Winter Stars in the Northern Hemisphere

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Venus Evening Star or Morning Star?

Posted on April 25, 2020 by Jack Case

venus

Venus sometimes appears as an evening star above the western horizon shortly after sunset and sometimes appears as a morning star above the eastern horizon shortly before sunrise. In primitive times, people regarded the evening and morning stars as two different heavenly bodies but in the sixth century BC, Pythagoras suggested that they might be one and the same body.

Ptolemy believed that the Earth was the centre of the Universe and that the Sun moved in a circular orbit around the Earth which was stationary. He also believed that Venus orbited the Sun as the Sun itself orbited the Earth and that this explained why Venus appeared from Earth as an evening star during part of its orbit and as a morning star during another part. The Ptolemaic hypothesis can be explained with the aid of diagram 1 below:

According to the Ptolemy, if Venus were at a point in its orbit somewhere on the semicircle ABC, say at B, then the line joining the Earth to Venus would be to the left of the line joining the Earth to the Sun. Therefore, looking from Earth, Venus would appear to be to the left of the Sun so, when the Sun sets in the west, Venus would be seen for some time after sunset above the western horizon as an evening star. In a similar way, if Venus were at a point somewhere on the semicircle ADC, say at D, it would appear from Earth to be to the right of the Sun and it would therefore appear above the eastern horizon just before sunrise as a morning star.

We now know, thanks to the work of Copernicus, that both Earth and Venus orbit the Sun, as do all of the planets in our solar system. According to the Copernican hypothesis, the reason that Venus sometimes appears as an evening star and sometimes as a morning star can be explained with the aid of diagram 2 below:

The average distance of Venus from the Sun is 108 million Km. while the average distance of Earth from the Sun is 150 million Km. Because the circumference of the orbit of Venus is smaller than that of the Earth, we can safely assume that Venus takes less time to complete its orbit than the Earth does. Whereas the Earth moves through 360^o in 365.25 days, Venus completes its orbit in 225 days. For this reason Venus gains 0.6^o per day on the Earth and overtakes it at intervals of approximately 600 days.

From the diagram it will be seen that, after Venus has overtaken the Earth (that is after it has passed the point V) it becomes a morning star. It increases its angular distance to the right of the Sun until it reaches 44^o at point X. After passing X, the angular distance to the right of the Sun decreases until it reaches point Z, which is behind the Sun. After passing Z, it makes its appearance to the left of the Sun and is now an evening star. As it approaches Y, its angular distance left of the Sun increases until it reaches 44^o at point Y. After passing Y on its way to V, its angular distance left of the Sun decreases until, at V, it is zero when it will pass either slightly above or below the Sun. On extremely rare occasions, it crosses in front of the Sun and this is known as a ‘transit of the planet’.

To sum up, Venus overtakes the Earth at intervals of approximately 600 days. During approximately 300 of these days, it is a morning star and for the other 300 days, it is an evening star. The maximum angular distance right or left of the Sun, is roughly 44^o.

What about Mercury? In a similar way, Mercury is also an evening and a morning star. Mercury is the closest planet to the Sun and because it has the most elliptical orbit, its distance from the Sun ranges from 29 million Km. to 47 million Km. Because the circumference of its orbit is comparatively small, it gains 3^o on the Earth per day and overtakes it on average every 120 days. For 60 of these days it will be a morning star and for the other 60 it will be an evening star. Its maximum angular distance left or right of the Sun is roughly 24^o.

So Venus and Mercury are both morning stars and evening stars but it is quite easy to tell them apart for the following reasons: Venus usually appears much higher in the sky than Mercury and is far brighter; in fact, Venus is the third brightest object in the sky after the Sun and the Moon.

In-depth coverage of this and similar topics can be found in the book Astro Navigation Demystified.

Books of the Astro Navigation Demystified Series:

Celestial Navigation. Theory and Pract

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The Retrograde and Prograde Motions of Mars and Jupiter

Posted on September 15, 2019 by Jack Case

Mars
We can best understand the retrograde and prograde motions of Mars by considering the planet’s motion relevant to distant stars. To do this, we must study the effect of the motions of both the Earth and Mars around the Sun. Firstly, the circumference of Mar’s orbit is 1.6 times greater than that of the Earth and with an orbital speed of 24 km/s, it takes 686.98 Earth days to complete each orbit while Earth travels at 30 km/s taking 365.25 days. Thus, the Mars’ year is 1.8 times as long as the Earth year.

Secondly, the orbit of the Earth is closer to the Sun than that of Mars, and this puts Earth on the inside track, so to speak. Taking this into account along with Earth’s greater speed, it is easy to see that the Earth will overtake Mars from time to time; in fact it laps Mars every 26 months.

The following diagram represents the orbits of the Earth and Mars around the Sun with the Earth on the ‘inside track’ and Mars on the outside.

