and G draw the lines F A, GD, parallel to the second bearing CB, and meeting CA and CD in the points A and D; join AD, and it will represent the ship's track; through C draw C K, parallel to AD, and the arch SK will be the measure of the ship's course. From C let fall the perpendicular CH upon the line A D, produced if necessary; and from H let fall the perpendicular HI upon the line FG, produced also, if necessary; then the measure of CI will give the interval between the time of the second bearing and that when the ship was nearest to the observer.

Make A L equal to the difference between the perpendiculars A F and DG; then, in the right angled triangle ALG, given the perpendicular AL and the base LD; to find the angle LAD, which is evidently equal to the angle a CK; to this let the inclination of CB to a parallel be applied, and the result will be the apparent course of the ship.

Example.

At 120 past noon a ship, sailing upon a direct course, was observed to bear N.W. b. N.; at 210", she bore N. W.; and at 325", the bearing was N.E. b. E.; required the apparent course steered by that ship, and the time when she was nearest to the observer?

Solution. The circle being described and quartered, and the three given bearings laid down as above directed, through C draw FG perpendicular to the second bearing CB; make FC equal to 50 minutes, the interval between the first and second bearings, and CG equal to 75 minutes, the interval between the second and third bearings: these may be taken from any scale of equal parts. Then proceed with the other parts of the construction, agreeably to the rule; now, the ship's apparent course, represented by the angle SCK, being applied to the line of chords, will be found to measure 721 degrees: hence the course is S. 72:30. E., or E. b. S. S. nearly. The perpendicular GD, being applied to the scale of equal parts from which the intervals were taken, will be found to measure 40, and the perpendicular FA 93; the difference between which 53, is the measure of A L. Then CI, measured upon the same scale, gives 26 minutes; which is evidently, by the construction, past the time of the second bearing: hence the time of the ship's nearest approach to the observer at C, is 210" 26" 2:36 past noon. Now, the figure being thus completed, the required parts may be obtained by trigonometrical calculation, in the following manner :

=

14

In the right angled triangle A F C, given the angles and the base FC 50 minutes; to find the perpendicular FA. Thus, since the straight line AC falls upon the two parallel straight lines CB and FA, it makes the alternate angles equal to one another (Euclid, Book I., Prop. 29): therefore the

angle F A C is equal to the angle ACB; but the angle ACB is given, being equal to 24 points, viz., the difference between N.W. b. N. and N. W. hence the angle FAC is also equal to 2 points.

:

In the same manner it may be shown (in the right angled triangle D G C, where the angles and the base CG 75 minutes are given; to find the perpendicular GD,) that the angle GDC is equal to the angle BCD; and since BCD is given, being equal to 5 points, viz., the sum of W., therefore the angle GDC is also equal to 5

N.E. b. E., and N. points. Hence,

To find the Perpendicular GD :—

90:

As radius =
Log. co-secant = 10.000000
Is to the base CG = 75" Log. =

1.875061

So is the angle GDC=54 points, Log. co-tangent = 9.727957

So is the angle FAC 2 points, Log. co-tang. = 10.272043

=

To the perpendicular FA=93.54 Log. . . .
Perpendicular GD

Difference =

=

40.09

1.971013

53.45, which is equal to the part A L.

In the right angled triangle ALD, given the base LD FG 125 minutes, and the perpendicular AL 53.45; to find the angle LAD: therefore,

Now, since CK is parallel to AD, and Ca to AL, the angle a CK is equal to the angle LAD; but the angle LAD is found, by computation, to be 66:50:54; wherefore the angle a C K is also equal to 66:50:54": to this let the angle a CS = the angle NCB 0 point, or 5:37:30", be added; and the sum 72:28:24" = the angle S C K is the apparent course of the ship between the south and the east, viz., S. 72:28:24" E, or E. b. S. S. nearly.

We have now to determine the measure of the base CI, in the right angled triangle CIH; to do which, we must first find the value of the hypothenuse AC in the right angled triangle A FC, and that of the base CH in the right angled triangle AHC. Thus,

=

90:
50

Log. co-secant=10.000000

Log.

=

1.698970

So is the angle FAC 2 points, Log. co-secant=10.326613

To the hypothenuse A C = 106.07 Log. =

As radius =

To find the Base CH:

2.025583

Is to the hypothenuse A C=106. 07 Log.=. ... 2.025583
So is LAD-FAC=CAH=38:43:24"Log. sine 9.796269

To the base CH=

66.35 Log.

= •

1.821852

Now, in the right angled triangle CIH, given the hypothenuse CH = 66.35 minutes, and the angle CHI; to find the base CI. The measure of the angle CHI is thus determined. In all quadrilateral or four-sided figures, the sum of the four angles is equal to four right angles, or 360 degrees. Now, in the quadrilateral figure AHIF, since three of the angles are given, the remaining or obtuse angle AHI is known by subtracting the sum of the given angles from 360 degrees: thus, the angle HIF 90+ IFA 90: + FAH 66:50:54% = 246:50:54"; and 360: -246:50:54% = 113:9:6", is the measure of the angle AHI; from which take away the right angle AHC 90°, and the remainder = 23:9'6" is the absolute measure of the angle CHI. Hence CI may be readily found; as thus:

To the interval or base CI= 26.09 minutes. Log. =
Time of second bearing = 210.0

Note. This interesting problem is thus worked at length, trigonometrically, with the view of adapting it to the use of mariners in general; though, indeed, in such cases, calculation need not be resorted to, since the solution deduced from geometrical construction will always be sufficiently near the truth.

SOLUTION OF PROBLEMS IN PRACTICAL GUNNERY.

Gunnery is the art of projecting balls and shells from great guns and mortars; of finding the ranges and times of flight of shot and shells; and of determining the different degrees of elevation at which those bodies should be projected, so as to produce the greatest possible effect.

PROBLEM I.

Given the Diameter of an iron Ball; to find its Weight.

RULE.

The diameter of an iron ball of 9 lbs. weight is 4 inches, very nearly; and, since the weights of spherical bodies, composed of the same materials, are as the cubes of their diameters, (Euclid, Book XII., Prop. 18,) it will be,-as the cube of 4, is to 9 lbs. ; so is the cube of the diameter of any other iron ball, to its weight. Hence the following rule :—

To thrice the logarithm of the diameter of the given ball, add the constant logarithm 9. 148063; and the sum (abating 10 in the index,) will be the logarithm of the required weight in lbs.

Example 1.

Required the weight of an iron ball, the diameter of which is 6.7 inches?

Required the weight of an iron ball, the diameter of which is 5.54 inches?

Note. The constant logarithm used in this problem is expressed by the arithmetical complement of the logarithm of the cube of 4, added to the logarithm of 9.

PROBLEM II.

Given the Weight of an iron Ball; to find its Diameter.

RULE.

This problem being the converse of the last, we obtain the following logarithmical expression :

To the logarithm of the weight of the given ball, add the constant logarithm 0.851937; divide the sum by 3, and the quotient will be the logarithm of the required diameter.

Note. The constant logarithm given in this rule is expressed by the arithmetical complement of the logarithm of 9, added to the logarithm of the cube of 4.

Example 1.

Given the Diameter of a leaden Ball; to find its Weight.

RULE.

A leaden ball of 1 inch in diameter, weighs of a lb.; which, reduced to a decimal fraction, is . 2143, very nearly: and, as the weights of spherical bodies are as the cubes of their diameters, it will be,-as the cube of 1, is to. 2143; so is the cube of the diameter of any other leaden ball, to its weight in lbs. Whence the following logarithmical rule:—