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Volume 11, 1878
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– 132 –

Art. IX.—On the Calculation of Distances by means of Reciprocal Vertical Angles.

[Read before the Philosophical Institute of Canterbury, 12th September, 1878.]

The distance between any two points on the earth's surface may be found, if the angle subtended by those points at the centre of the earth is known,

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as it is then only necessary to multiply the number of units in the given angle by the value of one unit at the earth's surface, in order to find the distance, or “Contained Arc,” as it is generally called.

The method of deducing the subtended angle, or rather the length of the “Contained Arc,” when the vertical angle from each station to the other is given, may be shown as follows, dividing the problem into two cases—first, when one angle is an elevation; and, secondly, when both are depressions.

Case 1.—When one angle is an elevation, and the other a depression.

To investigate a method of ascertaining distances by means of reciprocal vertical angles:—Let A and B in fig. 1 represent the two stations, and C the centre of the earth. Draw A F a horizontal line at A, and B G a horizontal line at B, and B H parallel to A F. Then A B G is the true angle of depression at B, and B A F is the true angle of elevation at A, and G B AH B A (or B A F) = C (the contained arc). Thus we see that the difference between the true angles of elevation and depression is equal to the “contained arc,” and taking the mean value of 1″ on the earth's surface = 101.4 feet or 153.6 links, we could thus obtain the distance between the two stations in feet or links.

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Fig. 1.

But the observed angles are not the true angles, as they are both affected with refraction.

Let A K and B K represent the apparent direction of each station from the other.

Let D represent the true angle of depression, G B A.

Let E represent the true angle of elevation B A F.

Let C represent the angle of A C B or contained arc; and

Let R represent the angle of refraction = K A B or K B A.

G B K will be the apparent angle of depression = DR.

And K A F will be the apparent angle of elevation = E + R, and the difference between the observed angles of depression and elevation will be (DR) − (E + R) = DE − 2R = C − 2R.

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Now, assuming R to be 1/15 of the contained arc, C− 2R will be C − 2/15 C = 13/15 C.

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Therefore the difference between the observed angles of elevation and depression will be 13/15 C; so, by multiplying the number of seconds in 13/15 C by 15/13 × 153.6, we shall get the number of links in the contained arc, or the distance between A and B. (Note 15/13 × 153.6 = 177.3). If the distance between A and B is required in feet instead of links, then multiply by 15/13 × 101.4 = 117.

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Case 2.—When both angles are depressions.

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Using the same notation as before, except that D and d represent the true angles of depression, and DR, dR the observed angles of depression; then D + d + F = 2 right angles, also C + F = 2 right angles ∴ D + d = C; and DR + dR = C− 2 R. That is, the sum of both angles of depression = C − 2 R = C − 2/15 C = 13/15 C, and 13/15 C × 15/13 × 153.6 (or sum of observed depressions in seconds multiplied by 177.3) = distance in links between A and B. If the distance between A and B is required in feet, then multiply by 117 instead of 177=3.

The above results expressed in words give the following

Practical Rule.

Take the sum of the observed vertical angles when both are depressions; or their difference when one is an elevation, and reduce this sum or difference to seconds; multiply by 177.3, and the result will be the approximate distance between the two stations in links. Note.—If the distance be required in feet, then multiply by 117.

Or the following general rule will apply to all cases:—Subtract 180° from the observed zenith distances, reduce the remainder to seconds, and multiply by 177.3, the result will be the approximate distance between the two stations in links.

In the preceding investigation, I have assumed the mean value of 1″ on the earth's surface = 101.4 feet, and I shall now show what is the greatest error that can be introduced in any case by this assumption.

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The radius of curvature on the meridian varies with the latitude from a minimum at the Equator (=P2/E) to a maximum at the Pole (E2/P).

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And the radius of curvature of the Prime Vertical also varies with the latitude from a minimum at the Equator (= E.) to a maximum at the Pole (=E2/P).

Also, the radius of curvature in any latitude varies with the Azimuth from a minimum on the meridian to a maximum on the prime vertical.