When Mars is at point X and Earth is at point Y, Mars and Earth are at their closest and although they are both moving from right to left of the diagram, because Mars is travelling 6 km/s slower, it will appear, to an observer on the Earth, to be moving left to right; this is retrograde motion. However, if the Earth is at point Z, it will appear that Mars is moving in the opposite direction, that is right to left. The speed at which the two planets will be moving in opposite directions will be equal to their combined orbital speeds; i.e. 54 km/s. This is called prograde motion’.

These changes in the apparent motion of Mars from retrograde to prograde and vice versa are not sudden changes. Before a change in direction, the planet seems to slow down and then pause for about a week before starting to move in the new direction.

To summarize, Mars will sometimes appear to be moving from left to right with respect to the background stars, sometimes it will seem to move in the opposite direction and in between these changes in, it will appear to pause.

Jupiter

Just as the movement of Mars appears to change between retrograde and prograde, so does that of Jupiter. Jupiter moves across the sky in a very predictable pattern, but every now and then it reverses direction in the sky, making a tiny loop against the background stars – this is Jupiter in retrograde.

The following diagram shows that, as Jupiter is overtaken by the Earth, its apparent motion across the sky appears to describe a loop as its direction changes from prograde to retrograde and then back to prograde again.

At position 1, Jupiter appears to be moving from west to east in prograde motion. At positions 2 and 3, its direction appears to have changed from prograde to retrograde so that it is now moving from east to west. At position 4, it appears to have resumed prograde motion as it moves from west to east again.

Note. Sky maps can be very confusing because they are not drawn in the conventional way with east on the right and west on the left. They are drawn as they would appear in the sky if we were lying down with our legs pointing to the south and looking upwards so that east would be on our left and west on our right.

Jupiter’s retrograde periods last for 4 months and are then followed by periods of nine months of prograde motion before going retrograde again. So the time from the beginning of one retrograde movement to the beginning of the next is approximately 13 months.

The relatively slow movement of Jupiter across the sky makes it very easy for the navigator to locate. Its orbit takes 11.85 earth years to complete and during that time, it appears to move from one constellation to another every 13 months describing its retrograde loop as it pauses in each one. Its path leads it through Leo, Virgo Libra, Scorpius, Sagittarius, Capricorn, Aquarius, Pisces, Aries, Taurus, Gemini, Cancer and then back to Leo to begin the sequence all over again. This predictable path across the sky together with the fact that Jupiter is the fourth brightest celestial body in the sky explains why it is such an important navigational planet.

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Making An Ocean Plotting Sheet

Posted on September 6, 2019 by Jack Case

Ocean plotting sheets are nautical charts designed for use at specific latitudes. They are used for ocean navigation where features such as shorelines, water depths and local tidal information are of no importance. Plotting sheets are particularly useful for plotting observed positions obtained by astro navigation.

Purchasing ocean plotting sheets can be an expensive outlay particularly when they are required for navigation courses or for practising plotting skills so why not make your own, all you need is a sheet of graph paper.

A nautical chart is a graphic representation of a sea area and is based on the principle of the Mercator projection which represents curved lines on the Earth’s surface as straight lines on the chart.

Chart-Scale. This is the ratio between distance on the ground and the corresponding distance on the chart. The example below shows part of an ocean plotting sheet. In the example, you will see that three different scales seem to have been used; one for distance, one for longitude and one for latitude. The reasons for this are explained below.

Nautical Mile. The international nautical mile (n.m.) is closely related to the geographical mile and is a unit of length corresponding to approximately 1 minute of arc along any meridian of longitude. It is defined as exactly 1852 metres. .

Longitude. Meridians of longitude are by definition great circles and an angle of 1 minute at the Earth’s centre will subtend an arc of 1 nautical mile along the surface of a great circle. From this it follows that 1 minute of arc along a meridian of longitude is equal to a distance of 1 nautical mile and 1 degree of arc is equal to 60 nautical miles. Since the chart scale in the example below is 1mm: 1 nautical mile and since 1 minute of latitude when measured along a meridian of longitude = 1 n.m. the scale used for latitude is 1 mm : 1′ of lat.

Latitude. A parallel of latitude is a small circle that is parallel to the Equator. We know that the distance between adjacent meridians of longitude when measured along the Equator is 60 n.m. This is because the Equator is a great circle; however, parallels of latitude are small circles and this presents us with a problem. Moving from the Equator towards the poles, the meridians of longitude gradually converge and the circumference of the parallels of latitude gradually become smaller. Because of this, the distance between adjacent meridians of longitude when measured along a parallel of latitude has to be calculated for each parallel.

Calculating the distance between adjacent meridians of longitude along parallel of latitude 53^oS in the example below. As explained in my book ‘Applying Mathematics To Astro Navigation’, the formula for calculating the distance along a parallel of latitude is “difference in distance (Ddist) = difference in longitude (Dlong) x Cos(Lat)” Therefore, the distance between adjacent meridians of longitude when measured along parallel of latitude 53^oS is calculated as follows: Ddist. = Dlong x Cos(Lat) = 1^o x Cos(53) = 60’x Cos(53) = 60 x 0.6 = 36 n.m. Therefore, since the chart-scale is 1mm : 1 n.m., the scale distance between adjacent meridians of longitude along latitude 53^oS in the example is 36 mm.

