f Divisions from beginning of Scale of Latitudes.

| D. Parts | D. Parts | D. Parts | |----------|----------|----------| | 1 | 247 | 22 | | 2 | 493 | 23 | | 3 | 789 | 24 | | 4 | 984 | 25 | | 5 | 1228 | 26 | | 6 | 1470 | 27 | | 7 | 1711 | 28 | | 8 | 1949 | 29 | | 9 | 2186 | 30 | | 10 | 2419 | 31 | | 11 | 2650 | 32 | | 12 | 2879 | 33 | | 13 | 3104 | 34 | | 14 | 3325 | 35 | | 15 | 3543 | 36 | | 16 | 3758 | 37 | | 17 | 3969 | 38 | | 18 | 4176 | 39 | | 19 | 4378 | 40 | | 20 | 4577 | 41 | | 21 | 4773 | 42 |

Distance of Divisions from beginning of Scale of Hours.

| Hours and Minutes | Parts | Hours and Minutes | Parts | Hours and Minutes | Parts | |------------------|-------|------------------|-------|------------------|-------| | XII. 0 | 0-0 | II. 0 | 517-7 | IV. 0 | 896-5 | | 5 | 30-3 | 5 | 534-1 | 5 | 913-1 | | 10 | 59-1 | 10 | 550-4 | 10 | 930-0 | | 15 | 87-0 | 15 | 566-4 | 15 | 947-1 | | 20 | 113-8 | 20 | 582-4 | 20 | 964-5 | | 25 | 139-6 | 25 | 598-3 | 25 | 982-0 | | 30 | 164-5 | 30 | 614-0 | 30 | 1000-0 | | 35 | 188-8 | 35 | 629-6 | 35 | 1018-3 | | 40 | 212-0 | 40 | 645-2 | 40 | 1036-9 | | 45 | 234-6 | 45 | 660-8 | 45 | 1055-8 | | 50 | 256-7 | 50 | 676-3 | 50 | 1075-2 | | 55 | 278-0 | 55 | 691-7 | 55 | 1095-0 | | I. 0 | 298-8 | III. 0 | 707-1 | V. 0 | 1115-4 | | 5 | 319-2 | 5 | 722-5 | 5 | 1136-2 | | 10 | 339-0 | 10 | 737-9 | 10 | 1157-5 | | 15 | 358-4 | 15 | 753-4 | 15 | 1179-6 | | 20 | 377-3 | 20 | 769-0 | 20 | 1202-2 | | 25 | 395-9 | 25 | 783-6 | 25 | 1225-4 | | 30 | 414-2 | 30 | 800-2 | 30 | 1249-7 | | 35 | 432-2 | 35 | 816-0 | 35 | 1274-6 | | 40 | 449-8 | 40 | 831-9 | 40 | 1300-4 | | 45 | 467-1 | 45 | 847-8 | 45 | 1327-2 | | 50 | 484-8 | 50 | 863-8 | 50 | 1355-1 | | 55 | 501-1 | 55 | 880-1 | 55 | 1383-9 | | VI. 0 | 1414-2 | | | | |

Construction of a Horizontal Dial by the Scales or the Tables.

36. Draw two parallel straight lines ac, bd, at a distance equal to the thickness of the style of the dial (Plate CLXXXV. fig. 4), for the double meridian line, and across them, at right angles, draw fabe, the six o'clock hour line. Supposing the dial is to be for London, in latitude 51°, make be and af each equal to the distance from 0 to 51° on the scale of latitudes (fig. 9), and place between the points e, f and the meridian lines, straight lines ed, fc, each equal to the whole length of the scale of hours, viz. from XII. to VI.; make cd and df each equal to the distance from XII. to I.; and through the points I, I draw axI. and dl for the eleven and one o'clock hour lines. In like manner, make c2 and d2 equal to the distance XII. II., and draw the hour lines ax, bII. and so on, for the remaining hour lines; the quarters and lesser divisions are to be laid down in the same way. The lines for the hours before VI. in the morning, and after VI. in the evening, are to be drawn as directed in art. 26.

37. If a common scale of equal parts and the tables be used, then make be and af each equal to 871-5 equal parts, this being the mean between the numbers for 51° and 52° in the table for the scale of latitudes; the lines ed and ac must in all cases be each 1414-2 from the same scale; and the distances d1 and c1 each 2998-1; and d2, c2 each 5177-7, these being the distances for 1 and 2 from the beginning of the scale (see Table). The distances of 3, 4, 5 from c and d must be 7017, 896-5, and 1154-4 equal parts respectively, as appears by the table; and in this way the points in de, cf, through which the hour lines pass, also the lines for every fifth minute, may be readily found, and the corresponding divisions transferred to the circumference of the dial.

