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EXAMPLE.

A map whose area is 126A. 3R. 16P. being given; to find the scale?

Suppose this map wast cast up by a scale of 20 perches to an inch, and the content thereby produced be 31A. 2R. 34P.

As the area found, 31A. 2R. 34P.-5074P.

Is to the square of the scale by which it was cast up, that is, to 20×20=400,

The given area of the map 126A. 3R. 16P. =20296P.

To the square of the scale by which it was laid down.

5074: 400 20296: 1600 the square of the required scale. Hence 40 perches to an inch is the scale required.

PROB. XVIII.

How to find the true content of a survey, though it be taken by a chain that is too long or too short.

Let the map be constructed, and its area found as if the chain were of the true length.

As the square of the true chain

Is the content of the map,

And it will be

:: The square of the chain you surveyed by

EXAMPLE.

If a survey be taken with a chain which is 3 inches too long; or with one whose length is 42 feet 3 inches, and the map thereof be found to contain 920A. 2R. 20P. Required the true content.

As the square of 42F. OIn. the square of 504 inches =254016

Is to the content of the map 920A. 1R. 20P.= 147260P.

The square of 42F. 3In. the square of 507 inches-250749

A. R. P.

To the true content, 931. 1. 19.

SECT. V.

Third Method for determining the Areas of rightlined Figures universally, or by Calculation.

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1. MERIDIANS ERIDIANS are north and south lines, which are supposed to pass through every station of the survey.

2. The Bearing of a line, is the angle it makes with the meridian passing through either end of it.

3. The difference of latitude, or the northing or southing of any stationary line, is the distance that one end of the line is north or south from the other end; or it is the distance which is intercepted on the meridian, between the beginning of the stationary line and a perpendicular drawn from the other end to that meridian. Thus, if N. S. be a meridian line passing through the point A of the line AB, then is

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4. The departure of any stationary line, is the nearest distance from one end of the line to a meridian passing through the other end. Thus Bb is the departure or easting of the line AB: but if CB be a meridian, and the measure of the stationary distance be taken from B to A; then is BC the difference of latitude, or northing, and AC the departure or westing of the line BA.

Cor. Hence it appears that the stationary distance, difference of latitude, and departure, constitute the three sides of a right angled plane triangle: the distance being the hypothenuse, the difference of latitude and departures, the two legs, and the angle opposite the departure, the bearing or course; consequently, any two of these four parts being given, the other two can be found.

5. The meridian which passes through the first station, is sometimes called the first meridian; and sometimes it is a meridian passing on the east or west side of the map, at the distance of the breadth thereof, from east to west, set off from the first station.

6. The meridian distance of any station is the distance thereof from the first meridian, whether it be supposed to pass through the first station, or on the east or west side of the map.

THEO. I,

In every survey which is truly taken, the sum of the northings will be equal to that of the southings; and the sum of the eastings equal to that of the westings.

Let a, b, c, e, f, g, h, represent a plot, or parcel of land. Let a be the first station, b the second, c the third, &c. Let NS be a meridian line, then will all lines parallel thereto, which pass through the several stations, be meridians also; as ao, bs, cd, &c. and the lines bo, cs, de, &c. perpendicular to those, will be east or west lines, or departures.

The northings ei+go+hq=ao+bs+cd+fr the southings: for let the figure be completed; then it is plain, that go+hq+rk-ao+bs+cd, and ei-rk-fr. If to the former part of this first equation ei―rk be added and fr to the latter, then go+hq+ei=ao+bs+cd+fr; that is the sum of the northings is equal to that of the southings.

The eastings cs+qa=ob+de+if+rg+ah, the westings. For aq +yo (az)=de+if+rg+oh, and bo=cs—yo. If to the former part of this first equation, cs-yɔ be added, and bo to the latter, then cs Laq ob+de+if+rg +oh; that is, the sum of the eastings is equal to that of the westings. Q. E. D.

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