We change the subject of the formula above to calculate the difference in longitude for 1 n.m. measured along lat. 53^oS as follows: Dlong = Ddist ÷ Cos(Lat) = 1 ÷ Cos(53) = 1 ÷ 0.6 = 1.66….. So 1 nm = 1.66’….. of long when measured along parallel of latitude 53^oS.

Summary. The scales used in the example plotting sheet are: 1mm : 1 n.m. 1 n.m. = 1′ of Lat so 1mm : 1′ of Lat 1 n.m. = 1.66…’ of long along latitude 53^oS. 1^o Long = 36 n.m. along latitude 53^oS

Note. The image of the example plotting sheet below has been reduced to fit the page and so it does not reflect the actual chart-scale.

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The Purpose of Altitude and Azimuth Angle in Position Finding at Sea

Posted on June 19, 2019 by Jack Case

To fully understand how the azimuth angle and the altitude of a celestial body help us to establish our position, we need to consider them in relation to the celestial sphere.

Consider the diagram below:

The celestial sphere is drawn in the plane of the observer’s meridian with the observer’s zenith (Z) at the top.

Point O represents both the observer and the Earth.

The arc PZQSP’ represents the observer’s celestial meridian.

The arc NAS is the celestial horizon and QRQ’ represents the celestial equator.

ZXAZ’ is a vertical circle running through the position of the celestial body (X). (A vertical circle is a great circle that passes through the observer’s zenith and is perpendicular to the celestial horizon).

The Azimuth Angle is the angle PZX (that is, the angle between the observer’s celestial meridian and the vertical circle through the celestial body).

The Altitude is the angle AOX (that is the angle from the celestial horizon to the celestial body measured along the vertical circle).

The Zenith. Point Z in the diagram represents the observer’s zenith which is an imaginary point on the celestial sphere directly above the observer. It is the point where a straight line drawn from the geocentric centre of the Earth, through the observer’s position and onwards, intersects with the celestial sphere.

The Zenith Distance. The Zenith Distance is the angular distance from the zenith to the celestial body measured from the Earth’s centre. In the diagram above, it is the angular distance ZX which is subtended by the angle XOZ.

Relationship between Altitude and Zenith Distance Since the celestial meridian is a vertical circle and is therefore, perpendicular to the celestial horizon, it follows that angle AOZ is a right angle and angles AOX and XOZ are complementary angles. From this we can deduce that:

Zenith Distance = 90^o – Altitude and Altitude = 90^o – Zenith Distance

Calculating the Zenith Distance.

Consider the next diagram.

PZX TRIANGLE

The diagram shows that the angular distance AU on the Earth’s surface is equal to the angular distance ZX in the spherical triangle PZX.

X represents the position of a celestial body on the celestial sphere,

Z represents a point on the sphere which coincides with the zenith of the DR position (A),

P represents the projection of the North Pole onto the celestial sphere,

PX = NU = (90^o – the declination of the body),

PZ = NA = (90^o – the latitude of the DR position),

ZX = AU = (90^o – the altitude of the body).

We can see that the triangle NAU on the Earth’s surface can be solved, in effect, by solving the triangle PZX in the celestial sphere.

Local Hour Angle (LHA)

In the PZX triangle diagram, LHA is the angle ZPX; that is the angle between the observer’s celestial meridian and the meridian of the celestial body.

Relationship between LHA and Azimuth Angle. Consider the next diagram.

This diagram is drawn in the plane of the celestial horizon. Imagine that you are looking down on the celestial sphere from a position directly above the observer’s zenith which is in the centre of the circle.

The circle WANESW represents the celestial horizon.

NZS represents the observer’s celestial meridian.

WQE represents the celestial equator,

P is the celestial pole,

X is the position of the celestial body,

PXR represents part of the meridian of the celestial body which cuts the Equator at R.

ZPX is the LHA.

PZX is the Azimuth angle.

When the LHA (ZPX) is less than 180^o, the celestial body lies to the west of the observer’s meridian and when the LHA is greater than 180^o it lies to the east. (Remember LHA is measured westwards from the observer’s meridian from 0^o to 360^o).

It follows that if the celestial body is to the west of the observer’s meridian, the azimuth angle must be west and when to the east, the azimuth angle must be east.

So we have the rule:

LHA 0^o to 180^o= Azimuth Angle West

LHA 180^oto 360^o = Azimuth Angle East

In Astro navigation. It can be seen that by measuring the altitude of a celestial body, we are able to easily calculate the zenith distance which will give us the distance in nautical miles from the observer’s position to the geographical position of the body. The azimuth angle will give us the direction of the GP from the observer’s position. This explains why calculating the altitude and azimuth angle are the first steps in determining our position in celestial navigation.

