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PROBLEM III. Suppose ABC a Triangular piece of ground, which, by an old Survey, is found to be thus; AB 260, AC, 160, and BC 150 rods. The bound at C is destroy

A.

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Fig. 43.
C

B

ed, and no remnants of the Lines AC and BC are to be found; the Line AB only remaining. What Angles must be set off from A and B to run new Lines exactly where the old ones were? Fig. 43. Here are the Sides given to find the Angles. It is performed by Case 4, Oblique Trigonometry, or Problem 6. Sec. 2.

Angle at A 32°
Ans. Angle at B 34°

The Point of Intersection, of the Lines AC and BC, will be the place for the bound at C.

PART II.

SURVEYING.

SECTION 1.

A brief description of some of the Instruments, used in Surveying.

CIRCUMFERENTOR.

THIS Instrument is a Circular Box, covered with a glass lid, generally about five or six inches in diameter, in the centre of which is a steel pin, on which is placed a needle, which, being constructed with a magnetic power, always points nearly to the North and South points of the Horizon, when the Instrument is Horizontal, and the needle at rest. On the North and South points of the Box, is an index,to the ends of which are screwed perpendicular brass sights. In each sight is a large and small aperture, one over the other; the small aperture in one, being opposite the large one, in the other. In the middle of the large opertures, is placed a horse hair, or fine silk

thread. The Instrument has a Socket, in which, being placed the head of a Staff, it is supported while used. To the Socket is generally a Ball, that the Instrument may be readily fixed in a Horizontal position.

The Circle is divided into 360°, marked under, or at, the ends of the Needle, thus; from the North to the East into 90, and from the North to the West into 90; from the South to the East into 90, and from the South to the West into 90. By this Instrument the Course or Bearing of the Boundary Lines of a Field are determined.

To ascertain whether the Needle be correct, and in good order for use, set the Compass in some place, where it is not affected by Iron or Steel; when the Needle is at rest, apply to one end of it a piece of Iron or Steel, to attract it from its place; then remove the Iron or Steel to a distance, and if the Needle settles at the same Point as before, in may depended on as correct.

Other Instruments, governed by the Needle, are sometimes used, as the Theodolite, Plain Table, Semicircle, &c. As they are, in many respects, similar to the Circumferentor, a description of them is omitted.

THE CHAIN.

A Four Pole Chain consists of 100 Links, each Link beng 7,92 Inches in length, but the Chain, commonly used in New-England, is two Rods in length, consisting of 50 Links. In the middle of the Chain and at every ten Links, is, usually, a piece of Brass. By this Instrument, the Distances of the Boundary Lines of a Field are measured.

Distances, in this country, are generally stated in Rods and Links, in Deeds, and other Instruments, where a description of Land is necessary.

It may be proper here to observe, that Inclined Surfaces, as the, sides of hills, are measured Horizontally, and not on the Plane or Surface of the hill. To effect this, in ascending a hill, the hinder end of the Chain must be raised, Perpendicularly, over the Stick, left by the forward Chainman, till the Chain is in a Horizontal position; at which time, the forward Chainman must place his Stick in the ground, at the end of the Chain, the Chain being straightly drawn. The Perpendicular position may be determined by a Plummet and Line. But in descending a hill, the same must be observed by the forward Chainman, respecting the point where he must place his Stick.

PROTRACTOR.

This Instrument is a Semicircle, usually made of brass, and 4 or 5 Inches in Diameter, the Arch of which is divided into 180 Degress, and numbered both ways. It is used with a Scale to delineate, or draw a Map or Plan, of a piece of land, from the Field Book.

GUNTER'S SCALE.

