EOT Calculations

Demonstrate an understanding of the term 'equation of time' (apparent solar time - mean solar time) and perform simple calculations

The questions you may be asked in the exam will ask you to make calculations based on the formulae for EOT, Apparent and Mean solar time. They may also ask you these type of questions:

• The culmination of the Sun (when the Sun reaches its highest point).
• Working out the above based on the rising of a star
• The time difference between time zones.
• Working out the longitude and latitude of a location

Let's start with a few questions and answers. To help you there is a statistics chart and graph on the right. If you want to know what the EOT is on 5th May, look up the date and you will find it is +3.

Question 1

If it is 12:00 on a Sundial in Greenwich inLondon, we can work out that a watch will read (mean solar time)?

EOT = apparent solar time – mean solar time

EOT = +3 (Sundial Fast)

EOT = 12:00 – 11.57

Question 2

If a Sundial reads 14:30 GMT on 15th July. What time will it be on an accurate clock?

Mean solar time = apparent solar time – EOT

Mean solar time = 14.30 - (-6)

Mean solar time = 14.36

When a negative is subtracted it becomes positive. When a positive is subtracted it becomes a negative.

Question 3

If a clock at Greenwich reads 10:45 on 25th October, what will a sundial read?

Apparent solar time = Mean solar time + EOT

Apparent solar time = 10:45 + (+)16

Apparent solar time = 11.01

When a positive is added to another positive it is a positive.

Question 4

An observer sees the Sun culminate at 11.56 GMT. The EOT on that date is -9. At what longitude is the observer located?

Apparent solar time = Mean solar time + EOT

Apparent solar time = 11.56 + -9 = 11.56 - 00.09
=11.47

4 minutes = 1 degree of longitude so 1 minute = 0.25 degree of longitude 13 minutes = 3.25 degrees

As the amount is negative (greater than the apparent solar time) it is East. Positive would be West.
= 3.25 degrees East

Question 5

From a longitude, an observer sees the star Betelgeuse rise at 21.20 GMT on 15th November. What time will the observer see the star rise on 15th December?

Star rises 4 minutes earlier each day

Difference of 30 days

30 x 4 = 120 minutes. Equivalent to 2 hours

21.20 – 02:00 = 19.20

Wasn't that fun? No? What do you mean?

Summary

EOT = apparent solar time – mean solar time

Mean solar time =
apparent solar time – EOT

Apparent solar time =
Mean solar time + EOT

Equation of Time Dates

Figures are rounded to minutes and are approximate.

Month

5th

15th

25th

January -5 -9 -12
February -14 -14 -13
March -12 -9 -6
April -3 0 +2
May +3 +4 +3
June +2 0 -2
July -4m -6 -6
August -6 -5 -2
September 0 +5 +8
October +11 +14 +16
November +16 +15 +13
December +10 +5 0
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