Spherical Trigonometry

Spherical Trigonometry distances : angles : declination or latitude : sunset position : sundial Introduction Spherical Trigonometry deals with triangles drawn on a sphere. The subject originated in the Islamic Caliphates of the Middle East, North Africa and Spain during the 8th to 14th centuries. It arose to solve an apparently simple problem: Which direction is Mecca? The development of this subject lead to improvements in the art of navigation, stellar map making, geographic map making, the positions of sunrise and sunset, and improvements to the sundial.

In the figure above, a triangle, ABC, is drawn on a spehere. Each line of the triangle is a Great Circle. These are circles drawn on a sphere with the same radius as the sphere. Great circles cover the shortest distance between two points. The capital letters (A, B, C) denote the angles between the great circle arcs of the triangle as measured on the surface of the sphere. The small latters (a, b, c) represent the lengths of the great circle arcs measured as angles from the centre of the sphere. A spherical triangle, differs from a plane triangle in that the sum of the angles is more than 180 degrees.

The Cosine Rule There is a Cosine Rule for spherical triangles: Cos(a) = Cos(b) × Cos(c) + Sin(b) × Sin(c) × Cos(A) Cos(b) = Cos(a) × Cos(c) + Sin(a) × Sin(c) × Cos(B) Cos(c) = Cos(b) × Cos(a) + Sin(b) × Sin(a) × Cos(C) The Cosine Rule allows the length of one of the arcs of a spherical triangle to be evaluated if the other two arcs and the angle opposite the arc are known.

The Sine Rule The Sine Rule for spherical triangles is: Sin(a) / Sin(A) = Sin(b) / Sin(B) = Sin(c) / Sin(C) The Sine Rule can be used to find an angle if two sides and an angle are known OR to find a side if two angles and a side are known.

Tables and Data This is a table of Latitudes and Longitudes of selected cities (in alphabetical order) for the worked examples. City Country Latitude Longitude Greece 38.00°N 23.44°E Athens Baghdad

Iraq

33.20°N

44.26°E

Beijing

China

39.55°N

116.26°E

Buenos Aires

Argentina

34.40°S

58.30°W

Cape Town

South Africa

33.56°S

18.28°E

Chicago

USA

41.50°N

87.45°W

Jakarta

Indonesia

6.08°S

106.45°E

London

UK

51.30°N

0.10°W

Mecca

Saudi Arabia

21.26°N

39.49°E

Mexico City

Mexico

19.25°N

99.10°W

Nairobi

Kenya

1.17°S

36.50°E

Australia 16.40°S 139.30°E Sydney This is a table of Declinations and Right Ascensions of selected stars (in alphabetical order) for the worked examples. Star Name Declination Right Ascension -60.5° 14h 40m Alpha Centauri +38.5° 18h 40m Vega The diameter of the Earth is assumed to be 12,756 km giving a circumference of 40,074 km. To convert kilometers to miles, divide by 1.609

Distance Between Two Points Definitions The Cosine and Sine Rules can be used to solve the basic problems of navigation on the surface of the Earth.

In the figure above, the points B and C are two points on the surface of the Earth. We can define the following:       

Point A is the North Pole. The great circle joining points B and C is the shortest distance between them. The great circle (in blue) joining B' and C' is the Equator (Latitude 0°). The great circle (red) joining ABB' is a line of Longitude. It is the Longitude of B. The great circle joining ACC' is another line of Longitude. It is the Longitude of C. The length of the great circle arc B'B is the Latitude of point B. The arc B'A is 90° (Equator to Pole). The length of the great circle arc C'C is the Latitude of point C. The arc C'A is also 90°.

Latitude (λ) is measured in degrees (°) measured from the Equator Northwards (marked N) or Southwards (S). The Latitude of the North Pole is 90°N and the Latitude of the South Pole is 90°S. Southern Latitudes are considered negative (by convention). The Equator is a natural line on the Earth as it represents the great circle bisecting the Earth's axis of rotation. Longitude (L) is measured in degrees East (E) or West (W) of the Line of Longitude passing through Greenwich Observatory, in a suburb of London (UK). This is called the Prime Meridian of the Greenwich Meridian. It is not a natural line and has been chosen by convention.