Relationship Between Azimuth Angle and Azimuth.

Azimuth is a specific type of bearing which measures the direction of an object in relation to true north, in the horizontal plane, clockwise from 0^o to 360^o.

azimuth and azimuth angle update

Azimuth Angle. In astro navigation, when we calculate the azimuth of a celestial body, what we actually calculate is the azimuth angle. Azimuth angle is measured from 0^o to 180^o either westwards or eastwards from either north or south. If the observer is in the northern hemisphere, the azimuth is measured from north and if in the southern hemisphere, it is measured from south.

For example, if the true azimuth of an object is 225^o, the azimuth angle for an observer in the northern hemisphere will be N135^oW but for an observer in the southern hemisphere, it will be S045^oW.

Summary Of The Discussions Above. The relationships discussed above illustrate the importance of altitude and azimuth angle in position finding at sea. It can be seen that from the altitude of a celestial body, we are able to easily calculate the zenith distance which will give us the distance in nautical miles from the observer’s position to the geographical position of the body. From the calculated azimuth angle we can find the true azimuth and this will give us the direction of the GP from the observer’s position. This explains why determining the altitude and azimuth angle are the first steps in determining our position in astro navigation.

In the next post, we will discuss sight reduction which is the process of reducing the data gathered from an observation of a celestial body down to the information needed to establish an astronomical position line.

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Note. 1 minute of arc at the Earth’s centre will subtend a distance of 1 nautical mile at the Earth’s surface.

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Planning Star and Planet Sights

Posted on May 11, 2019 by Jack Case

Fixes from Sightings of Stars and Planets. There are 59 navigational stars and 4 navigational planets which we can use to achieve position fixes. However, there are only two short periods during the day in which we can do this because we need it to be dark enough to see the bodies in the sky yet light enough to see the horizon. In other words, we are restricted to taking star and planet sights during the times of morning and evening nautical twilight.

Because nautical twilight gives us only a short period of time to make observations, advanced planning is essential. We need to establish what our estimated position at nautical twilight will be in advance and select the stars and planets that we intend to use for the fix. To be sure of taking three reliable sights for a three point fix, we should select at least four or five bodies for our observations.

Having selected our stars and/or planets, the next step is to calculate what the altitude and azimuth of each of them will be at the estimated position at the time of the planned observation. In this way, we will know what the approximate altitude and azimuth of each of the bodies will be when we come to make the sightings from the true position and this will cut the time we take to locate them in the sky.

There is also the added advantage that by making the calculations for the estimated position in advance, most of the work required to calculate the intercept will have been completed by the time the observation is made. This will of course, speed up to process of calculating the position lines and the ultimate fix.

Sight Reduction. This is the process of reducing the data gathered from observations of celestial bodies down to the information needed to establish an astronomical position line. The two essential items of data that we need to begin the process of sight reduction are the azimuth and the altitude of the celestial body in question. Methods of sight reduction usually fall under two categories, tabular and formula. Tabular methods such as Rapid Sight Reduction involve interpolating large tables of data by entering latitude, declination and LHA to extract altitude and azimuth. Formula methods involve mathematically calculating the altitude and azimuth from the same input data.

Using Spherical Trigonometry For Sight Reduction. With spherical trigonometry, we have the tools to quickly calculate the altitude and azimuth of selected celestial bodies at the estimated position without being encumbered by large tables of data.

Why Spherical Trigonometry? At first sight, the term ‘spherical trigonometry’ might seem quite daunting but with the knowledge of just two simple formulas and with a little practice of the methods demonstrated below, it will be found to be quick and easy to apply. The method is comprehensively taught in my book ‘Celestial Navigation – Theory and Practice’.

Why Calculate Azimuth? The true azimuth and the azimuth angle provide exactly the same directional information albeit in different formats. This begs the question: “why go to the trouble of calculating the azimuth angle and then converting it to the true azimuth when it is easier just to measure the true azimuth directly with a compass?” However, we calculate the azimuth angle by finding the angle PZX from the values of the sides PZ, PX and ZX in the spherical triangle ZPX and these values are derived from data relating to the DR position. If we measure the azimuth by compass, we can only do so from the true position. At the time of taking the altitude, we would not know where the true position is so our aim must be to find the direction of the true position from the DR position and we can only do this by calculating the azimuth angle at the DR position. There is also the point that, unless you are in the fortunate position of having a gyro compass, you must take magnetic compass readings and these have to be corrected for variation and deviation; so you might just as well calculate the azimuth in the first place.

There is another important reason to be able to use spherical trigonometry for this task as the following statement in the ‘International Maritime Organization Regulations’ makes clear. “The provision of trigonometric tables onboard and regular practice with them by all officers and navigation related staff is compulsory”. In the event of GPS and other electronic navigation systems failure, it would be irresponsible of a ship’s master if his ship were to go dangerously off course simply because trigonometric calculations could not be made”.

