which placedover Z S will complete the dial. The arcs X «, N b, N e, See. maybe easily calculated trigonometrically since in triangles X P a,X p l, Isi p, ike. there is given the side X, P and the angle at P,the angle at N being a right angle. We proceed to the calculation asfollows.

Dialing by Spherical Trigonometry .

The construction of sun-dials on all planes whatever may beincluded in one general rule: intelligible, if that of a horizontaldial for any given latitude be well understood. For there is noplane, however obliquely situated-with respect to any given place,but what is parallel to the horizon of some other place ; and there-fore if we can find that other place by a problrm on the terrestrialglobe, or by a trigonometrical calculation, and construct a horizon-tal dial for it; that dial applied to the plane where it is to servewill be a true dial for that place.—Thu;, an erect direct S. dial in514 degrees X. latitude, would be a horizontal dial on the samemeridian, 90° southward of 51-j degrees of X. latitude: which fallsin with 38 j degrees of S. latitude. But if the upright plane de-clines from lacing the S. at the given place, it would still be ahorizontal plane 9*>° Irom that place, hut fora different longitude,which would alter the reckoning of the hours accordingly. Tocalculate the angles which the hour-lines of a horizontal dial makewith the meridian or twelve o’clock line: bee fig. 18. Let X ES W represent the horizon of any place, P S X the meridian, andP the north pole of the sphere: let KPII be any hour-circle, forexample the circle which makes with the meridian an angle of 15degree.-, then it is evident from what has been said in the pre-ceding part of this article, that Nil, the ate of the horizon inter-cepted between N, the north, and PII the hour-circle in whoseplane the sun is at XI or 1 o'clock, measures the angle containedby the substile of the dial, and the hour-lines corresponding to thesehours. In the spherical triangle P X PI, right-angled at X, thereare given the side PX w;hirh is the elevation of the pole above thehorizon end the angle X P II which is contained by the meridianand hour-circle, to find XII the arc of the horizon opposite thatangle. By spherical trigonometry, radius is to the sine of P X asthe tangent ol X P II to the tangent of X II the side required,lienee v.e have this practical rule. To find the angle which anvhour-line of a horizontal dial makes with the meridian, or wbieliis the same,, to find the angle which the hour-lines on any dialmake with the subside :—'I o the logarithmic sine of the latitudeot the place for which the dial is made, add the log. tangent of thesun’s distance from the meridian for the hour required, the sum,abating 10, is the log. tangent of the angle required.

'Example. To find the angles which the hour-lines of XI or Imake w ith the meridian of a horizontal dial for the latitude ot Lon-don , which is 51-j°

To log. sine of 51° 30' 9.S9354

Add log. tan. of 15° 9-4'>05

Sum, rejecting 1.0, is *>.'32159

■which is the tangent of 11° 5 F nearly. In like manner it will hefound, that the hour-linos of X and 11 make each \viih the meridianan auale of 24“ IS'.Vc. And by computing in this-manner, witht.ie sine of the latitude, and the tangents of 30, 45, Oil, and 75°, fortic 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 dis-tances irom XII to he 24° lb', 38° 3', 53° 35', and 71° whichare all that there is occasion to compute for.—And ttiese distancesmay he set off from XII by a line of chords ; or rather, by takinglOOl) from a scale of equal parts, and setting that extent as a ra-dius from C to XII, fig. 19, and then, taking 209 of the sameparts (which are the natural tangents of li c iff),. and setting them

from XII to XI and I, on the line hi, which is perpendicular toC XII: and so for the rest of the hour-lines, which in the tableof natural tangents, against the above distances, are 452, 782, 1355,and 2920, of such equal paits from XII as the radius C XII con-tains 1000. And, lastly, set offl25, (the natural tangent of 5 I°30')for the angle of the stile’s height, which is equal to the latitude ofthe place.

