The system of rules and operations by which the relative position of any number of points in a tract of country may be determined, so that it may be delineated on a plane surface, is called surveying. When the extent of country is not great, the subject involves little difficulty; but when a kingdom, such as Britain or France, is to be surveyed, in addition to some of the more profound theories of pure mathematics, the aid of astronomy, and other branches of natural philosophy, is required; but such an extensive view of the subject will not be expected in this work.
As a surveyor has perpetual occasion for calculation, it is necessary that he be thoroughly master of arithmetic, and understand the nature of logarithms, the use of logarithmic tables, and algebraic notation. As it is his business to investigate and measure lines and angles, and to describe them on paper, he should be well acquainted with the elements of geometry and trigonometry, and with the application of their principles to the mensuration of heights, distances, and surfaces. In particular, he should be familiar with the best practical methods of solving the ordinary geometrical problems, and should be expert in drawing lines and describing figures. He should know something of the principles of geology, optics, and magnetism, and possess some skill in drawing and painting.
In a survey, the most remarkable objects, such as the summits of hills, spires, towers, &c., must be chosen as stations, and, if necessary, marked by signals. These must be considered as joined by straight lines forming a chain of triangles, each connected with all the others. The angles of the triangles should neither be very acute nor very obtuse; their sides should be as long as possible, so as to admit of two of the stations, at the angles, being seen from the third; the nearer each triangle approaches to the equilateral form, the better. The three angles of each triangle should be measured, if possible, for although two of them be sufficient to determine the third, yet it will conduce to accuracy if all the three be taken; and if their sum hardly differ from $180^\circ$, it may be supposed that they have been correctly determined. The principal points should also be intersected as much as possible by lines from different stations, to ascertain whether different data give the same position.
A theodolite is the most convenient instrument for measuring the angles, because it gives them at once reduced to the plane of the horizon. When a sextant is used, the angles, if out of that plane, must be reduced to it by calculation.
Supposing a proper disposition of the triangles to have been made, when their angles are known, if a side of any one of them were also known, then the sides of all the others might be found by calculation, and a plan of the country constructed. Therefore, a side of one of the triangles must be taken as a base, and measured with great care, for upon this the just determination of the absolute length of all the other lines depends; and if the survey be extensive, it will be proper to assume another base, called a base of verification, and compare its measured with its calculated length. The base should be as long as possible, and lie on a flat surface; if it is not perfectly level, its slope must be measured, and its horizontal length calculated. In ordinary cases, the base may be measured by a chain or tape, but more accurately by stretching a rope, 100 yards or more, tight in its direction, and applying repeatedly to it a twenty-feet deal-rod. A pin may be stuck in the rope at the fore-end of the rod, each time it is removed, and before the rope is removed to a new position: a point directly under the end of the last rod may be marked on the top of a peg fixed in the ground. In the trigonometrical survey of Britain, which has been going on ever since 1784, under the direction of the Board of Ordnance, an original base of about five miles was measured three times over, first with twenty-feet deal-rods, then with glass rods of the same length, and at a later period by a hundred feet steel chain of a particular construction. Several bases of verification have Surveying also been measured. The labour and ingenuity exerted in these operations have been very great.
As, generally speaking, all bodies expand by heat and contract by cold, a rod or chain varies in absolute length as it varies in temperature. A steel rod at the temperature of water boiling is the one thousandth part longer than at the temperature of water freezing. Hence, when great accuracy is required, the temperature of the measure must be noted, that its precise length may be known.
When the sides of the primary triangles are known, they serve as bases by which the situation of objects near them may be determined. In this way, the position of any number of points in a road, or the course of a river, also the situation of towns, country-seats, the boundaries of enclosures, and every feature of a country, may be found, and laid down in a plan or map; and to construct it with accuracy, the principal points should be laid down by their calculated distances from each other, rather than by the measures of the angles formed by the lines joining them, because a point may be more correctly laid down by the intersection of two arcs of circles, than by that of lines which make given angles with a third line.
A small extent of the earth's surface may be regarded as a plane, and lines perpendicular to it as parallel to one another. However, in an extensive survey, such as that of Britain, the curvature of the earth must be taken into account, and then its figure and magnitude enter as elements into all the calculations.
This connection between the figure of the earth and the magnitude and position of lines traced on its surface affords, reversely, the means of determining the former when the latter are known; so that such surveys, besides their immediate object, are applicable to the solution of the still more sublime problem of finding the magnitude and figure of the earth itself.
Of Terrestrial Refraction.
It is a matter of experience, that the rays of light proceeding from the heavenly bodies are bent from their original rectilineal direction in passing through the atmosphere. A ray in its progress may be considered as passing through a very great number of thin strata of air, which are denser the nearer they are to the earth; and these, by their action on the particles of light, bend the ray downwards, so that, in fact, its path is a curve line concave towards the earth, and situated in a vertical plane passing through the luminous object and the eye of the observer.
