**Greenwich Mean Time**** (GMT****)** is the local mean time anywhere on the meridian of Greenwich. In other words it is the Local Hour Angle of the Mean Sun on the meridian of Greenwich.

Since the Greenwich meridian is used as the base meridian from which the longitude of all places on Earth are identified, it provides the link between the LMT of a place and the LMT at Greenwich (or GMT).

**Example.** In the diagram below, imagine that we are looking down on the North Pole.

P represents the North Pole

G represents the position of Greenwich on the Earth’s surface.

GP represents part of the Greenwich Meridian (0^{o}).

MP represents part of the meridian of longitude 45^{o}W and M is a point on that meridian.

AP represents part of the meridian of longitude 30^{o} E and A is a point on that

meridian.

The meridian of the Mean Sun, for a very brief instant, coincides with the meridian 45^{o}W. and so, at that instant, the Local Mean Time at point M is noon.

At the same instant, the Local Hour Angle of the Mean Sun at Greenwich is 45^{o}. Therefore, the LMT at Greenwich must be 3p.m. since the time difference for 45^{o} is 3 hours and Greenwich is to the East of M.

It follows that the Greenwich Mean Time must also be 3p.m. (since GMT is equal to the LMT at Greenwich).

The LMT at point A must be 2 hours after GMT (since the time difference for 30^{o} is 2 hours and A is to the East of Greenwich).

Therefore, the LMT at point A must be 5p.m.

**Greenwich**** Hour Angle**** (GHA**). The angle between two meridians of Longitude can be expressed as an hour angle. The hour angle between the Greenwich Meridian and the meridian of a celestial body is known as the Greenwich Hour Angle.

The Local Hour Angle between an observer’s position and the geographical position of a celestial body can be found by combining the observer’s longitude with the GHA.

In the diagram below , O represents the longitude of an observer;

X represents the meridian of a celestial body;

G represents the Greenwich Meridian.

Because in this case, the observer’s longitude is east and because LHA is measured westwards from the observer’s meridian to the meridian of the celestial body, LHA is equal to the longitude plus the GHA.

For a fuller explanation of GHA and LHA click here

**Converting GMT to GHA.**

Because GMT is measured westwards from the reciprocal of Greenwich (i.e. 180** ^{o}**) and GHA is measured westwards from the Greenwich meridian (i.e. 0

**) we convert GMT to GHA as follows: If GMT, when converted to arc, is less than 180**

^{o}**then add 180**

^{o}**; if GMT is greater than 180**

^{o}**then subtract 180**

^{o}**). Examples:**

^{o}**Example 1.** Convert 0840 GMT to GHA.

**Step 1.** Convert GMT to arc.

8^{h} = 8 x 15 = 120** ^{o}** 0’ 0”

40^{m} = 40 ÷ 4 = __ 10__^{o}__ 0’ 0”__

** = ** 130** ^{o}** 0’ 0”

**Step 2.** Convert to GHA.

GHA =** **130** ^{o}** 0’ 0” + 180

**0’ 0” = 310**

^{o}**0’ 0”**

^{o}**Example 2.** Convert 1530 GMT to GHA.

**Step 1.** Convert GMT to arc.

15^{h} = 15 x 15 = 225** ^{o }** 0’ 0”

30^{m } = 30 ÷ 4 = __ 7__^{o}__ 30’ 0”__

= 232** ^{o }**30’ 0”

**Step 2.** Convert to GHA.

GHA = 232** ^{o }**30’ 0” – 180

**0’ 0” = 52**

^{o}**30’ 0”**

^{o}**Note.** Because GHA relates to apparent solar time and GMT relates to mean solar time, we must take the equation of time (EOT) into account when converting GMT to GHA. Therefore the next example includes a calculation for EOT.

**Example 3.** Convert 0415 GMT to GHA. E0T = +1^{ m}

Equation of Time = mean solar time – apparent solar time

**∴ **apparent solar time = mean solar time – equation of time

**∴ **GHA = GMT – EOT.

**Step 1.** Convert 0415 GMT to arc.

4^{h} = 4 x 15 = 60** ^{o}** 0’ 0”

15^{m } = 15 ÷ 4 = __ 3__^{o}__ 45’ 0”__

** ** = 63** ^{o }**45’ 0”

– EOT __ – 1’ 0” __ (correction for EOT)

** = **63** ^{o }**44’ 0”

**Step 2.** Convert to GHA.

GHA =** **63** ^{o }**44’ 0” + 180

**0’ 0’’ = 243**

^{o}**44’ 0”**

^{o}A fuller treatment of this topic can be found in the book Astro Navigation Demystified