third number in the first Departure Column. To this add 9.56 the Easting of the fourth Course, and you have 84.72 the length of the Line 1 E, which represents the Sum of the Eastings made by the four first Courses, and is the fourth number in the first Departure Column. These two, viz. the Lines 3 D 75.16 and 1 E 84.72 added together make 159.88 the fourth number in the second Departure Column; which being multiplied by 49.15 the length of the Line 3, 1 which represents the Southing made by the fourth Course, will give double the Area of the Trapezoid 1 ED 3. The number thus produced is 7858.1020, which is to be placed for the first number in the Column of South Areas.

The fifth Coarse being due South, it is evident the Sum of the Eastings will remain the same as at the end of the fourth Course; That is, the Line 4 F equals the Line 1 E, which is 84.72. These added make 169.44 the fifth number in the second Departure Column. This being multiplied by 54.10 the length of the Line EF, which is the Southing of the fifth Course as corrected in balancing, and the same as the Line 1, 4-will give do uble the Area of the Parallelogram 1 EF 4, which is 9166 .7040 the second number in the Column of South Areas.

From the Line AF 84.72 subtract 39.95 which is a West Course, and it leaves 4G 44.77 the Sum of the Eastings, or the Meridian Distance, at the end of the sixth Course, and the sixth number in the first Departure Column. From this subtract 23.75 the Westing made by the seventh Course, and you have 21.02 the length of the Line 5H, which is the Meridian Distance at the end of the seventh Course, and the seventh number in the first Departure Column. The Line 4G 44.77 added to the Line 5H 21.02 make 65.79 the seventh number in the second Departure Column. This being multiplied by 32.21 the length of the line 4, 5-which is the Southing of the seventh Course, will give double the Area of the Trapezoid 4GH5, which is 2119.0959 the third number in the Column of South Areas.

The Line H5, 21.02 is the Westing of the last Course, and the last number in the second Departure Column. This being multiplied by 26.65 the length of the Line 5 A, and the Northing of the last Course, produces 560

.1830, which is double the Area of the Triangle A5H, and the last number in the Column of North Areas.

Note. it will be observed that against the third and sixth Courses there are no Areas; the reason is that these Courses being one East and the other West, there is no Northing or Southing to be multiplied into them; regard can therefore be had to them only in forming the Departure Columns.

By inspecting the Figure, and attending to the preceding illustrations, it will be seen that the three North Areas represent double the Area of the Triangle A2B, the Trapezoid 2BC3, and the Triangle A5H, all of which are without the boundary Lines of the Field: Also, that the three South Areas represent double the Area of the Trapezoid 3DE1, the Parallelogram 1EF4, and the Trapezoid 4GH5; and that these include not only the Field but also what was included in the North Areas Therefore the North Areas subtracted from the South, the Remainder will be double the Area of the Field, contained within the black Lines.

Additional Directions and Explanations.

The Northings and Southings may be added and subtracted instead of the Eastings and Westings; then there will be two Latitude Columns instead of Departure Columns; and the numbers in the second Latitude Column must be multiplied into the Eastings and Westings, and you will have East and West Areas.

When the Course is directly North or South, the Distance must be set in the North or South Column; When East or West, in the East or West Column. There will therefore sometimes be no number to be added to or subtracted from the number last set in the Latitude or Departure Column; then the number last placed in the Column must be brought down and set against such Course; as in EXAMPLE I. at the 5th Course. It may also sometimes be the Case that there will be no number to multiply into the number in the second Latitude or Departure Column; then that number must be omitted, and against such Course there will be no Area as in

When the Northings or Southings, Eastings or Westings, beginning at the top, will not admit of a continual addition of the one and subtraction of the other, without running out before you get through the several Courses, you may begin at such a Course as will admit of a continual addition and subtraction; and when you get to the bottom go the top, and you will end in Cypher at the Course next above that where you began; as in EXAMPLE II. which begins at the 9th Course to add the Eastings and subtract the Westings.

