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3. Required the area of a square, in acres, the side of which is 132 yards. 132 x 132

3.6 acres = 3a. 2r. 16p. the area 4840 4. What is the area of a rectangle whose length is 2470 Links, and its breadth 1114 links?

2470 x 1114 = 27.51580 = 27a. 2r. 2 p. the arca. 5. The length of a rectangular field is 324 yards, and its breadth 235 what is its content in acres ?

Ans. 15a. 2r. 37p 6. What is the area of a rectangular pleasure ground, the length of which is 960 and its breadth 125 links?

Ans. la. Or. 32p.

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PROBLEM II.

TRIANGULAR FIELDS. 1. Let A B C be a triangle, of which the survey, plan, and content are required.

Set up poles or marks at the angles A, B, and C, and meagure from A towards B, and when at or near D, try with the cross for the place of the perpendicular CD; plant the cross, and turn it till the marks A

с and B can be seen through one of the grooves; then look through the other groove, nnd, if the mark at C can be seen through it, the cross is in the right place for the perpendicular; if not, move

D the cross backward or forward till the three marks can be seen as before directed. Suppose the distance A D to be 625 links, and the whole A B, to be 1257 links; return to D, and measure the perpendicular DC, which suppose to be 628 links, thus completing the survey of the triangle.

CONSTRUCTION. From a scale of equal parts, or plotting scale, lay off the base A B, = 1257 links; on which take A D, 625 links; at D) erect the perpendicular DC, which make = 628 links; join AC, CB, then A B C is the plan of the triangle.

A

B

TO FIND THE CONTENT. RULE. Multiply the base by the perpendicular, and half the product will be the area.

EXAMPLES.

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1. The dimensions being the same as found above, required the content.

Ans. 1257 x 628 = 2= 3.94698 acres = 3a. 3r. 32p. 2. The distance from the beginning of the base to the place of the perpendicular is 375 links, the whole base 954, and the perpendicular 246; what is the area of the triangle?

954 x 246 ; 2= 1•17342 la. Or. 27p. the content. 3. Measuring the base of a triangle the place of the perpendicular was found at 863 links, and its length 645; the whole base was 1434 links; required the plan and area.

Area. 4a. 2r. 20p.

PROBLEM III.

FIELDS IN THE FORM OF TRAPEZIUMS. Fields in this form are usually divided into two triangles by a diagonal, which is a base to both the triangles.

Let ABCD be a field in the form of a trapezium, the plan and area of which is required.

Measure from A towards C; and let the place of the perpendicular m B be at 5.52, and its length 3.76, also let the place of the perpendicular n D be at 11•82, and its length 3•44, and the length of the whole diagonal AC be 13:91 chains, which completes the survey : but it is usual also to measure the other diagonal B D for a proof line, which is found to be 9:56 chains.

Note 1. The construction of each of the two triangles, forming the trapezium, is the same as the construction given to the first example in PROB. II.

Note 2. The longer of the two diagonals should always be selected for the base of the two triangles, forming the trapezium, for sometimes the perpendicular will not fall on the shorter diagonal, without its being prolonged; and when this is the case with both diagonals, one of the sides may be taken for a base, or two of the sides, if necessary.

TO FIND THE CONTENT. RULE. Multiply the sum of the two perpendiculars by the diagonal, and half the product will be the content.

EXAMPLES. 1. Let the measurement of a trapezium be as above found, required the content.

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1.21600

Ans. 5a. Or. Ip. 2. From the following notes, plan and find the content of a field. Perpendiculars

Base or

Perpendiculars Station Line.

on right. to OC

on left.

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Begin
at OA

and range West.

Content. 37a. 3r. 2p. 8. Give the plan and area of a field from the following notes.

AC

872 B 652

731

423 545 C Begin at O A and range

E.

Area. 5a. Or. 35p. ANOTHER METHOD. A four-sided field may frequently be surveyed by dividing it into two triangles and a trapezoid, by perpendiculars on its longest side.

TO FIND THE CONTENT. RULE. Multiply the sum of the two perpendiculars by their distance on the base line for the double area. The double areas of the two triangles must be found as in PROB. II., and both be

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added to the double area of the trapezoid ; the sum being divided by 2, and five figures marked off for decimals, will give the content required.

EXAMPLES

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595 } perp.

1. Required the survey and area of the following field.

Measure the base A B, and put down in the field book the dis. tances of P and Q, where the perpendiculars rise, &c., as below.

Trapezoid PCDQ.
to o D.

352
AB=1097
QD=595 A Q=

= 745
PC= 352 AP= 110

947 sum. Perpen. From o A go E.

635 =PQ.
Triangle A CP. Triangle Q D B. 4735
352
595
2841

D
110
352

5682

C
38720
1190

631345
2975

38720 1785

209440

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O A

go E.

Area. 4a. Or. 393p.

39.6032 2. Required the plans and areas of two fields from the following notes. A B

A B 1169

1448 E 615 339 D. D 513 1102 E. G 234 461 C. C 683

436 G. From O A go W.

From

PROBLEM IV.
TO SURVEY FIELDS CONTAINED BY MORE THAN FOUR SIDES.

Fields or plots of ground bounded by more than four sides, may be surveyed by dividing them into trapeziums and triangles.—Thus, a field of five sides may be divided into a trapezium and a triangle; of six sides, into two trapeziums; of seven sides into two trapeziums and a triangle; &c.

TO FIND THE CONTENT.
RULE. By the two last Problems, find the double areas of

each trapezium and triangle in the field; add all the double areas together, and half their sum will be the content.

EXAMPLES.

1. Lay down a field and find its area from the following dimensions.

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CONSTRUCTION. The above field is divided into trapeziums ABCG, GDEF and the triangle GCD.-Draw the diagonal A C, which make = 550 links; at 135 and 410 set off the perpendiculars mG= 130, and nB = 180 links respectively; join AB, BC, CG, and GA, and the first trapezium will be completed. Then on CG,

, lay off Cq = 152, and draw the perpendicular q D = 230 ; join CD, DG, and the triangle is completed. Lastly, with centre G and radius o G= 120 describe an arc; and with centre D and radius o D = 314 (= 520 – 206) describe another arc, intersecting the former in o: through o draw the diagonal DF = 520 links, upon which, at 288 links, draw the perpendicular rE; join DE, EF, F G, and the figure will be completed. 130 440 120

Double areas.
180
230
80

170500 trap. A BCG.

101200 tri. CDG.
13200

104000 trap. D E F G.
88
5.20

200

310 550

2)3-75700 15500 101200 104000 1550

1•87850 = la. 3r. 3034

170500

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