Vertical North and South Dials.

38. These dials are represented in fig. 1 and 2 of Plate CLXXXVII. It has been explained in art. 15 that south and north dials for a given latitude would be horizontal Dialling.

Geometrical Construction of Vertical East and West Dials.

42. On the east and west vertical plane (fig. 8, 9) draw the horizontal line HR, and at an assumed point c in that line draw a straight line acb, so as to make with it an angle equal to the latitude of the place, and directed, in each dial, towards the north. This line is to serve as the six o'clock hour line, over which the rod or wire AB is to be placed on its supports, Aa, Bb, as the style of the dial. In the line acb take cd equal to Bb, the height of the style. About b as a centre, with a radius equal to bc, describe a semicircle, and divide each quadrant into six equal parts; draw EcQ perpendicular to bc, and from b, through the points of division of the circle, draw straight lines, producing them until they meet the line EQ. Through the points of intersection draw perpendiculars to EQ, which will be the hour lines; against these the hours are to be written as in the figures. At the points a and b erect the supports of the style perpendicular to the plane of the dial, and each equal in length to the line bc, and over them place the rod AB, and the dial is finished.

43. The east dial will show the hours from sunrise until noon, and the west dial from a little after noon until sunset; but neither can indicate the exact time of noon by a shadow, because then it goes off parallel to their planes.

Polar Dial.

44. This dial is shown in fig. 3, Plate CLXXXVII. It is described on a plane perpendicular to the meridian, and passing through the poles. It has a great affinity with east and west dials; for if a prism with six rectangular faces were placed with its axis directed to the pole, and two of its faces due east and west, the remaining two would form planes for polar dials, and the face directed to the pole would serve as the plane of an equinoctial dial.

45. The style of a polar dial, like that of an east or west dial, must be parallel to its face, and may have the same form (see fig. 10). To construct it, draw ab in the plane of the meridian for the twelve o'clock hour line, and cross it at right angles by the horizontal line HR. The hour lines are to be found exactly as in the east and west dials, and marked with the hours as in the figure.

A polar dial may, if it be of sufficient extent, show time from a little after six in the morning to a little before six in the evening. At the hours of six the sun is in the plane of the dial, and the shadow parallel to its face.

Putting x to denote the hour from noon, and y for the distance of the hour line from the meridian line, and d for the height of the style above the plane of the dial; the formula for a polar dial is

\[ y = d \tan x \]

Vertical Declining Dials.

46. It seldom happens that an upright wall faces exactly one of the cardinal points; therefore, in general, a dial described on the plane of a wall will be a vertical declining dial.

47. The declination of a plane is an arch of the horizon between the plane and the prime vertical; or it is the arch between the meridian and a plane perpendicular to the dial plane, and is always reckoned from the south or north.

48. The meridian of a plane is the meridian perpendicular to the plane of the dial. This differs from the meridian of the place, which is the meridian that is perpendicular to the horizon.

The substyle of a dial is the common section of its plane and the plane of its meridian, or it is the line in which perpendiculars drawn from every point in the axis of the dial meet its plane. In horizontal and in vertical south and north dials, the substyle coincides with the twelve o'clock hour line; but not in declining dials. The difference of longitude of a dial plane is the angle which the plane of its meridian makes with the meridian of the place.

The latitude of any dial plane is the angle which the axis makes with the plane; which is also the latitude of the place where the dial would be a horizontal one.

49. In the adjoining figure let \( AB \) be a line drawn on a vertical plane, a wall, for instance. It may have any aspect, but, to fix our ideas, let us suppose it to face some point between the south and west; let \( CE \) be the style or axis fixed at \( C \) in the wall in the direction of the earth's axis; draw the vertical line \( CD \) on the plane, and from \( E \), the extremity of the style, draw \( ED \) perpendicular to \( CD \); the plane of the triangle \( CDE \) will manifestly coincide with the meridian, and \( CD \) will be the twelve o'clock hour line on the dial.

Let \( NCF \) be a horizontal line passing through \( C \) due east and west; conceive a plane to pass along \( NF \) and the vertical line \( CD \); this will coincide with the prime vertical in the heavens. Let us now suppose that at some hour, for example two in the afternoon, the horary plane (that is, the plane passing through the sun and the axis \( CE \)) meets the plane of the dial in the line \( CH \), and the plane of the prime vertical in \( Ch \); the first of these lines will be the hour line of two on the dial to be constructed, and the second the line of the same hour on the prime vertical, that is, on a vertical south dial. Now by the theory of that dial (art. 38) the angle \( DCA \) will be known; it is, however, the angle \( DCH \) that is required in order to construct the dial under consideration.

