Find the area of the part IAB C, according to SECTION III. page 64, as follows: set the latitude and departure of the three first sides, I A, A B, and B C, in their proper columns, in a traverse table; and place as much southing, viz. 109.1, equal to the line CK, and as much westing, viz. 71.7, equal to the line KI, as will balance the columns. This southing and westing will be the latitude and departure made by the line CI. The area of IABC will be found to be 8722 rods, which is less than half the area of the whole field by 3470 rods, the quantity to be contained in the triangle I C N. Find the bearing and distance of CI by RIGHT ANGLED TRIGONOMETRY, CASE IV. as follows:*

As C K, the southing of CI, 109, nearly

: radius

::KI, the westing of C I, 71.7

2.037426 10 000000 1.855519

11.855519
2 037426

: tangent course S. 33° 20′ W.

9.818093

The mode given above is undoubtedly the most correct, but the use of the traverse table will save many figures. From that table the

NOTE. In this way the course and distance may be found from any. angle of a field to another.

Having found the line C I, divide 3470, the number of rods to be contained in the triangle IC N, by one half the line CI, viz. 65.25; the quotient will be the length of the perpendicular P N, viz. 53.18.

Now, by the bearings of CI and CD, it appears that they form an angle of 60° 20′; wherefore, in the triangle CP N, are given the side P N 53.18, and the angle at C 60° 20′, to find the hypothenuse C N.

Thus the dividing line must go from I to a point on the line C D, which is 61.2 rods from C. The bearing and distance of this line may be found by the directions given above for finding the bearing and distance of the line CI. Or, they may be found by oblique trigonometry, CASE III.

ANOTHER MEthod of finding the diSTANCE CN.

Having ascertained the latitude and departure of the line CI, set them down in a traverse table; find the latitude and departure of the line C D, and place them in the table; the difference between the northing of the line IC, and the southing of the line C D, will be the southing of the line DI, viz. 6.6; and the sum of the eastings of those lines, as they are both easterly, will be the westing of the line DI, viz. 123.9. Proceed to calculate the area of the triangle ICD, which will be found to be 6522 rods, nearly.

NOTE. As in this triangle two sides and their contained angle are

Having found the area of this triangle, proceed to find CN according to PROB. II. page 75, as follows:

As the area of the triangle; is to CD the base; so is the quantity to be contained in the triangle ICN; to CN, its proportion of the base.

As 6522: 115:: 3470: 61.19.

A THIRD METHOD OF FINDING THE DISTANCE CN.

To the logarithm of double the area to be contained within the triangle I C N, add radius; from this sum substract the logarithmic sine of the angle at C; and from the remainder substract the logarithm of the side IC; the last remainder will be the logarithm of the side C N.

The double area of the triangle ICN, is 6940; the angle at C is 60° 20'; the side IC is 130.

Double area 6940

3.841359

NOTE. Radius may be added by placing a unit before the index of the logarithm for the double area, without the trouble of setting down the ciphers.

BY NATURAL SINES.

Divide the double area by the natural sine of the given angle, and that quotient by the given side; the last quotient will be the side C N.

Nat. sine of the angle at Ĉ 60o 20′ 0.86892

6940+0.86892=7986.92

7986.92+130.5=61.2

The log. of IC, as found by calculation on page 77, is employed, instead of taking from the tables, that of 130.5, which is not the exact

From the above the following general rule may be drawn. TO FIND THE SIDE OF A TRIANGLE WHEN THE AREA IS GIVEN, AND ALSO ONE OF THE SIDES, AND THE ANGLE CONTAINED BETWEEN THE GIVEN SIDE AND THE SIDE REQUIRED.

TO THE LOGARITHM OF DOUBLE THE AREA ADD RADIUS;

FROM THIS SUM SUBSTRACT THE LOGARITHMIC SINE OF THE GIVEN ANGLE, AND FROM THE REMAINDER SUBSTRACT THE LOGARITHM OF THE GIVEN SIDE; THE LAST REMAINDER WILL BE THE LOGARITHM OF THE SIDE REQUIRED.

OR, BY NATURAL SINES: DIVIDE THE DOUBLE AREA BY THE NAT. SINE OF THE GIVEN ANGLE, AND THAT QUOTIENT BY THE GIVEN SIDE; THE LAST QUOTIENT WILL BE THE SIDE REQUIRED.

CONCLUDING REMARKS.

Other methods of surveying fields are taught by some authors on this subject. The preceding, however, will be found most useful in actual practice. Other instruments besides those mentioned in this book are sometimes used; such as the plain table, semicircle, perambulator, theodolite, &c. But of these instruments very little use is made in New England; and they are not often to be met with. For general practice, none will be found more useful than a common chain, and a compass upon Rittenhouse's construction.

The com

It will be observed that in this work there is no descriptions of mathematical and surveying instruments. piler omitted such descriptions, from a belief that nothing which can be written on the subject will enable a person to understand them, without an actual inspection of the instruments themselves, and some instruction from those acquainted with them.

The general principles here taught may be applied to the

APPENDIX.

OF THE DECLINATION AND VARIATION OF THE MAGNETIC NEEDLE, AND OF THE ATTRACTIONS TO WHICH IT IS SUBJECT.

THE declination of the needle is the number of degrees it deviates from the true north, either east or west. This differs in different places, and in the same place at different times. [At Hartford, Conn., the declination was, in 1829, 6o 3' west of the true meridian of the earth; and was increasing by an annual variation of about 3'.]

The following method of ascertaining the variation, by the north star, has been adopted by many surveyors, as the most eligible to be practised on land. It was communicated to the compiler by Moses Warren, Esq., of Lyme, an experienced surveyor, with permission to publish it.

The star, commonly called the north star, is not directly north, but revolves round the pole in a small circle, once in 24 hours.* It can therefore be due north only twice in that period; and that is within a very few minutes of the time, when a star, called Alioth, in the constellation of Ursa major, or the great bear, is directly over or under it. There is also another star nearly in an opposite direction from the pole, called Gamma, in the constellation of Cassiopeia. When these three stars are vertical, the north star is very near the meridian; and when they are horizontal, it is at its greatest elongation, that is, at its greatest distance east or west of the pole, and on the same side as the star in Cassiopeia. The variation may be calculated when the star is on the meridian, or when at its greatest elongation; more accurately, however, at the latter period, because its motion be

* More exactly, 23 hours, 56 minutes, and 4 seconds.-ED