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chain at A, and carry it round until it falls on the line AE at F. Then place a staff at F, and ADF will be a right angle, being an angle in a semi-circle.

97. There is a very simple instrument, used exclusively in laying off right angles on the ground, which is called the

SURVEYING CROSS.

Pl. 2. Fig. 1. This instrument consists of two bars, AB and CD, permanently fixed at right angles to each other, and firmly attached at E to a pointed staff, which serves as a support. Four sights are screwed firmly to the bars, by means of the screws a, b, c, and d.

As the only use of this instrument is to lay off right angles, it is of the first importance that the lines of sight be truly at right angles. To ascertain if they are so, let the bar AB be turned until its sights mark some distinct object; then look through the other sights and place a staff on the line which they indicate: let the cross be then turned until the sights of the bar AB come to the same line: if the other sights are directed to the first object, the lines of sight are exactly at right angles.

The sights being at right angles, if one of them be turned in the direction of a given line, the other will mark the direction of a line perpendicular to it, at the point where the instrument is placed.

PROBLEM II.

From a given point without a straight line, to let fall a perpendicular on the line.

98. Let C be the given point,

and AB the given line.

From C measure a line, as CA,

E

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as AF, and at Ferect FE perpendicular to AB. Having stationed a person at A, measure along the perpendicular FE until the forward staff is aligned on the line AC: then measure the distance AE. Now, by similar triangles, we have

in which all the terms are known except AD, which may,

therefore, be considered a wound. laid off from A, the point D, at CD meets AB, becomes known.

The distance AD being which the perpendicular If we wish the length of

the perpendicular, we use the proportion

AE : EF :: AC: CD,

in which all the terms are known, excepting CD: there. fore, CD is determined.

PROBLEM III.

To determine the horizontal distance from a given point to an inaccessible object.

99. Let A be an inaccessible object, and E the point from which the distance is to be measured.

F

A

B

E

At E lay off the right angle AED, and measure in the direction ED, any convenient distance to D, and place a staff at D. Then measure from E, directly towards the object D A, a distance EB of a convenient length, and at B lay off a line BC perpendicular to EA. Measure along the line BC, until a person at D shall range the forward staff on the line DA. Now, DF is known, being equal to the difference between the two measured lines DE and CB. Hence, by similar triangles,

DF FC: DE: EA, in which proportion all the terms are known, except the fourth, which may, therefore, be regarded as found: hence, EA is determined.

SECOND METHOD.

100. At the point E lay off EB perpendicular to the line EA, and measure along it any convenient distance, as EB.

At B lay off the right angle B EBD, and measure any distance in the direction BD. Let a person at D align a staff on DA,

E

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while a second person at B aligns it on BE: the staff will thus be fixed at C. Then measure the distance BC. The two triangles BCD and CAE being similar, we have, BC BD :: CE: EA,

in which all the terms are known, except the fourth, which may, therefore, be regarded as found.

THIRD METHOD.

101. Let B be the given point, and A the inaccessible object, it is required to find BA.

Measure any horizontal base line, as BC. Then, having placed staves at B and C, measure any convenient distances BD and CE, such that the points D, B and A, shall be in the same right line, as also, the points E, C and A; then measure the diagonal lines DC and EB.

D

E

Now, in the triangle BEC, the three sides are known, therefore, the angle ECB can be found. In the triangle CDB, the three sides are also known, therefore the angle CBD can be determined. These angles being respectively subtracted from 180°, the two angles ACB and ABC become known; and hence, in the triangle ABC, we have two angles and the included side, to find the side BA.

PROBLEM IV.

To find the altitude of an object, when the distance to the vertical line passing through the top of it is known.

102. Let CD be the altitude required, and AC the known distance.

From A, measure on the

line AC, any convenient. distance AB, and place a

B

C

the object D, and let the point, at which the line AD cuts the staff BE, be marked. Measure the distance BE on the staff; then say,

As ᎯᏴ : ᏴᎬ :: ᎯᏟ : CD,

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then, CD becomes known.

If the line AC cannot be measured, on account of intervening objects, it may be determined by calculation, as in the last problem, and then, having found the horizontal distance, the vertical line is readily determined, as before.

CHAPTER III.

Of the area or content of ground.-Of laying out and
dividing land.

103. We come next to the determination of the area or content of ground.

The surface of the ground being, in general, broken and uneven, it is impossible, without great trouble and expense, to ascertain its exact area or content. To avoid this incon venience, it has been agreed to refer every surface to a horizontal plane: that is, to regard all its bounding lines as horizontal, and its area as measured by that portion of the horizontal plane which the boundary lines enclose.

For example, if ABCD were a

piece of ground having an uneven surface, we should refer the whole to a horizontal plane, and take for the measure of the area that part of the plane which is included between the bounding lines AB, BC, CD, DA.

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In estimating land in this manner, the sum of the areas of all the parts into which a tract may be divided, is equal to the area estimating it as an entire piece: but this would not be the case if the areas of the parts had reference to the actual surface, and the area of the whole were calculated from its bounding lines.

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104. The unit of a quantity is one of the equal parts of which the quantity is composed (Arith. In. VI). Thus, a line of three feet in length is made up of three single feet, and of this line, 1 foot is the unit. The unit of a line may be 1 foot, 1 yard, 1 rod, 1 chain, or any other known distance. If, on the unit of length, a square be described, it will form the unit for computing areas.

1 foot.

Thus, is 1 square foot,

1 square yard, or 9 square feet,

1 square chain, or 16 square rods. .

1 yard 3 feet.

1 chain 4 rods.

Thus it is seen that there are two kinds of quantity to be considered, viz. lines, and areas or surfaces; and each kind has its own unit of measure.

When, therefore, the linear measures of ground are feet, yards, rods, or chains, the superficial measures will be square feet, square yards, square rods, or square chains; and the number expressing the area will be nothing else than the number of times which the unit of superficial measure is contained in the land measured.

It has been already observed (Art. 83), that Gunter's chain of four rods or 66 feet in length, and which is divided into 100 links, is the chain in general use among surveyors. We shall, therefore, take the length of this chain for the unit

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