Example. Using Spherical Trigonometry To Calculate the altitude and azimuth of the star Alioth at the estimated position at the planned time of observation using the data provided in the scenario.

Scenario.

Estimated Position: Lat. 50^oN Long. 45^oW

Data from Nautical Almanac re. Alioth: SHA = 166. Declination = 56^oN GHA Aries = 300

Step 1. Estimated position at planned time of observation: Lat. 50^oN Long. 45^oW

Step 2. From the Nautical Almanac, extract data for planned time of observation as follows: Alioth. SHA = 166. Declination = 56^oN GHA Aries = 300

Step 3. Calculate LHA.

SHA Alioth 166

GHA Aries 300

. 466

Long -45 (subtract if long is west, add if long is east)

. 421 (subtract 360 if LHA is greater than 360)

. -360

LHA Alioth 61^o(West since LHA less than 180^o)

Step 4. Calculate PZ, PX and ZPX

PZ = 90^o – Lat = 90^o – 50^o = 40^o. PX = 90^o – Dec = 90^o – 56^o = 34^o. ZPX = LHA = 61^o

Step 5. Calculate Zenith Distance (ZX).

Formula to Calculate Azimuth (ZX):

Cos (ZX) = [Cos(PZ) . Cos(PX)] + [Sin(PZ) . Sin(PX) . Cos(ZPX)]

Enter values of PZ, PX and ZPX in the formula: (decimals to 3 places).

Cos (ZX) = [Cos(PZ) . Cos(PX)] + [Sin(PZ) . Sin(PX) . Cos(ZPX)]

= [Cos(40^o) . Cos(34^o)] + [Sin(40^o) . Sin(34^o) . Cos(61^o)]

= [0.766 x 0.829] + [0.643 x 0.559 x 0.485]

= 0.635 + 0.174

Cos (ZX) = 0.809

∴ ZX = Cos^-1 (0.809) = 36^o

Step 6. Calculate Altitude.

Altitude = 90^o – ZX = 90^o – 36^o = 54^o.

Step 7. Calculate Azimuth.

Data previously calculated: PZ = 40^o PX = 34^o ZX = 36^o

Formula to Calculate Azimuth (PZX):

Cos PZX = [Cos(PX) -(Cos(ZX) x Cos(PZ))] / [Sin(ZX) . Sin(PZ)]

Enter values of PZ, PX AND ZX into the formula: (decimals to 3 places).

Cos PZX = [Cos(34) – (Cos(36) x Cos(40))] / [Sin(36) . Sin(40)]

= [0.829 – (0.809 x 0.766)] / [0.588 x 0.643]

= [0.829 – 0.620] / 0.378

= 0.21 / 0.378

∴ Cos(PZX) = 0.557

∴ PZX = Cos^-1(0.557) = 56.15 ≈ 56^o

∴ Azimuth angle = N56^oW (True Azimuth: 304^o) (Lat. North and LHA less than 180^o)

Summary: The altitude and azimuth of the star Alioth at the estimated position (EP or DR) at time of planned observation have been calculated in advance to be:

Altitude: 54^o. Azimuth: 304^o

This data serves two purposes:

Firstly it helps the navigator to quickly locate the position of the star at the planned observation time by providing its approximate altitude and azimuth. Secondly, by calculating the altitude and azimuth from the EP or DR in advance, the navigator will have all the data necessary to quickly calculate the intercept.

Calculating the Intercept. Ho is observed altitude. Hc is calculated altitude.

To calculate intercept (p): p = Ho – Hc

If p is positive, intercept is from DR position towards azimuth.

If p is negative, intercept is from DR position away from azimuth (i.e. towards reciprocal).

Calculating the Intercept in the Alioth Example. Suppose the actual measurements of altitude and azimuth of Alioth at the true position at the time of observation were as follows: Altitude: 53.785^o Azimuth: 304^o

We would calculate the intercept in the following way:

Calculated altitude (Hc) = 54^o Observed altitude (Ho) = 53.785^o

p = Ho – Hc = 53.785^o – 54^o = –0.125^o = –7.5′ = -7.5 nautical miles

Therefore intercept = 7.5 n.m from 304^o i.e. 7.5 n.m. towards 124^o (reciprocal).

The Rapid Sight Reduction method is taught in my book ‘Astro Navigation Demystified’.

The trigonometric method of sight reduction is taught in my book ‘Celestial Navigation – Theory and Practice’.

Books of the Astro Navigation Demystified Series:

email: astrodemystified@outlook.com

Posted in astro navigation, astronomy, celestial navigation, Marine Navigation, navigation | Tagged astro navigation, celestial navigation, navigation, Planning star and planet observations | Leave a comment

The Importance of Morning and Evening Star Sights – Part 2

Posted on March 25, 2019 by Jack Case

The ‘Where To Look’ Method. Devised by Jack Case HOME

In part 1 of this series, we discussed how we can establish whether or not a selected star or planet will be visible during nautical twilight. In this post, we discuss how we can use the ‘Where To Look’ method to calculate its approximate azimuth and altitude.

