Meridian PassageAND Calculation of Latitude... Today’s lesson Aim is:-To introduce you to the theory of Meridian Passage, why we obtain it and how we calculate the observer’s latitude..
Trang 1Meridian Passage
AND Calculation of Latitude
Trang 2Today’s lesson Aim
is:-To introduce you to the theory of Meridian Passage, why we obtain it and how we
calculate the observer’s latitude
Trang 3The Objective
being:-That by the end of this two hour lesson YOU will be able
to:-Explain the theory of Meridian Passage and position lines
Explain why navigators use this theory to obtain position lines and latitude
Demonstrate how to extract the LMT of Meridian Passage from the Nautical Almanac
Calculate the GMT of Meridian Passage by applying longitude
Demonstrate how to obtain Declination and necessary corrections from the Nautical Almanac
Calculate a meridian passage latitude using the supplied pro-forma layout.
Trang 4Meridian Passage
When we talk about Meridian Passage, the Meridian we are
referring to is the Observer’s Celestial Meridian
Please remember that
Trang 5Meridian Passage
Meridian Passage occurs when the celestial body, in it’s
movement across the heavens crosses the observers
celestial meridian
At this point the bearing of the body is either due north or
due south of the observer and the altitude of the body is at it’s maximum
Trang 6Observers Meridian
Body bearing North or South
Here we have the OOW at 10 0 North Latitude observing a body on his/her meridian which has a Declination of 10 0 S
Pn
Equinoctial
E W
Ps
X
Z
Equinoctial
Q
Trang 7OBSERVERS MERIDIAN
Trang 8Meridian Passage
As we already know, taking an altitude of a celestial body
allows us to calculate it’s bearing and as a result, a position line
Observing the Altitude at Meridian Passage provides a quick
Observing the Altitude at Meridian Passage provides a quick
and easy method of obtaining a position line which, as the bearing of the body is either due north or south then the
position line will run in an east/west direction
This same position line in effect will then become the
observer’s latitude
Trang 9Observers Meridian
Here we have a 2/O at 10 0 North Latitude observing a body with a Declination of 10 0 S
Pn
Z
Pn
Z
P
Equinoctial
E W
Ps
X
Ps X
Dec
Lat
Equinoctial
Q
TA Dec
Trang 10Meridian Passage
In order to make the observation at the correct time we
need to know the time of meridian passage
We also need to know the GMT of meridian passage in order
to extract the declination of the body from the Nautical
Almanac
In order to find the time of meridian passage for the sun &
planets we enter the Nautical Almanac daily pages and
extract the time of Mer Pass as
Trang 13follows:-Meridian Passage
These times are the LMT of the local Meridian Passage of the body
We must apply longitude (in time) to this time in order to obtain the
GMT of Meridian Passage at the local meridian before we can extract Declination from the Nautical Almanac
Trang 14A little problem for you!
• On September 23rd
The Sun is observed
on the meridian in DR
position Lat 120 15’ N,
0
• LMT Mer Pass: 23d 11h 52m
• GMT Mer Pass 23d 18h 03m
0920 45’ W Find the
time of meridian
passage and the
declination
Trang 15Procedure for Altitude correction to derive
Latitude by Meridian Altitude.
• Sextant Altitude : 450 25.5’
• Index Error (I.E.) : 00.2’
• Observed Altitude : 450 25.7’ So tell me, was the IE on or off the arc?
• Dip : 8.7’ + or - ??
• Apparent Altitude : 450 17.0’
• Total Correction : 000 15.3’ Lower limb + or - ??
• Total Correction : 000 15.3’ Lower limb + or - ??
• True Altitude : 450 32.3’
• ~900 00.0’ : 900 00.0’
• ZX : 440 27.7’
• + or – Dec : ??0 ??.?’
• Latitude : ??0 ??.?’ N/S Position Line runs 0900 / 2700
Trang 16Applying Declination to ZX
Z
Pn
Let us imagine we are on the Celestial Sphere, looking down on a 2/O on the Equator
with Sextant in hand
Pn
N
What does this line also represent?
Z
Ps
W
Ps
What does this line
also represent?
N S
Trang 17Applying Declination to ZX
Observers Meridian
Let us now imagine the 2/O moves to
10 0 North and observes a body with Declination of 10 0 S
Pn
Z Q
Lat
ZX
Pn
E W
Ps
X
Z
Equinoctial
Q
TA
TA
X Lat Dec
Trang 18Applying Declination to ZX
Observers Meridian
Let us now imagine the 2/O moves to
10 0 North and observes a body with Declination of 10 0 S
Pn
Z
P
Equinoctial
E W
Ps
X Z
X
Dec
Lat
Equinoctial
Q
TA
Trang 19Applying Declination to ZX Let us now imagine the 2/O moves to
20 0 North and observes a body with Declination of 10 0 N
Equinoctial
Observers Meridian Pn
Z P
E
Q
TA
Z
X
X
Q
ZX
Dec
Lat
Trang 20Going back to our original example
Z
P
• Sextant Altitude : 45 0 25.5’
• Index Error (I.E.) : 00.2’
• Observed Altitude : 45 0 25.7’
• Dip : 8.7’
• Apparent Altitude : 45 0 17.0’
• Total Correction : 00 0 15.3’
X
Dec
Lat
• Total Correction : 00 0 15.3’
• True Altitude : 45 0 32.3’
• ZX : 44 0 27.7’
• + or – Dec : ?? 0 ??.?’
• Latitude : 20 0 00.0’N
+ or – Dec : 24 0 27.7’S