To understand the nature of this effect, let AB (fig. 1) represent a portion of the earth's surface, and DFG the upper boundary of the atmosphere: a succession of particles of light from the sun or a star S, proceed on a straight line, until they arrive at F; afterwards, in their passage through the atmosphere, their path is gradually bent into the curve FE, so that when they reach the eye at E, the object S appears as if it were at S', in the direction of a straight line which touches the curve at E.
This bending of the ray is called refraction; it increases the apparent elevation of the heavenly bodies above the horizon, except when they are in the zenith, and in that position there is no refraction.
The rays of light, by which terrestrial objects are rendered visible, are, in their passage through the atmosphere, bent downwards, exactly as those proceeding from the heavenly bodies; so that the apparent elevation of a remote object is always greater than its true elevation. This incursion is called terrestrial refraction; in the case of the heavenly bodies, it is astronomical refraction.
Terrestrial refraction varies with the state of the atmosphere, so that an object appears more elevated at one time than at another: the displacement, however, is always in a vertical plane, but never sensibly in a horizontal direction. How it may be found for any case will appear from the following problem.
Prob. I.
Having given the apparent positions of two remote stations, as seen from each other at the same instant, to determine the error produced by refraction.
Let A and B be the stations (fig. 2), and C the earth's centre; draw the lines CA, CB, and produce them towards Z and V, the zeniths of the stations; join AB; then the true zenith distance of B, as seen from A, is the angle ZAB, and the true zenith distance of A, as seen from B, is the angle VBA. These, however, cannot be directly measured, for, by refraction, the point B, as seen from A, appears elevated to the position b; and the point A, as seen from B, is elevated to the position a. The errors, then, produced by refraction are the angles bAB and aBA, and these will be nearly equal, if the angles be observed at the same instant, which may be done by setting two watches to the same time, or making a signal at one station so as to be seen from the other.
Put the greater apparent zenith distance ZAb = d, the lesser VBa = d', the refraction bAB = aBA = r, the angle C at the earth's centre = C.
Then ZAB = ABC + C, and VBA = BAC + C (Geom. 23, 1), therefore ZAB + VBA = ABC + BAC + C + C.
But ABC + BAC + C = 180°, therefore ZAB + VBA = 180° + C.
Again, ZAB = ZAb + bAB = d + r, VBA = VBa + aBA = d' + r, therefore ZAB + VBA = d + d' + 2r.
Put the two values of ZAB + VBA equal to each other, and we have
\[d + d' + 2r = 180° + C,\] and \(r = 90° + \frac{1}{2}C - \frac{1}{2}(d + d').\)
Ex. In the British survey, on Wisp Hill and Cross Fell are two stations in a triangle, which connects the north of England with the borders of Scotland. Their distance is computed at 235,018.6 feet = 44.511 miles, this corresponds on the surface of the earth to an arc of 38° 33' 7". From Cross Fell, Wisp Hill was seen depressed 30° 48" below the horizon, and from the latter place the former was found to have a depression of 2° 31". Here we have
\[d = 90° 30' 48",\] \[d' = 90° 2° 31",\] \[C = 38° 33' 7";\] hence
\[r = 90° + \frac{1}{2}C - \frac{1}{2}(d + d') = 2° 37' 8".\]
In this case the error produced by refraction is nearly \(\frac{1}{12}\)th of the arc intercepted between the stations. Dr Maskelyne reckoned it to be \(\frac{1}{15}\)th. Delambre and Legendre, French mathematicians, estimated it, the former at \(\frac{1}{12}\)th, and the latter at \(\frac{1}{11}\)th; and Col. Mudge, by numerous and correct observations, found its medium value to be about \(\frac{1}{12}\)th. In the examples of the mensuration of heights, the vertical line to be measured has been supposed to stand on the horizontal plane of the station, where the angle of elevation was taken; but in estimating the difference between the heights of two stations which are at a considerable distance from each other, this is not exactly true: in such a case it is necessary to correct the observed vertical angle, on account of the earth's curvature.
**Prob. II.**
Having given the distance between two stations, and the elevation or depression of the one as seen from the other, to determine the correction to be made in the vertical angle on account of the earth's curvature.
Let A and B be the two stations (fig. 3), and C the earth's centre. Draw the horizontal line AH in the plane of the triangle ABC, then BAH will be the apparent angle of elevation or depression of B, according as it is above or below the horizon of A.
Take CD = CA, and join AD; the line BD is the difference between the heights of the stations A and B, and to determine BD, the vertical angle BAD must be known; now this angle differs from the apparent elevation BAH by the angle HAD, therefore HAD is the correction of the vertical angle, depending on the earth's curvature.
In the isosceles triangle ACD, the sum of its angles, that is, 2CAD + C, is equal to two right angles (Geom. II and 24, I), and therefore CAD + \(\frac{1}{2}C\) is one right angle; but CAH is a right angle, therefore CAH = CAD + \(\frac{1}{2}C\), and taking away the common angle CAD, there remains HAD = \(\frac{1}{2}C\). Hence the correction on account of the earth's curvature is half the arc intercepted between the stations.
**Cor.** If the station B be above the horizon, as seen from A, the corrected vertical angle will be the sum of its apparent elevation and half C, the angle contained by the verticals passing through the stations; but if B be below the horizon, then it will be the excess of half C above the apparent depression.