Still the limits of variation are so small, compared with the ordinary errors of observation, that in general practice it is sufficient to assume 101.4 feet as the mean value of 1″ on the surface of the earth for New Zealand.

The following are the precise values for latitudes 39° and 44°, taking 39° as the mean latitude of the North Island of New Zealand, and 44° as the mean latitude of the South Island.

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Feet.
Taking Bessel's value of the equatorial radius (E) = 20923597
And Bessel's value of the polar semi-axis (P) = 20853654
The value of 1″ on the meridian at lat. 39° = 101.164
The value of 1″ on the prime vertical at lat. 39° = 101.575
∴ The mean value of 1″ at all azimuths in lat. 39° = 101.370
Again, the value of 1″ on the meridian at lat. 44° = 101.252
And the value of 1″ on the prime vertical at lat. 44° = 101.604
∴ The mean value of 1″ at all azimuths at lat. 44° = 101.428
And the mean value of 1″ at all azimuths at lat. 39° = 101.370
∴ The mean value of 1″ at all azimuths for both Islands of N.Z. = 101.399
Or say, 101.4 feet.

It will thus be seen that, by using this mean value, the results would be sometimes slightly in excess of the true values, and sometimes slightly in defect; but in any case the difference would only amount to about ⅕ per cent., and may therefore in ordinary practice be neglected.

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With regard to the co-efficient of refraction which I have adopted, it may be thought that 1/15 is too small, as in most works on surveying it is stated to be from 1/13 to 1/14.

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The reason I have used 1/15 is because I find it more in accordance with actual observations in hilly country in New Zealand.

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The factor 177.3, as stated above, is obtained by taking the value of 1″ on the earth's surface as 153.6 links, and the refraction as 1/15 of the contained are; but if it is required to obtain the distance in any other denomination, such as feet, metres, miles, etc., for any other values of terrestrial curvature and refraction, this may easily be done by means of the following formula:—

Let v = value of 1″ on the earth's surface, in the given denomination

Let m = co-efficient of refraction

Let F = the factor required;– then

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F=v/1−2m

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Example. Suppose v = 30.89 metres and m = .071 then v/1−2m = 30.89/.858 = 36, the factor required.

It must be borne in mind that this method is only approximate, as the observed vertical angles are liable to an error of 2″ or 3″ even when an 8-inch theodolite is used, and a mean of several observations taken.

Supposing the average error of each double observation to be 5″ or 6″ then the error in the calculated distance would be 5 or 6 times 177 links, say about 10 chains. This would be 1 per cent. in a distance of 1000 chains, which is the usual distance between geodesical station in New Zealand.

The chief advantage of this method is that the observations are not subject to a ratio of error in proportion to the distance. Most approximate methods, by telemeters, etc., although tolerably correct for short distances, fail altogether when applied to long distances; but this method gives pro-

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tionately better results, the longer the distance, as I estimate it as subject to an average error of 5″ or 6″ which is equivalent to about 10 chains, and this error is the same for all distances. Thus, in finding the distance between two hills 50 miles apart, this would only introduce an error of 2½ links per 10 chains, thus nearly approaching in accuracy to a chained measurement, besides being free from accidental errors and omissions which all chained measurements are liable to.

But although the errors of observation do not affect the results in proportion to the distance, still, any error in the estimated refraction will do so; therefore this method is only suitable for hilly country, where other methods are not available; as, whenever the line of sight between the two stations passes for any considerable distance close to the surface of water or level land, the refraction is generally very variable and uncertain, and the results obtained by this method will then be unreliable.

In my own practice, using an 8-inch transit theodolite, reading to 10″, and noting the level readings at each observation, the distances found by this method have an average error of half-a-chain to the mile.

For instance, in a circuit of 50 miles between two known points, average distance of stations 10 miles apart, the error was found to be 23 chains, or less than half-a-chain per mile. In another case, there was an error of 31 chains in 60 miles, or about half-a-chain per mile.

It is requisite, in this method, to use only the corrected vertical angles, that is, they must be corrected for the height of the eye and object.

