Capt.S.S.Chaudhari

Due to the earth rotation the spin axis of a free gyroscope will trace out a circular path as it remains pointed in a fixed direction in space. The requirement of a gyro compass is that the spin axis should point in a fixed direction on earth and the chosen direction is true north.
To convert a free gyroscope into a gyroscope the free gyroscope must be made.

1. North seeking i.e. seek the meridian.
2. North settling i.e. must point in or near the meridian.

Stage 1, Free Gyroscope: The north as well as south axle follow fixed points in space (stars), the period being 23^h56^m4.1^s

Stage 2, Gravity Control: The gravity control causes the period of declinations circle to reduce to 84½ minutes. This period is also to cause Schuler tuning. Gravity control is caused by making the gyro top heavy or bottom heavy.

In northern latitudes the axle would tend to follow the star on observers inferior meridian. (The meridian from North pole till horizon is the inferior meridian). Such stars would cross inferior meridian from west to east.  As the axle drifts off the meridian to east, it will tend to rise in altitude or tilt upwards.  This tendency is due to tilting upwards in east. [T_g = 0 when axle is on meridian & downwards in west].  T_g depends on the latitude & the azimuth.  By azimuth here what we mean is the angular separation wrt meridian.  As axle drifts eastwards.  The following may be observed.

1. Steady rate of eastwards drifting.
2. T_g increasing as axle moves away from meridian.
3. Rate of Control Precession increasing with increase in tilt.

At a certain point rate of precession becomes equal & opposite to rate drifting. When this happens the axle stops going any more east & in fact has only upwards motion due to T_g.  With more & more tilt the westwards rate of precession becomes more & more. Axle now moves in upwards & west wards direction.

As the axle moves towards meridian, T_g gradually decreases & diminishes to 0 on meridian.  On meridian there is no N-S forces, hence represents northern most peak. Axle overshoots here to the west of meridian. Once the axle has come on west side it attains down wards T_g like any star which is on west side of observer’s meridian. 

Thus, a westward rate of precession higher than eastwards drifting rate, along with downwards T_g moves the axle to west as well as down, till a point is reached where west wards rate of precession equals rate of eastwards drifting.  At this point the axle stops going anymore to west. Any further down wards movement will cause westwards precession to become smaller than eastwards drifting. Axle then comes back to meridian & points in the original direction.

T_g (Rate of tilting) a Azimuth or angular separation from meridian. (T_g = 0 at meridian, upward to east & downward to west)

C_p (Control Precession) a Tilt or angular separation from Horizon, C_p = 0 at horizon.

D (Rate of Drifting) can be assumed for a given latitude, to be permanent throughout the trace.

Position (1) is on the meridian on horizon. Because of D axle is driven out of meridian to east. As the axle is off the meridian T_g acts on the axle, giving the axle an upward motion. Gradually the tilt increases.

Position (2) is off the meridian and above the horizon giving rise to tilting and control precession respectively.

Position (3) has a tilt at which C_P becomes equal to D. At this point T_g is maximum in upward direction.

Position (4) is the northernmost point of the trace on meridian. At (4)T_g is 0. C_Pis maximum. Above the major axis of the trace C_P is more than D.

Position (5) has forces same as at position (3), except that T_g is downwards.  

Position (6) has forces same as at position (2), except that T_g is downwards.

The path of axle of free gyroscope & undamped gyro respectively on the observer’s R.H. would be as follows.

Stage 3, Damping: Damping is a type of precession much less than the gravity controlled precession and caused to act in a direction that will either be towards the horizontal (damping in tilt) or towards the meridian (damping in azimuth). In gyros damped in tilt very small fraction of control precession acts towards the horizontal, perpendicular to control precession. Whereas, in the gyros damped in azimuth the damping precession has a tendency to take the axle towards the meridian.  The path of axle becomes spiral, eventually settling the axle slightly off the meridian in the gyros damped in tilt and on the meridian in the gyros damped in azimuth.

The arrangements are made accordingly to cause couple in a direction 90º off the required precession.  Such couple & hence precession must reverse in sign when crossing the horizontal line downwards or upwards.  Since the damping must be proportional to (a function of) tilt it is also proportional to (a function of) control precession. Thus, damping precession = τC_P.

Damping in tilt

Position (1) is on the meridian on horizon. Because of D axle is driven out of meridian to east. As the axle is off the meridian T_g acts on the axle, giving the axle an upward motion. Gradually the tilt increases.

Position (2) is off the meridian and above the horizon giving rise to tilting and control precession respectively. A small part of C_P is given as damping precession, perpendicular to C_P, acting towards horizontal plane. D_P = ρC_P.

Position (3) has a tilt at which C_P becomes equal to D. At this point T_g is maximum in upward direction. T_g>D_P.  Axle moves upwards.

Position (4) is the northernmost point of the trace on meridian. At (4)T_g = D_P. C_P  as well as D_P is maximum. This  northernmost peak occurs to the east of meridian.

Position (5) has forces same as at position (3), except that T_g is downwards joins D_P,  bringing the axle down rapidly. 

Position (6) is southernmost peak, east of meridian, with T_g = D_P & D>C_P.

Position S is the settling position with opposite forces balancing each others.

ρC_p = T_g = 15 Cos lat Sin Az. But C_P = D = 15 Sin lat Cos alt.
Substituting value of C_P we get,  ρ. 15Sin lat Cos alt = 15 Cos lat Sin Az.
Since tilt always is v v small, Cos alt will approx. be 1. We get ρ.Sin lat = Cos lat Sin Az

Rearranging we get Sin Az = ρ. Tan lat. For small angles of Az, Sin Az = Az/57.3⁰. Az = ρ.57.3 tan lat. This (Az), is the damping error, easterly in NH & westerly in SH. This error is there only in case of gyros damped in tilt.

Why  does the axle of gyro damped in tilt not settle in meridian? Why does it settle to east in NH & west in SH?
Settling occurs on a tilt where C_P = D to balance horizontal forces. On this tilt C_P bound to occur, which means D_P  is bound to occur. To balance this D_P, T_g must be there. For this axle can’t be in meridian.

In NH an eastward D must be balanced by westward C_P, which will give rise to downward D_P, which must be balanced by an upward T_g, which will be available to east of meridian. Similarly in SH a westward D must be balanced by eastward C_P, which will give rise to upward D_P, which must be balanced by a downward T_g, which will be available to west of meridian.

Damping in azimuth
Additional precession about vertical axis out of phase with normal precession caused. Direction of this precession is such that movement of a´le towards meridian is encouraged.

Position (1) is the start position withhh axle horizontal & in meridian. D drives axle off the meridian.

Position (2) is in comparison to ellipse is  the major axis. This is the easternmost point. Easternmost limit is reached at smaller Ð of tilt, compared to ellipse or the trace of gyros damped in tilt because C_p & D_p add up to equal D.

Position (3) is the northernmost point & lies in meridian where T_g is 0.C_P>D.

Position (4) is in comparison to ellipse is above the major axis. This is the westernmost point. Westernmost limit is reached at a higher Ð of tilt, compared to ellipse or the trace of gyros damped in tilt because D & D_p add up to equal C_P.

Position (5) is the second extreme east position. Vectors are similar to position 2. T_g however, being smaller but D & C_P bigger in size than at (2).

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