**Ex.** In the British trigonometrical survey, the distance between the stations on Wisp Hill and Cross Fell was found to be 235,018-6 feet, which, reckoning 6094\(\frac{1}{2}\) feet to a minute, corresponds to 38° 33' 57" on the earth's surface. From Wisp Hill, Cross Fell appeared depressed below its horizon 2° 31' 3"; but, by a corresponding observation at the other station, the error arising from refraction was estimated at 2° 37' 3". Hence the difference between the heights of the stations is required.
Let A be Wisp Hill, B Cross Fell. The observed depression HAB was 2° 31' 3"; but as the line AB was elevated 2° 37' 3" by refraction, the correct value of HAB was 2° 31' 3" - 2° 37' 3" = 5' 58" 3". This subtracted from HAD = \(\frac{1}{2}C\) = 19° 16' 8", leaves 14° 8' 5" for the vertical angle BAD. And because in the triangle BAD the angle D is almost a right angle, therefore BD will be found by this proportion; rad.:tan.BAD(14°8'5"):AD(235,018'6):BD=966-8feet, the height of the station on Cross Fell above that on Wisp Hill.
On a survey, it will sometimes happen that the instrument cannot be conveniently placed at the very centre of a station, in order to determine the angles subtended by remote objects: it must then be placed at a point near the station, and the angles taken at that point must be corrected by calculation, so as to reduce them to the centre. How this may be done is shown in the next problem.
**Prob. III.**
To reduce an angle taken out of a station to the centre of the station.
Let C be the centre of a permanent station (fig. 4), where the angle ACB, subtended by two remote objects A and B, is to be determined; let O be a given point at a little distance, where the instrument is placed, and the angle AOB actually measured; then, having given the distance CO, and the angles COA, COB; also the distances CA, CB, or at least their values nearly, it is required to find the difference between the angles ACB, AOB.
Let I be the intersection of AC and BO; and because the angle AIB is the sum of ACB and CBO, also the sum of AOB and CAO (Geom. 24, I); therefore
\[ ACB + CBO = AOB + CAO, \] and \( ACB - AOB = CAO - CBO. \)
Now, in the triangles COA, COB,
\[ CA : CO :: \sin. COA : \sin. CAO, \] \[ CB : CO :: \sin. COB : \sin. CBO. \]
From these proportions, the angles CAO, CBO may be found, and their difference, which is also the difference of the angles ACB, AOB will be known.
It is useful in practice to have a formula that expresses the difference of the angles BCA, BOA in minutes of a degree. For this purpose, put the angles ACB = C, AOB = O, BOC = v; then COA = O + v; also put the lines CA = m, CB = n, OC = d; then the two proportions become
\[ m : d :: \sin.(O + v) : \sin. CAO ; n : d :: \sin. v : \sin. CBO ; \]
hence sin. CAO = \(\frac{d}{m}\) sin. (O + v), sin. CBO = \(\frac{d}{n}\) sin. v.
But small angles being almost proportional to their sines, it follows that the number of minutes in the angle CAO will be \(\frac{\sin. CAO}{\sin. 1'}\) nearly; and in like manner the number of minutes in the angle CBO will be \(\frac{\sin. CBO}{\sin. 1'}\) nearly; therefore
\[ CAO = \frac{d \sin.(O + v)}{m \sin. 1'}, \quad CBO = \frac{d \sin. v}{n \sin. 1'}; \]
hence, since CAO - CBO = C - O, we have
\[ C - O = \frac{d \sin.(O + v)}{m \sin. 1'} - \frac{d \sin. v}{n \sin. 1'}. \]
This is the correction of the angle O expressed in minutes of a degree; in its application, attention must be paid to the signs of \(\sin.(O + v)\) and \(\sin. v\), agreeably to the rules laid down in *Arithmetic of Sines*, Algebra (225).
A different and more simple expression for the correction of the angle O may be found as follows:
Let a circle be described about the triangle ABC, meeting BO in H, and join CH, AH. Let the angles BCA, BOA, BOC, and the lines CA, CB, CO, be denoted by the same letters as before; and in addition, put the angle BAC = A. Surveying. Because
\[ C - O = BCA - BOA = BHA - BOA \] (Geom. 24, I),
and \( HAO = BHA - BOA \),
therefore \( C - O = HAO \).
In the triangle \( OHc \), \( \sin CHO = \sin CHB : \sin HCO : CO : HO \);
and in the triangle \( OHA \), \( HA : HO : \sin HOA : \sin HAO \).
Remarking now that \( CHB = CAB = A \), and that \( HCO = CHB - COB = A - v \), also that \( HA \) and \( CA = m \) are nearly equal, these proportions may be expressed thus:
\[ \frac{\sin A}{m} : \frac{\sin (A - v)}{HO} :: d : HO, \]
\[ \frac{\sin O}{m} : \frac{\sin O}{\sin HAO}; \]
hence \( HO = \frac{d \sin (A - v)}{\sin A} \),
and \( \sin HAO = \frac{HO \times \sin O}{m} = \frac{d \sin (A - v) \sin O}{m \sin A} \).
