CAPT. MANALO , M.E BSMT
NAV 4 – COMPASS ERROR CALCULATION
GYRO ERROR by AMPLITITUDE of the Decl. = S 11° 50.5
d corr = - 0.5
Sun Decl = S 11° 50.0’
Example:
5. Solve for the value of Amplitude:
On Feb. 17, 2025, @ 1735 Lt, in Position
Latitude: 13° 45' 36'' N / Longitude: 121° 03' 00'' Sin Amplitude = Sin Dec
E, the ships on Gyro Course 315°, and Magnetic Cos Lat
Compass heading of 318°, the OOW took a
sunset observation of the sun Gyro bearing of Amplitude = Sin-1 ( sin 11° 50.0’/ Cos 13° 45'
the sun was 258 ᵒ T. Determine the gyro error 36'')
= Sin-1 (0..97130)
by amplitude if the variation of the area is 1ᵒE.
Amp = S 12.2° W
To solve for Amplitude you will need:
To get the Azimuth (ZN or True Bearing) do the
UTC operation as per the following quadrants:
Gyro Bearing N
Latitude
Declination + 270 -90
1.Solve for UTC
W E
Local Time = 1735
ZD (-E) = 8 -270 + 90
GMT = 0935H
2.Get the value of declination from the nautical
almanac corresponding to the date and the
whole number of time of GMT. Take note also of Azimuth (Zn) = 270 - 12.2°
the value of “d” at the bottom left of the table. (Zn) = 257.8 T
UT DECLINATION: “d”
09 S 11°50.5’ 0.9’ HOW TO SOLVE GYRO ERROR:
3. Get the value of the correction corresponding True Bearing
to “d” from the “Increments and correction Gyro Bearing
table” of the nautical almanac. Locate first the Gyro Error
minutes of GMT. At the column “v and d” Gyro Course
correction, find the value corresponding to “d”.
True Course
Magnetic Course
UT “d” v and d correction
Magnetic Error
0935 0.9’ 0.5’
Variation
Deviation
4. Based on whether the declination is
ascending or descending, add or subtract the v
and d correction. (based on the given example,
declination is descending, so we subtract.)
NAVIGATION 4 CELESTIAL NAVIGATION
, CAPT. MANALO , M.E BSMT
NAV 4 – COMPASS ERROR CALCULATION
TB = 257.8
Always Subtract the smaller value from the bigger value
GB = 258
GE = 0.2 W Name the gyro error by : compass Best error WEST / compass least error EAST
GC = 315 Apply gyro error to the gyro course ( East plus(add) / West minus (subtract) to
TC = 315.2 get the True Course.
MC = 318 Always Subtract the smaller value from the bigger value (TC and MC)
ME = 2.8 W Name the compass error by : compass Best error WEST / compass least error
EAST
Var = 1 E (W) (Reverse the name of variation then: if same name with ME add Var, if different
name with ME subtract Var
Dev = 3.8 W Takes the name of the higher Value
CHECKING
C = 318 “C” = Magnetic Compass Course
D = 3.8 W “D” = Deviation (East plus / West minus from “C”)
M = 314.2 “M” = Corrected compass heading
V = 1E “V” = Variation (Use the original name then EAST plus / WEST minus)
T = 315. 2 “T” = True Course (This should be the same value with the TC (true
course) above. If it is not the same, something wrong my frend.)
NAVIGATION 4 CELESTIAL NAVIGATION
NAV 4 – COMPASS ERROR CALCULATION
GYRO ERROR by AMPLITITUDE of the Decl. = S 11° 50.5
d corr = - 0.5
Sun Decl = S 11° 50.0’
Example:
5. Solve for the value of Amplitude:
On Feb. 17, 2025, @ 1735 Lt, in Position
Latitude: 13° 45' 36'' N / Longitude: 121° 03' 00'' Sin Amplitude = Sin Dec
E, the ships on Gyro Course 315°, and Magnetic Cos Lat
Compass heading of 318°, the OOW took a
sunset observation of the sun Gyro bearing of Amplitude = Sin-1 ( sin 11° 50.0’/ Cos 13° 45'
the sun was 258 ᵒ T. Determine the gyro error 36'')
= Sin-1 (0..97130)
by amplitude if the variation of the area is 1ᵒE.
Amp = S 12.2° W
To solve for Amplitude you will need:
To get the Azimuth (ZN or True Bearing) do the
UTC operation as per the following quadrants:
Gyro Bearing N
Latitude
Declination + 270 -90
1.Solve for UTC
W E
Local Time = 1735
ZD (-E) = 8 -270 + 90
GMT = 0935H
2.Get the value of declination from the nautical
almanac corresponding to the date and the
whole number of time of GMT. Take note also of Azimuth (Zn) = 270 - 12.2°
the value of “d” at the bottom left of the table. (Zn) = 257.8 T
UT DECLINATION: “d”
09 S 11°50.5’ 0.9’ HOW TO SOLVE GYRO ERROR:
3. Get the value of the correction corresponding True Bearing
to “d” from the “Increments and correction Gyro Bearing
table” of the nautical almanac. Locate first the Gyro Error
minutes of GMT. At the column “v and d” Gyro Course
correction, find the value corresponding to “d”.
True Course
Magnetic Course
UT “d” v and d correction
Magnetic Error
0935 0.9’ 0.5’
Variation
Deviation
4. Based on whether the declination is
ascending or descending, add or subtract the v
and d correction. (based on the given example,
declination is descending, so we subtract.)
NAVIGATION 4 CELESTIAL NAVIGATION
, CAPT. MANALO , M.E BSMT
NAV 4 – COMPASS ERROR CALCULATION
TB = 257.8
Always Subtract the smaller value from the bigger value
GB = 258
GE = 0.2 W Name the gyro error by : compass Best error WEST / compass least error EAST
GC = 315 Apply gyro error to the gyro course ( East plus(add) / West minus (subtract) to
TC = 315.2 get the True Course.
MC = 318 Always Subtract the smaller value from the bigger value (TC and MC)
ME = 2.8 W Name the compass error by : compass Best error WEST / compass least error
EAST
Var = 1 E (W) (Reverse the name of variation then: if same name with ME add Var, if different
name with ME subtract Var
Dev = 3.8 W Takes the name of the higher Value
CHECKING
C = 318 “C” = Magnetic Compass Course
D = 3.8 W “D” = Deviation (East plus / West minus from “C”)
M = 314.2 “M” = Corrected compass heading
V = 1E “V” = Variation (Use the original name then EAST plus / WEST minus)
T = 315. 2 “T” = True Course (This should be the same value with the TC (true
course) above. If it is not the same, something wrong my frend.)
NAVIGATION 4 CELESTIAL NAVIGATION
