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Art. LIII.—*On the Deflection of the Plumb-line from the Vertical due to the Spheroidal Form of the Figure of the Earth, and its Effect on the Latitude of Stations above the Mean Sea-level*.

[*Read before the Otago Institute, 22nd September, 1914*]

It is now well known that the attraction of mountains and elevated regions has a sensible influence on the plumb-line. These local deflections have been the subject of numerous investigations since 1854, when Archdeacon Pratt computed the effect of the attraction of the Himalaya Mountains on the plumb-line in North India (Phil. Trans., 1854).

In modern geodetic surveys the degree of precision in the determination of the apparent astronomical latitude by means of the zenith telescope is so great that the station error, as it is called, or the difference between the observed and computed latitudes, due to local causes, has become a part of the routine work.

The deflection of the plumb-line by the attraction of mountains or by the variation of the density of the underlying strata will not be further discussed, the object of the present investigation being to determine the

amount of deflection and its direction that a plummet will deviate from the vertical line, or normal, at its point of suspension, in the plane of the meridian, for any length of the suspending cord, due to the spheroidal form of the figure of the earth.

### Statics of the Problem of a Plummet suspended from a Fixed Point.

Let PEP_{1}E_{1}, fig. 1, represent the meridian ellipse of the earth, PP_{1} the axis of rotation or polar axis, and EE_{1} the equatorial axis.

We can infer from consideration of the symmetry of the figure that a plumb-line suspended at the Poles or at the Equator does not deviate from the vertical, and that at all other latitudes there is a deviation.

Let A be a point at the mean level of the sea, the latitude of which is φ', and suppose *p* a plummet suspended from the point A. Such a suspension can be actually obtained by means of a mine-shaft for small depths, but for the statement of the problem the existence of an opening in the earth need not be considered, for the solution of the problem gives the direction that a shaft takes when sunk by means of the plumb-line.

Let C be the centre of the earth, F and F_{1} the foci.

The forces acting on *p* are the attraction of the earth and the tension of the suspending cord, the weight of the cord being neglected.

Through *p* let a confocal ellipse be described, and let it generate a spheroid by revolving about the axis PP_{1}. It is a well-known theorem in attractions that the potential within a spherical or ellipsoidal shell is constant, and therefore the force acting on *p* is the attraction of the above-generated spheroid, the surface of which is equipotential (Routh's “Analytical Statics,” ii, p. 104).

A second theorem in attraction is: If there be any resultant force, due to any attracting mass, this force acts along the normal to that equipotential surface on which the point lies (Pierce's “Newtonian Potential Function,” p. 38).

Since the tension of the cord is the resultant force, it is normal to the confocal ellipse, and therefore bisects the angle F*p* F_{1}.

Thus for all positions of *p* the direction of the resultant is towards the point A, and the angle F*p*F_{1} is bisected by A*p* produced.

### Geometry of the Problem.

The problem to be solved can be stated as follows, in its most general form: Given the three sides of a triangle ABC, it is required to find the locus of a point, such that the line from the angle A to the point bisects the angle included between the lines from the angles B and C to the point.

The bisected angle contains the point A, and no restriction is placed on the size of the angle except that it is less than four right angles.

The deviation of the plumb-line is a particular case of the general form.

Let ABC, fig. 2, be the given triangle, and D a point such that the angles ADB, ADC are equal.

Take A as the origin of co-ordinates and the line bisecting the angle BAC as the axis of X.

Let *r*, θ be the polar co-ordinate of D. Denote the equal angles ADB, ADC by *χ*.

To obtain the equation to the locus of D, we have from the triangle ABD

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r/C=sin(A/2+θ-χ)/singχ ∴cotχ=r+c cos (A/2+θ)/c sin (A/2+θ) (1)

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Again, from the triangle ACD we find cotχ=r + b cos (A/2-θ)/b sin (A/2-θ) (2)

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Eliminating cot χ we obtain sin 2θ-mr sing θ+nr cos θ=0 (3)

Wherein *m* has the value b+c/bc cos A/2, and *n* the value b-c/bc sin A/2, or 2 sin θ - *mr* sin θ + *nr* = *o*, since θ is small. ∴ sign θ=2nr/2-mr (4)

The amount of the deflection is *r* sin θ, and is towards the Equator, since θ is positive.

To transform the equation to rectangular co-ordinates we substitute in (3) the values of the variable in terms of *x* = *r* cos θ, *y* = *r* sin θ, and *x*^{2} + *y*^{2} = *r*^{2}, and we have nx^{3}-myx^{2}+2xy+nsy^{2}-my^{3} = 0 (5)

When B and C are the foci of an ellipse the triangle ABC is isosceles if A is at the extremity of the minor axis.

Then *m* = 2/b cos A/2 and *n* = *o*.

Substituting these values of the constants in equation (5) we obtain
2xy-2/bcos A/2yx^{2}-2/bcoxa/2y^{3} = 0

Therefore *y* = *o* (6) and *y*^{2} + (x-b/2secA/2) = (b/2secA/2)^{2} (7)

Equation (6), *y* = *o*, represents the axis of X, and for a plummet suspended at the Pole of the earth the axis of X is the vertical, and consequently there is no deviation from the vertical when the plummet is suspended from the Pole of the earth considered as an oblate spheroid.

