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SOLUTION. 1. In the triangle EDC, the EDC = 30.27', ECD=1759.23'. DEC=1°.10', and the side oc=1809 5 feet, hence ce=5348.08 feet. 1

2. In the triangle GCE, the LGCE=2°.43', and ce=5348.08 feet, hence cg=AH=5342:07 feet, and GE=253.48 feet.

3. In the triangle KCE, the 2 KCE=0°.55', CEK=92°.43', CKE =86o.22', and CE=5348.08 feet, hence EK=8573,1 feet.

4. In the triangle BAI, AB = 1809 5 feet, and the angle BAIS 19.54', hence Bi=59.994 feet.

Lastly, if to the height GE, the height Ac, of the instrument be added, it will give HE, the height of the hill,

EXAMPLE XLIII.

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At the distance of 25 miles from a tower its top just appeared in the horizon; required its height. The diameter of the earth being 7964 miles, and its circumference 25019.7024 miles.

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ECOA SOLUTION. 25019-7024 miles : 3600::25 miles : 211.34"= LDAB. In the right-angled triangle ABD, AB and the DAB are given, to find ad=3982-078446 miles, from which take AC=AB, the remainder dc=•078446 miles=414.19 feet, the height of the tower.

EXAMPLE XLIV. Supposing it were possible to see a light-house or other object D, in the horizon, at the distance of 200 miles, it is required to find its height CD, the diameter Ec of the earth being 7964 miles, and its circumference 25019.7024 miles.

Answer. The Ľ DAB = 29.52.39”, AD = 3987.0272 miles, and dc=5.0272 miles=26543•616 feet, being 5910.616 feet higher than Chimboraço, the highest of the Andes.

EXAMPLE XLV.

The Peak of Teneriffe is said to be 2 miles above the level of the sea; at what distance can it be seen, supposing the radius of the earth to be 3982 miles?

Artswer. VED X DC=DB (Euclid 36 of III.)=141:12 miles,

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EXAMPLE XLVI.

If the Peak of Teneriffe be 24 miles high, and the angle cdB made by a plumb-line, and a line DB meeting the surface of the sea in the farthest visible point B, be equal to 870.58'.13", it is required to find the diameter ce of the earth, supposing it to be a perfect sphere, and the utmost distance de that can be seen from the top D of the mountain. Answer. (rad. +sine CDB) DC

=DB=141.12 miles, and CE cosine CDB =7964 miles.

CHAP. IV.

OBSERVATIONS ON THE ADMEASUREMENT OF A BASÉ LINË.

(G) Where the ground is perfectly level, the manner of measuring a straight line from one object to another appears to be simple and easy; yet, on account of the curvature of the earth, no two points on its surface can be exactly situated in the same horizontal line; the chord of the arc, and not the arc itself, being the horizontal distance. Now the radius of one circle is to the radius of any other circle, as any arc of the former is to a similar arc of the latter. If we take, for instance, the base line measured on Hounslow-heath, (D. 72.) 27404.2 feet, the radius of the earth 3982 miles, or 21024960 feet, we shall have 21024960 feet : 27404 2 feet::1:. 274042

De feet, the length of the measured arc in terms of the 21024960 radius 1. But the difference between any arc and its chord, the radius being 1, is n'y of the cube of the length of the arc *;

27404:2 13 hence

xat="000000000092258 will express the dif21024960 ference in terms of the radius 1, which multiplied by 21024960 feet, the radius of the earth, produces .001939, the extent in feet by which a terrestrial arc of 27404.2 feet exceeds the chord of the same arc, a difference scarcely worth notice, even where the greatest accuracy is required.

(H) When the ground on which a base line is to be measured is sloping, it will be necessary, in some cases, to reduce it to a horizontal level. Thus, after having determined the direction of the base Art, by poles LA, IM, Hn, go, pointed at one end and fixed perpendicularly in the ground by means of a plumb)

See Chapter V, following. i + The point A, and the suinmits of the hills in, n, o, F, should be connected, so as to form a regular slope, af,

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А

B

C

line; the sum of the horizontal distances Lm, in,

K..... HO, GF will evi

LA dently be equal to the whole horizontal distance AE; and if the heights AL, MI, NH, Go be successively measured, their sum will give the whole height EF.

(I) If the ground be irregular, or if it rise and descend alternately, it is evident that the difference between the heights of the poles must be added when ascending, and subtracted when descending, in order to determine the different elevations and depressions of the ground.

(K) Surveyors generally ascertain the altitudes of irregular hills by the assistance of a spirit-level, and perpendicular poles placed at convenient distances from each other. This practice is called levelling.

(L) A base line on a sloping ground may likewise be mea-, sured by taking angles at its extremities with a theodolite. Thus, let im represent a theodolite, al a pole fixed perpendicular to the horizon and equal in length to the height of the instrument; also, let ki be a horizontal line (which may be ascertained by the bubble of air in the spirit-level of the telescope resting in the middle) and kil the angle of depression between the top of the pole Al and the horizontal line ki.