50. Suppose a horizontal plane to pass along \( ED \), and meet the horary plane in the line \( EHA \), the plane of the dial in \( DH \), and the prime vertical in \( DA \); the plane \( EDA \) may now be regarded as that of a horizontal dial, of which \( EC \) is the axis, \( E \) the centre, and \( ED \) the meridian line; and on this dial \( EH \) will be the hour line of two; but for any given hour, the angle \( DEH \) will be known by formula 1, art. 23. And because the horizontal lines \( DE, DA \) lie, one in the plane of the meridian \( CDE \), and the other in the plane of the prime vertical \( CDH \), which is perpendicular to the former, the angle \( EDH \) will be a right angle; now the angle \( HDH \), or its equal \( BCE \), is the declination of the dial plane (art. 46), and therefore is given, or may be found; therefore the angle \( EDH \), its complement, is known; and hence

51. All the angles of the triangle \( DEH \) are known.

Let \( L \) denote the latitude of the place for which the dial is to be made; \( D \), the angle \( HDH \) or \( BCE \), the declination of the plane; and \( E \) the angle made by the meridian \( ED \) of the assumed horizontal dial, and \( EH \) the hour line; then, in the triangle \( DEH \) we have the angle at \( E \) (denoted by \( E \)), the angle \( EDH = 90 - D \), and therefore the third angle \( DHE = 180 - (E + 90 - D) = 90^\circ - (E - D) \).

In the right-angled triangles \( CDE, CDH \) (which have \( CD \), one of the sides about the right angle, common to both), by trigonometry,

\[ \frac{DE}{DH} = \tan. DCE \text{ or } \cot. DEC : \tan. DCH. \]

But in the triangle \( DEH \),

\[ \frac{DE}{DH} = \sin. DHE : \sin. DEH = \cos. (E - D) : \sin. E; \]

therefore, \( \cos. (E - D) : \sin. E = \cot. DEC : \tan. DCH. \)

Now the first three terms of this proposition are known, because the angles \( E \) and \( D \) are given, and also the angle \( DEC \), which is the latitude; therefore the fourth term, viz. the tangent of \( DCH \), is known; and hence the angle \( DCH \), between the hour line \( CH \) on the dial and the meridian line \( CD \), is known. Thus may all the angles made by the hour lines and the meridian of the dial be determined by plane trigonometry.

52. This way of finding the hour lines requires two operations for each, viz. one to find the angle \( DEH \) at the centre of the auxiliary horizontal dial; and a second to find the angle \( DCH \) at the centre of the dial to be constructed. We shall now investigate a formula which gives each angle by a single operation, when two subsidiary quantities common to them all have been found.

For any vertical declining dial let

\( L \) = latitude of the place;

\( D = BCE \), the declination of the dial plane reckoned from the east towards the south;

\( x \) = the variable horary angle described by the sun since noon;

\( y \) = the corresponding angle \( DCH \) at the centre of the dial; and, as before, put \( E \) for the angle described at the centre of a horizontal dial for the same latitude in the time \( x \).

From what has been shown in the preceding articles, we have

\[ \tan. y = \frac{\cot. L \sin. E}{\cos. (E - D)} \quad \ldots \quad (1) \]

\[ \tan. E = \sin. L \tan. x \quad \ldots \quad (2) \]

In formula (1), instead of \( \cos. (E - D) \), put its equal \( \cos. E \cos. D + \sin. E \sin. D \) (Algebra, art. 239), then divide the numerator and denominator by \( \cos. E \), and lastly, put \( \tan. E \) instead of \( \frac{\sin. E}{\cos. E} \); the formula will then be transformed to this,

\[ \tan. y = \frac{\cot. L \tan. E}{\cos. D + \sin. D \tan. E}. \]

Now put \( \sin. L \tan. x \) instead of \( \tan. E \); and again \( \frac{\sin. x}{\cos. x} \) instead of \( \tan. x \); we then have

\[ \tan. y = \frac{\cot. L \sin. L \sin. x}{\cos. x \cos. D + \sin. L \sin. x \sin. D} \quad \ldots \quad (3) \]

Let \( P \) denote a subsidiary angle, such that

\[ \tan. P = \sin. L \tan. D = \frac{\sin. L \sin. D}{\cos. D}; \]

then \( \cos. D = \frac{\sin. L \sin. D \cos. P}{\sin. P} \);

this value of \( \cos. D \) being substituted, instead of it, in formula 3, we get

\[ \tan. y = \frac{\cot. L \sin. P}{\sin. D} \cdot \frac{\sin. x}{\cos. x \cos. P + \sin. x \sin. P} \quad \ldots \quad (4) \]