Declining Dials. I.<( us suppose that an upright plane atLondon declines 3b degrees westward from facing the south, andthat it is required to find a place on the globe to whose horizon lhesaid plane is parallel; and also the difference of longitude betweenLondon and that place. Let N KS W be the horizon of London ,fig. 21, whose zenith is Z, and P the north pole of the sphere; andlet ZA be the position of a vertical plane at Z, declining west-ward from S (the south) by an angle of 30 degrees; on which planean erect dial for London at Z is to lie described. Make the semi-diameter ZD perpendicular to Z/i; and it will cut the horizon inD, 36 degrees west of the south S. Then a plane, in the tangentHD, touching the sphere in D, will lie parallel to the plane Z/i;and the axis of the sphere will be equally inclined to both theseplaces. Let W Q E be the equinoctial, whose elevation above thehorizon of Z (London ) is 38-4 degrees ; and PHD be the meridianof the place D, cutting the equinoctial in II. Then it is evident,that the arc RD is the latitude of the place D (where the plane / Awould he horizontal; and the arc RQ is the difference of longitudeof the planes Z/f and Dll. 1. In tire spherical triangle W D R,the arc \V D is given, for it is the complement of the plane’s de-clination from S the south ; which complement is 54° (viz. 90° —.36°:) the angle at R, in which the meridian of the place D cutsthe equator, is a right angle; and the angle II W' D measures theelevation of the equinoctial above the horizon of Z, namely 384degrees. Say therefore, As radius is to tiie co-sine of the plane’sdeclination from the south, so is the co-sine of the latitude of Zto the sine of RD the latitude of I): which is of a different de-nomination from the latitude of Z, because Z and D are on differ-ent sides of the equator.

As radius...10.00000

To co-sine 3b° 0'=RQ *>.90796

So co-sine 51° 30' = QZ 9.79415

To sine 30° 14' = DR (9.70211 ) = the latitudeof D, whose horizon is parallel to the vertical plane V.h at Z. 2.To find RQ. the difference of longitude oi the places D and Z, sar,A- radius is to 1 he cosine of II \V 1) 38 j degrees, the height of tireequinoctial at Z, so is the co-tangent of l)\V 36 degrees, the plane’sdeclination, to the co-tangent of RQ the difference of longitudes.Thus,

To the logarithmic sine of 51° 30' 9-89354

Add to the logarithmic tang, of 54° 0' 10.13874

Their sum rejecting 10.10.03228 is thenear

est tangent of 47° 8'= \VR; which is the co-tangent of 42° 52'=RQ, the difference of longitude sought. W hich difference, beingreduced (o time, is 2 hours 51 i minutes. 3. And thus havingfound the latitude and longitude of the place D, to whose horizontiie vertical plane at Z is parallel, we proceed to the constructionof a horizontal dial for tiie place D, whose latitude is 30° 14' south ;but anticipating the time at D by 2 hours 51 minutes (neglectingfiie-4 min. in practice,) because D is so far westward in longitudefrom the meridian of London ; and this will he a true vertical dialat London , declining westward 36 degrees. Assume any rightline CSL, fig. 19, for the substile of the dial, and make the angleKCP equal to the latitude of the ulace (viz. 30° 14'), to whosehorizon the plane of the dial is parallel; then CUP will he tiie axisof the stile, or edge that casts the shadow on the hours of the day,in the dial. This done, 'draw the contingent line FQ, cutting thesuhslilarliqe at right angles in K ; and Irom K make KR perpen-dicular to the axis CUP. Then KG (=KR) being made radius,that is, equal to the chord of 60° or tangent of 45°, on a good sec-tor take 4J° 52' (the difference of longitude of the places Z andID) from the tangents, and having set it from K to M, draw CMfor the hour-line of XII. Take KN, equal to the tangent of anangle less by 15 degrees than KM; dial is, tiie tangent of 27° 52':anil through the point X draw CN for the hour-line of 1. Thetangent of 12° 52' (which is 15° less than 27° 52'), vt off the samewav, will give a point between K and X', through which the houn-

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