Then, because ki is parallel to AB, the angle Kil is equal to the angle MAB; then (0.44.), rad. : Am:: cosine 2 KIL : AB, or AB= Am x cosine _ KIL

rad. (M) If Am=400 yards, and < KIL=4°, AB will be 399.025 yards, hence the difference between am and AB is less than 1 yard. It appears from this example, that when the measured base is inclined to the horizon in a small angle, a reduction of this kind will be unnecessary, except in cases where great accuracy is required. OF THE ERRORS WHICH OCCUR IN TAKING ANGLES OF ELEVA

TION AND DEPRESSION WITH A THEODOLITE. (N) When the observer is at a considerable distance from the object, the altitude taken with a theodolite will require correction. In the first place the horizon of the observer and that of the object observed are not the same. Let c be the centre: of the earth, the summit of a mountain, and HPOB the horizon of the observer. Through draw EDF perpendicular to DC, and it will be the horizon of the point D.

Now so will be the true height of the mountain above the horizontal line ps; DPS

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PBS

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(0) The French academicians measured an inclined base at Peru, the length pd of which was found to be 6274:057 toises *, the angle of elevation DPB was 19.5'.43" (the effect of refraction being deducted). Now, rad. : PD :: sine DPB : DB,

PD x sine DPB hence DBS

119.93 toisés. In order to cal

rad. culate Bs; pd may be used instead of PB, and cs and CB may be considered as equal to each other without sensible error. And as (2cs + BS) X BS=PB2 (Euclid 36 of III.), it will follow that 2cs X BS=PB”, and hence Bs=

2cs (P) If the diameter of the earth be taken=6543373 toises, as deduced from the admeasurements in Lapland, Paris, and Peru;

6274-057 BS will be found

6•0158 toises. By adding BS

6543373 to bd the true height so of the mountain=125.9458 toises. Had the height os been determined from the triangle dps, by the most exact calculation, it would have been=125.97 toises. Hence it appears that in most cases the angle at c may be rejected, that PB may be taken =PD, the PBD=a right angle, and cs=CB without material error:

(Q) Angles of elevation or depression taken with a theodolite may be corrected thus: Let o be the place of the telescope when the theodolite stands on the vertical line cd: p the situation of the telescope on the vertical line cp. Then if the tele

* Bouguer, Figure de la Terre, A toise=6 French feet, and 107 French feet=

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2

scope at P be directed to an object at p, the elevation of that object above the horizontal line HPB is the LDPB; and when the telescope is at D and directed to an object at P, the angle of depression, below the horizontal line EDI, is the FDP.

Now, bcause PB touches the circle, and is cuts it, the LBPS (measured by half the arc Ps *)= Lc; therefore 2 DPS=LC + the L of elevation DPB,

Through p draw DG parallel to sp, then the GDPE LDPS (Euclid 29 of I.), hence L GDP = { 2c + the L of elevation DPB; and because the triangles cps and cdG are isosceles, and the Lc is common to both of them, the L CPS LCDG, but the angles CPB and cor are right angles, therefore the BPS )=1L c)= FDG.

Again, LFDPS LGDP+FdG=} Lc+the 2 of elevation DPB + ] Zc; therefore _ FDP= Lc+the Lof elevation DPB; to each of these equals add the 2 DPB, then the 2 of depression Fort the L of elevation DPB = LC + twice that of elevation DPB; hence (Zof depr. FDB + L.of elev. DPB)— LC

- the true Ľ of elevation.

When the angles are both elevations or both depressions, their difference must be diminished by the Lc, and half the remainder will be the true 2 of elevation of the higher of the two objects.

The Lc is generally very small, and where the measured base does not exceed six or seven hundred yards, it may

be rejected.

EXAMPLE XLVII. (R) Suppose D and p to be two objects fixed exactly at the same height above the ground as the height of the telescope of the theodolite; now if the 2 FDP of depression be 26', and the 2 DPB of elevation 14', what will be the error in observation? The arc Ps, or distance of the stations, being 8000 feet.

The length of a degree in latitude 510.9ʻis 364950 feet t; 364950 feet : 60'::8000 feet : 1'.19" nearly = L C. Then (26' + 14')-1.19"

=19.20"£ the true 2 of elevation DPB, hence 2

The angle formed between the tangent of any arc and its chord, is measured by kalf that arc.

The 2TBF (Plate I, fig. 1.) is measured by half the arc Bp. For the LTBF is equal to the 2 fbb in the alternate segment (Euclid 33 of III.), and the Z FØB = LPCB (Euclid 20 of III.) therefore the LTBF=Z FCB; but the FCB is measured by the arc By, therefore the LTBF is measured by half the arc Bs.

it Trigonometrical Survey of England and Wales, Vol. II. Part II.

Q. E. D.

page 113.

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