Using the Cosine Rule The Cosine Rule can be used to determine the distance between points B and C (the arc, a) as follows: Cos(a) = Cos(b) × Cos(c) + Sin(b) × Sin(c) × Cos(A) where

  

A is an angle measured in degrees. It is the difference in Longitude between points B and C. The great circle arc, b, is 90° minus the Latitude of C. This is called the Polar Distance. The great circle arc, c, is 90° minus the Latitude of B.

Example 1: Find the distance bewteen London (UK) and Baghdad (Iraq). If we use the diagram below then B is London while C is Baghdad. From the table above London has a Latitude of 51.30°N and a Longitude of 0.10°W. Baghdad has a Latitude of 33.20°N and a Longitude of 44.26°E.

To use the Cosine Rule, we need to list the other terms needed: 



 

We seek a, the angular distance along a great circle between London and Baghdad. This is the shortest distance between these two cities along the surface of the Earth. It is shown in red in the diagram. A (the difference in Longitude between London and Baghdad) is 44.26°E + 0.10°W = 44.36°. The two cities lie on different sides of the Greenwich Meridian (just) so we add the Longitudes. b is the Polar Distance of Baghdad given by 90° minus the Latitude of Baghdad (90° 33.20° = 56.80°). c is is the Polar Distance of London: 90° minus the Latitude of London (90° - 51.30° = 38.70°).

By the Cosine Rule: Cos(a) = Cos(b) × Cos(c) + Sin(b) × Sin(c) × Cos(A) putting in the values:

Cos(a) = Cos(56.80°) × Cos(38.70°) + Sin(56.80°) × Sin(38.70°) × Cos(44.36°) which gives: Cos(a) = 0.5476 × 0.7804 + 0.8368 × 0.6252 × 0.7150 Cos(a) = 0.4273 + 0.3741 = 0.8014 Therefore, the great circle arc joining London and Baghdad (a) is 36.74°. The Earth's circumference is 40,074 km which is equivalent to 360°. The value 36.74° is 36.74 / 360 of the Earth's circumference. Therefore the distance between London and Baghdad is given by 40,074 km × 36.74 / 360 = 4089 km

Cosine Rule With Latitudes In the section on Trigonometry, we saw that for any angle, X, in degrees, Cos(90 - X) = Sin(X) and Sin(90 - X) = Cos(X) so we can re-write the Cosine Rule to use (the more readilly available) Latitudes instead of Polar Distances. There are two forms of this rule depending on if the values of the two Longitudes. If the Longitudes are both on the same side of the Greenwich Meridian, (i.e both E or both W), the formula is given by: Cos(DistBC) = Sin(λC)Sin(λB) + Cos(λC)Cos(λB)Cos(LC - LB) If the Longitudes are on different sides of the Greenwich Meridian (i.e. One is E and the other is W), the formula is given by: Cos(DistBC) = Sin(λC)Sin(λB) + Cos(λC)Cos(λB)Cos(LC + LB) In either form:     

DistBC is the angular distance between B and C along an arc of a great circle. This is the shortest distance between the two points. λB is the Latitude of B (positive for N, negative for S). λC is the Latitude of C (positive for N, negative for S). LB is the Longitude of B. LC is the Longitude of C.

Example 2: Find the distance bewteen Chicago (USA) and Mexico City (Mexico). If we set B to Mexico City and C to Chicago in the diagram below and we use the Cosine Rule With Latitudes:

Cos(DistBC) = Sin(λC)Sin(λB) + Cos(λC)Cos(λB)Cos(LC [+-] LB)

    

DistBC is the required angular distance between Mexico City and Chicago. This is a on the diagram above. λB is the Latitude of Mexico City (Angle BB' = 19.25°N). λC is the Latitude of Chicago (CC' = 41.50°N). LB is the Longitude of Mexico City (99.10°W). The great circle ABB' is this Longitude. LC is the Longitude of Chicago (87.45°W). Great circle ACC'.

