04-22-2023, 04:49 PM
Paul,
An example of finding the AP longitude in western longitudes including adjusting the AP lon to be within 30.0' of the DR lon.
A sight of the sun's lower limb was taken on July 15, 2022 at corrected watch time time of: 15h 01m 53s
DR position at 15 00 hrs was 37° 47.5 N 123 26.2W
The local time is Pacific Daylight Time so it is +7 hours to get UTC
UTC time was 22 01 53 on July 15, 2022
GHAsun for 22 00 00 148° 29.3'
Inc for 00 01 53 1° 28.3'
GHAsun 22 01 53 148° 57.4' <= the minutes that you append to the DR lon for the AP lon.
AP lat 38° N AP long 123° 57.4' <= subtract to get LHA
LHA 25° 00.0'
Now we check to see if the AP longitude is within 30.0' of longitude to the DR
AP lon 123° 57.4
DR lon 123° 26.2 <= subtract
lon. diff. 0° 31.2' <= too big for the right answer on the exam !
So now we adjust our AP lon by 1° to bring the AP lon to within 30.0' of the DR and check again
DR lon 123° 26.2'
Updated AP lon 122° 57.4 <= subtract
lon. diff. 0° 28.8' <= less than 30.0' and what the examiner would use
Last adjust the LHA
GHAsun 22 01 53 148° 57.4'
AP lat. 38° N AP long 122° 57.4' W <= subtract to get LHA
LHA 26° 00.0' <= the LHA the examiner would use
Remarks:
If you did the entire sight reduction all the way through to plotting with 25° LHA instead of 26° you would still get a useful line of position, however your intercept and azimuth would not match the exam answers.
Also with slightly different numbers you might have to add a degree to the AP lon. to make it work out to less than 30.0' longitude from the DR position.
Peter
An example of finding the AP longitude in western longitudes including adjusting the AP lon to be within 30.0' of the DR lon.
A sight of the sun's lower limb was taken on July 15, 2022 at corrected watch time time of: 15h 01m 53s
DR position at 15 00 hrs was 37° 47.5 N 123 26.2W
The local time is Pacific Daylight Time so it is +7 hours to get UTC
UTC time was 22 01 53 on July 15, 2022
GHAsun for 22 00 00 148° 29.3'
Inc for 00 01 53 1° 28.3'
GHAsun 22 01 53 148° 57.4' <= the minutes that you append to the DR lon for the AP lon.
AP lat 38° N AP long 123° 57.4' <= subtract to get LHA
LHA 25° 00.0'
Now we check to see if the AP longitude is within 30.0' of longitude to the DR
AP lon 123° 57.4
DR lon 123° 26.2 <= subtract
lon. diff. 0° 31.2' <= too big for the right answer on the exam !
So now we adjust our AP lon by 1° to bring the AP lon to within 30.0' of the DR and check again
DR lon 123° 26.2'
Updated AP lon 122° 57.4 <= subtract
lon. diff. 0° 28.8' <= less than 30.0' and what the examiner would use
Last adjust the LHA
GHAsun 22 01 53 148° 57.4'
AP lat. 38° N AP long 122° 57.4' W <= subtract to get LHA
LHA 26° 00.0' <= the LHA the examiner would use
Remarks:
If you did the entire sight reduction all the way through to plotting with 25° LHA instead of 26° you would still get a useful line of position, however your intercept and azimuth would not match the exam answers.
Also with slightly different numbers you might have to add a degree to the AP lon. to make it work out to less than 30.0' longitude from the DR position.
Peter