The denominator of this last fraction is manifestly \( \cos. (P - x) \). Hence we have the following simple formula for computing the angle which the shadow of the axis of any vertical declining dial describes in any time before or after noon.

\[ \tan. y = \frac{\cot. L \sin. P}{\sin. D} \cdot \frac{\sin. x}{\cos. (x - P)} \]

This formula gives the afternoon hours on a vertical Having determined the constants, the hour line angles Dialling may be found as follows:

Calculation of hour line angle of XI. A. M.

\[ \begin{align*} \sin(x = 15^\circ) & = 941300 \\ \tan(Q) & = 979080 \\ \cos(x + P = 45^\circ 36') \text{ ar. comp.} & = 0.15511 \\ \tan(y = 12^\circ 52') & = 935891 \\ \end{align*} \]

Calculation of hour line angle of I. p. m.:

\[ \begin{align*} \sin(x) & = 941300 \\ \tan(Q) & = 979080 \\ \cos(x - P = 15^\circ 36') \text{ ar. comp.} & = 0.01630 \\ \tan(y = 9^\circ 26') & = 922010 \\ \end{align*} \]

The following table exhibits at one view the elements of the dial and the hour line angles for its construction:

| Hours | x | x + P | x - P | y the hour angle | |-------|---|-------|-------|-----------------| | IX. a.m. | 45° | 75° 36' | 60° 24' | | | X. | 30 | 60 36 | 32 10 | | | XI. | 15 | 45 36 | 12 52 | | | XII. | 0 | 30 36 | 0 | | | I. p.m. | 15 | 15° 36' | 9 26 | | | II. | 30 | 0 36 | 17 10 | | | III. | 45 | 14 24 | 24 17 | | | IV. | 60 | 30 24 | 31 34 | | | V. | 75 | 44 24 | 39 52 | | | VI. | 90 | 59 24 | 50 30 | | | VII. | 105| 74 24 | 65 44 | | | VIII. | 120| 89 24 | 88 52 | |

If the hour lines of this dial were traced on a transparent plane and extended, and if the style were produced through the dial, the reverse would show a north dial declining eastward 36°.

Ex. 2. Suppose a vertical south dial decline east 49° in the latitude 51° 30'; to determine the hour line angles. (See fig. 4 of Plate CLXXXVII.)

In this case L = 51° 30'; D = 49° east. We now apply formula B of (3). The data of the dial, the constants, and hour line angles, are exhibited in the following table:

| Hours | x | x - P | x + P | y | |-------|---|-------|-------|---| | III. a.m. | 135° | 93° | 95° 59' | | | IV. | 120 | 78 | 71 12 | | | V. | 105 | 63 | 56 19 | | | VI. | 90 | 48 | 46 5 | | | VII. | 75 | 33 | 39 5 | | | VIII. | 60 | 18 | 32 42 | | | IX. | 45 | 3 | 26 32 | | | X. | 30 | 12 | 19 49 | | | XI. | 15 | 27 | 11 35 | | | XII. | 0 | 42 | 0 | | | I. p.m. | 15 | 57 | 18 32 | | | II. | 30 | 72 | 48 47 | |

In the calculation for the hour line angle of III. a.m. the cosine of \(x - P = 93°\) is negative; this makes the sign of \(y\) negative, and therefore \(y\) an angle between 90° and 180°.

58. If this dial were traced on a plane, and the hour Dialling lines extended and continued through it, and if the style also were produced through the plane, the reverse would be a north vertical dial, declining to the westward 49°.

In general, to make a north declining dial, we have only to make a south declining dial whose declination is the same and lies the same way, and then turn it upside down, and it will be the dial required; but the hours must be numbered the contrary way. Therefore these two examples, duly considered, will serve for examples of all declining dials.

59. It has been already observed (art. 15), that a dial on any plane whatever, given in position at a given place, will be a horizontal dial at some other place, which may be found. This principle gives another method of constructing a declining vertical dial; for if the latitude of that place, and also the difference between its longitude and that of the place where the dial is to show the time, be found, the former will be the angle which the style must make with the plane of the dial; that is, the angle it makes with the substyle; and the latter will give the time the shadow takes to pass between the twelve o'clock line and the substyle, from which the angle they contain may be found. These three elements being known, viz. the latitude and longitude of the place where the dial would be horizontal, and the angle contained by the meridian line and substyle, the construction is reduced to that of a horizontal dial.

60. These elements may be found by spherical trigonometry as follows: let SZNz be the meridian, in which Z and z are the zenith and nadir, and P, p the poles: Let SEN be the horizon, S and N being the south and north points, and E the east; let ZE be a vertical plane or great circle of the sphere on which the dial is to be drawn; let this plane cut the horizon in F, and the plane of the meridian in the straight line Zz; and let it be cut perpendicularly in the line Aa by a circle PAp which passes through the poles. Then Cp or CP will be the axis of the dial, according as it faces the south or the north; Cz or CZ the twelve o'clock hour line, and Ca or CA the substyle (art. 48).