There are various methods and devices that can be used for this purpose such as celestial navigation star-globes, star Identifiers, ABC tables and navigation software systems. However, the ‘Where To Look’ method (WTL Method) solves the problem without the need to clutter the chart table with cumbersome globes, computers and large books of tables, some of which are quite expensive and complicated. It is a particularly advantageous method for use in the close confines of a small-boat chart table and it is cost-free.

It is emphasized that it is not intended to be a method of accurately pin-pointing the position of a star or planet but a simple and quick method of establishing whether or not the body is likely to be visible at the time an observation is required and if so, what its approximate position in the sky will be.

The Method in a Nutshell. Although the following explanation and proof of the WTL method is necessarily quite lengthy, once it is fully understood, it all boils down to the application of this very simple formula: Approx. Alt. = 90^o – (Lat ∆ Dec)and the mental image of this very simple diagram:

Demonstration Example. For those who prefer to simply learn and apply a method without bothering to study lengthy explanations and proofs, the following example will adequately demonstration the WTL method. For those who would rather see the proof before deciding whether or not to apply the method, a full explanation is given after the demonstration example.

Brief Demonstration Example. To Estimate The Approximate Azimuth Angle And Altitude Of The Star Alioth by the WTL method.

Scenario: A navigator intends to use the star Alioth for a position fix during morning nautical twilight. Before he can do this, he needs to locate Alioth in the sky. This he does this by using the WTL method to calculate its approximate azimuth angle and altitude as demonstrated below: He bases his calculations on the following data:

DR Position of observer: Lat. 30^oN Long. 45^oW

SHA of Alioth = 166. Declination = 56^oN

GHA Aries = 250

Step 1. Calculate LHA of Alioth.

SHA Alioth 166

GHA Aries 250

. 416

Long -45 (subtract if long is west, add if long is east)

. 371

-360 (subtract 360 if LHA is greater than 360)

LHA Alioth 11 = 11^o

Step 2. Calculate whether or not Alioth is above the celestial horizon.

The rules for this are:

Body is above western celestial horizon if: LHA = 0^o TO 90^oor LHA – 360^o = 0^o TO 90^o

Body is above eastern celestial horizon if: LHA = 270^o to 360^o or 360^o – LHA = 0^o TO 90^o

Conclusion. The LHA of Alioth is 11^oW and so we can conclude that it will be above the celestial horizon to the west.

Step 3. Is Alioth above the celestial horizon to the north and south?

The rules for this are:

Latitude North: visible range = 90^oN to (90^o – Lat)S.

Latitude South: visible range = 90^oS to (90^o – Lat)N.

DR Position of observer: Lat. 30^oN so the visible ranges are: 90^oN to (90^o – 30^o)S = 90^oN to 60^oS

Declination of Alioth: 56^oN which is within the visible range.

Step 4. Estimate Azimuth Angle. Having calculated the LHA of Alioth, we can now make a rough approximation of its azimuth angle by plotting its LHA and declination in relation to the observer’s DR position using the simple diagram shown earlier. It is necessary to draw this simple diagram in order to explain the method but in practice, the navigator would visualize it in his mind’s eye without the need to draw it.

In the drawing, we have plotted the position of Alioth in terms of its LHA and declination which are as follows: LHA 11^oW, Dec. 56^oN

If we draw a line joining the position of Alioth to the position of the observer, we will see that the approximate azimuth angle is: N15^oW which, in terms of true azimuth, is: 345^o

It is reiterated that, in practice, it would not be necessary to draw the diagram; by picturing the diagram in the mind’s eye and visualizing the positions of the observer’s latitude and the declination and LHA of the body, we can easily estimate the approximate azimuth.

Of course, as discussed earlier, we cannot find an accurate solution to a spherical problem with a straight-line, two dimensional drawing such as this which does not take account of the fact that the Earth is not a perfect sphere and does not allow for phenomena such as refraction and parallax . However, at this stage we only need an approximate answer to the question “in what direction do we look”? It is also stressed that the drawing above is for illustrative purposes only and therefore, it is not drawn to scale and angles are not accurately drawn.

Step 5. Calculate Approximate Altitude.

The formula for calculating approximate altitude is: Approx. Alt. = 90^o – (Lat ∆ Dec)

Reminder of data gathered to date: Observer’s Lat: 30^oN Dec Alioth: 56^oN

We can apply the formula to calculate the approximate altitude of Alioth as follows:

Approx alt. Alioth = 90^o – (30^o ∆ 56^o) = 90^o – 26^o = 64^o

Summary of the Demonstration Example Results. Firstly, we used the WTL method to establish that Alioth will be above the observer’s horizon during the predicted times of nautical twilight. Secondly, we used the method to calculate the approximate azimuth and altitude of Alioth as follows: Approx Az. = N15^oW (345^o), Approx Alt: 64^o

Full Explanation of the Method. To find if a star or planet will be above the horizon at our position at the time of the planned observations, we need to take two things into account, its local hour angle and its declination.