But, on account of the smallness of the angle \( HAO \), the number of minutes it contains will be \( \frac{\sin HAO}{\sin I'} \) nearly;
therefore, because \( C - O = HAO \), we have, in minutes of a degree,
\[ C - O = \frac{d \sin (A - v) \sin O}{m \sin A \sin I'}. \]
This expression for the difference of the angles \( C \) and \( O \) is not quite so accurate as the former, yet, in practice, it is near enough the truth. It requires that approximate values of the distance \( AC \) and the angle \( BAC \) be known; the distance \( d \) should, however, be accurately determined.
It is obvious that if the instrument were placed at \( H \), in the circumference of a circle passing through \( A, B, C \), the observed angle \( AHB \) would then be equal to the angle \( C \) at the station. This may be done by moving it along \( OB \) until the angle \( CHB \) is found to be equal to \( CAB \), and then no correction is wanted.
Ex. In a survey, \( A, B, C \) are three stations, and \( AB = 6196 \) feet. At \( A \), the angle \( BAC = 63^\circ 44' \); and at \( O \), a point 115 feet from the centre of the station \( C \), the angle \( AOB = 45^\circ 42' \); the angle \( COB = 50^\circ 18' \). Hence it is required to find the angle \( ACB \).
The angle \( ACB \) will be nearly equal to \( AOB = 45^\circ 42' \), hence \( CBA = 70^\circ 34' \) nearly. In the triangle \( CAB \) we have now approximate values of all the angles, and a side \( AB \); hence (by the first case of oblique-angled triangles) \( BC = 7763 \), and \( AC = 8164 \) feet.
To apply the formulae for reducing to the centre, we have
\[ \begin{align*} m &= 8164 & O &= 45^\circ 42' & O + v &= 96^\circ 0' \\ n &= 7763 & v &= 50 18 & A - v &= 13 26 \\ d &= 115 & A &= 63 44 \end{align*} \]
to find \( C - O \). The logarithmic calculation by the first formula may stand thus:
\[ \begin{align*} \frac{d}{\sin (O + v)} &= 1.060698 \\ \frac{m}{\sin O} &= 9.997614 \\ \frac{n}{\sin A} &= 6.088097 \\ \frac{d}{\sin I'} &= 3.536274 \end{align*} \]
Hence \( C - O = 4^\circ 816 - 3^\circ 918 = 0^\circ 898 = 54'' \) nearly; this gives \( C = 45^\circ 42' 54'' \).
Calculation by the second formula.
\[ \begin{align*} \frac{d}{\sin (A - v)} &= 1.060698 \\ \frac{m}{\sin O} &= 9.366075 \\ \frac{n}{\sin A} &= 9.854727 \\ \frac{d}{\sin I'} &= 6.088097 \\ \frac{d}{\sin A} &= 0.047331 \\ \frac{d}{\sin I'} &= 3.536274 \end{align*} \]
\( C - O = 0^\circ 898 = 54'' \)
The correction is positive, because the angle \( A - v \) is less than 180°, and therefore \( \sin (A - v) \) is a positive quantity.
When a theodolite is employed in surveying, the angles are taken at once in a horizontal plane; but when a sextant is used, the angles are measured in the planes of the objects, and if they are oblique, the corresponding horizontal angles are found by calculation.
**Prop. IV.**
Having given the inclination of each of two lines to the horizon, and the oblique angle they contain, to find the corresponding horizontal angle.
Let \( AB, AC \) be the straight lines, which contain between them the given oblique angle \( BAC \); in \( AV \), a perpendicular to the horizon, take any point \( H \), and let a horizontal plane pass through \( H \) and meet \( AB, AC \) in \( B \) and \( C \); join \( HB, HC, BC \); then \( BHC \) is the horizontal angle corresponding to the given oblique angle \( BAC \), and \( HAB, HAC \) the complements of the inclinations of the lines \( AB, AC \) to the horizontal plane.
Put the angles \( HAB = b, HAC = c \), the given oblique angle \( BAC = A \), and its corresponding horizontal angle \( BHC = H \).