In the spherical triangle ZPA, right angled at A, PZ, the complement of the latitude of the place where the dial is to show time is given; so also is the angle PZA, for it is measured by FZ, the arc of the horizon of which the complement is the declination of the dial (art. 47); from these, AP, the measure of the angle contained by PC the axis, and AC the substyle, that is, the latitude of the dial, (art. 48), also AZ, the measure of the angle contained by the substyle AC and the vertical or twelve o'clock hour line, lastly, the angle ZPA, the difference of longitude of the planes ZPC, APC (art. 73), may be all found.

61. By the principles of spherics (see SPHERICAL TRIGONOMETRY),

\[ \begin{align*} \text{rad.} : \sin PZ & = \sin Z : \sin AP \\ \text{rad.} : \cos PZ & = \tan Z : \cot P \\ \text{rad.} : \tan PZ & = \cos Z : \tan AZ. \end{align*} \]

Let L be the latitude of the place where the dial is to serve,

D the declination of the dial (art. 47),

l the latitude of the dial, that is, of the place where it would be horizontal (48),

a the longitude of the dial,

b the angle between the substyle and the vertical.

Then from the above proportions there is got

\[ \begin{align*} \frac{\cos L}{\sin l} \cdot \frac{\cos D}{\cot a} & = \alpha \\ \frac{\sin L}{\sin l} \cdot \frac{\cot D}{\cot b} & = \beta \\ \frac{\cot L}{\cot b} \cdot \frac{\sin D}{\tan b} & = \gamma \end{align*} \]

These formulae, when L and D are known, give the three elements for the construction of the dial, which is now reduced to that of a horizontal dial. The hour line angles must, however, be so found that one of them shall fall on the vertical or twelve o'clock hour line.

62. Let us again take Ex. I of art. 56, and suppose that a vertical south dial, declining 36° to the west, is to be constructed, the latitude being 54° 30' (fig. 5 of Plate CLXXXVII.). In this case L = 54° 30', D = 36°.

Calculation of l, of a, of b.

| cos L | sin L | cot L | |-------|-------|-------| | 9-76395 | 9-91069 | 9-85327 | | cos D | cot D | sin D | | 9-90795 | 10-13874 | 9-76922 |

\[ \begin{align*} \sin b & = 9-67190 \\ \cot b & = 10-04943 \\ l & = 28° 1' \\ a & = 41° 45' \\ b & = 22° 45' \end{align*} \]

The construction of our south declining dial for latitude 54° 30' is now reduced to that of a horizontal dial for latitude 28° 1'; for the sake of brevity, let us call the former place A and the latter B. The earliest hour that can be shown at A, on the dial, is about IX. a.m., that is, 45° of an hour angle from noon. Now when it is 45° from noon at A, because of the difference of longitude, it will be 45° + a = 86° 45' at B; the hour line angle with the meridian corresponding to this will, by the formula for a horizontal dial, be found by this proportion:

As rad. .......................................................... 10-00000 to sin. latitude (28° 1') ........................................... 9-67191 so is tan. hour angle from noon (86° 45') .................... 11-24577

to tan. hour line angle with meridian (83° 6') ............. 10-91768

Here we have found that at the place B, when it is nine in the morning at A, the hour line will make with the meridian line there an angle of 83° 6'; this is the angle which the hour line of IX. at A will make with the substyle. In the same manner may the angles which the remaining hour lines on the dial make with the substyle be found. The whole are shown in this table.

| Hours at A. | Hour Angles at B. | Hour Line Angles with Meridian at B. | Hour Line Angles with Meridian at A. | |-------------|------------------|-------------------------------------|-----------------------------------| | IX. | 86° 45' | 83° 6' | 60° 21' | | X. | 71° 45' | 54° 56' | 32° 11' | | XI. | 56° 45' | 35° 37' | 12° 52' | | XII. | 41° 45' | 22° 45' | 0° 0' | | I. | 26° 45' | 13° 19' | 9° 26' | | II. | 11° 45' | 5° 35' | 17° 10' | | Substyle. | | | | | III. | 3° 15' | 1° 32' | 24° 17' | | IV. | 18° 15' | 8° 49' | 31° 34' | | V. | 33° 15' | 17° 7' | 39° 52' | | VI. | 48° 15' | 27° 45' | 50° 30' | | VII. | 63° 15' | 42° 59' | 65° 44' | | VIII. | 78° 15' | 66° 7' | 88° 52' | The angles which the hour lines make with the vertical or meridian line at A are found from the like angles at B, by taking the difference between them and the angle \( b = 29^\circ 45' \) for all those on one side of the substyle, and the sum of each and that angle when they lie on the other side. The agreement of the results in this table with those in article 56 will appear by inspection.