Rules for calculating the correction are given in most books on surveying, but the following blank form will be convenient when the difference of heights of the eye and object is given in feet and inches, and the distance between the stations in links:—

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Blank Form.
Difference of height of eye and object in inches log
Distance between stations in links colog
Colog tang 1″—log 7.92 = constant log 4 . 4 1 5 7 0
Correction in seconds of arc = log

Note.—When the height of the eye exceeds the height of the object, the correction is to be added to an elevation or subtracted from a depression.

When the height of the object exceeds the height of the eye, the correction is to be added to a depression, or subtracted from an elevation; Or the rule for applying the corrections may be simplified thus:

Mark angles of elevation + mark angles of depression.
Mark height of eye + mark height of object.

Then take the algebraical sum of the heights of the eye and object, to compute the correction, to which prefix the same sign; then the algebraical sum of this correction, and the observed vertical angle, will give the true vertical angle.

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In order to compute this correction by the above rules, the distance between the stations is required to be known; but as in all cases where this method is used the distance between the stations is not known, we must proceed as follows:—

With the observed vertical angles, as they stand in the field-book, compute the distance between the stations; and with this approximate distance, compute the eye and object correction. Then, with the corrected angles, again compute the distance, and in most cases no further calculation will be required; but in cases where the second calculation gives a result differing greatly from the first approximation, it may be advisable to repeat the calculation.

Instead, however, of neglecting the eye and object correction altogether, in calculating the first approximation, it will be sometimes advantageous to ascertain the correction roughly, and take it into account. This may be done as follows:—

As 1 inch subtends 1″ at 26044 links or 3 ¼ miles nearly, we can easily ascertain the angle subtended by any number of inches, at any number of miles distance, by the following rule:—

Multiply the inches by 3 ¼ and divide the product by the number of miles, the quotient will be the number of seconds subtended. The distance in miles can generally be estimated to within 10 per cent. or so, and calculating the first approximate correction in this way will often save time.

Example.

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Bryant's Hill to Barker's Hill. Elev. 1° 14 13″
Bryant's Hill to Barker's Hill. Dep. 1° 22′ 50″
FT. 1N.
Bryant's Hill to Barker's Hill. Height of eye = 3 1
Bryant's Hill to Barker's Hill. " "object = 0 0
IN.
Eye exceeds object 3 1 = 37
3 ¼
Distance, say 10 miles 10)120
12.0″
FT. IN.
Barker's Hill to Bryant's Hill. Height of eye = 2 4
Barker's Hill to Bryant's Hill. " " eye object = 7
IN.
Eye exceeds object = 1 9 = 21
3 ¼
10)68
6.8

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Bryant's Hill to Barker's Hill. Elev., 1° 14′ 13″ + 12″ = 14′ 25″
Bryant's Hill to Barker's Hill. Dep., 1° 22′ 50″ −; 6.8″ = 1 22 43.2
8 18.2 = 498″.2
3771
49820
34874
3478
To compute the eye and object corrections 149
First approximation = 88330
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Inches
37 = log 1.56820
88330 = colog 5.05389
constant log 4.41570
Correction 10.”9 = log. 1.03779
Inches.
21 = log 1.32222
88330 = colog 5.05389
constant log 4.41570
Correction 6.”2 = log 0.79181

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Corrected Angles.
Elev. 1° 14′ 13″ + 10″.9 = 1° 14′ 23″ .9
Dep. 1° 22′ 50″ − 6. 2 = 1 22 43.8
8 19.9 = 499″.
3771
49990
34993
3499
150
∴ Distance from Bryant's Hill to Barker's Hill = 88632 links
True distance as found by Triangulation = 89197 " "
Difference = 565 " "

Which is about half a chain per mile.

Having found the contained arc, or distance between the stations, in links, by the rules given above, the difference in altitude may be obtained in the usual way, viz., by converting the links into feet and then multiplying the distance in feet between the stations by the tangent of the true angle of elevation or depression. (Note.—The true angle of elevation or depression is half the sum of the observed reciprocal angles, when one is an elevation; or half the difference when both are depressions; or, generally, if zenith distances are used, the true vertical angle is equal to half the difference of the reciprocal zenith distances;—of course supposing the eye and object corrections to have been applied.)

