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In this example we find,
The area of OEFGHO = 5270.5
Consequently of

HOMH = 826.0
Dif. lat. of the line HO=HV 35.2
Departure of ditto

=QV = 38.2 As HI happens to be a meridian, the area of HOMH divided by half OV (19.1) quotes HM (43.23) without finding the area of HOIH, as we did of ICDI in example 2d. and HM-HV = VM= 8.03 = dif. lat. of OM, which with its dep. Vo=38.2. gives the bearing and distance as before.


PL. 12. fig. 4.

A trapezoidal field ABCD, bounded as under specified, is to be divided into two equal parts by a right line EF parallel to AB or CD; required AF or BF?

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In the triangle CBG are given BC and all the angles (known by the bearings) to find BG, and thence the area by prob. 9. sect. 4. which+half the area of ABCD area of EFG. then as the area of CBG to that of EFG, so is the square of BG to the square of FG, and FG-BG=BF.

Operation at large.
Angle G 39° 30', log. S, Co. Ar. 0.19649
Side BC 60 per. log.

1.77815 Sadd Angle C 40° 30', sine


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Ans. BF= 16.46 per.

1.89055 By the application of this method a tract of land may be divided accurately, in any proportion, by a line running in any assigned direction.

Note. When the practitioner would wish to be very accurate, it will be much better to work by four-pole chains and links than by perches and tenths; one tenth of a perch square being equal to 62 square links.


The following Field-Notes (from A. Burns) are's of a piece of land, which is proposed, as an example, to be divided into three equal parts by tno rightlines running from the sixth and sevenih stations ; and proved, by calculating the content of the middle part.

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PL. 8. fg. 5.

The plot ABCDEFGHA is proposed to be divided, geometrically, in the proportion of 2 to 3, by a right line from a given point in any boundary or angle thereof, suppose the point D.

Reduce the plot to the triangle cDe, as already taught; divide the base ce in the point N, so that eN be to Nc in the ratio of two or three, by prob. 14. page 53; draw DN, and it is done.


PL. 12. fig. 3.

Example 2d may likewise be performed geometricaliga

Produce CD both ways for a base, and reduce the whole to a triangle, making the vertical point; then bisect the base in N, and draw IN. But,

Notwithstanding this geometrical method is demonstrably true in theory, it is not as safe, on practical occasions requiring accuracy, as the calculation, even when performed with the greatest care ; for which reason we will not enlarge on it here.


Suppose 864 acres to be laid out in forma of a right-angled prerallel

ogram, of which the sides shall be in proportion as 5 to 3; required their dimensions ?

For the greater side, multiply the area by the greater number of the given proportion, and divide


by the less, or, for the less side, multiply by the less number, and divide by the greater; the square root of the quotient will be the side required : thus,

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If it be required to lay out any quantity of ground, suppose 47A. PR. 16P. inform of a parallelogram, of which the length is to exceed the breadth by a given difference, for instance 80 perches, then add the square of half this difference to the area, and take the square-root of the sum; to which add half the difference for the greater side, and subtract it therefrom for the less ; thus,


47A. 2R. 16P.=7616 perches.



V 9216=96

1600 half diff. add and subt.--40

the length

= 136

the breadth Any proposed quantity of ground may be laid out or inclosed in the form


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