Errors Gyro Compass

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Errors of Marine Gyrocompass

OPERATIONAL ERRORS OF THE GYRO: 1. Latitude error (Damping Error or Settling Error) 2. Course, Latitude & Speed Error (Steaming Error) 3. Ballistic Deflection 4. Ballistic Tilt 5. Rolling Error 6. Inter cardinal Rolling Error

Latitude Error (Damping Error or Settling Error): This error is due to the eccentricity of the damping weight (i.e. offset of the mercury ballistic cone bearing). The spin axis reaches equilibrium and settles in a position at which drifting is counteracted by control precession & the damping precession counteracts tilting. Since for any latitude except the equator there will always be a drifting given by 15o Sin (Lat) per hour. This Dg is counteracted by control precession (Pc) which comes into action only with tilt. In a given proportion damping precession (Pd) also accompanies, which counteracts the tilting (Tg). This results in tilt being reduced which in turn will reduce the Pc. This means that some sustained increase in Tg is required which compensates for this. But Tg can occur only when spin axis is off the meridian, as Tg at meridian is nil.

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Therefore the spin axis of the gyro settles off the meridian & slightly tilted, where all the forces balance out. That is Dg = Pc and Tg = Pd In N lat this position is slightly above the horizon and east of the meridian and in S lat the spin axis settles slightly below the horizon and west of meridian. This error can be calculated for given latitude and applied manually. In Sperry compass this error is allowed for by moving lubber line by means of auxiliary latitude corrector. Tilt is very slight and can be ignored. This error occurs only in gyro compasses damped in tilt and not in compasses damped in azimuth Formula for Latitude error: At the settling position: Drifting (Dg) = Control precession (Pc) Tilting (Tg) = Damping precession (Pd) Or, 15 Sin lat = Pc and, ---------------- (1) 15 Cos lat. Sin Az. = Pd -------------- (2) Dividing (2) by (1), we have: Cos lat. Sin Az = Pd = Pc Sin lat Pc 40 Pc

(In Sperry compass, Pd = Pc/40)

Or, Sin Az = K Tan lat (Where, K = 1/40, in Sperry compass) Thus: Sin Az = Tan lat/40 (in Sperry compass eccentricity = 1/40) Since azimuth is very small, we have: Az (radians) = Tan lat/40 Or, Az (degrees) = 57.3 Tan lat/40 = 1.43 Tan lat Thus Azimuth (DE or SE or LE in degrees) = K Tan Lat (where K=constant, about 1.43 in Sperry Compass) Thus students can use either of the two formulas: 1. Sin Az = 1/40.Tan lat. 2. Az in degrees = 1.43 Tan lat. Where Az is Damping error (or Settling error or Latitude error) Course & Speed error (Steaming error): Gyrocompass is basically a fast spinning gyrosphere which is controlled and damped so as to make it point and keep pointing true north. We know that all


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meridians, which represent true north, are perpendicular to equator and all parallels of latitudes. This means that the gyro spin axis points at right angle to the direction of rotation of the earth. This is alright as long as gyro is placed on a stationary vessel. But on a moving vessel, gyro senses the resultant of earth’s E-W motion and vessel’s own motion as the actual direction of rotation of the earth and aligns the axis along the perpendicular to this resultant. This is called gyro north and its displacement from true north is called gyro error. A little imagination will tell straightaway that no error would be caused on easterly or westerly course and maximum error would be caused on northerly or southerly course. On other courses, error will lie between zero and the maximum value. Error is also dependent on latitude (i.e. length of earth’s motion vector) and speed of vessel (length of ship’s motion vector). Further, error is high or west on northerly courses and low or east on southerly courses. This error is independent of the design of the compass and is the same for all types of compasses for a given course & speed for a particular Latitude. Tan (Error) =

V Cos co . (900 Cos lat ± V Sin co)

