CASE I.

In the annexed triangle, the side AC is given N. 400 W. 50 rods. The latitude and departure or the northing and westing are required.C

AB is a meridian or a north and south line. The angle at A contains 40°, and that at C, 50°, as appears by their respective arches. BC is an east and west linę, of course the angle at B contains 90°, or it is a right angle. AC is made radius. From the point at A, the

Fig. 1.

40'

500

arch Ca is described. From the point at C, the arch Az is described. BC is the sine of the angle at A, or of the arch Ca, and AB is the co-sine. AB is the sine of the angle at C, or of the arch Az, and BC is the co-sine. Sines and cosines lie within their respective arches. Each side is proportioned to its opposite angle, and each angle to its opposite side.

TO FIND THE REQUIRED SIDES TAKE THE FOLLOWING RULE.

For the latitude, make radius, the angle at B, the first term, the logarithm of the side AC 50, the second; the sine of the angle at C, the third, and the fourth will be the logarithm of the latitude AB. For the departure, take the same first and second terms, the sine of the angle at A for the third, and the fourth will be the logarithm of the departure BC.

Required the latitude and departure of S. 28° W. 88 rods.

As radius,

10.000000 As radius,

: distance 88, logarithm, 1.944483: distance, 88 rods,

10.000000 1.944483

:: cosine of course 28° 9.945935: : sine of the course 28° 9.671609

Required the latitude and departure of N. 75°, E. 245 rods.

As radius,

10.000000 As radius,

10.000000

: distance 245,

2.389166: distance, 245,

2.389166

: : co-sine of course, 75° 9.412996: sine of the course,

9.984944

The logarithm of the distance may be set down in two places; and the sine of the course may be placed under one, and the co-sine under the other, and when they are added, the unit at the left hand, in each case, may be cancelled, which will be the same as subtracting radius.

CASE II.

Suppose you run north from A to B, 220 B rods; then east 150 rods to C. What is your course and distance from C to A? First find the course. In this case, make the side AB radius. With your dividers on the point at A, descibe the arch Ba, then on the point at C, describe the arch Bz. In describing the last arch BC is made radius. Each arch contains as many degrees as its opposite angle.

Fig. 2.

150

As the sides AB and BC, both lie without their respective arches, they are tangents. BC is the tangent of the angle at A; or of the arch Ba, and AB is the co-tangent. AB is the tangent of the angle at C; or of the arch Bz, and BC is the

NOTE. When an angle is required, a side must be the first term. When a side is required, an angle must be the first term.

To find the course from C to A, Fig. 2, adopt the following rule.

Suppose yourself standing at the angle A; make the side AB radius, then state; as the log. of the side AB 220; is to radius; so is the log. of the side BC, 150 to the tangent of the angle at A. See the example worked.

As log. of AB,

220-2·342423 : R.10.000000 :: log. BC, 150--2.179160 2.176091

12.176091

2.342423

: Tang, of A 34° 17′ 9.833668

In the column of tangents, under 34° against 17', you will find the last term or that which is nearest to it. The course from A to C is N. 34° 17′ E.or from C to A, S. 34°, 17', W. Next find the distance.

CASE III.

TO FIND THE DISTANCE.

Make the side AC radius. From the point at A describe the arch Cx. In this case, the side BC is a sine as it lies within the arch. Take the following rule.

As the sine of the angle at A, is to the side BC, so is radius to the side AC. See the example worked.

220

Suppose you begin at a pine and run north 125 rods to a hemlock, then east 216 rods to a spruce, what is the course and distance from the spruce to the pine? By CASE II. find the course, and by CASE III. the distance. Suppose yourself standing at the pine, and making the northing radius, the proportion will be,

The course from the spruce to the pine is S 60° W.

NOTE. As the easting is greater than the northing, the course is over 45°. therefore look at the bottom of the page to find it.

To find the distance from the spruce to the pine, make that side radius, and the proportion will be, as the sine of the angle at the pine; is to the easting; so is radius to the distance required; or as the sine of the angle at the spruce; is to the northing; so is radius to the distance from the spruce to the pine.

As sine at the pine 60° 9.937531 As sine at the spruce 30°,9.698970 : the easting 216.5, 2.335458: northing, 125, 10.000000 : radius,

:: radius,

12.335458

9.937531

2.096910

10.000000

12.096910

9.698970

required dist. 250 rods, 2.397927: required dist. 250 rods, 2.397940

The following courses and distances of a survey are given, and the course and distance of the closing line are required.

Beginning at a maple tree, thence running as follows:S. 80 W. 90 rods, N. 15° W. 95 rods, N. 85° E. 45 rods, S. 10° E. 40 rods, N. 85° E. 42 rods to a beach tree. First, arrange the given courses and distances in a table, having a blank line for the course, distance, latitude and departure of the closing line. Find by the traverse table the latitude and departure of each course and distance, and insert them in

Subtract the sum of the southings from that of the northings, and 44.32 will remain to fill the blank in the column of southings. Also, subtract the sum of the eastings from that of the westings, and 19.60 will remain to fill the blank in the column of eastings.

These numbers or distances are the legs of a right angled triangle, the hypothenuse of which is the closing line. You are now standing at the beach tree.

Make the

southing radius and to find the course, the proportion will be,

:: tangent of the course, 23° 51′ 9.645656

The course of the closing line is S. 23°, 51', E. which may fill the blank in the column of courses.

To find the distance make the line from the beach to the maple, or the closing line radius, and the easting will be a