ways according to the view that is taken of the subject. They may all, however, be deduced from the formula investigated in art. 21, namely, that radius is to the sine of the latitude, as the tangent of the norary angle described by the sun between any hour and noon, is to the tangent of the angle which the hour-line on the dial makes with the meridian. From this formula we inmediately derive the following results.

Method 1.

27. Let CMO, C'M'O (fig. 8) be the meridian line on the dial, the space between CM, C'M' being left for the thickness of the stile, and CC' its centre, and 6 C 6 the six o'clock hour-line.

Make a right-angled triangle cmo, fig. 9, of any magnitude, having one of its acute angles c equal to the latitude of the place.

In the meridian, take CM and C'M' equal to cm, the hypothenuse of the triangle, and MO and M'O' equal to mo, the side opposite to the angle c.

Through M, M' draw PQ perpendicular to CO.

On O and O', as centres, with OM as a radius, describe quadrants MH, M'H'.

Divide each quadrantal arc into six equal parts.

Through the points of division draw the lines O 1, O 2, O 3, &c. also, O' 11, O' io, O' 9, &c. meeting PQ in v, u, x, &c. and in r, s,

t, &c.

From the points C, C' draw lines C 1v, C2u, C 3 x, &c. to the points v, u, x, &c. and C' 11r, C 10 s, Cg t, &c. to the points r, s, t, &c. and these will be the hour-lines of the dial, viz. C1 and C 11 will be the hour-lines of I in the afternoon and XI in the forenoon, and C 2, C 10 the hour-lines of II and X, and so on.

The hour-lines before six in the morning, and after six in the evening, are to be found from the adjoining intermediate hours, as directed

in art. 22.

The demonstration of this construction is obvious; for in the rightangled triangles OMv, CMv, we have

CM: Mv :: rad, : tan. MCv,

and Mv MO:: tan. MOv: rad.

Therefore, ex æquo inv. CM: MO :: tan. MOv: tan. MCv, but CM MO:: cm: mo :: rad. : sin. lat.; hence, rad. : sin. lat. :: tan. MOV: tan. MCv.

Therefore, the angle MCv is rightly determined, and the demonstration applies alike to all the hour-lines.

This construction, although very simple, is rather inconvenient in practice, because the lines 04, 05, and 08, 07, may go off the surface on which the dial is to be delineated, before they meet the line PQ. The next construction has not this defect.

4

Method 2.

28. Let CM, C'M' be the double meridian line (fig. 10), and 6C6 the six o'clock hour-line, and let cmo (fig. 9), be a right-angled triange, constructed as directed in the first operation of Method 1.

On C, C' the centres of the dial, with a radius equal to cm, the hypothenuse of the triangle com (fig. 9), describe semicircles on opposite sides of the meridian.

On the same centres, with a radius equal to om, (the side opposite to the apple which is the latitude) describe other two semicircles on opposite sides of the meridian,

Divide each quadrant of the two semicircles into six equal parts, at the points of division 1, 2, 3, &c. 11, 10, 9, &c. and let the numbers be written at the points of division, in the same order, in respect to the meridian, as the characters for the hours are to be placed on the dial.

Then, to find the position of any hour-line, as, for example, that for three in the afternoon: let D be the third point of division on the inner circle, and E the third point of division on the outer circle, reckoned from the meridian on the quadrant through which the afternoon hour-lines are to pass. Draw EBA perpendicular to the meridian, and DB parallel to it, meeting the perpendicular in B.

Draw a straight line from C through B, and the line CB will be the hour-line for III in the afternoon, as required.

And in the very same way may all the other hour-lines be drawn on the dial.

To prove the truth of this construction, let EB meet the meridian in A, and join EC, which will evidently pass through D. Because BD is parallel to AC, CE: CD :: AE: AB; but by the construction, CE: CD :: rad. : sin. lat. ; and, by trigonometry, AE AB :: tan. ACE: tan. ACB; therefore, rad. : sin. lat. :: tan. ACE: tan. ACB; How ACE is equal to the horary angle which the sun describes in three hours; therefore CB is the hour-line for three in the afternoon. (Art. 39.)

Construction of Dialling Scales.

29. There is another very elegant geometrical construction for the hour-lines, by which scales may be made for the construction of dials, which save the labour of dividing circles.

To construct these scales, divide AB (fig. 11), a quadrant of a circle, into six equal parts. Draw the line ba to touch the middle of the arc at G. Draw lines from the centre through A and B, the extremities of the arc. to meet the tangent in a and b, and also through the divisions, to meet the tangent in the points against which the numerals VI, V, IV, &c. are placed. Then the line between the extreme points a and b is the scale of hours.

Next, divide EF, a quadrant of the same circle, into 90 equal parts, (only every tenth division is marked in the figure). From the points of division draw perpendiculars to OF, the radius. Draw lines through E and the bottoms of the perpendiculars, and produce them, until they meet the circumference again in the points 10, 20, 30, &c. transfer the chords of the arcs D 10, D 20, D 30, &c. (also the chords of the intermediate arcs not distinguished in the figure) to a straight Iine df, numbering them as in the figure; and the line df will be the scale of latitudes.

