the first station which is on the west side of the figure, of course, the eastings are added and the westings substracted. The whole is illustrated by this figure.

Area in rods 160)4135.665

Dist. D. Dist.

Merid. D. M. Merid II. D. N. Areas. S. Areas.

16.90

No.

from 3 to n. their sum is 22.11, the meridian distance of 3, or the distance which is the meridian distance of 2, or the distance from 2 to the easting of the first line 16.90, into the head of the column, To form the first column marked at the top Merid. Dist. set To this number, add the easting of the second line, and To this number add the easting of the third

S.

line, and their sum is 80.07, the meridian distance of 4, or the distance from 4 to m. From this number, substract the westing of the fourth line, and 73.82 remains, the meridian distance of 5, or the distance from 5 to a. From this number, substract the westing of the fifth line, and 70.33 remains, the meridian distance of 6, or the distance from 6 to z. From this number, substract the westing of the closing line, and 00.00 remains, which proves the work correct so far. When the last substraction does not end in a cipher, an error has been committed somewhere.

In forming the column of double mean distances, take the following directions :

Add together two opposite sides of each figure formed by the meridian distances. Set the first meridian distance into the head of the next column, or that marked at the top D. M. D., and it gives one side of the parallelogram 1s-2c. This figure is divided into two equal triangles by the line 1-2. The meridian distance multiplied by the northing of the line 1-2, gives the area of the whole parallelogram, or the double area of one of the triangles. Let this be consid ered, that which lies without the field. To the line 2s, add 3n, and their sum is 39.01, the length of the parallelogram nxos, which is divided into two equal trapezoids by the line 2-3. The double mean distance 39.01, multiplied by the northing of the line 2-3, gives the area of the parallelogram, or the double area of one of the trapezoids.

Let this be considered that which lies without the field.

To the line 3n, add 4m, and their sum is 102.18, the length of the parallelogram mien, which is divided into two equal trapezoids by the line 3-4. The double mean distance 102.18, multiplied by the northing of the line 3-4, gives the area of the parallelogram, or the double area of one of the trapezoids. Let this be that which lies without the field. To the line 4m, add 5a, and their sum is 153.89, the length of the parallelogram mdfa, which is equally divided by the line 4-5. The double mean distance 153.89, multiplied by the southing of the line 4-5, gives the area of the parallelogram, or the double area of one of the trapezoids.

Let this be that which abuts on the meridian, being partly within and partly without the field.

To the line 5a, add 6z, and their sum is 144.15, the length of the parallelogram auvz, which is equally divided by the line 5-6. The double mean distance 144.15, multiplied by the southing of the line 5-6, gives the area of the parallelo

be a5-6z, partly within and partly without the field. To the line 6z, nothing is to be added, of course, it is one side. of the parallelogam 26-w1, which is equally divided by the closing line. The distance 70.33, multiplied by the southing of the closing line, gives the double area of the triangle 26-1, partly within and partly without the field.

The three products in the column of north areas, all lie without the field. Those in the column of south areas lie partly within and partly without the field, and they include all the space which is covered by the products in the column of north areas.

Hence it is plain, that when the sum of the former is substracted from that of the latter, the double area of the field will remain. In this method, when the column of meridian distances and that of double mean distances are correct, the sum of the latter will be double to that of the former.

PENNSYLVANIA METHOD.

If the writer has been correctly informed, this improvement was made by Dr. Rittenhouse.

Only one column is used for meridian distances, but the final results are the same as when two columns are used.

This method is not so easily explained to the learner, but is perhaps preferable in practice, because an error may be committed in forming the column of double mean distances which may not be discovered; but in this method, an error cannot be committed in the meridian distances, without being detected. A full explanation of this method would be tedious and perplexing to the learner, without being of the least use in performing practical operations, and probably, if made, but little attention would be paid to it; therefore, only direc tions how to form the column, and how to apply the numbers in their practical use, will be given.

The operation of this method is substantially the same as that of the other, and the diagram given in that will answer for this. The north and south products are precisely the same in both methods. A further investigation of this method will be left for the learner when he shall be more experienced.

When the two columns in the other method are proved by addition, the process is longer than in this.

The second column in the foregoing example, marked at the top, meridian distances, is the method here treated of.

each station. Set the first easting 16.90 into the head of the column at the upper place. Add it to itself, and the sum is 33.80 in the lower place. To this number add the next easting, and they make 39.01 in the upper place. To this number add the same easting again, and they make 44.22 in the lower place. To this number add the third easting, and they make 102.18 in the upper place. To this number add the same easting again, and they make 160.14 in the lower place.

From this number substract the first westing, and 153.89 remains in the upper place. From this number substract the same westing again, and 147.64 remains in the lower place.

From this number substract the second westing, and 144.15 remains in the upper place. From this number substract the same westing again, and 140.66 remains in the lower place. From this number substract the last westing, and 70.33 remains in the upper place.

From this number substract the same westing again, and 00.00 remains. The upper number against each station in this column, is the same as the double mean distance which stands against it; and the remainder of the process may be performed as before directed. Another column may be formed as the eleventh in this example, marked H. D. at the top, which, for distinction, is here called half distance. It contains half the sum of the numbers in the double mean column. These numbers, when multiplied by their respective northings or southings, give the simple areas of the different figures.

This method is preferable in practice, as the multiplications are greatly diminished. When the last decimal in the double mean distances is an odd number, a unit may be taken off, and take half the remainder, rather than annex another decimal: perhaps this would not make the difference of a rod in a survey of three hundred acres. Or the odd numbers in the last place of decimals may be balanced by sometimes adding a unit.

If the numbers are diminished a trifle, it may be remarked, that, on account of uneven surfaces, there is danger of making the distances too much, rather than falling short of the true measure. There are other methods by which surveys may be calculated arithmetically, but none are more correct, and perhaps none are so simple and so easy to be understood as the methods here treated of. These methods may be applied in every case in practical surveying, which depends on a calculation from courses and distances.