Then, supposing \( AH \) to be the radius of a circle, and \( I = 1 \), it is evident that in the right-angled triangles \( AHB, AHC \),
\[ AB = \sec b, HB = \tan b, AC = \sec c, HC = \tan c. \]
Now, in the triangles \( ABC, HBC \), by Trigonometry,
\[ AB^2 + AC^2 = BC^2 + 2AB \times AC \times \cos A, \]
\[ HB^2 + HC^2 = BC^2 + 2HB \times HC \times \cos H; \]
Hence, by subtracting equals from equals, and observing that \( AB^2 - HB^2 = AH^2 \), and \( AC^2 - HC^2 = AH^2 \), we have
\[ 2AH^2 = 2AB \times AC \times \cos A - 2HB \times HC \times \cos H, \]
that is,
\[ 1 = \sec b \sec c \cos A - \tan b \tan c \cos H. \]
From this expression, after substituting \( \frac{1}{\cos b} \) and \( \frac{1}{\cos c} \) for \( \sec b \) and \( \sec c \), also \( \frac{\sin b}{\cos b} \) and \( \frac{\sin c}{\cos c} \) for \( \tan b \) and \( \tan c \), we get
\[ \sin b \sin c \cos H = \cos A - \cos b \cos c. \]
Now \( \cos H = 1 - 2 \sin^2 \frac{1}{2} H \) (Algebra, 248), hence, by substituting and transposing, we get this other expression,
\[ 2 \sin b \sin c \sin^2 \frac{1}{2} H = \cos b \cos c + \sin b \sin c - \cos A. \]
Again, \( \cos b \cos c + \sin b \sin c = \cos (b - c) \) (239),
and \( \cos (b - c) - \cos A = 2 \sin \frac{A + (b - c)}{2} \sin \frac{A - (b - c)}{2} \) (240); therefore,
\[ 2 \sin b \sin c \sin^2 \frac{1}{2} H = 2 \sin \frac{A + (b - c)}{2} \sin \frac{A - (b - c)}{2}. \]
If we put \( \frac{A + b + c}{2} = s \), then \( \frac{A + (b - c)}{2} = s - c \),
\[ \frac{A - (b - c)}{2} = s - b, \]
and
\[ \sin b \sin c \sin^2 \frac{1}{2} H = \sin (s - b) \sin (s - c), \]
and hence
\[ \sin \frac{1}{2} H = \sqrt{\frac{\sin (s - b) \sin (s - c)}{\sin b \sin c}}. \]
Note. In this formula, the sines are supposed to be computed to a rad. = 1, but in the table of log. sines, the rad. is a number of which the log. is 10; therefore, to adapt the formula to the table, we must divide each sine Surveying, to rad. = 1 by this number, so that calling the number R, we have
\[ \sin \frac{1}{2} H = R \sqrt{\frac{\sin (s - b) \sin (s - c)}{\sin b \sin c}} \]
When the lines which contain the given oblique angle have their inclinations to the horizontal plane less than 2° or 3°, the general solution does not conveniently apply. In this case, instead of seeking the horizontal angle directly, it is better to find its difference from the oblique angle.
To investigate a rule, let the horizontal triangle BHC (see last figure) be applied upon the plane of the oblique triangle BAC, so that they may still have a common base BC; join AH, producing it to K, and draw HP, HQ perpendicular to AB, AC. And because BHK = BAH + ABH, and CHK = CAH + ACH, therefore BHC = BAC + ABH + ACH, and putting \( d \) for the difference BHC - BAC, we have
\[ d = ABH + ACH. \]
Put \( r \) for AH, \( m \) for the angle HAP, and \( n \) for the angle HAQ; then \( HP = r \sin m, HQ = r \sin n, AP = r \cos m, AQ = r \cos n \), and, by the formulae of Art. (246) Algebra,
\[ \begin{align*} HP + HQ &= \sin m + \sin n \\ AQ + AP &= \cos n + \cos m = \tan \frac{1}{2}(m + n) = \tan \frac{1}{2}A, \\ HP - HQ &= \sin m - \sin n \\ AQ - AP &= \cos n - \cos m = \cot \frac{1}{2}(m + n) = \cot \frac{1}{2}A; \end{align*} \]
therefore \( HP + HQ = (AQ + AP) \tan \frac{1}{2}A, \)
\[ \frac{HP - HQ}{2} = (AQ - AP) \cot \frac{1}{2}A; \]
and, by adding and subtracting,
\[ \frac{HP}{2} = \frac{AQ + AP}{2} \tan \frac{1}{2}A + \frac{AQ - AP}{2} \cot \frac{1}{2}A, \]
\[ \frac{HQ}{2} = \frac{AQ + AP}{2} \tan \frac{1}{2}A - \frac{AQ - AP}{2} \cot \frac{1}{2}A. \]
Supposing \( b \) and \( c \) to denote the same things as in the general solution, put \( b' \) for the difference between \( b \) and a right angle, and \( c' \) for the difference between \( c \) and a right angle; then \( b' \) and \( c' \) are the inclinations of the sides of the given oblique angle to the horizontal plane.