Local Hour Angle (LHA). For a star or planet to be visible, its meridian must be within 90^o east or west of the observer’s longitude at the time of the planned observations.

If a body’s LHA is greater than 0^oand less than 90^oor if LHA is greater than 360^o and LHA – 360^o is less than 90^o then body will be above the western celestial horizon.
If a body’s LHA is greater than 270^obut less than 360^othen 360^o – LHA will be less than 90^o and the body will be above the eastern celestial horizon.

We can formulate the above statements as follows:

Body is above western celestial horizon if: LHA = 0^o TO 90^oor LHA – 360^o = 0^o TO 90^o

Body is above eastern celestial horizon if: LHA = 270^o to 360^o or 360^o – LHA = 0^o TO 90^o

The following diagram demonstrates what has been discussed above.

visiblehorizon LHA

Point O is the position of an observer on the Earth’s surface at latitude 50^oN.

NS is the meridian of the observer and in terms of LHA, is 0^o.

WE is the celestial equator.

The limits of the observer’s western and eastern celestial horizons are at LHA 90^o and LHA 270^o which are both 90^o from the observer’s meridian.

Suppose that point A in the diagram represents the position of a celestial body whose LHA is 40^o and that point B represents the position of another celestial body whose LHA is 295^o. Since the LHAs of these bodies is either less than 90^o or greater 270^o, they will be visible above the celestial horizon to the west and the east respectively.

Bodies whose LHAs are greater than 90^o but less than 270^o would be below the celestial horizon and therefore would not be visible.

Calculating LHA From the above, we can see that our first step in determining the position of a celestial body is to calculate its LHA.

LHA of A Star. The LHA of stars is not listed in the Nautical Almanac so we must calculate these ourselves. The following method can be used to calculate the LHA of a star.

From the nautical almanac daily pages, find the Greenwich Hour Angle (GHA) of Aries (to the nearest degree) at the planned time of the observation.
From the ‘Index to Selected Stars’ in the Nautical Almanac, find the Sidereal Hour Angle (SHA) of the star to the nearest degree. (A list of navigational stars can be found on page 170).
Calculate your estimated longitude (to the nearest degree) at the planned time of the observation.
Combine the SHA, GHA Aries and the estimated longitude to find the approximate LHA.

Examples. We will use three celestial bodies, stars X and Y and planet V to demonstrate the method outlined above: DR Position of Observer: Lat : 50^oN, Long: 135^oW

Details of Star X: SHA = 182 Dec = 65^oN

GHA Aries = 335

Longitude of observer = 135^oW

To Calculate LHA of Star X

SHA X 182

GHA Aries 335

. 517

Long -135 (subtract if long is west, add if long is east)

. 382

-360 (subtract 360 if LHA is greater than 360)

LHA X = 22^o

Since the LHA of star X is between 0^o and 90^o it will be visible above the horizon to the West. Therefore, LHA = 22^oW

Details of Star Y: SHA = 130 Dec = 15^oN

GHA Aries = 335

Longitude of observer = 135^oW

To Calculate LHA of Star Y

SHA Y 130

GHA Aries 335

. 465

Long -135 (subtract if long is west, add if long is east)

LHA Y = 330^o

Since the LHA of Y is between 270^o and 360^o it will be visible above the celestial horizon to the East. However, from a practical point of view, when the LHA is from 270^o to 360^o, we need to know how far this is to the east of the observer’s meridian. We do this by subtracting the LHA from 360^o which in this case is 360^o – 330^o = 30^oE. Therefore, LHA = 30^oE

LHA Of Planet V. Unlike the stars, the positions of the planets in the celestial sphere are not fixed and for this reason the four navigational planets are listed in the daily pages of the Nautical Almanac by their GHA and declination instead of by their SHA. The Navigational planets are: Mars, Saturn, Venus and Jupiter.

The method used to calculate the LHA of a planet is the same as that for a star except that because the GHA is given in the Nautical Almanac, we do not need to calculate it. The following calculations for planet V demonstrate the method.

Details of Planet V: GHA = 354^oDec = 20^oS

Longitude of observer = 135^oW

To Calculate LHA of Planet V:

GHA 325

Long -135

LHA V = 190

Because the LHA of planet V is greater than 90^o but less than 270^o it will be below the celestial horizon and therefore it will not be possible to use it for position fixing at the observer’s position.

Azimuth and Azimuth Angle. The difference between azimuth and azimuth angle is demonstrated in the following diagram.

azimuth and azimuth angle update

In astro navigation, when we calculate the azimuth of a celestial body, the result is expressed as an azimuth angle. Whereas True Azimuth is measured clockwise from 0^o to 360^o in relation to true north, azimuth angle is measured from 0^o to 180^o either westwards or eastwards from either north or south. If the observer is in the northern hemisphere, the azimuth angle is measured from north and if in the southern hemisphere, it is measured from south.

