No. IV.

ON RECTANGULAR SURVEYING.

Rectangular surveying is a name given to the method here treated of by the late Governor Treadwell.

In the operation of it the whole survey is reduced to squares and parallelograms, and each of these figures is divided into two right angled triangles of equal shape, or into two trapezoids of equal size and shape, by a boundary line of the field running obliquely through it, except when the boundary line is directly north and south, then the figure is divided into two squares or parallelograms.

The calculations are made from a meridian drawn either at the eastern or at the western extremity of the map; and the areas are produced by multiplying the latitudes by the longitudes made from this meridian.

All the spaces lying between the field and the meridian, and between the parallels of latitude of the northern and southern extremities of it, also large spaces lying without the figure on the opposite side of it, and between those parallels, are included in the calculation; but that which lies without, is eventually excluded.

Parallels of latitude are drawn from each angle to the meridian, which are called meridian distances.

In forming the column of meridian distances, when the meridian is drawn at the eastern extremity, the westings are added, and the eastings substracted; but when the meridian is laid on the west side, the eastings are added and the westings substracted.

The meridian distances proceeding from each end of a line are added together, to form the column of double mean distances, which the compiler of the foregoing work called second departure.

Meridian distances and double mean distances,† are more proper terms or names for these columns, than first departure and second departure.

The meaning of the term meridian distance, is the distance made from any meridian.

The following survey is calculated from the meridian of

* Rectangular calculation is as proper.

+ The compiler of the foregoing work would have made use of these names, but he was afraid of infringing on a certain copy-right

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.

70-33

Area in rods 160)4135.665

Merid. D. M. Merid H. D. N. Areas. S. Areas.
Dist. D. Dist.

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 this number, add the easting of the second line, and To form the first column marked at the top Merid. Dist. set 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 considered, that which lies without the field. To the line 2s, add 37, 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 z6-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 z6-1, 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 directions 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.