The lines BA, BH are nearly equal, because the one is the secant and the other the tangent of an angle nearly = 90°; and the same is true of the lines CA, CH; hence the points A and H will be near each other, and the angles ABH, ACH will be small, and BP = BH nearly, also CQ = CH nearly; and since
\[ BA = \sec b = \csc b' = \frac{1}{\cot \frac{b'}{2} + \tan \frac{b'}{2}}, \]
\[ BH = \tan b = \cot b' = \frac{1}{\cot \frac{b'}{2} - \tan \frac{b'}{2}} \]
(249);
therefore, by subtracting, \( AP = BA - BH = \tan \frac{b'}{2}. \)
In like manner it appears that \( AQ = CA - CQ = \tan \frac{c'}{2}. \)
These values of \( AP \) and \( AQ \) being substituted in the values of \( HP \) and \( HQ \), we have
\[ \frac{\tan \frac{c'}{2} + \tan \frac{b'}{2}}{2} \tan \frac{1}{2}A + \frac{\tan \frac{c'}{2} - \tan \frac{b'}{2}}{2} \cot \frac{1}{2}A, \]
\[ \frac{\tan \frac{c'}{2} + \tan \frac{b'}{2}}{2} \tan \frac{1}{2}A - \frac{\tan \frac{c'}{2} - \tan \frac{b'}{2}}{2} \cot \frac{1}{2}A. \]
Now, in the triangles HBP, HCQ, because HB = cot \( b' \)
\[ \frac{1}{\tan \frac{b'}{2}} \text{ and } HC = \cot \frac{c'}{2} = \frac{1}{\tan \frac{c'}{2}}, \]
we have
\[ \sin HBP = \frac{HP}{HB} = HP \times \tan \frac{b'}{2}, \]
\[ \sin HCQ = \frac{HQ}{HC} = HQ \times \tan \frac{c'}{2}; \]
therefore, by adding, and observing that the sum of the sines of the angles HBP, HCQ, is, by reason of their smallness, almost equal to twice the sine of half their sum, that is, to twice the sine of \( \frac{1}{2}(H - A) = \frac{1}{2}d \), we have
\[ 2 \sin \frac{1}{2}d = HP \times \tan \frac{b'}{2} + HQ \times \tan \frac{c'}{2}. \]
In this expression, substitute the values already found for HP and HQ, and the result will be
\[ \left[ \frac{1}{2}(\tan \frac{c'}{2} + \tan \frac{b'}{2})(\tan \frac{c'}{2} + \tan \frac{b'}{2}) \tan \frac{1}{2}A, \right] \]
\[ - \frac{1}{2}(\tan \frac{c'}{2} - \tan \frac{b'}{2})(\tan \frac{c'}{2} - \tan \frac{b'}{2}) \cot \frac{1}{2}A. \]
The approximation will be sufficiently accurate, and more simple, if we put the arcs instead of the sine and tangents, to which they are almost equal. The formula then becomes
\[ d = \left( \frac{c' + b'}{2} \right) \tan \frac{1}{2}A - \left( \frac{c' - b'}{2} \right) \cot \frac{1}{2}A. \]
In this formula, the arcs or angles \( d, b', c' \) are expressed in parts of the radius. If we suppose the same letters to express the number of minutes in each, then, observing that an arc of 1° = 0.0029888, in parts of the radius, we must substitute \( d \times 0.0029888 \), or, which is the same thing, \( \frac{d}{3438} \) instead of \( d \), and similarly, \( \frac{b'}{3438} \) and \( \frac{c'}{3438} \) instead of \( b' \) and \( c' \).
This being done, and the quantities \( \tan \frac{1}{2}A \) and \( \cot \frac{1}{2}A \) being divided by \( R \), the radius, for the convenience of logarithmic calculation, we have
\[ d = \left\{ \frac{(c' + b')^2 \tan \frac{1}{2}A}{R} - \frac{(c' - b')^2 \cot \frac{1}{2}A}{R} \right\} \frac{1}{3438}. \]
Note. We have supposed the angles \( b', c' \) to be both elevations. If one be a depression, the formula will still hold true, provided the arc of depression be regarded as a negative quantity. This may be inferred from the position of the lines, also from the rules for the signs of the cosine and sine of an arc (Algebra, 225).
Ex. 1. The angles of elevation of two straight lines, which contain an oblique angle of 64° 10', are 6° 20' and 8° 46': Find the corresponding horizontal angle.
In this case \( A = 64° 10' \). Log. calcu. by the 1st formula.
\[ \begin{align*} b &= 90° - 6° 20' = 83° 40' \sin b \\ c &= 90° - 8° 46' = 81° 14' \sin c \\ \end{align*} \]
Ar. Co. \( \left\{ \begin{array}{l} 0.002659 \\ 0.005104 \\ \end{array} \right\} \)
\[ \begin{align*} s &= 114° 32' \\ s - b &= 30° 52' \sin (s - b) = 9710153 \\ s - c &= 33° 18' \sin (s - c) = 9739590 \\ \end{align*} \]
\[ 2)19-457506 \]
\[ \sin \frac{1}{2}H = 32° 22' 42'' = 9728758 \]
The horizontal angle \( H = 64° 45' 24'' \), the answer.
As the ar. comp. of two logarithms are added, we ought to reject 20 from the sum, or 10 from half the sum; but the formula requires that 10, the log. of rad., be added to half the sum. As these operations compensate each other, both are here omitted.
Ex. 2. As an example of the second formula, let \( A = 97° 36' \); \( c' \), an angle of elevation, = 1° 30'; \( b' \), an angle of depression, = 1° 6': Find the horizontal angle \( H \).