This Instrument is a Rule two feet in length; it is generally made of wood. On one side are Lines of Numbers, Sines, and Tangents, by which the different statements in the Cases of Trigonometry may be solved; also Lines of Sine and Tangent Rhumbs, Versed Sines, &c., the use of which it is unnecessary here to describe. On the other side is a Line of Chords, for measuring and laying off Angles, and answers the purpose of a Protractor. At the left end are two Scales of equal parts, one of an Inch, and the other of half an Inch; at one end of the large Scale is an Inch, divided into ten equal parts; at the other end of the small Scale is half an Inch, divided also into 10 equal parts; both of which are Diagonally divided, by Lines drawn slantwise across the Scale. This part of the Scale is used for taking Distances with the Dividers, for the purpose of drawing a Plan, and is thus performed. If it be required to draw a Plan of 20 Perches to an Inch, then the extent of one Inch, in the Dividers, will represent 20 Perches, and one Division on the Diagonal Inch, 2 Perches; and, proceeding downwards, on the Diagonal Line, each Division is two tenths of a Perch.

A thorough knowledge, of the Instruments here described, cannot be obtained without some practice, and instruction from persons acquainted with their use.

SECTION II.

Introductory Problems, for Reducing the Measures used in Surveying.

The usual Measure of land is the Acre: 40 Square Rods make a Rood, and 4 Roods, or 160 Square Rods, Perches, or Poles, make an Acre.

PROBLEM I. To reduce Two Rod Chains to Rods and Decimal Parts.

Multiply the Chains by 2 for the Rods, and the Links by 4 for the Decimal. If the Links exceed 25, add one to the Rods, and multiply the remainder of the Links by 4 for the Decimal. If the Links do not exceed 2, a Cipher must be prefixed to the left hand.

1. In 19 Chains 21 Links, how many Rods, &c.?

19-21
2 4

Ans. 38,84 Rods.

2. In 15 Chains 27 Links how many Rods, &c.?

Ans. 31,08 Rods.

PROBLEM II. To reduce Two Rod Chains to Four Rod Chains.

Divide the Chains by 2, to which annex the Links if any. If the given Chains be an odd number, call the remainder 50 Links, which must be added to the given Links.

In 17 Two Pole Chains 42 Links, how many Four Pole Chains and Links?

PROBLEM III.

Ch. L. 2)17-42

Ans. 8-92

To reduce Four Rod Chains and Links to

Rods and Decimal Parts.

Multiply the Chains and Links by 4, the Product will be Rods and Hundredths.

In 13 Chains and 64 Links, how many Rods and Decimal Parts?

13-64
4

Ans. 54,56 Rods.

PROBLEM IV. To reduce Rods and Links to Four Rod Chains and Links.

Divide the Rods by 4, to the Quotient annex the Links, adding thereto 25 for every Unit in the remainder.

In 53 Rods 17 Links, how many Chains and Links?

4)53

-17

Ans. 13 Cha. 42 Links.

PROBLEM V. To reduce Square Chains to Acres.

Divide the Chains by 10, or which is the same, cut off the Right hand figure; the Quotient will be Acres and Decimals: Thus 846 Square 4 Pole Chains make 84, 6 Acres.-Multiply the Decimal by 4, and cut off from the Right hand of the Product, one figure; the figure at the left will be Roods; multiply the figure cut off by 40, cutting off as before, and the figures at the left will be Rods.

PROBLEM VI. To reduce Square Rods to Acres.

Divide by 160 for the Acres, and the remainder by 40, if it exceed that number, for the Roods, or Quarters of an Acre; the last remainder will be Square Rods.

In 656 Square Rods how many Acres?

Ans. 4 Acres 16 Rods.

SECTION III.

To calculate the Area of Plain Rectilinear Figures and Circles.

PROBLEM I. To find the Area of a Square.

Multiply the length of one Side by itself; the Product is the Area.

How many Acres in a Square piece of land, the length of one Side being 40 Rods?

Ans. 40X40-1600-160-10 Acres.

PROBLEM II. To find the Area of a Parallelogram.
Multiply the longest by the shortest Side.

How many Acres in a piece of land, 63 Rods long and 28 broad?

gle.

Ans. 63×28-1764÷160—11 Acres 4 Rods,

PROBLEM III. To find the Area of a Right Angled Trian

Multiply the Base by half the Perpendicular, or the Perpendicular by half the Base, the Product is the Area; or, multiply the Base and Perpendicular together, and half the Product is the Area.

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