The two Longitudes are both West of the Greenwich Meridian, so we use the following Cosine Rule With Latitudes: Cos(DistBC) = Sin(λC)Sin(λB) + Cos(λC)Cos(λB)Cos(LC - LB) Putting in the values and taking the difference between the two Longitudes we have: Cos(DistBC) = Sin(41.50°)Sin(19.25°) + Cos(41.50°)Cos(19.25°)Cos(11.65°) This gives: Cos(DistBC) = 0.6626 × 0.3297 + 0.7490 × 0.9441 × 0.9794 = 0.9110 Therefore: DistBC = 24.36° The distance between Chicago and Mexico City is given by: 40,074 km × 24.36 / 360 = 2712 km Example 3: Find the distance bewteen Buenos Aires, Argentina and Athens, Greece.

Set B to Buenos Aires and C to Athens and we use the Cosine Rule With Latitudes, Cos(DistBC) = Sin(λC)Sin(λB) + Cos(λC)Cos(λB)Cos(LC [+-] LB)     

DistBC is the required angular distance between Buenos Aires and Athens (a). λB is the Latitude of Buenos Aires (B'B = 34.40°S). Being South of the Equator this can be written as -34.40°. λC is the Latitude of Athens (CC' = 38.00°N). LB is the Longitude of Buenos Aires (58.30°W). Great circle AB'B. LC is the Longitude of Athens (23.44°E). Great circle ACC'.

The two Longitudes are on different sides of the Greenwich Meridian, so we use the following Cosine Rule With Latitudes: Cos(DistBC) = Sin(λC)Sin(λB) + Cos(λC)Cos(λB)Cos(LC + LB) Put in the values. Remember we are using a negative Latitude for Buenos Aires: Cos(DistBC) = Sin(38.00°)Sin(-34.40°) + Cos(38.00°)Cos(-34.40°)Cos(81.74°) From the graphs of sines and cosines in the Trigonemtric Equations essay we know that: Sin(-X) = -Sin(X) and Cos(-X) = Cos(X) Therefore, Cos(DistBC) = -Sin(38.00°)Sin(34.40°) + Cos(38.00°)Cos(34.40°)Cos(81.74°) Which gives: Cos(DistBC) = -(0.6157 × 0.5650) + 0.7880 × 0.8251 × 0.1437 = -0.2544

Therefore: DistBC = 104.74° The distance between Buenos Aires and Athens is given by: 40,074 km × 104.74 / 360 = 11,659 km

Direction Between Two Points Definitions In this section we will look at the direction between the two points. If we are travelling from B to C (in the diagram below) then we must travel in a certain direction given by angle ABC. This angle is called the Heading or Course and is measured in degrees clockwise from North.

Using The Sine Rule The Sine Rule can be used to determine the direction between points B and C (the angle, B) as follows: Sin(a) / Sin(A) = Sin(b) / Sin(B) = Sin(c) / Sin(C) where  

The lower case letters (a, b, c) are the arcs of great circles. The upper case letters(A, B, C) are the angles between the great circles making up the triangle. The angle A is the difference in Logitudes between points B and C. The

Longitudes are added if they are on different sides of the Greenwich Meridian and subtracted if they are on the same side of the Greenwich Meridian. Example 4: What direction is Mecca (Saudi Arabia) from Jakarta (Indonesia). Using the diagram below, set C to Jakarta and B to Mecca.

We are looking for angle C. This is the angle between the line of Longitude (b) which faces due North and a, the arc of the great circle joining point C to point B. The arcs of the great circles (c and b) are known. They are simply the Polar Distances of points B and C respectivly (90° minus the Latitudes of these points). Angle A is also known: it is the difference in Longitudes between points C and B. The only way that the Sine Rule can be used is as follows: Sin(c) / Sin(C) = Sin(a) / Sin(A) Rearranging for C, gives Sin(C) = Sin(A) × Sin(c) / Sin(a) where    

C is the angle required. It will be West of North (Angle ACB on the diagram). A is the Longitude difference between Mecca and Jakarta. In this case the value is 66.96°. c is the Polar Distance of Mecca (90° - 21.26° = 68.74°). a is the Distance between Mecca and Jakarta. This is found by using the Cosine Rule and has a value of 71.08° (corresponding to a distance of 7912 km).

