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part of the figure above this line will be found, eleven acres and thirty-four rods, leaving a deficiency of one acre and a half, and thirty-three rods, of making one half of the whole figure.

This quantity is to be laid in a triangle on the line from to 5, and on that from 5 to 6, running to a point at 2.

The question is, what distance from 5 towards 6, will be required to give the contents of the triangle? To answer such a question, the area, the contained angle, and one side of the triangle must be given, to find the length of the contiguous side, including the angle. Having these data, take the following rule.

To the sine of the given angle, or to its supplement if obtuse, add the logarithm of the given side; substract radius from this sum, and substract the remainder from the logarithm of the double area: the last remainder will be the logarithm of the side required. In this case, the angle formed by the line from 5 to 2, and by that from 5 to 6, contains 750°25', the given side is 57.72 rods, the area of the triangle is 273 rods, and the double area is 546 rods. See the rule worked. Sine of 750 25'

9.985778 logarithm of the given side 57.72 1.761369 cancel the first figure, and R. is substracted

1,1.747147 logarithm of the double area

2.737193 remainder brought down

1.747147 logarithm of the side required 9.77 rods

0.990046

be

Measure from 5 towards 6, 9 rods and 194 links to a beech tree, from thence on a straight line to 2, and the figure will be equally divided. The true position of the divisional line being established, the course and distance of it may calculated as before directed.

The course and distance from 2 to 5 are N. 800 25' E. 57.72 rods : from 5 to the beech tree, S. 5° W., 9.77 rods. On the first, 9.62 rods of northing and 56.92 of easting are made. On the second, 9.74 rods of southing and .85 of a rod of westing are made.

The northing of the first course being substracted from the southing of the second, and the westing of the second from the easting of the first, .12 of aerod of northing, and 56.07 rods of westing, are given as data by which the course and distance from the beech tree to 2 may be calculated.

The

Suppose it is next required to lay out 15 rods in width beow and parallel with the dividing line.

As the eastern and western boundary lines both turn obliquely from the dividing line, the question is, what distance will be required on each to give the required width?

Make the distance on each boundary line the hypothenuse of a right angled triangle, and 15 rods, the width required, one of the legs or the base. The angle at 2, formed by the boundary and dividing lines, contains 114° 53'. Substract 90° from this, and 24° 53' remains, being the angle formed by the given leg (15) and the hypothenuse.

Take the following proportion to find the hypothenuse.

As the co-sine of 24° 53'; is to 15, the given side; so is radius to the distance required. See it worked. Co-sine of 24° 53'

9.957687 : given side 15 rods

1.176091 : : radius

10.

11.176091
9.957687
1.218404

: side required 16.54 rods

The dividing line and the eastern boundary, at the beech tree, form an angle of 85° 7'. Substract this from 90°, and 4° 53' remains, being the angle by which the east side must be worked. Co-sine of 4° 53'

9.998421 : given side 15 rods

1.176091 : : radius

10.

11.176091
9.998421

: side required 15.05 rods

1.177670 The distances required are 15 rods 11 link on the eastern, and 16 rods 133 links on the western boundary.

Measure from 2 towards 1, 16 rods 131 links to a spruce tree, and from the beech tree towards 6, 15 rods 14 link to a pine tree, and calculate the distance from the pine to the spruce, and the division will be completed. This calculation must be made by a traverse from the courses and distan. ces already known.

From the spruce to 2, N. 25° E. 16.54 rods; from 2 to the beech, S. 89° 53' E. 56.07 rods; from the beech to the pine in their proper columns as follows, and calculate the latitude and departure of each, having a blank line for the last.

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15.11|15.11/63.0563.05 Having calculated the latitudes and departures, substract the northing of the first course from the sum of the southings of the two last, and .12 of a rod remains to be entered in the blank column of northings; also, substract the westing of the third from the sum of the eastings of the two first, and 61.74 remains to fill the blank column of westings. This balance of northing and of westing is the latitude and departure of the line from the pine to the spruce. The distance is 61.74 rods, and the course agrees with that on the opposite side.

Suppose it is next required to lay out four acres adjoining and parallel with the line from the pine to the spruce.

As the figure of this division is not regular, no rule can be given by which it can be calculated by one operation, but it must be done by approximation.

Reduce the quantity of the division to rods, and divide that sum by the length of the known side, and the quotient will be 10.36 rods, but as the south side will be longer than the north, the width cannot be so much.

Assume 10 rods as the width, and calculate the distance on each end by the same process that you did those on the last. The angles in this are the same as in that. Co-sine of 24° 53'

9.957687 : given side 10 rods

1.000000 : : radius

10.

11.000000
9.957687

Co-sine of 40 53'
: given side 10 rods
: : radius

9.998421 1.000000 10.

11.000000
9.998421

: side required 10.04 rods

1.001579 The distances required are 10 rods 1 link on the east, and H rods of a link on the west boundary.

Measure from the spruce towards 1, 11 rods of a link to a hemlock, and from the pine towards 6, 10 rods 1 link to a maple, and calculate the distance from the latter tree to the hemlock.

From the hemlock to the spruce N. 25° E. 11.03 rods; from the spruce to the pine, S: 890 53' E. 61.74 rods; from the pine to the maple S. 5° W. 10.04 rods.

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10.12 10.12166.40166.40 The northing and westing in the blank line is the latitude and departure of the line from the maple to the hemlock. The distance required is 65.53 rods. To this distance, add that of the north side, and the sum of both is 127.27 rods. Multiply half of this sum by 10 rods, the width of the division as it now stands, and the product is 636.3 rods. Substract this from 640, and 3.7 rods remain, being a deficiency. To this deficiency annex three ciphers, and divide that sum by the distance from the maple to the hemlock, and the quotient will be nearly 6, which is .06 of a rod, or 11 link. To the quantity already found, add this width on the south side, and the division will be completed.

The south division of the whole figure contains twelve acres, three quarters, and twenty-seven rods. This division The first contains five acres, two quarters and three rods; the second, four acres, and if the foregoing calculations are correct, the third contains three acres, one quarter, and twenty-four rods. See the work proved.

There now remains between the southeast corner and the last subdivision, 5.08 rods, and between the southwest corner and the same division, 12.37 rods, to be calculated on.

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As the magnetic needle cannot be relied on in renewing lost boundaries, it is of the first importance that good, substantial bounds, be kept up. In divisions, or distributions of lands, surveyors ought to see that such bounds are erected. It is their business to cause them to be made, and such bounds ought to be described in deeds, or in the instruments by which lands are conveyed, and to be a part of the record. By proper care and attention to this part of the business, many disputes may be prevented.

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