Star's apparent altitude 10: 0 0 Log, secant 10.006649
Refraction Table VIII.

5.15

Star's true altitude .. 9.54.45 Log. cosine = 9.993467

Correction of the Logarithmic Difference when the Moon's Distance from a Planet is observed.

The arguments of this Table are, the apparent altitude of a planet in the left or right-hand marginal column, and its horizontal parallax at top; in the angle of meeting stands the corresponding correction, which is to be applied by subtraction to the logarithmic difference deduced from Table XXIV., when the moon's distance from a planet is observed. Hence, if the apparent altitude of a planet be 20 degrees, and its horizontal parallax 21 seconds, the corresponding correction will be 16 subtractive, and so on. This Table was computed by the rule in page 51, under which the correction corresponding to the sun's apparent altitude in Table XXV. was obtained, as thus :

Let the apparent altitude of a planet be 23 degrees, and its horizontal parallax 21 seconds; required the correction of the logarithmic difference?

TABLE XXVII.

Natural Versed Sines, and Natural Sines.

Since the methods of computing the true altitudes of the heavenly bodies, the apparent time at ship or place, and the true central distance between the moon and sun, or a fixed star, are considerably facilitated by the application of natural versed sines, or natural sines, this Table is given; which, with the view of rendering it generally useful and convenient, is extended to every tenth second of the semicircle, with proportional parts corresponding to the intermediate seconds; so that either the natural versed sine, natural versed sine supplement, natural co-versed sine, natural sine or natural cosine of any arch, may be readily taken out at sight. The numbers expressed in this Table may be obtained in the following

manner :

A

E

Let ABC represent a quadrant, or the fourth part of a circle; and let the radius CB =unity or 1, be divided into an indefinite number of decimal parts: as thus, 1.0000000000, &c. Make BD the radius CB; and since the radius of a circle is equal to the chord of 60 degrees, the arc BD is equal to 60 degrees: draw DM, the sine of the arc BD, and, at right-angles thereto, the cosine DE: bisect the arc BD in F, and draw FN and FG at rightangles to each other; then will the former represent the sine, and the latter the cosine of the arc BF 30 degrees: bisect BF in H; then HO will express the sine, and HI the cosine of the arc BH = 15 degrees. Proceeding in this manner, after 12 bisections, we come to an arc of 0:0:52"44"3""45"", the cosine of which approximates so very closely to the radius C B, that they may be considered as being of equal value. Now, the absolute measure of this arc may be obtained by numerical calculation, as follows, viz.

G

T

I

M

N OB

Because the chord line BD is the side of a hexagon, inscribed in a circle, it is the subtense of 60 degrees, and, consequently, equal to the radius C B (corollary to Prop. 15, Book IV., of Euclid); wherefore half the radius B S = BM, will be the sine of 30 degrees FN, which, therefore, is .5000000000. Now, having found the sine of 30 degrees, its cosine may be obtained by Euclid, Book I., Prop. 47: for in the right-angled triangle FNC, the hypothenuse FC is given the radius, or 1.0000000000, and the perpendicular FN = half the radius, or .5000000000, to find the base CN the cosine GF; therefore

✓ F C × F C − FN × FN = CN .8660254037, or its equal GF; hence the sine of 30 degrees is . 5000000000, and its cosine. 8660254037. Again,

In the triangle FN B, the perpendicular FN is given = .5000000000, and the base CB-CN=NB. 1339745963, to find the hypothenuse BF: but half the side of a polygon, inscribed in a circle, is equal to the sine of half the circumscribing arc; therefore its half, BT HO, will be the sine of the arc of 15 degrees: hence FN × FN + NB × NB =BF.5176380902; the half of which, viz., .2588190451, is therefore equal to BT, or to its equal HO, the sine of 15 degrees, and from which its cosine HI may be easily obtained; for, in the triangle COH, the hypothenuse C H is given = 1.0000000000, and the perpendicular HO =.2588190451, to find the base CO the cosine HI. Now,

✓CH X CH-HO× HO CO.9659258263 the cosine HI; hence the sine of 15 degrees is. 2588190451, and its cosine .9659258263. Thus proceeding, the sine of the 12th bisection, viz., 52"44"3":"45"", will be found = .0002556634. And because small arcs are very nearly as their corresponding sines, the measure of 1 minute may be easily deduced from the sine of the small arc, or 12th bisection determined as above; for,

As the arc of 52"44"3""45"" is to an arc of 1 minute, so is the sine of the former to the sine of the latter: that is, as 52"44"3""45"" : 1 :: .0002556634: .0002908882; which, therefore, is the sine of 1 minute, the cosine of which is . 9999999577; but this approximates so very closely to the radius, that it may be esteemed as being actually equal to it in all calculations; and hence, that the cosine of 1' is 1.0000000000.