The Course, Latitude & Speed errors are corrected by various means as per the design: 1. To be allowed for by the navigator from the supplied tables or by calculation. 2. This error can be allowed for by a corrector mechanism which can be adjusted for ship’s speed and latitude. The correction is made to the position of the LL and is made to vary as the cosine of ship’s course by means of a cam which runs in a cosine grove cut beneath the compass card. 3. In Arma Brown compasses this error is eliminated by injecting a signal into the azimuth servo motor system so that the twist is produced in the vertical torsion wires. The resultant tilt of the gyro ball in tilt is equal and opposite to the rate of tilting due to N-S component of ship’s speed and the tilting sensed by the pendulum is that due to earth’s rotation only. The strength of the signal injected into the azimuth servo motor is determined by setting a speed control and by an input from the transmitter, which varies the signal as the cosine of ship’s course.

Derivation of the formula: Derivation for vessel’s course in each quarter is given separately. In each of the following cases, following notations are used: AD = BC = E = 900 Cos lat (vector representing earth’s west to east rotation). At equator, E =900 (as Cos 0 = 1) and it reduces as lat. Increases. AB = DC = V (vector representing vessel’s motion)


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AC = R (vector representing resultant of E and V) CP is perpendicular drawn to vector E TN is true north (perpendicular to E) GN is gyro north (perpendicular to R) Angle ‘a’ is gyro error. It is high (or west), if GN is west of TN. It is low (or east), if GN is east of TN Angle ‘b’ is the acute angle which V makes with E Angle ‘co’ is the vessel’s quadrantal course. In all cases ‘co’ = 90 ─ ‘b’ Courses in NE quarter:

Tan a =

V Sin b . 900 Cos lat + V Cos b

Or, Tan a =

V Sin (90-co) . 900 Cos lat + V Cos (90-co)

Or, Tan a =

V Cos co . 900 Cos lat + V Sin co Note: Course has northerly component. GN is west of TN. Hence error is high (west)


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Courses in SE quarter:

Tan a =

V Sin b . 900 Cos lat + V Cos b

Or, Tan a =

V Sin (90-co) . 900 Cos lat + V Cos (90-co)

Or, Tan a =

V Cos co . 900 Cos lat + V Sin co

Note: Course has southerly component. GN is east of TN. Hence error is low (east).


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Courses in SW quarter:

Tan a =

V Sin b . 900 Cos lat ─ V Cos b

Or, Tan a =

V Sin (90-co) . 900 Cos lat ─ V Cos (90-co)

Or, Tan a =

V Cos co . 900 Cos lat ─ V Sin co

Note: Course has southerly component. GN is east of TN. Hence error is low (east).


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Courses in NW quarter:

Tan a =

V Sin b . 900 Cos lat ─ V Cos b

Or, Tan a =

V Sin (90-co) . 900 Cos lat ─ V Cos (90-co)

Or, Tan a =

V Cos co . 900 Cos lat ─ V Sin co

Note: Course has northerly component. GN is west of TN. Hence error is high (west)


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IMPORTANT NOTES REGARDING STEAMING ERROR: 1. Error is directly proportional to vessel’s speed (Vector V). 2. Error is directly proportional to north/south component of course; i.e. closer the course to north/south greater the error. (Cos 0=1, Cos 180= -1) 3. Error is inversely proportional to earth’s speed of rotation. At equator the speed is maximum (900 nm/hr) and the error is the least. At higher latitudes speed is less (given by 900 Cos lat) resulting in smaller denominator and thus greater error. 4. On courses having easterly component (NE and SE) perpendicular CP falls outside on extended vector E. The vector V Cos b is thus added to vector E. 5. On courses having westerly component (SW and NW) perpendicular CP lies inside, on vector E. The vector V Cos b is thus deducted from vector E 6. Error is independent of the name of latitude i.e. independent of which hemisphere vessel is in. 7. In the final formula, term V Sin co is quite small compared to 900 Cos lat. Hence this term can be ignored and the formula is further simplified as below: Tan a =

V Cos co . 900 Cos lat

For small values of ‘a’, Tan a = ‘a’ radians Thus, a (in radians) =

V Cos co . 900 Cos lat

Thus a (in degrees) = =

V Cos co . x 180 900 Cos lat π V Cos co . (Approximate formula) 5 π Cos lat

However, in case of high speed crafts and in higher latitudes, V Sin co will become significant and 900 Cos lat will become smaller. In such a case V Sin co will not be small enough to be ignored and it would be advisable to use the main formula and not the approximate one.

