Page images
PDF

vey, we shall state a very few particulars, besides what is already mentioned in art. 12.

General Roy, who first measured the base on HounslowHeath, measured another on the flat ground of RomneyMarsh in Kent, near the southern extremity of the first series of triangles, and at the distance of more than 60 miles from the first base. The length of this base of verification, as actually measured, compared with that resulting from the computation through the whole series of triangles, differed only by 28 inches.

General Mudge measured another base of verification on Salisbury-Plain. Its length was 36574.4 feet, or more than 7 miles; the measurement did not differ more than one inch from the computation carried through the series of triangles from Hounslow-l leath to Salisbury-Plain. A most remarkable proof of the accuracy with which all the angles, as well as the two bases, were measured

The distance between Beachy. Head in Sussex, and Dun. nose in the Isle of Wight, as deduced from a mean of four series of triangles, is 339397 feet, or more than 64; miles. The extremes of the four determinations do not differ more than 7 feet, which is less than 1; inches in a mile. Instances of this kind frequently occur in the English survey *. But we have not room to specify more. We must now proceed to discuss the most important problems connected with this subject ; and refer those who are desirous to consider it more minutely to General Mudge and Colonel Colby’s “Account of the Trigonometrical Survey :” Mechain and Delambre, “Base du Système Métrigue Décimal;” Swanberg, “Ex. position des Opérations faites en Lapponie;” and Puissant's works entitled “Geodesie,” and “Traite de Topographie, d'Arpentage, &c.”

* Puissant, in his “Geodosie,” after quoting some of them, says, “Neanmoins, jusqu'à présent. rien n'égale en exactitude les opiations géodesigues qui ont servi de fondement à motre système motrique.” Ile, however, gives no instances. We have no wish to depreciate the la

hours of the French measures; but we cannot yield them the preference on there assertium.

SECTION II.

Problems connected with the detail of Operations in Extensive Trigonometrical Surveys.

PROBLEM I.

It is required to determine the most advantageous
conditions of triangles.

1. In any rectilinear triangle ABC, it is, from the propor. tionality of sides to the sines of their opposite angles, AB : Bc :: sin c : sin A, and consequently AB . sin A = Bc . sin c. Let AB be the base, which C is supposed to be measured without perceptible error, and which therefore is assumed as constant ; then finding the extremely A B

small variation or fluxion of the equation on this hypothesis,

it is AB. cos A. A = sin c. BC + Bc. cos c. c. Here, since we are ignorant of the magnitude of the errors or variations

expressed by A and c, suppose them to be equal (a probable supposition, as they are both taken by the same instrument), and each denoted by v : then will

AB cos A—nc cos c.

[ocr errors]

or, substituting ; for its equal o the equation will be.

- cos a cos c come bc = v × (Bc. in , T BC - i.); or finally, be a r. bc (cot A — cot c). This equation (in the use of which it must be recollected that t taken in seconds should be divided by R", that is, by

the length of the radius expressed in seconds) gives the error

Ic in the estimation of Bo occasioned by the errors in the angles A and c. Hence, that these errors, supposing them to be equal, may have no influence on the determination of po, we must have A = C, for in that case the second member of the equation will vanish.

2. But, as the two errors, denoted by A. and & which we have supposed to be of the same kind, or in the same direction, may be committed in different directions, when the equation will be bc = + v. Bc (cot A + cot c); we must inquire what magnitude the angles A and c ought to have, so that the sum of their cotangents shall have the least value

possible ; for in this state it is manifest that Bc will have its least value. But, by the formulae in chap. 3, we have

[ocr errors]

cos (A or c) + cos h"
* - 2 sin B

Consequently, Bc = + v. Bc. or
And hence, whatever be the magnitude of the angle B, the
error in the value of Bc will be the least when cos (A or c) is
the greatest possible, which is, when A = c.

We may therefore infer, for a general rule, that the most advantageous state of a triangle, when we would determine one side only, is when the base is equal to the side sought.