Now, having thus found the sine and cosine of one minute, the sines of every minute in the quadrant may be obtained by the following rule; viz.

As radius is to twice the cosine of 1 minute, so is the sine of a mean arc to the sum of the sines of the two equidistant extremes; from which let either extreme be subtracted, and the remainder will be the sine of the other extreme: as thus,

To find the Sine of the Arc of 2 Minutes.

As radius 1:2::.0002908882 to .0005817764, and .0005817764 -.0000000000.0005817764; which, therefore, is the sine of the

are of 2 minutes.

To find the Sine of 3 Minutes.

As radius = 1; 2 :: .0005817764 to .0011635528, and .0011635528 -.0002908882.0008726646; which, therefore, is the sine of an arc

of 3 minutes.

To find the Sine of 4 Minutes.

As radius=1: 2 :: .0008726646 to .0017453292, and .0017453292 -.0005817764 = .0011635528; which, therefore, is the sine of the arc of 4 minutes.

To find the Sine of 5 Minutes.

As radius = 1; 2::.0011635528 to .0023271056, and .0023271056 -.008726646 = .0014544407; which, therefore, is the sine of the arc of 5 minutes.

In this manner, the sines may be found to 60 degrees; from which, to the end of the quadrant, they may be obtained by addition only; for the sine of an arc greater than 60 degrees, is equal to the sine of an arc as much less than 60, augmented by the sine of the excess of the given arc above 60 degrees: thus,

All the sines being found to 60 degrees; required the sine of 61 degrees?

Solution.-Sine of 59.8571673, and sine of 1:= .0174524; their sum.8746197; which, therefore, is the sine of 61 degrees, as required. Again,

All the sines being found to 60 degrees; required the sine of 62 degrees?

Solution.-Sine of 58.8480481, and sine 2.0348995; their sum.8829476; which, therefore, is the sine of 62 degrees, as required.

Now, the natural sines being thus found, the natural versed sines, natural tangents, and natural secants, may be readily deduced therefrom, agreeably to the principles of similar triangles, as demonstrated in Euclid, Book VI., Prop. 4. Thus,

To find the Natural Versed Sine of 30 Degrees = N B, in the Diagram.

Since the versed sine of an arc is represented by that part of the diameter which is contained between the sine and the arc; therefore N B is the versed sine of the arc B F, which is the arc of 30 degrees; and since the versed sines are measured upon the diameter, from the extremity B to C continued to the other extremity, the natural versed sines under 90 degrees are expressed by the difference between the radius and the cosine, and those above 90 degrees by the sum of the radius and the sine: hence, the radius CB 1.0000000- the cosine FG, or its equal N C. 8660254 = NB. 1339746; which, therefore, is the natural versed sine of 30 degrees.

To find the Natural Tangent of 60 degrees = B Q, in the Diagram.

As the cosine C M is to the sine DM, so is the radius CB to the tangent BQ: that is,

As CM.5000000: DM.8660254 :: CB 1.0000000 : BQ = 1.7320508; which is the natural tangent of 60 degrees.

To find the Natural Secant of 60 Degrees CQ, in the Diagram.

As the cosine CM is to the radius CD, so is the radius C B to the secant CQ: that is,

As CM.5000000: CD 1.0000000 :: CB 1.0000000: CQ = 2.0000000; which is the natural secant of 60 degrees. Hence, the manner of computing the natural co-tangent A P, the natural co-secant CP, and the natural co-versed sine EA, will be obvious. The versed sine supplement of an arc is represented by the difference between the versed sine of that arc and the diameter or twice the radius: thus, the versed sine supplement of the arc BF is expressed by the difference between twice the radius C B, and the versed sine NB; viz., twice CB 2. 0000000 - NB . 1339746 = 1.8660254; which, therefore, is the natural versed sine supplement of the arc B F or the arc of 30 degrees, and so of any other.

Now, the natural sines, versed sines, tangents, and secants, found as above, being principally decimal numbers, on account of the radius being assumed at unity or 1; therefore, in order to render these numbers all affirmative, they are to be multiplied by ten thousand millions respectively; and then the common logs. corresponding thereto will be the logarithmic sines, versed sines, tangents, and secants, which are generally given in the different mathematical Tables under these denominations.