Ballistic Deflection: BD is a precession which results from accelerations imparted to compass by change in speed and/or course of the vessel. It is an error caused by the precession imparted to the compass by N-S change in speed and / or course of the vessel. If vessel going on North course alters course to 090, there will be rush of mercury from S pots to N pots, as governed by Newton’s first law of motion. As the rotor is


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spinning ACW, precession will be caused towards East, which is opposite of westerly CLSE when the vessel was initially going on Northerly course. On settling on new course the mercury regains its natural level, but as long as the acceleration exists the error also exists. This is known as BD and depends on free surface of mercury and the amount of change in the N-S component of vessel’s motion. It is independent of the latitude and thus can be made exactly equal to “change in CSLE”. But CSLE varies with latitude. Hence BD is usually made equal to “change in CSLE” for standard latitude, usually 45. It is found that for making BD = change in CSLE, the time period of undamped gyroscope i.e. the ellipse (given by T = 2π√ R/g (where R is radius of earth) T has to be 84.7 minutes. R= 6378388 m, g =9.81 m/s2 Alteration of course towards N or increase in northerly speed or decrease in southerly speed means: Northerly (positive) acceleration (means Hg flows to S pots) produces a westerly (negative) change in azimuth. Alteration of course towards S or increase in southerly speed or decrease in northerly speed means: Southerly (negative) acceleration (means Hg flows to N pots) produces an easterly (positive) change in azimuth. These accelerations cause a false vertical and hence false horizontal. The false vertical is the resultant of the acceleration due to gravity and that due to vessel’s change of motion. The control element remains in the true horizontal due to gyroscopic inertia, but the liquid senses false horizontal and flows to cause N or S heaviness. The control element remains in the true vertical due gyroscopic inertia but liquid associates itself with false vertical and flows to produce N or S heaviness. N heaviness gives easterly precession and vice versa. The rate of precession is proportional to acceleration causing it. The precession continues for as long as the acceleration continues, so that for a given acceleration of speed and/or course the total change in azimuth will be constant, irrespective of the rate at which the maneuver is carried out.  = tilt a (acceleration) 

False Horizontal

g (gravity)

False vertical

Tan  = a/g, for small angles Tan  = c i.e.  in radians Thus,  c = a/g We know that precession is given by P = B  c/H = a B gH


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(Where B and H are compass constants) And, if this precession exists for time t, then BD is given by: BD = P x t = a B t gH Both BD and CLSE are caused by change in N-S component of vessel’s velocity. In order to help spin axis settle at new settling meridian, BD is made equal to change in CLSE. Let the initial course-speed and final course-speed be V1-Co1 and V2-Co2 respectively. Initial CLSE = V1-Co1 . (R = radius of earth, ώ = earth’s angular velocity) Rώ Cos lat Final CLSE = V2-Co2 . Rώ Cos lat Note: Linear velocity = Angular velocity x radius Change of CLSE = V2-Co2 - V1-Co1 . = V2 Cos2 – V1 Cos1 Rώ Cos lat Rώ Cos lat Rώ Cos lat Equating BD and change of CLSE, we have: a B t = V2 Cos2 – V1 Cos1 gH Rώ Cos lat Now, acceleration a = V2 Cos2 – V1 Cos1 t Thus, we have: B = 1 gH Rώ Cos lat Or, R g

=

H B ώ Cos lat

Multiplying both sides by 2π and square rooting: 2π√R/g = 2π√H/B ώ Cos lat 2π√R/g is the period of a pendulum whose length is R (in this case radius of earth) and is equal to about 84.7 minutes. And, 2π√H/B ώ Cos lat is the period of undamped ellipse traced the spin axis of the gyro. Thus BD can be averted by making the time period of undamped ellipse equal to about 84.7 minutes.