Note. In the above EXAMPLE you might begin at the 4th Course to add the Westings and subtract the Eastings; or at the 6th Course to add the Northings and subtract the Southings; or at the 11th Course to add the Southings and subtract the Northings. So in every Survey some place may be found where you may begin to add and subtract, without running out before you get through all the Courses.

When a Field is very irregularly shaped, it will often happen that parts of the same Area will be contained in several different Products in the Columns of Areas; but in the final result, one column being subtracted from the other will leave what is included within the boundary Lines of the Field.

DEMONSTRATION. See PLATE III. Fig. 64. and ExAMPLE II.

where the calculation begins, is the Triangle I2K, all without the Field.

The Area against the 10th Course is the Trapezoid 2KL3, also without the Field.

The Area against the 11th Course is the Trapezoid 4ML3. This is a South Area, and contains a part of the Field and also part of the preceding North Area.

The Area against the 12th Course is the Trapezoid 5NM4, part within and part without the Field.

The Area against the 13th Course is the Trapezoid 6AN5, part within and part without the Field.

The Area against the 1st Course is the Trapezoid 6AB7, part within and part without the Field. This is a North Area and to be ultimately subtracted from the South Areas; but this includes a part of the preceding South Area, viz. the space nAso; it will however be seen hereafter that this same space is included in another South Area. This North Area contains also a part of the first North Area, viz. the space 6n07; but the same space is also included in another South Area.

The Area against the 2d Courses is also a North Area, and is the Trapezoid 7BC8. This Trapezoid contains the space sBCx, without the Field; the space osxw, within the Field; and the space 70w8, without the Field. But the space osxw will be contained in the next South Area; and the space 7ow8, which was contained in the two first North Areas, will be contained in the next South Area.

By examining the whole Figure in this manner, it will be seen that the North Areas contain all without the Field that is taken into the Calculation, and some of it twice over; they also contain part of the Area within the Field. The South Areas contain all within the Field, and all without the Field that is contained in the North Areas. They also contain, twice over, so much of the Field as is included in any of the North Areas; and likewise, twice over, that part without the Field which is contained twice in the North Areas. So that subtracting the North from the South Areas leaves double the Area of the Field.

This method of calculating the Area of a Field by the Northings, Southings, Eastings and Westings, divides the

Field, with a certain quantity of the adjoining ground, into Right Angled Triangles, Right Angled Trapezoids, Parallelograms, or Squares, as may be seen by the Figures. It may therefore with propriety be called RECTANGULAR SURVEYING.

A USEFUL PROBLEM.

To find the true Area of a Field which has been measured by a Chain too long or too short.

Calculate the Area as if the Chain was of a true length, then institute the following Proportion:

As the Square of the length of the true Chain;
Is to the Area, as found by the Chain made use of;
So is the Square of the length of that Chain;
To the true Area of the Field.

EXAMPLE.

Suppose a Field, measured by a Two Rod Chain 3 Inches too long, is found to contain 41 Acres 1 Rood and 33 Rods, what is the true Area ?

As the Square of 33 Feet, the true length of a Two Rod Chain; Is to 41 Acres 1 Rood and 33 Rods; So is the Square of 33 Feet 3 Inches, the length of the Chain used in the Survey; To 42 Acres and 13 Rods. 33 Feet 396 Inches. 396x396=156816 Square Inches. 41 Acres 1 Rood 33 Rods=6633 Rods. 33 Feet 3 Inches=399 Inches. square inches.

159201x6633-156316-6733 Rods.

399X399=159201

6733-160=42 Acres 13 Rods, the true Area.

PART II.

LAYING OUT Land.

PROBLEM I. To Lay out any number of Acres in the form of a Square.

Annex 5 Cyphers to the number of Acres, which will turn them into Square Links, the Square Root of which will be the Side of the Square in Links.

EXAMPLE. It is required to lay out 810 Acres in the form of a Square.

Answer. Each Side of the Square must be 9000 Links, or 90 Chains.

PROBLEM II. To lay out any number of Acres in the form of a Parallelogram, whereof one Side is given.

Divide the number of Acres, when turned into Square