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The Arca against the 13th Course is the Trape zoiu 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 Course 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 7ow8, without the Field. But the space osxw will be contained in the next South Area ; and the space 70w8, 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
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 ;
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.
33x123396 Inches. 396 X 396 =156816 Square Inches.
33—3X12=399 Inches. 399X399=159201 Square Inches.
159201x6633---156816=6733 Rods. 6733-160=42 Acres 13 Rods, the true Area.
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 Şide of the Square must be 9000
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 Links, by the given Side ; the Quotient will be the Side required.
Example. What must be the longest side of a Parallelogram, which is to contain 25 Acres, when the shortest side is 5 Chains and 50 Links !
Answer. 2500000-550=4545 Links for the longest Side.
PROBLEM III. To lay out any number of Acres in a Field, 3, 4, 5, 6, &c. times as long as it is broad.
Divide the Acres, when turned into Square Links, by the proportion between the length and breadth ; the Square Root of the Quotient will be the shortest Side.
EXAMPLE. It is required to lay out 100 Acres 5 times as long as it is broad,
Answer. 10000000-5=2000000 the Square Root of which is 1414 Links for the shortest Side, and the longest will be 7070 Links.
PROBLEM IV. To make a Triangle which shall contain a given number of Acres, being confined to a certain Base.
Double the given number of Acres, to which annex 5 Cyphers, and divide by the Base ; the Quotient will be the Perpendicular in Links.
EXAMPLE. Upon a Base of 40 Chains to lay out 100 Acres in a Triangular form.
Answer. 5000 Links or 50 Chains will be the length of the Perpendicular.
The Perpendicular may be erected from any part of the Base : Thus, the Triangle ABC. See Plate II. Fig. 55. is the same as ABE, each containing 100 Acres.
When the given Base is so situated that a Perpendicular of sufficient length cannot be erected therefrom, continue the Base as from B to D. Fig. 56. from which erect the Perpendicular DC, and complete the Triangle
DivIDING LAND. As different Fields are so variously, and many of them irregularly shaped, and as they are required to be divided in many different proportions, it is difficult to give Rules which will apply to particular cases.
The business of dividing Land must therefore be left, in a great measure, to the skill and judgment of the Surveyor ; who, if he is well acquainted with Trigonometry, and with measuring Land, will not find it difficult after a little practice, to divide a Field in such a manner às shall be desired. If he has before him a Plot of the Field, and knows the number of parts into which it is to be divided, and the proportion which each part is to bear to the others, he will readily find out where the dividing Lines are to be drawn.
A few Rules and Examples will be given for the general instruction of the Learner.
PROBLEMI. To cut off any number of Acres from a Square or Parallelogram.
Say, As the whole number of Acres in the Field ; Is to the length of the Square or length or breadth of the Parallelogram ; So is the number of Acres proposed to be cut off; To their proportion of the length or breadih.
PROBLEM II. To cut off any number of Acres by a Line proceeding from any Angle of a Triangle.
Measure the Base, or Side opposite the Angle from which the dividing Line is to be drawn: Then say, As the number of Acres in the whole Triangle; Is to the whole Base ; So is the given number of Acres; To their part
of the Base.
EXAMPLE. See PLATE II. Fig. 57. In the Triangle ABC, which contains 48 Acres, it is requred to cut off 18 Acres, by a Line proceeding from C to the Base AB, which is 40 Chains.
As 48 : 40 :: 18: 15 Lay 15 Chains on the Base from B to D, and draw the Line CD. . The Triangle will then be divided as
PROBLEM III. To take off any given number of Acres from a multangular Field.
EXAMPLE I. See Plate III. Fig. 65. Let ABCD, &c. be the Plot of a Field containing 11 Acres, from which it is required to cut off 5 Acres.
Join two opposite Corners of the Field as D and G, with the Line DG (which you may judge to, be near the partition Line) and find the Area of the part DEFG, which suppose may want 140 Rods of the quantity proposed to be cut off. Measure the Line DG, which suppose to be 70 Rods; divide 140 by 35 the half of DG, and the Quotient 4 will be the length of a Perpendicular whose Base is 70 and Area 140. Lay off 4 Rods from G to I, and draw the Line DI, which will be the dividing Line.
Example II. See Plate III. Fig. 60. Let ABCD, &c. be a Tract of Land, to be divi. ded into two equal parts, by a Line from I to the oppo. site Side CD: To find Arithmetically on what part of the Line CD the dividing Line IN will fall; or to find the Distance CN.
70.9 BC. S. 77 0 E. 91
GH. N. 36 OW. 47 CD. S. 27 O E. 115 HI. North,
64.3 DE. S. 52 o W. 58
IA. N. 62 15 W. 59 EF. S. 15 30 E. 76
Acres Rood Rods
Whole Area 152 · 1 25 Find the Area of the part IABCI, according to SecTION III. Page 59, as follows: Set the Latitude and Departure of the three first Sides IA, AB and BC 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 IABCI 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