To trace a Meridian Line on any Plane.

63. In constructing a dial it is always necessary to determine the line in which the plane of the meridian meets the plane of the dial. On an assumed point in the meridian line as a centre describe several concentric circles with any distances in the compasses; at this point fix a wire truly perpendicular to the plane, and of such a length that when the sun shines its shadow shall extend beyond the circles. Watch now the instant when the extremity of the shadow exactly reaches some one of the circles in the forenoon, and mark the point in which it crosses the circle. In the afternoon, mark in like manner the point in which it again crosses the same circle; and any point that is equally distant from the two points so determined will be in a meridian line passing through the centre of the circle, which may now be drawn. By a like attention to the path of the extremity of the shadow when it crosses the other circles, points may be found in a meridian line passing through their common centre; and a mean position of all the meridian lines thus found will be sufficiently accurate for a dial.

The theory of this method is sufficiently obvious. At equal intervals of time each way from noon, the shadow is nearly of the same length, and makes nearly equal angles with the meridian line; the deviation from absolute equality arises chiefly from the continual change in the sun's declination. This, however, is but little in the course of a day near the solstices. Accordingly, greater accuracy will be obtained about midsummer or midwinter than nearer the equinoxes.

64. A good compass will give an approximation to the true position of the meridian; but the variation, which is not the same at all places and at all times, must be known. In 1823 it was about \( 24^\circ 10' \) west at London, and in the same year \( 27^\circ 48' \) west at Edinburgh. If the direction of the meridian be known at one place, the variation may be there determined; and thence the direction of the meridian at another place not very distant may be found.

65. If a good watch be set to true time, or its deviation from it be found, by taking equal altitudes of the sun with a sextant, or by a single altitude, or any other astronomical observation, a dial may thereby be set truly in the meridian.

66. The pole star is distant from the pole about \( 1^\circ 36' \); hence, knowing the latitude, its greater azimuth may be found. At London its greatest deviation from the meridian either way is about \( 2^\circ 34' \) in azimuth. If now two plummets be suspended at some distance from each other, and in such a position that the pole star may appear in the same plane with them, that plane will be nearly in the meridian. If one of the plummets be moveable, and it be shifted, following the change of position in the star, the two extreme positions of the plumb-line will, with the other plummet, determine two vertical planes, which deviate equally from the meridian in opposite directions. If lines were now traced on a horizontal plane in the directions of the vertical planes thus found, these would form with the meridian line sought equal angles on opposite sides of it, and the meridian would be found by bisecting the angle they contain. The pole star comes to the meridian about noon and midnight in the beginning of October. It may then be seen twice in the same night at its greatest distance on opposite sides of the meridian; but one observed maximum distance, and the computed deviation in azimuth, would serve to find the meridian at any time.

The time of the meridian passage of the pole star on any day of the year may be found as taught in works on astronomy. At this time, as shown by a watch, two plumb-lines may be placed in the meridian, or one, and the corner intersection of two walls truly perpendicular instead of another. Great accuracy in the time of the observation is not required; the motion of the star in azimuth being only 7 minutes of a degree in ten minutes of time at London.

67. A meridian line may be traced by observing when two stars which have the same right ascension, or which differ in right ascension by twelve hours, come into the same vertical plane; for then they are both in the meridian.

The pole star and the star (the first of the three in the tail) of the great bear have nearly this relative position. On the 1st of January 1834, their mean right ascension, as given in our table of the places of the fixed stars (Astronomy), will be,

| Hours | Min. | Sec. | |-------|------|-----| | Ursae Majoris | 12 | 46 | 41 | | Polaris | 1 | 0 | 34 |

The difference from twelve hours is about forty-six minutes of time. By the revolution of the heavens the star comes into the meridian under the pole; but the pole star is not then exactly in the meridian. Afterwards they come into the same vertical, and then the pole star is very near the meridian, and, for finding a meridian line, may be considered as exactly so. The azimuth of the pole star, at the instant when the stars are in the same vertical, may be found by calculation, and the approximate meridian thereby corrected.

There are other pairs of stars which come into the meridian nearly at the same time. The stars \( \beta \) Draconis and \( \alpha \) Ophiuchi are such. The stars \( \delta \) of the lesser bear, and \( \alpha \) of the lyre, are also well adapted to the end in view, in latitudes exceeding the polar distance of the most remote star, which is about \( 51^\circ \).