Equation (7) represents a circle whose radius is b/2 sec A/2, and is therefore the circle circumscribing the triangle ABC.

The co-ordinates of the centre are *y* = *o*. *x* = b/2 sec A/2.

When the point A coincides with the extremity of the major axis of the ellipse the angle A vanishes, and we find
*m* = b+c/bc and *n* = *o*. These values of the constants substituted in equation (5) reduce it to the following two results:—y=0 (8) y^{2}+(x-bc/b+c)^{2} = (bc/b+c)^{2} (9)

The equation (8) represents the axis of X, and is the solution of the problem when the plummet is suspended at the Equator, from which we infer that for places situated on the Equator there is no deviation of the plumb-line from the vertical.

Equation (9) represents a circle of radiu bc/b+c, the co-ordinates of the centre being *y* = *o*. *x* = bc/b+c

If in fig. 1 the inner ellipse represents the surface of the earth and A a point at an elevation above the surface, from which a plummet is suspended by a cord A*p*, the line A*p* produced bisects the angle F*p*F_{1}, and is normal to the surface of the earth at *p*; and the point A is therefore vertically above *p*. This condition holds for any length of the suspending cord, and the plumb-line at the earth's surface suspended from a point above it always coincides with the vertical.

The level surface at *p* is at right angles to A*p*, and the angle EH*p* is the geographic latitude.

The level surface at A is at right angles to the line AK, which bisects the angle FAF_{1}, and the geographic latitude of A is the angle EKA. The difference of these angles is the variation of the latitude of *p*, due to the spheroidal form of the earth, and the elevation *p*A.

The level surfaces become more spherical as we recede from the earth, the point K moves along the major axis and ultimately coincides with the centre.

If φ_{1} denotes the latitude of *p* and φ_{2} the angle EKA, then φ_{1} - φ_{2} is the angle θ in equation (3), where A is the origin, and the line bisecting FAF_{1} the initial, or the axis of X.

Let δ_{0}=HAF. δ=AF_{p}. δ′_{0}=HAF_{1}. δ′=AF_{1p}.

F_{p}F_{1}=ψ. AP=r.

F_{p}=f_{1}. F_{1}P =f.

It is now proposed to express AF, AF_{1}, and δ_{o}, δ'_{o} in terms of *p*F, *p*F_{1}, and *r*.

The triangle AF*p* gives
AF^{2}f_{1}^{2}+r^{2}+2f_{1}r cosψ/2 (10) f_{1} sin δ = r sing δ_{0} (11) δ + δ_{0} = ψ/2 (12)

Similar expressions are obtained from the triangle AF_{1}*p* for AF_{1}, δ‘_{o}, and δ’.

From (11)
sin δ = r/f_{1} sin δ_{0} = p_{1} sin (ψ/2 - δ), where p_{1} = r/f_{1} (13)

For even the highest mountain-tops the values of *p*_{1} and δ are small, and equation (13) can be developed by Maclaurin's series, thus: δ = tan ^{1} (p_{1}sin ψ/2/1+p_{1} cos ψ/2) = p_{1} sin ψ/2-½ p_{1}^{2} sin ψ +&c (14)

Similarly, if *p* = p=r/f δ′ = p sin ψ/2 - ½ p^{2} sing ψ + &c (15)

To take a numerical example, suppose that in latitude 45° S. it is required to find, the difference in latitude between the top of a hill, 490 metres high, and the sea-level vertically below it, using Helmert's spheroid.

The value of θ can be computed by equation (3), but it can be found directly from (12), (14), and (15), as under.

ψ/2 = δ +δ_{0} = δ′+δ′_{0} δ′_{0}-δ_{0} = δ -δ′ (16) Now, δ′_{0} = A/2 +θ. δ_{0} = A/2-θ. ∴δ′_{0}-δ_{0} = 2θ. or θ ½ (δ′_{0} - δ_{0}) (17)

And by the aid of (14) and (15) θ in seconds of arc is
θ″=½ p_{1}-p/sin 1″ sin ψ/2 - ¼. p_{1}^{2}-p^{3}/sin 1″ sing ψ +&c (18) p_{1}-p = r/f_{1} - r/f = r (f-f_{1})/f f_{1} = 10 -^{6}x 893.

Computing the first term of (18) from the known value of sin ψ/2 we find θ =0″.053

Thus the latitude of the hilltop is
φ_{2}= φ_{1} - θ. = 44° 59′ 59″.947.

These differences of latitude are so small that in most cases they will be masked by the greater differences due to local conditions, which deflect the plumb-line; they possess the advantage that they can be easily computed, and the observed value can be corrected or reduced to the sea-level.

A summary of the three principal results is as follows:—

(1.) |
A plummet suspended from a point at the sea-level deviates from the vertical in the plane of the meridian, except at the Poles or the Equator; the amount of the deviation depends on the length of the suspending cord, and the deviation is towards the Equator. |

(2.) |
A plummet suspended from a point above the surface of the sea-level, when the suspending cord just reaches the surface, always coincides with the vertical for any length of the suspending cord. |

(3.) |
At heights above the surface the observed latitude is less than its corresponding value at sea-level. |