By the Sine Rule,

Sin(C) = Sin(A) × Sin(c) / Sin(a) = Sin(66.96°) × Sin(68.74°) / Sin(71.08°) which gives Sin(C) = 0.9202 × 0.9319 / 0.9460 = 0.9066 This gives a value for angle C, C = 65.04° To face Mecca from Jakarta, one must face North and turn anti-clockwise through an angle of 65.04°. Example 5: An airplane is to fly from Cape Town in South Africa to Beijing in China. If the craft flies in a straight line, what Heading should the pilot follow. Using the diagram below, set B to Cape Town and C to Beijing.

We are looking for angle B. This is the angle between the line of Longitude (c) which faces due North and a, the arc of the great circle joining points B and C. Again, we use the Sine Rule: Sin(a) / Sin(A) = Sin(b) / Sin(B) Rearranging for B, gives Sin(B) = Sin(A) × Sin(b) / Sin(a) where

   

B is the angle required. It will be East of North. Heading is measured clockwise from North therefore this angle will be the Heading. A is the Longitude difference between Cape Town and Beijing. In this case the value is 97.98°. b is the Polar Distance of Beijing (90° - 39.55° = 50.45°). a is the Distance between Cape Town and Beijing. This is found by using the Cosine Rule and has a value of 116.18° (corresponding to a distance of 12,933 km).

By the Sine Rule, Sin(B) = Sin(A) × Sin(b) / Sin(a) = Sin(97.98°) × Sin(50.45°) / Sin(116.18°) which gives Sin(B) = 0.9903 × 0.7711 / 0.8974 = 0.8509 This gives a value for angle B, B = 58.31° To fly from Cape Town to Beijing, one must keep to a Heading of 58.31°.

Declination or Latitude from Stars Definitions In the essay about Coordinate Systems, it was shown that there are two local coordinates that can be used to fix the position of a star from a particulat location. In the diagram below, point O is the observer and S is a star:  

The angle NOT is the Azimuth, α, (measured in degrees clockwise from North). The angle TOS is the Altitude, measured in degrees from the horizon. In practice it is easier to measure Zenith Distance, z, which is 90° - Altutude. The Zenith is the point directly overhead.

A star map (or a table of star positions) is a very useful aid in navigation. The Celestial Sphere has a coordinate system analagous to that on the Earth. The Right Ascention is analagous to Longitude on the Earth. It is found by measuring times and will not be discussed here. The Declination of a star is the number of degrees North or South of the Celestial Equator. Once these two coordinates are known, the star can be plotted accurately onto a star map or tabulated for use in navigation.

Finding Declinations If a star's Azimuth (α) and Zenith Distance (z) are measured at the same time from a location whos Latitude (λ) is known, the star's Declination (given the symbol, δ) can be found by using the Declination Formula below. Sin(δ) = Sin(λ)Cos(z) - Cos(λ)Sin(z)Cos(α) Stars have fixed positions (in a human lifetime) but objects like the Sun, Moon, planets and comets change their positions in the sky so this process can also be used to track their movements against the background stars and plot their paths on a star map. Example 6: A new comet is found in Sydney, Australia. Its Azimuth (α) is measured as 57.40° while its Zenith Distance (z) is 83.70°. What is its Declination. Before using the formula we list the values to be used:    

The Declination (δ) required. The Azimuth (α) is 57.40°. The Zenith Distance (z) is 83.70°. The Latitude of Sydney (λ) is 16.40°S (which can be written as -16.40°).

Using the Declination Formula,

Sin(δ) = Sin(λ)Cos(z) - Cos(λ)Sin(z)Cos(α) which gives Sin(δ) = Sin(-16.40°) × Cos(83.70°) - Cos(-16.40°) × Sin(83.70°) × Cos(57.40°) Remembering that Sin(-X) = -Sin(X) and Cos(-X) = Cos(X), this can be re-written as Sin(δ) = -Sin(16.40°) × Cos(83.70°) - Cos(16.40°) × Sin(83.70°) × Cos(57.40°) This gives Sin(δ) = -(0.2823 × 0.1097) - (0.9593 × 0.9940 × 0.5388) = -0.5447 Therefore, The Declination (δ) = -33.00° On a star map, the comet lies 33° South of the Celestial Equator.