68. In whatever way a meridian line has been found on a horizontal plane, two plummets hung over it, or so that their lines may pass through it, will indicate the position of the plane of the meridian in space. The line in which it cuts any other plane, as a vertical wall not far from the meridian line, may be found by placing the eye in the meridional plane, with the plummets between it and the wall, and noting points on the wall which their lines cover from the eye. The ingenious dialist may now, with a little dexterity in practical geometry, fix the style of his dial in the wall or other plane, with its edge in the meridian plane, and making with a vertical line the proper angle, so that it may point to the pole. A perpendicular drawn now from the end of the style on the plane, will give a point in the substyle, which may now be drawn to the centre. Thus two important elements in a dial will be determined.

To find the Declination of a Vertical Plane.

69. Place a board in a position truly horizontal, with a straight side in contact with the plane; trace a meridian line on the board; and the angle made by this line and the line in which the board meets the vertical plane will be the inclination of the plane to the meridian.

To find the Inclination of any Plane to the Horizon.

70. Extend a surface truly level until it meet the plane; Dialling.

the line of their intersection will be horizontal. Draw perpendiculars to this line on the two planes from any point in it; and the angle which these form will be the inclination of the plane to the horizon.

If the plane were made to form one side of a temporary trough into which water was poured, the surface of the water would accurately mark the horizontal line. The inclination of the plane to the surface of the water, which is horizontal, might be found by various expedients too obvious to require being pointed out.

When the position of the meridian and the axis of the dial are truly determined, the finding of the inclination and declination of the plane are problems in practical geometry which may be resolved by the application of levels, or plumb-lines and squares, and those simple principles with which most workmen are familiar.

Inclining Dials.

71. We come now to consider inclining dials, or those which stand oblique to the horizon, either projecting forwards from the perpendicular, or retiring backwards.

From what has been taught, any dial plane being given in position, a meridian line may be traced on it; also, at a point in that line an axis may be fixed pointing to the pole of the world. Suppose now that in the following figure, OB is a meridian line drawn on an inclining dial plane, and OC

the axis directed to the pole; the angle COB will be the complement of the latitude of that place where the dial, being vertical, would show true time; this, therefore, is known. If, in addition, we knew the declination of the dial, the hour line angles might be found by the formulae of art. 55. Now it is easy to find the declination, for since the horizon must be perpendicular to the vertical OB, if CB be drawn in the plane of the meridian, and HB in the dial plane, both perpendicular to OB, these lines will be in the plane of the horizon of the place where the dial, being vertical, would show true time; and the angle CBH will be the inclination of the plane of the dial to the meridian, which is the complement of the declination (art. 47). If now we make BH = Bk of any length, and join CH, CA, we have, by trigonometry,

\[ \cos. CBH = \frac{(Ch + CH)(Ch - CH)}{4BH \cdot BC} \]

Hence, by measuring the lines on a scale of equal parts, the inclination may be found.

72. Thus, having the position of the meridian, and the axis, on a dial plane making a given angle with the horizon, we may find the latitude of the place where it would be a vertical declining dial, and also its declination at that place; and with these data the angles which the hour lines make with the meridian may be found by the formulae for vertical dials, art. 55; or else we may find the latitude and longitude of that place where the dial would be horizontal, and construct it by the rules for a horizontal dial.

73. To begin with the first of these methods, let SZN be the meridian (see fig. in next column), Pp the axis of the sphere, Z the zenith of a place where the dial is to stand, SEN the horizon; S, E, and N, the south, east, and north points respectively; also let HFI be a circle of the sphere on the plane of which the dial is to be constructed, and let it meet the horizon in F, and the plane of the meridian Dialling, in the line HA; the line POP will be the axis of the dial, and HOh the meridian line. The arch EF between the east point of the horizon and the plane will be its declination, and the spherical angle HFN its inclination to the horizon, or the complement of its declination from the vertical position.

Now, let SEN be the horizon of a place having H for its zenith, which of course is the place where the dial would be a vertical dial, and let it cut the dial plane circle in f; because the two horizons are perpendicular to the meridian, their intersection will be in the east point in both.

Let L = PN, the given latitude of the place where the dial is to be made;

D = FE, the declination of the dial;

R = complement of angle FEF, its declination;

l = the latitude of the place where the dial would be vertical, which is sought;

d = Ef, its declination there, also sought;

then l - L = Na, the measure of the angle FEF.

