Imágenes de páginas
PDF
EPUB

6. From a station P there can be seen three objects A, B and C, whose distances from each other are known : viz. AB=800, AC=600, and BC

= 400 yards. Now, there are measured the horizontal angles

APC=330 45' and BPC=22° 30': it is required to find the three distances PA, PC, and PB.

Ans.

PA=710.193 yards.
PC=1042.522
PB=934.29 1

OF MEASUREMENTS WITH THE TAPE OR CHAIN ONLY.

92. It often happens that instruments for the measurement of angles cannot be easily obtained; we must then rely entirely on the tape or chain.

We now. propose to explain the best methods of determining distances, without the aid of instruments for the measurement of horizontal or vertical angles.

PROBLEM I.

To trace, on the ground, the direction of a right line, that shall be

perpendicular at a given point, to a given right line.

FIRST METHOD.

D 93. Let BC be the given right line, and A the given point. Measure from A, on the line BC, two equal distances AB, AC, one on each side of the point A. Take a B A portion of the chain or tape, greater than AB, and place one extremity at B, and with the other trace the arc of a circle on the ground. Then remove the end which was at B, to C, and trace a second arc intersecting the former at D. The straight line drawn through D and A will be perpendicular

SECOND METHOD.

94. Having made AB=AC, take any portion of the tape or chain, considerably greater than the distance

B between B and C. Mark the middle point of it, and fasten its two extremi. ties, the one at B and the other at č. Then, taking the chain by the middle point, stretch it tightly on either side of BC, and place a staff at D or E: then will DAE be the perpendicular required.

THIRD METHOD.

95. Let AB be the given line, and © the point at which the perpendicular is to be drawn. From the point C measure a distance CA equal to 8.

A

B With C as a centre, and a radius equal

с to 6, describe an arc on either side of AB : 'then, with A as a centre, and a radius equal to 10, describe a second arc intersecting the one before described at E: then draw the line EC, and it will be perpendicular to AB at C.

REMARK. Any three lines, having the ratio of 6, 8 and 10, form a right-angled triangle, of which the side corresponding to 10 is the hypothenuse

FOURTH METHOD.

96. Let AD be the given right line, and D the point at which the perpendicular is to be drawn. Take any distance on the tape or chain, and place one extremity at D, and fasten the other at some point as E, between the two lines which are to form the right angle. Place a staff at E. Then, having stationed a person at D, remove the extremity of the chain and carry it round until it ranges on the line 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 perpen

dicular on the line. 98. Let C be the given point, and AB the given line.

From C measure a line, as CA, to any point of the line AB. From

-B A, measure on AB any distance

F D as AF, and at F erect 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 as found. The distance AD being laid off from A, the point D, at which the perpendicular CD meets AB, becomes known. 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: therefore, CD is determined.

PROBLEM III.

[ocr errors]

E

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.

At E lay off the right angle AED, and measure in the direction ED, any convenient distance to D, and

B В place a staff at D. Then measure from E, directly towards the object D F 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

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.

C

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.

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.

[ocr errors]

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

с

« AnteriorContinuar »