Finding Latitude This is the reverse of finding Declinations. However, two stars must be used. Both must be of known Declination. To find the Latitude, the navigator requires a table of Declinations and an instrument to measure the Zenith Distance and Azimuth of the selected stars. The method is to measure the Zenith Distance and Azimuths of two known stars at the same time. The formula is long so it is broken down below: Cos(λ) = -P / Q where P = Sin(δ1)Cos(z2) - Sin(δ2)Cos(z1) and Q = Cos(z1)Sin(z2)Cos(α2) - Sin(z1)Cos(z2)Cos(α1) The symbols are defined as follows:    

λ is the required Latitude. z1, z2 are the Zenith Distances of the two selected stars. α1, α1 are the Azimuths of the two selected stars. δ1, δ1 are the Declinations of the selected stars (from tables or a star map).

In the following example, angles are measured to only 1 decimal place for brevity. As a side note, the Longitude is found by knowing the local time and comparing it with Greenwich Mean Time. Each difference of one hour corresponds to a difference of 15° in Longitude. This is not covered in this essay. Example 7: On a ship, the following measurements are made of the stars Alpha Centauri (Azimuth: 183.0°; Zenith Distance: 74.0°) and Vega (Azimuth: 51.0°; Zenith Distance: 54.5°). Find the Latitude. Firstly we list the items required for the formula.    

λ is the required Latitude. z1, z2 are the Zenith Distances of Alpha Centauri (74.0°) and Vega (54.5°). α1, α1 are the Azimuths of Alpha Centauri (183.0°) and Vega (51.0°). δ1, δ1 are the Declinations of Alpha Centauri (-60.5°) and Vega (38.5°) taken from the table above.

We can now begin to insert the values into the formula, begining with P: P = Sin(δ1)Cos(z2) - Sin(δ2)Cos(z1) = Sin(-60.5°) × Cos(54.5°) - Sin(38.5°) × Cos(74.0°) giving P = -(0.8704 × 0.5807) - (0.6225 × 0.2756) = -0.6770 Now, Q: Q = Cos(z1)Sin(z2)Cos(α2) - Sin(z1)Cos(z2)Cos(α1) = Cos(74.0°) × Sin(54.5°) × Cos(51.0°) - Sin(74.0°) × Cos(54.5°) × Cos(183.0°) giving Q = (0.2756 × 0.8141 × 0.6293) - (0.9613 × 0.5807 × -0.9986) = 0.6987 Finally, Cos(λ) = -P / Q = - (-0.6770) / 0.6987 = 0.9690 This gives us a value for the Latitude, λ = 14.3°

Sunrise and Sunset Points Definitions The Sun's Declination changes throughoiut the year from +23.5° on 21 June to -23.5° on 21 December. These two dates are called the Solstices. This movement from north of the Equator to south of the Equator causes the seasons. It also changes the length of daylight and the position of local sunrise and sunset throughout the year. On the Equinoxes (21 March and 23 September) the Sun's Declination is 0° and the Sun rises exactly in the East (Azimuth 90°) and sets exactly in the West (Azimuth 270°). At other times of the year the Sun rises and sets North or South of the East or West point as its Declination changes.

In the diagram above (set in the Northern Hemisphere), the red line is the daily path of the Sun on the Equinox. It rises in the East (Azimuth = NOE) passes through the South (when it is at its highest) and sets in the West (Azimuth = NOW). The blue line is the daily path of the Sun during the December Solstice. The Sun rises South of East (Azimuth = NOR), passes through the South (lower than at the Equinoxes) and sets to the South of West (Azimuth = NOT).

In the diagram above, the green line is the daily path of the Sun during the June Solstice. The Sun rises North of East (Azimuth = NOR), passes through the South (higher than at the Equinoxes) and sets to the North of West (Azimuth = NOT). This formula calculates the rising and setting points of the Sun (or any other object) if its Declination is known for a given Latitude. Cos(α) = Sin(δ) / Cos(λ) The symbols are defined as follows:   

α is the required Azimuth. δ is the Declinations of the Sun (or other object). λ is the Latitude of the observer.