Here \( c' = + 90' \), \( b' = - 66' \), \( c' + b' = 90' - 66' = 24' \), \( c' - b' = 90' + 66' = 156' \). The calculation by logarithms may be thus: Here 20 is subtracted from each sum, viz. 10 because of the ar. co. of the log. that is added, and 10 because the rad. is a divisor of each term of the formula. The calculation gives \( d = 0^{\circ}48 - 1^{\circ}549 = -1^{\circ}501 = -1^{\circ}30' \), a negative quantity, and the required horizontal angle is \( 97^{\circ}36' - 1^{\circ}30' = 97^{\circ}34'30'' \).
**Scholium.** If a station be taken at a known height above a horizontal plane, and the oblique angles subtended by any number of objects be measured, also their depressions, the corresponding horizontal angles may be found by this problem. Their horizontal distances may also be found, and their places laid down in a map of the country.
When a tract to be surveyed has been covered with a series of triangles, so as to connect the principal points, and all the angles of each, and a side of one, are known, the sides of all the triangles may be found, and a plan made, by constructing the triangles on the sides of each other; but in a plan constructed by this method, an error in the position of one triangle must necessarily affect the positions of all the others that depend on it.
To avoid this inconvenience, we may determine the position of a side of one of the triangles in respect of a meridian, or line drawn due north and south, by a compass, or more accurately by astronomical observations, and then calculate the distances of all the stations from the meridian, and also the distances from each other reduced to the direction of the meridian: from these, the position of each station may be laid down in the plan independently of the others, and also the direct distance between any two points may be easily found. The manner of proceeding will appear from the following example.
Let A, B, C, D, E, F, be six stations connected by four triangles ABC, BCD, BDE, EDF; the angles are
\[ \begin{align*} BAC &= 79^{\circ}20' & CBD &= 39^{\circ}20' \\ ABC &= 51^{\circ}31' & BCD &= 69^{\circ}28' \\ ACB &= 49^{\circ}9' & BDC &= 71^{\circ}12' \\ \end{align*} \]
sum \(= 180\) sum \(= 180\)
\[ \begin{align*} DBE &= 45^{\circ}28' & EDF &= 62^{\circ}3' \\ BDE &= 72^{\circ}3' & DEF &= 52^{\circ}25' \\ BED &= 62^{\circ}29' & DFE &= 65^{\circ}32' \\ \end{align*} \]
sum \(= 180\) sum \(= 180\)
A side AB of one of the triangles is 4213 yards, and it makes with NS, the meridian, an angle of \(62^{\circ}52'\) at the point A. Find the points in which perpendiculars from the stations cut the meridian, and the length of each perpendicular.
Draw Bb, Ce, Dd, Ea, Ef perpendicular to the meridian, and Bu, Dq, parallel to it, forming the right-angled triangles ABb, BCm, BDu, EDp, FDq. Because the angles of the four triangles ABC, BCD, BDE, DEF are given, and also AB, a side of one of them, the five lines AB, BC, BD, DE, DF may be found from AB and each other (by case I, oblique-angled triangles). Their logarithms will be
AB \(= 624591\), BD \(= 733559\), DF \(= 575833\),
BC \(= 738255\), DE \(= 638690\).
In the right-angled triangle ABb, the side AB (or its log.) and the angle BAb \(= 62^{\circ}52'\) are known; hence we find
\[ \begin{align*} Ab &= 1921^{\circ}4, Bb &= 3749^{\circ}3 \text{ yards}. \end{align*} \]
If from ABm \(= 117^{\circ}8'\) (the sup. of BAb) the angle ABC \(= 51^{\circ}31'\) be taken, there remains CBm \(= 65^{\circ}37'\). Surveying. Therefore, in the right-angled triangle CBm, the angles and the log. of BC are now known; hence we find Bm \(= 2259^{\circ}6\), Cm \(= 4985^{\circ}2\).
And if from CBn \(= 65^{\circ}37'\), CBD \(= 39^{\circ}20'\) be subtracted, there remains DBn \(= 26^{\circ}17'\). Therefore, in the right-angled triangle DBn, the angles, and the log. of BD, are known; and hence
\[ \begin{align*} Ba &= 4854^{\circ}7, Dn &= 2397^{\circ}6. \end{align*} \]
From BDq \(= 153^{\circ}43'\) (the sup. of BDn) subtract BDE \(= 72^{\circ}3'\), and there remains EDp \(= 81^{\circ}40'\); then, in the right-angled triangle EDp, the angles, and the log. of DE, are known, and hence Dp \(= 630^{\circ}7\), Ep \(= 4306^{\circ}1\).
Lastly, subtracting EDF \(= 62^{\circ}31'\) from EDp \(= 81^{\circ}40'\), there remains FDq \(= 19^{\circ}37'\); and hence, in the right-angled triangle FDq, we find Dq \(= 3569^{\circ}1\), Fq \(= 1272^{\circ}1\).
