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the decimal point or separatrix being removed one figure to the left, which leaves a figure to the right to spare.

If the distance be any number of chains or perches, and the decimal of a chain or perch, the lat. and dep. must be taken out at two or more operations, by taking out the lat, and dep. for the chains or perches in the first place; and then for the decimal parts.

To save the repeated trouble of additions, a judicious surveyor will always limit his stations to whole chains; or perches and lengths, which can commonly be done at every station, save the last.

1. In order to illustrate the foregoing observations, let us suppose a course or bearing to be S. 35o. 21' E, and the distance 79 four-pole chains. Under 35o. 15', er 35 degrees; and opposite 79, we find 64. 52 for the latitude, and 45. 59 the departure, which signify that the end of that station differs in latitude from the beginning 64. 52 chains, and in departure 45. 59 chains.

Note. We are to understand the same things if the distance is given in perches or any other measures, the method of proceeding being exactly the same in every case.

Again, let the bearing be 54. degrees, and distance as before; then over said degrees we find the same numbers, only with this difference, that the lat. before found, will now be the dep. and the dep. the lat. because 543 is the complement

of 351 degrees to 90, viz. lat. 45. 59. dep. 64.

52.

Here we

2. Suppose the same course, but the distance 7 chains 90 links, or as many perches. find the same numbers, but the decimal point must be removed one figure to the left.

Thus, under 351 and in a line with 79 or 7.9,

are

Lat. 6. 45
Dep. 4. 56.

the 5 in the dep. being increased by 1, because the 9 is rejected ; but over 54 we get

Lat. 4. 56
Dep. 6. 45.

3. Let the course be as before, but the distance 7.79, then opposite

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THEOREM II.

When the first meridian passes through the map.

If the east meridian distances in the middle of

each line be multiplied into the particular southing, and the west meridian distances into the particular northing, the sum of these products will be the area of the map.

Plate X. fig. 1.

Let the figures abkm be a map, the lines, ab, bk to the southward, and km ma to the northward, NS the first meridian line passing through the first station a.

The meridian) dd X gọ
Distances east) tu x ox (by)

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The meridian efx gr

=Area Distances west 5 hh x ga (my)

These four areas am + ow + xp + gl will be the area of the whole figure cmswiprlc, which is equal to the area of the map abkm. Complete the figure.

The parallelograms am and ow, are made of the east meridian distances dz and tu, multiplied into the southings ao and ox. The parallelograms xp and gl are composed of the west meridian dis

tances ef and hh, multiplied into the northings xg and ga (my) but these four parallelograms are equal to the area of the map; for if from them be taken the four triangles marked Z, and in the place of those be substituted the four triangles marked 0, which are equal to the former; then it is plain the area of the map will be equal to the four parallelograms. Q. E. D.

THEOREM III.

If the meridian distance when east, be multiplied

into the southings, and the meridian distance when west be multiplied into the northings, the sum of these less by the meridian distance when west muliiplied into the southings, is the area of

the survey.

Plate X. fig. 2.

Let a b c be the map.

The figure being completed, the rectangle af is made of the meridian distance eq when east, multiplied into the southing an ; the rectangle yk is made of the meridian distance xw, multiplied into the northings ca or ya. These two rectangles, or parallelograms, af tyk, make the area of the figure dfnyikd, from which taking the rectangle oy, made of the meridian distance tu when west, into the southings oh or bm, the remainder is the area of the figure dfohikd, which is equal to the area of the map.

Let bou –Y, urih L, ric=0, wrc=Z, akio =K, and efb= B, ade=A. I say that Y+Z+B =K+L+A.

Y=L + O, add Z to both, then Y + Z=L+0 +Z; but Z+0=K, put K instead of Z + O : then Y+Z=L+ K, add to both sides the equal triangles B and A, then Y + 2 +B=L+K+ A. If therefore B + Y + Z be taken from abc, and in lieu thereof we put L + K + A, we shall have the figure dfohikd=abc, but that figure is made up

of the meridian distance when east, multiplied into the southing, and the meridian distance, when west, multiplied into the northing less by the meridian distance, when west, multiplied in the southing. Q. E. D.

COROLLARY.

Since the meridian distance (when west) mulplied into the southing, is to be subtracted, by the same reasoning the meridian distance when east, multiplied into the northing, must be also subtracted.

SCHOLIUM.

From the two preceding theorems we learn how to find the area of the map, when the first meridian passes through it; that is, when one part of the map lies on the east and the other on the west side of that meridian. Thus,

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