Example 8: Show that on 21 March (when the Sun's Declination is 0°), the Sun rises exactly due East in London (UK), Nairobi (Kenya) and Sydney (Australia). First we list the values to be used:   

α is the required Azimuth. δ is the Declinations of the Sun (0° on 21 March). λ is the Latitude of London, Nairobi or Sydney. These values 51.30°N, 1.17°S and 16.40°S respectively.

Using the formula, Cos(α) = Sin(δ) / Cos(λ)

It is clear that, regardless of the value of λ, when δ = 0, Cos(α) = Sin(0°) / Cos(λ) = 0 / Cos(λ) = 0 therefore, α = 90° (due East) and 270° (due West) This proves that the Sun rises and sets due East and West respectively on 21 March, for the three cities. Example 9: Find the sunrise position in Nairobi on 21 June and 21 December. First we list the values to be used:   

α is the required Azimuth. δ is the Declinations of the Sun on these two days (+23.5° on 21 June; -23.5° on 21 December). λ is the Latitude of Nairobi (1.17°S or -1.17°).

Using the formula, Cos(α) = Sin(δ) / Cos(λ) For Nairobi on 21 June, Cos(α) = Sin(23.5°) / Cos(-1.17°) = 0.3987 / 0.9998 = 0.3988 For Nairobi on 21 December, Cos(α) = Sin(-23.5°) / Cos(-1.17°) = -0.3987 / 0.9998 = -0.3988 This gives the two sunrise points required: α = 66.50° (23.5° North of East) and 113.5° (23.5° South of East) Nairobi is close to the Equator (λ = 1.17°S). The difference in sunrise position on the two Solstice days is 47°, which is the difference in the Sun's Declination between the two dates. Example 10: Find the sunrise position in London on 21 June and 21 December. First we list the values to be used:  

α is the required Azimuth. δ is the Declinations of the Sun on these two days (+23.5° on 21 June; -23.5° on 21 December).



λ is the Latitude of London (51.30°N).

Using the formula, Cos(α) = Sin(δ) / Cos(λ) For London on 21 June, Cos(α) = Sin(23.5°) / Cos(51.30°) = 0.3987 / 0.6252 = 0.6377 For London on 21 December, Cos(α) = Sin(-23.5°) / Cos(51.30°) = -0.3987 / 0.6252 = -0.6377 This gives the two sunrise points required: α = 50.38° (39.62° North of East) and 129.62° (39.62° South of East) London is far North of the Equator (λ = 51.30°N). The difference in sunrise position on the two Solstice days is over 79°.

The Sundial History and Description The earliest time keeper was the sundial. These originally consisted of a stick (called a gnomon) placed vertically in the ground. From the position of the shadow, an idea of the time of day could be obtained. The simple gnomon was used by all the major ancient civilisations including the Babylonians, Egyptians, Indus Valley, Chinese, Greek and Roman. One problem with a vertical gnomon is that the Sun's Declination changes throughout the year. This affects the Sun in two ways:  

The Sun is at different Altitudes for a given time of day at different times of the year. The Sun's apparent path in the sky changes throughout the year.

These two effects mean that the movement of the Sun's shadow during a day in June is different to the movement of the shadow in December. An hour in June, as measured from the Sun's shadow at the foot of a gnomon, is a different length to an hour measured in December. Each month requires a different scale at the foot of the gnomon for telling the time accurately.

The Modern Sundial During the 10th century AD, the Moorish rulers of Spain solved this problem by making the sundial at an angle.

The diagram above shows the Moorish sundial which is now a common ornament in UK gardens. The angle OQP is equal to the Latitude of the place where the sundial will be located. A sundial for Baghdad will not work in London or Athens. The sundial is positioned (in the Northern Hemisphere) so that the line QP faces the point directly above the North Pole in the sky (The North Celestial Pole). In other words, the sundial is aligned along the North-South meridian. This makes the line QP parallel to the Earth's axis of rotation. This simple change has the effect of making all hours, measured by the Sun's shadow, the same length throughout the year, regardless of the Sun's apparent path in the sky. The Sun's shadow at local noon will face due North. For each hour after local noon, the shadow of the sundial (called the Shadow Angle and shown by OTQ in the diagram) will face a different direction. There will be a regular movement of the shadow for each time unit and this can be calculated. Once the sundial has been built with the correct angle and has been positioned properly, the graduations for measuring the time can be found with the following formula. Tan(S) = Sin(λ) × Tan(15 × H) where