If the chords of all the arcs from 0° to 90° of the quadrant EF be transferred to another straight line ef, a scale of chords will be formed, which is frequently wanted in making dials.

Construction of a Horizontal Dial by the Scales.

30. Let CM, C'M' be the meridian, and 6 C'C' 6 the six o'clock hour-line (fig. 12.)

From the scale of latitudes take the extent from the beginning of the scale to the division corresponding to the latitude of the place for which the dial is to be made, and set it off from C to a, and from C to a'.

From the points a, a', place lines a b, a' b', each equal to the whole length of the scale of hours, to terminate at band ' in CM, C'M', the meridian line.

Transfer the divisions of the scale of hours to the lines a b, a' b', numbering them as in the figure.

From the points C, C' draw the lines C 1, C2, C3, &c. also C' 11, C' 10, C9, &c. and these will be the hour-lines of the dial.

The morning hours before VI, and evening hours after VI, are found as explained in the other constructions. And the stile is to be formed in all respects as has been described.

To demonstrate the truth of this construction, let the latitude for which the dial is made be equal to the number of degrees in the arc Ep, (fig. 11.) Then, pq being drawn perpendicular to OF, and Eq drawn meeting the circle in r, and Dr joined, it is manifest, from the construction of fig. 11 and fig. 12, that the triangle DrE (fig. 11) is in all respects equal to the triangle aCb (fig. 12), so that Dr=Ca, TE=C, and DE=ab; and since, in fig. 13, rad. : sin. lat. :: EO: Oq :: Er rD; therefore, in fig. 14, rad. : sin. lat. :: ¿C: Ca.

Let H (fig. 12) be the point in which any one of the hour-lines (for example, that for IV in the afternoon) meets ab. In the six o'clock line, place CN equal to Cb; join ¿N, and through H draw KHL parallel to CN, meeting the meridian in K, and the line 6N in L; and join CL. And because No and ab are similarly divided at L and H, and aH and Hb, in fig. 12, are respectively equal to a IV and IV b in fig. 11; therefore, No in fig. 12, and ab in fig. 11, are similarly divided at L and IV. Now the triangles NC (fig. 12) and Ob (fig. 11) are manifestly similar; therefore it is easy to see that the angle UCL in fig. 12, must be equal to OIV in fig. 11; and hence bCL in fig. 12, must be equal to the horary angle described by the sun between noon and IV in the afternoon."

Now LK=UK: HK:: tan. LCb: tan. HCv. But K: HK bC: Ca rad. : sin. lat.; therefore, rad. : sin. lat. :: tan. hor. ang. : tan. HCb. Hence it follows that the angle which the hour-line EC, or IVC, makes with the meridian, is of the proper magnitude: and the same may be proved in like manner of all the others.

Construction of Horizontal Dials by a Globe.

31. The construction of a horizontal dial, and indeed of any dial whatever, as will appear farther on, may be very naturally deduced from the doctrine of the sphere. For, let aPBp (fig. 13) represent the earth, which we may suppose transparent, and let its equator be divided into 24 equal parts by meridian circles a, b, c, d, e, &c. one of which is the geographical meridian of any given place, as Edinburgh, which we may suppose at the point a. If now the hour of 12 were marked at the equator, both upon that meridian and the opposite one, and all the rest of the hours in order on the other meridians, they will be the hour circles of Edinburgh, and the sun will move from one of them to another in an hour.

Now, if the sphere had an opake axis, terminating at the points Pp, the shadow of the axis, which is in the same plane with the sun and each meridian successively, would fall upon every particular meridian, and hour, when the sun came to the opposite meridian, and would therefore show the time at Edinburgh, and all other places on the same meridian. If the sphere were now cut through the middle, by a plane ABCD, in the rational horizon of Edinburgh, one half of the axis would be above the plane, and the other half below it; and if straight lines were drawn from the centre of the plane to those points where its circumference is cut by the hour circles of the sphere, those lines would be the hour circles of an horizontal dial for Edinburgh; for the shadow of the axis would fall upon each hour-line of the dial when it fell on the like hour-circle of the sphere.

32. It appears, then, that to construct a horizontal dial by the terrestrial globe, we must place the globe in such a position, that the arc of the brazen meridian between the pole and horizon may be equal to the latitude of the place, and that any one of the meridians on the globe may coincide with the brazen meridian; and then the arcs of the horizon between its north point and its intersections with the 24 meridians on the globe will be the measures of the angles which the hour-lines on the dial must make with the meridian line.

33. From the same principles we may derive immediately the formula already investigated. For, let PHp be any hour circle which cuts the horizon in H, then in the right-angled spherical triangle PBH, there are given PB, one of its sides adjacent to the right angle B, equal to the latitude, and the angle HPB at the pole, which is equal to the hour angle from noon, to find HB, the arc of the horizon between the meridian and hour circle, passing through the sun, which arc is the measure of the angle at the centre of the dial contained by the meridian and hour-line corresponding to that hour circle.

By the principles of spherics (see Spherical Trigonometry), in any right-angled spherical triangle, radius is to the sine of either of the sides about the right angle, as the tangent of the adjacent angle to the tangent of the other side about the right angle; that is, in the pre