To determine the stations, we have now
\[ \begin{align*} \text{Yards.} & \quad \text{Yards.} \\ Ab & \quad = 1921^{\circ}4, Bb & \quad = 3749^{\circ}3 \\ Ac & \quad = Ab + Bm & \quad = 4181^{\circ}0, Cc & \quad = Cm - Bb = 1235^{\circ}9 \\ Ad & \quad = Ab + Bn & \quad = 6776^{\circ}1, Dd & \quad = Bb - Dn = 1851^{\circ}7 \\ Ae & \quad = Ad + Dp & \quad = 7406^{\circ}8, Ea & \quad = Dd + Ep = 5657^{\circ}8 \\ Af & \quad = Ad + Dq & \quad = 10345^{\circ}2, Ef & \quad = Dd + Fq = 2629^{\circ}8. \end{align*} \]
By these numbers the position of each station may be laid down with great accuracy in a plan, independently of the others: also the distance between any two may be readily found; for example,
\[ CE = \sqrt{(Cc + Ee)^2 + (Ae - Ac)^2} \quad (\text{Geom. 13, 2}). \]
**Of Surveying with a Compass.**
When a tract of country is to be surveyed with expedition, and no great degree of accuracy is required, the compass, notwithstanding its imperfections, is preferable to any other instrument. Even in a correct survey, it may be applied with advantage in filling up the less important parts of a plan.
The principle which renders the compass applicable to surveying is generally known. The instrument should admit of being fixed in a horizontal position on the top of a staff, and it should be furnished with two sights diametrically opposite to each other. The circumference of a circle immediately under or opposite to the point of the needle should be divided into degrees, and it may be numbered both ways, beginning from the extremities of the diameter that passes through the sights; or it may be numbered all one way from \(0^{\circ}\) to \(360^{\circ}\), beginning at a point immediately under one of the sights.
The use of the instrument is to determine the bearings of objects as seen from each other, or the angles which the lines joining them make with the magnetic meridian. Thus, if the perimeter of a polygon is to be surveyed and laid down on paper, the instrument must be placed at one of its angles, so that an object at the next angle may be seen through the two sights. The needle of the compass being then allowed to turn freely on its pivot, it will settle in the magnetic meridian, and point out on the divided circle the number of degrees the line deviates from the north or south, and on which side of the meridian it lies. If this be done at each angle, the positions of all the lines in respect of the magnetic meridians passing through their intersections will be determined. The sides of the polygon may be measured in passing from one angle to another; and then, as the meridians may be supposed parallels, sufficient data will be obtained for constructing a plan of the figure, and verifying the accuracy of the survey; for any angle of the figure will manifestly be equal to the sum or difference of the angles which the lines that contain it make with the meridian passing through their intersection, according as they lie on opposite sides or the same side of that meridian. All the angles of the figure may therefore be found; and, if they have been accurately observed, their sum, together with four right angles, will make twice as many right angles as the figure has sides (Geom. 21, 1). If the survey begin and end at the same point, the positions of the beginning and end ought to coincide in the plan.
If the lines which join the objects do not form an enclosed figure, still the manner of proceeding may be the very same; and the positions of many points may be determined by observing their bearings from other points having given positions in the plan, without the trouble of measuring distances on the ground; and, in the whole process, the compass serves merely as an instrument for measuring angles.
In practice, the figure may be constructed in the field, or else a mode of registering the observations may be easily devised, from which a plan may be afterwards made; and it may prevent mistakes, if the bearings of all the objects be reckoned from the north quite round the circle in one direction, either to the right or left. Thus, proceeding to the right, we may consider all lines on the east side of the meridian between due north and due south as making with it angles between $0^\circ$ and $180^\circ$, and all lines on the west side between due south and due north as making angles between $180^\circ$ and $360^\circ$.
It must be recollected that the magnetic does not coincide with the true meridian, and that the angle between them, called the variation of the compass, changes. It varies at different times and in different places (see Magnetism). In 1823 it was $27^\circ 48'$ at Edinburgh, and $24^\circ 10'$ at London; the deviation was westerly.
This mode of surveying was applied by Richard Norwood, an English mathematician of the seventeenth century, to determine the magnitude of the earth. Having found the latitudes of London and York, he travelled from the one place to the other, measuring along the road with a chain, and taking the bearings with a compass. He says, "When I measured not, I paced; and I believe the experiment has come within a scantling of the truth."
In our article Mensuration, we have explained the principal problems by which heights and distances are determined by trigonometry: these find continual application in surveying. The theory of levelling is a branch of the same subject; this has been discussed in a particular article (see Levelling). Maritime surveying requires the solution of this problem: Three stations being given, and the angles which the lines joining them subtend at a fourth station in their plane; to determine the position of that fourth station. This is Problem V. in our article Mensuration of Heights and Distances, and is there resolved in the usual way. Since that was printed, Dr Wallace, emeritus professor of mathematics in the University of Edinburgh, has published a work entitled "Geometrical Theorems and Analytical Formulae," in which various new formulae and constructions are given for this and other geodetical problems.
On extensive trigonometrical surveying, see the following works: Delambre, Base du Système Métrique Décimal, 3 vols. 1806–10; Biot et Arago, Recueil d'Observations Géodésiques, &c. 1821; Dalby, Mudge, Colby, &c. Trigonometrical Survey of Britain, 1799–1811; Puissant, Traité de Géodésie, 2 vols. 1819.
SURVIVORSHIP. See Annuities and Mortality.