and G draw the lines F A, G D, parallel to the second bearing C B, and meeting C A and C D in the points A and D; join A D, and it will represent the ship's track; through C draw C K, parallel to A D, and the arch S K will be the measure of the ship's course. From C let fall the perpendicular CH upon thè line A D, produced if necessary; and from H let fall the perpendicular HI upon the line F G, 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 AL equal to the difference between the perpendiculars AF and D G ; then, in the right angled triangle A L G, given the perpendicular AL and the base LD; to find the angle LA D, which is evidently equal to the angle a C K ; 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 3:25", 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 F G 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 co nstruction, agreeably to the rule; now, the ship's apparent course, rep resented by the angle SC K, being applied to the line of chords, will be found to measure 72 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 upori 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 2:10" +26" = 2:36" past noon. Now, the figure being thus completed, the required parts naay be obtained by trigonometrical calculation, in the following man ner :—

In the right angled triangle AFC, given the angles and the base

FC 50 minutes; to find the perpendicular FA. Thus, since the straight line A C 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 AC B; 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 F A C is also equal to 2 points.

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

In the right angled triangle ALD, given the base LD FG 125 minutes, and the perpendicular A L 53. 45; to find the angle LAD: therefore,

Now, since C K is parallel to A D, and C a to AL, the angle a C K is equal to the angle LA D; but the angle LA D 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 N CB 0 point, or 5:37:30, be added; and the sum 72:28:24′′ = the angle SCK 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 C I, in the right angled triangle CIH; to do which, we must first find the value of the hypothenuse AC in the right angled triangle AFC, and that of the base C H in the right angled triangle A H C. Thus,

To the hypothenuse A C = 106.07 Log. =

To find the Base C H.

1.698970

2.025583

So is LAD FAC CAH=38:43:24" Log. sine =

9.796269

Now, in the right angled triangle CIH, given the hypothenuse CH66.35 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 A HIF, 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: +F A H 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 96 and the remainder = 23:9:6" is the absolute measure of the angle CII I. Hence C I may be readily found; as thus :

To the interval or base CI=26 .09 minutes Log.
Time of second bearing=2:10". 0

Sum =

2:36.09; which is the time of the ship's

nearest approach to the observer.

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.

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

Given dianteter 6.7; thrice its log. = 2.478225

Note. The constant logarithm used in this Problem is expressed by he 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.

Given the Diameter of a Leaden Pall; 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

T T