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Ballistic Tilt: BT is a byproduct of BD. While BD is precession in azimuth, BT is precession in tilt. Rush of mercury also causes a torque about vertical axis because of the eccentricity of the damping weight (offset of mercury ballistic cone bearing) and causes the spin axis to tilt as well. A southerly acceleration causes north end to precess upwards. A northerly acceleration causes north end to precess downwards.

Ballistic tilt caused by southerly acceleration In the figure shown above, TN is true north, GN1 is the initial gyro north (before the maneuver) and GN2 is the new gyro north at the end of the maneuver. Because of the offset of bearing of the mercury ballistic the spin axis does not go to M directly, but instead goes to G. This shift is indicated by vector FG. This vector consists of two components: Vector FM representing the BD towards east Vector FN representing the BT upwards Clearly the vessel is undergoing southerly acceleration. After reaching at G, the spin axis traverses a small spiral path, shown in green dotted curve, and finally settles at M, on settling level corresponding to the latitude of the vessel. A similar figure is shown below which shows the shift and the subsequent spiral path of the spin axis when vessel undergoes northerly acceleration.


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Ballistic tilt caused by northerly acceleration Thus the axis arrives at the new settling meridian slightly displaced in tilt from the settling position. This causes the spin axis of the gyro to execute a small damped spiral until the axis finally settles. BT causes maximum error of no more than 1.5 degrees at the end of a first quarter of the cycle i.e. after about 20 minutes from the completion of maneuver. Damping Error and BT are both due to the eccentricity of the pivot of the cone bearing of the mercury ballistic. By reducing eccentricity both DE and wander due to BT can be reduced. For this reason damping %age is kept within limits so as to reduce BT. Damping %age of 66.67% is chosen with this in mind. This keeps BT within tolerable limits and no further attempt is made to tackle it. ROLLING ERROR: If an unsymmetrical pendulum ( e.g. a ring) is tied as a bob and set oscillating, it would be found that it tends to align so as to have the maximum MOI lying in the plane of the swing. The MOI of a ring is greater about an axis than about a diameter. Thus when vessel is rolling and the gyroscope is swinging like a pendulum in gimbals, a torque is produced about the vertical axis tending to turn the plane of the plane of the vertical ring into the plane of the swing. This T will cause P in tilt and a subsequent wander of the compass. In Sperry compasses this is prevented by compensator weights, called quadrantal weights attached on each side of the vertical ring so that the MOI of the rotor is same in all directions about the vertical axis.


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In AB compasses control and damping is by electrical signals and torsion wires and no gravitational controls are attached to the gyroscope. A further source of rolling error can develop if sensitive element has unequal MOIs about N-S and EW axes. This is prevented by spherical shape of the gyro ball. INTERCARDINAL ROLLING ERROR: It is a combination of two effects. When vessel rolls on an E-W course, gyroscope swings in N-S plane in its gimbals and vice versa. Swing in N-S plane causes Hg to surge to and fro between the Hg pots, though inertia keeps the rotor and pots stable with respect to the horizontal. Since the surge is equal and opposite with the alternate N and S swing, horizontal P produced are also equal and opposite on each successive swing and no error is allowed to accumulate. When on N-S course the swing is in E-W plane. There is no surge of Hg but the link attachment (between rotor casing and MB) shifts alternately between eastwards and westwards. We know that the link is deliberately offset to provide the damping T about the vertical axis. The alternate E and W swing thus results in the DT being alternately greater than and less than the desired value. Here also, the average value is not affected and the settling position is not disturbed. Now consider vessel rolling in NE course. The swing will be in NW and SE plane. Both the effects will be seen now. Hg will surge and the link also will be shifted. On NW swing Hg will surge to N pots and link will be carried westwards and vice versa on SE swing. This error is approximately corrected by restricting the bore of the tubes connecting the Hg bottles, so that the surge of Hg lags about a quarter of period behind the roll.


Errors Gyro Compass

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