Of Dialing.
227 -
R U L E I.
¥0 find the angks ivhicb tbe hour-lines on any dial make with the subfiile .
To the logarithmic sine of the given latitude, or of the stile' ’selevation above the plane of the dial, add the logarithmic tangentof the hour * distance from the Meridian, or from the -j- fubstile; andthe sum minus radius will be the logarithmic tangent of the anglesought.
For, in Fig. 2. KC is to KM in the ratio compounded of the ratioof KC to KG (=KR) and of KG to KM-, which, making CKthe radius 10,000000, or io,cooo, or 10, or I, are the ratio ofjo,000000, or of 10,0000, or of 10, or of 1, to KG x KM.
Thus, in a horizontal dial, for latitude 51 0 30', to find the angulardistance of XI in the forenoon, or I in the afternoon, from XII.
To the logarithmic sine of 51° 30' 9.89354 J
Add the logarithmic tang. of 15 0 o' 9.42805
The sum — radius is - 9.32159 = the logarithmic
tangent of n° 50', or of the angle which the hour-line of XI or Imakes with the hour of XII.
And by Computing in this manner, with the sign of the latitude,and the tangents of 30, 45, 60, and 75 0 , for the hours of II, III, IIII,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 » anc ^ 7 1 * ^ ; which are all that there is occasion to computefor.——And these distances may be set off from XII by a line ofchords; or rather, by taking 1000 from a scale of equal parts, andsetting that extent as a radius from C to XII; and then, taking 209of the fame parts (which, in the tables, are the natural tangent ofii» 50') and setting them from XII to XI and to I, on the line ho ,which is perpendicular to < 7 X 11 : and so for the rest of the hour-lines
* That is, of 15, 30, 45, 60, 75*, for the hours of I, II, III, IIII, V in theafternoon; and XI, X, IX, VIII, VII in the forenoon.
f In all horizontal dials, and erect north or fouth dials, the fubstile and Meridianare the fame: but in ali declining dials, the fubstile line makes an angle with theMeridian.
} In which case, the radius CK is supposed to be divided into 1000000 equal parts.
G g 2 which,