By spherical trigonometry, in the triangle EFE, right angled at f,

\[ \begin{align*} \text{rad. : sin. EF} &= \sin. F : \sin. Ef, \\ \text{rad. : cos. EF} &= \tan. F : \cot. E. \end{align*} \]

From these proportions we find

\[ \begin{align*} \sin. d &= \frac{\cos. R \cdot \sin. D}{\text{rad.}} \\ \cot. (l - L) &= \frac{\cot. R \cdot \cos. D}{\text{rad.}} \end{align*} \]

Hence the angles d and l are determined; and these being substituted in the formula of art. 55, instead of L and D, it will become a general expression for the angle y, which the shadow makes with the meridian line on the reclining dial.

Example.—To find the hour lines on a south dial plane FHb, that declines westward 25°, and reclines 15°, in latitude 54° 30'.

Here L = 54° 30', D = 25°, R = 15°.

To find d.

\[ \begin{align*} \text{rad.} &\quad 10-00000 \\ \cos. R &\quad 9-98494 \\ \sin. D &\quad 9-62595 \end{align*} \]

\[ \sin. (d = 24° 6') = 9-61089 \]

To find L.

\[ \begin{align*} \text{rad.} &\quad 10-00000 \\ \cot. R &\quad 10-87195 \\ \cos. D &\quad 9-95728 \end{align*} \]

\[ \cot.(l - L) = 16° 27' 10-52923 \]

Since l - L = 16° 27', and L = 54° 30', therefore l = 70° 57'. The dial declines to the west; therefore (art. 55),

\[ \tan. y = \frac{\tan. Q \cdot \sin. x}{\cos. (x + P)} \quad \text{for the forenoon hour lines,} \]

\[ \tan. y = \frac{\tan. Q \cdot \sin. x}{\cos. (x - P)} \quad \text{for the afternoon.} \]

In these tan. P = \(\frac{\sin. l \cdot \tan. d}{\text{rad.}}\). tan. Q = \frac{\cot l \sin P}{\sin d},

and hence \(P = 22^\circ 53'\), and log. tan. Q = 9.51759.

The elements of the dial and the results of the calculation are exhibited in this table.

| Hours | Hour Angle \(x\) | \(x + P\) | \(x - P\) | Hour Line Angles \(y\) | |-------|-----------------|-----------|-----------|----------------------| | VIII. a.m. | 60° | 82° 55' | 66° 37' | | IX. | 45 | 67 55 | 31 46 | | X. | 30 | 52 55 | 15 16 | | XI. | 15 | 37 55 | 6 10 | | XII. noon. | 0 | 22 55 | 0 0 | | I. p.m. | 15 | 7° 55' | 4 55 | | II. | 30 | 7 5 | 9 25 | | III. | 45 | 22 5 | 14 6 | | IV. | 60 | 37 5 | 19 40 | | V. | 75 | 52 5 | 27 22 | | VI. | 90 | 67 5 | 40 13 | | VII. | 105 | 82 5 | 66 35 |

The hour lines are to be drawn on the dial, so as to make with the XII. o'clock hour line the angles in this table. The dial is represented in fig. 7 of Plate CLXXXVII.

74. To construct the dial by finding the latitude and longitude of a place where it would be horizontal, we must find the angle which the substyle makes with the meridian, and also the latitude and longitude of the plane.

Resuming the figure of art. 73, let a great circle passing through the pole, and perpendicular to the plane of the dial, meet it in the line OA; this will be the substyle, and AOH the angle which it makes with the meridian or XII. o'clock hour line; AOP the angle which the axis makes with the plane of the dial, and the spherical angle APH its difference of longitude.

In the spherical triangle AHP right-angled at A, AHP = \(fhs\) is measured by the arc \(fs\); but this arc is the complement of the arc \(fE\), which we have denoted by \(d\), therefore the angle AHP is the complement of \(d\). Again, the arc HP is the complement of the arc Pa, which was expressed by \(l\). Now formulae have been given (in the preceding article) for the computation of \(d\) and \(l\), therefore the angle AHP and the side HP of the spherical triangle may be considered as known. The three remaining parts of the triangle may be found by spherical trigonometry by these proportions,

\[ \begin{align*} \text{rad.} : \cos H &= \tan PH : \tan AH; \\ \text{rad.} : \sin H &= \sin PH : \sin AP; \\ \text{rad.} : \tan H &= \cos PH : \cot P. \end{align*} \]

By substituting the symbols \(d\) and \(l\) in these proportions, they give

\[ \begin{align*} \tan AH (\text{the angle made by the substyle and meridian}) &= \frac{\sin d \cot l}{\text{rad.}}; \\ \sin AP (\text{the angle made by the axis and substyle}) &= \frac{\cos d \cos l}{\text{rad.}}; \\ \cot P (\text{the dif. of long.}) &= \frac{\cot d \sin l}{\text{rad.}}. \end{align*} \]

These formulae, applied to the example of last article, give

angle made by substyle and meridian = 8° 1'; angle made by axis and substyle = 17° 20';