3. Since, by this rule, the base should be equal to the side sought, it is evident that when we would determine two sides, the most advantageous condition of a triangle is that it be cquilateral. 4. It rarely happens, however, that a base can be commodiously measured which is as long as the sides sought. Supposing, therefore, that the length of the base is limited, but that its direction at least may be chosen at pleasure, we proceed to inquire what that direction should be, in the case where one only of the other two sides of the triangle is to be determined. Let it be imagined, as before, that AB is the base of the triangle ABC, and BC the side required. It is proposed to find the least value of cot A + cot c, when we cannot have A = c. Now, in the case where the negative sign obtains, we have

[ocr errors]

Bc. sin B ab. sin b T as ope, sin p" This equation again manifestly indicates the equality of Ah and Bc, in circumstances where it is possible : but if AB and BC are constant, it is evident, from the form of the denominator of the last fraction, that the fraction itself will be the least, or cot A — cot G the least, when sin R is a maximum, that is, when b = 90°. 5. When the positive sign obtains, we have cot A+cot c= s/(Roo-AH3 sin” o pc3 cot A + —T-- cot a + v (viz. – 1). Here, the least value of the expression under the radical sign, is obviously when A = 90°. And in that case the first term, cot A, would disappear. Therefore the least value of cot A+ cot c, obtains when A = 90° ; conformably to the rule given by M. Bouguer (Fig. de la Terre, p. 88). But we have Vol. II. 11

already seen that in the case of cot A–cot c, we must have B = 90 : whence we conclude, since the conditions A = 90°, B = 90°, cannot obtain simultaneously, that a medium result would give A = B. If we apply to the side Ac the same reasoning as to Bc, similar results will be obtained : therefore in general, when the base cannot be equal to one or to both the sides required, the most advantageous condition of the triangle is, that the base be the longest possible, and that the two angles at the base be equal. These equal angles, however, should never, if possible, be less than 23 degrees.

- PROBLEM II.

To deduce, from angles measured out of one of the stations, but near it, the true angles at the station.

When the centre of the instrument cannot be placed in the vertical line occupied by the axis of a signal, the angles observed must undergo a reduction, according to circumstances.

1. Let c be the centre of the station, B A.

P the place of the centre of the instru. ment, or the summit of the observed angle App : it is required to find c, the measure of ACB, supposing there to be known APB = P, BPC = p, cp = d, C P Bc = L, AC = R. Since the exterior angle of a triangle is equal to the sum of the two interior opposite angles (th. 16 Geom.) we have, with respect to the triangle IAP, AIB = P + IAP ; and with regard to the triangle bic, AIB = c + chP. Making these two values of AIB equal, and transposing IAP, there results c = P + IAP – cBP. But the triangles CAP, chP, give d. sin (p+p).

- - cp .
sin cap = sin IAP = I, sin Arc = R *

- _ CP - _d , sin p sin cap == . sin bec = −. And, as the angles CAP, CBP, are, by the hypothesis of the problem, always very small, their sines may be substituted for their arcs or measures : therefore

d. sin (r-Ep) d. in p
h L

Or, to have the reduction in seconds,

[ocr errors]

sin 17 I

[ocr errors]

The use of this formula cannot in any case be embarrassing, provided the signs of sin p, and sin (P + p) be attended to. Thus, the first term of the correction will be positive, if the angle (p + p) is comprised between 0 and 180° ; and it will become negative, if that angle surpass 180°. The contrary will obtain in the same circumstances with regard to the second term, which answers to the angle of direction p. The letter R denotes the distance of the object A to the right, L the distance of the object B situated to the left, and p the angle at the place of observation, between the centre of the station and the object to the left.

2. An approximate reduction to the centre may indeed be obtained by a single term; but it is not quite so correct as the form above. For, by reducing the two fractions in the second member of the last equation but one to a common denominator, the correction becomes

C - P =to (p +p) dr sing.
Lht - -
- - R. sin A R. sin A

But the triangle ABC gives L = Tsin a T snow-Ec)"
And because P is always very nearly equal to c, the sine of
A + r will differ extremely little from sin (A + c), and may
therefore be substituted for it, making L = #.
Hence we manifestly have

[ocr errors][ocr errors]

Which by taking the expanded expressions for sin (p + p), and sin (A + F), and reducing to seconds, gives

[ocr errors]

3. When either of the distances R, L, becomes infinite, with respect to d, the corresponding term in the expression art. 1 of this problem, vanishes, and we have accordingly d : sin p d. sin (p+ p)

[ocr errors]
[ocr errors]

The first of these will apply when the object A is a heavenly body, the second when b is one. When both A and B are such, then c — P = 0. But without supposing either A or B infinite, we may have c — P = 0, or c = P in innumerable instances : that is, in all cases in which the centre P of the instrument is placed in the circumference of the circle that passes through the three points A, B, c ; or when the angle bec is equal to the angle *Ac, or to BAc + 180°. Whence though c should be inaccessible, the angle AcB may commonly be obtained by ob

« PreviousContinue »