But instead of finding the distance between the stations in links, and then converting it into feet, it would be more simple to find the distance in feet at once, by using the factor 117 instead of 177.3, as before explained:—

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Example.
Bryant's Hill to Barker's Hill. Corrected Elev. 1° 14′ 23″.9
Barker's Hill to Bryant's Hill. " " Dep. 1° 22′ 43″.8
Diff. 8′ 19″.9 = 499″.9
Sum. 2° 37 07″.7
½ Sum. 1° 18′ 33″.8

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499.9 = log 2.698883
117 = constant log 2.068186
1° 18′ 33″.8 = tangent 8.359040
1337.0 feet = 3.126109
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If no logarithmic or trigonometrical tables are at hand, the difference of altitude may be found as follows:—

As .00000485 represents the value of sin 1″ arc 1″ or tang 1″ (true to the last figure), and as the tangents of small angles vary very nearly as the number of seconds contained in the angle, we may substitute for the tangent of the angle the number of seconds multipled by .00000485.

In practice, the operation may be shortened by combining the two multipliers together; thus, .00000485 × 117 = .0005675.

(Note.—In order to show how very nearly the sines, arcs, and tangents agree for the first two degrees, their values at two degrees are given, for the sake of comparison.

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Diff.
Thus sin 2° = .0348995 71 = 1 ½”
arc 2° = .0349066 71 =1½”
tang 2° = .0349208 142 = 3″

Therefore, the arc of 2° = sin 2° 00′ 01 ½”, and the tangent of 2° = arc of 2° 00′ 03″.

Also, in obtaining the tangent of 2° by multiplying .00000485 × 60 × 60 ×2, the result is .0349200, or just ⅙ of a second below the true value.

Similarly the tangent of 1°, found in the same manner, is .0174600, or just 1″ above its true value; but the value used for tang 1″, viz., .00000485, is slightly in excess of its true value, which is .0000048481368, etc.)

Then the difference of altitude may be found by the following rules:—

Case 1.—When one angle is in an elevation.

Rule.—Take the difference of the observed vertical angles, and also half the sum, both reduced to seconds; multiply them together, and their product by .0005675; the result will be the difference of altitude between the two stations in feet.

Case 2.—When both angles are depressions.

Rule.—Take the sum of the observed verticle angles, and also half the difference, both reduced to seconds, multiply them together, and their product by .0005675; the result will be the difference of altitude between the two stations in feet.

Or, if zenith distances are used, the following general rule will apply in all cases:—

Rule.—Subtract 180° from the sum of the observed zenith distances and reduce the remainder to seconds; then take half the difference of the observed zenith distances and reduce it to seconds; multiply the two quantities together, and the product by .0005675, and the result will be the difference of altitude between the two stations in feet.

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Example as before.
½ Sum. = 1° 18′ 33″.8 = 471″3.8
Diff. = 8′.9 = 499.9 9994 = 499″.9 reversed
18855
4242
424
42
2356300
.0005675 5765000 = 0005675 reversed
11782
1414
165
12
Diff. of altitude = 1337=3 feet

Even when a book of logarithms is available, the calculation by logs will be more expeditiously performed by using the logs of the above quantities than by using the log tangent.

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Example.
1° 18′ 33″.8 = 4713=8 log 3.673371
8′ 19″.9 = 499.9 log 2.698883
Constant log .0005675 = 6.753966
1337.3 feet = 3=126220

With regard to the actual results obtained by this method, I may mention that in the circuit of 50 miles previously referred to, the altitudes closed to 18 feet, and in the circuit of 60 miles, the error in closing was only 2 feet.

It is thus evident that this method is quite capable of giving reliable results in hilly country, and is well adapted for the topographical survey of a new country. A line of stations might be selected in the most accessible positions, and each line used as a base from which to extend triangles on either side, and as every line is determined independently, there would be no accumulation of error.

On the contrary, by observing to distant hills on either side, the distances found would check each other, and any erroneous result could be rejected.

In very level country, where the refraction is too uncertain to give reliable results by this method, other methods may be employed, such as chained lines, or triangulation from a measured base, etc.