 

S is the Shadow Angle for a given time (say 1pm). The Shadow Angle is measured in degrees from the position of the Sun's shadow at local noon. This is where the mark for that time will be placed on the scale at the foot of the sundial. λ is the Latitude of the sundial (which must equal angle OQP for it to work correctly). H is the hour required. It is multiplied by 15 in the formula because during a complete 24 hour solar day, the Sun appears to go once around the Earth (360°). This means that 360° is equivalent to 24 hours; making 15° equivalent to one hour (360 / 24 = 15).

Example 11: Find the Shadow Angle for 2pm for a sundial in London . The values needed are:   

S (the Shadow Angle). This is what we are looking for. λ is the Latitude of London (51.30°). H is the hour required, in this case, 2.

Using the formula, Tan(S) = Sin(λ) × Tan(15 × H) = Sin(51.30°) × Tan(15° × 2) = Sin(51.30°) × Tan(30°) This gives, Tan(S) = 0.7804 × 0.5774 = 0.4506 The Shadow Angle, S, is therefore S = 24.26° The Sun's daily motion is East to West (if facing South, the motion is left to right in the Northern Hemisphere). Therefore the Sun's shadow will move from its North position at local noon towards the East. The mark for 2pm will be put at 24.26° from the North. Since the movement of the Sun is symetrical, the mark for 10am (noon minus 2 hours) will also be at 24.26° but on the West side of North. In this manner all the hours (or half hours or even quarter hours) can be marked on the sundial scale.

Tables The following is a table of Shadow Angles from local noon (0h) to 5pm (5h) for sundials in London (UK), Athens (Greece) and Mexico City (Mexico) derived from the above formula. Of course, the same angles can also be used from 7am to local noon. London Athens Mexico City Time (λ = 51.30°) (λ = 38.00°) (λ = 19.25°) 0° 0° 0° 0h

1h

11.81°

9.37°

5.05°

2h

24.26°

19.57°

10.78°

3h

37.97°

31.62°

18.25°

4h

53.51°

46.84°

29.73°

71.05° 66.48° 50.90° 5h This table allows the building and marking of sundials in the three cities named above or in any place with the same Latitudes. From the table we can see that in all three cities the Shadow Angle is 0° at local noon (0h). The other angles all differ. For example, in London, the Sun's shadow will move 37.39° between local noon and 3pm. This is the Shadow Angle for 3h in the table. In Athens the same time period will produce a shadow movement of 31.62°. In Mexico City, the figure will be 18.25°. The closer the sundial is to the Equator, the less movement of the Sun's shadow between local noon and 3pm.

Sundial Corrections Before a sundial can be used for telling the time, three corrections must be made to the reading of the position of the shadow. The Sun is not, in fact, a perfect time keeper. This is because its movements have certain irregularities that can make the time read from a sundial differ by up to 17 minutes from clock time. A correction needs to be made (called the Equation of Time) and subtracted from the time read from a sundial to give the correct clock time. This correction depends only on the date. A diagram is shown below. The Equation of Time is zero on four days of the year.

Most countries use Time Zones. These are strips of territory, usually 15° wide. The clock time is set to the local time in the centre of the time zone. If the sundial is exactly on the centre of the time zone, no correction needs to be made. If, however, the sundial is located at the edge of the

time zone it may be 7.5° away from the centre. Since 1 hour is equivalent to 15°, this error will cause the sundial time to be up to half an hour out from the clock time. This correction depends on the location of the sundial and has a constant value at all times. It is equal to Time Zone Correction (in hours) = (LS - LT) × 15° where LS is the Longitude of the sundial and LT is the Longitude of the centre of the Time Zone. If the sundial is to the East of the time zone centre, the correction is subtracted, if to the West the correction is added. The third correction is Summer Time or Daylight Saving Time. If it is force, an hour must be added to the sundial time to give the clock time. Each country has different rules for when this extra hour is in force.