For example, in the diagram above, the true azimuth of an object is shown as 135^o but for an observer in the southern hemisphere, the azimuth angle is S45^oE. If the observer had been in the northern hemisphere, the azimuth angle would have been N135^oE.

Once we have calculated the LHA of stars X and Y, we are in a position to make a rough approximation of their azimuth angles by plotting their LHA and declination in relation to the observer’s DR position on the celestial horizon diagram that we previously used to explain LHA.

In the drawing below, we have added to the LHA diagram by plotting the positions of X and Y in terms of their LHA and declination which are as follows:

Star X: LHA 22^o, Dec. 65^oN

Star Y: LHA 330^o, Dec. 15^oN.

azimuth X and Y

If we join the positions of X and Y to the position of the observer, we will see that the approximate azimuth angles are as follows:

Star X: Approximate Azimuth Angle = N30^oW which in terms of true azimuth is 330^o

Star Y: Approximate Azimuth Angle = N140^oE which in terms of true azimuth is 140^o

Summary of data relating to stars X and Y which are represented in the diagram:

DR Position of observer: Lat. 50^oN Long. 135^oW

Star X: Declination = 65^oN, LHA = 22^o, Approximate Azimuth = 330^o

Star Y: Declination = 15^oN, LHA = 330^o, Approximate Azimuth = 140^o

Once you have become familiar with this diagram, there will be no need to draw it every time you wish to calculate the approximate azimuth. By picturing the diagram in the mind’s eye and visualizing the positions of the observer’s latitude and the declination and LHA of the body, it is an easy task to estimate the approximate azimuth.

Of course we cannot find an accurate solution to a spherical problem with a straight-line, two dimensional, drawing such as this which does not take account of phenomena such as refraction and parallax or the fact that the Earth is not a perfect sphere. However, at this stage we only need an approximate answer to the question “in what direction do we look”?

It is stressed that the diagram above is for illustrative purposes only, it is not drawn to scale and angles are not accurate.

Declination. As well as the LHA, we have to take into account the declination of a celestial body. For a celestial body to be visible above the horizon, its declination must be within 90^o of the latitude of our position. If our latitude is north, then the declination must be within the range 90^o north to (90^o – latitude) south. If the latitude is south, then the declination must be within the range 90^o south to (90^o – latitude) north.

We can formulate the above statements as follows:

Latitude North: visible range = 90^oN to (90^o – Lat)S.

Latitude South: visible range = 90^oS to (90^o – Lat)N.

We can explain the reason for this rule with the aid of another diagram which shows the positions of the bodies X and Y from the previous example in terms of their declinations. Please note that the diagram is for illustrative purposes only; for this reason, it is not drawn to scale and angles are not drawn accurately.

In the diagram , O is the position of the observer at latitude 50^oNorth and Z is the zenith at that position.

The line AB represents the observer’s celestial horizon.

The line PQ is horizontal to AB and represents the observer’s visible horizon at point O.

PQ is tangential to the circumference of the Earth at point O.

C is the center of the Earth and the line CZ is perpendicular to both the celestial horizon and the visible horizon.

The celestial horizon at point O will cut the circumference of the Earth at latitudes that are 90^o to the north and to the south of the observer’s latitude (in this example, latitude 40^oN and latitude 40^oS).

X is the position of the star X in the celestial sphere and X₁ is its geographical position on the Earth’s surface.

Y is the position of the star Y in the celestial sphere and Y₁ is its geographical position.

The diagram shows that although the geographical positions of X and Y are below the visible horizon and therefore out of sight to the observer, the bodies themselves are above the celestial horizon and therefore since they are within the limits for LHA, they will be visible above the horizon at the observer’s position.

Note. Because of the vast distances of the stars from the Earth, we can assume that, in their cases, the celestial horizon and the visible horizon correspond with very little error.

Calculating Approximate Altitude. We can make use of the previous diagram to demonstrate how we calculate the approximate altitude of a celestial body For this demonstration, we will continue with the example of stars X and Y but will ignore V because we have calculated that it will be below the celestial horizon.

Data previously determined:

Latitude of observer’s DR position = 50^oN

Star X: Declination = 65^oN. Approximate Azimuth = 330^o

Star Y: Declination = 15^oN. Approximate Azimuth = 140^o

Calculating approximate altitude of star X

. In the diagram , Angle, ECZ is equal to 50^o, the latitude of the observer.

Angle ECX is equal to 65^o, the declination of star X.

Angle BCZ is equal to 90^o and is the angle between the celestial horizon and the zenith of the observer.

Angle ZCX is 15^o and is the angle between the zenith of the observer and the zenith of star X measured at the Earth’s centre.

Angle ZOX is equal to angle ZCX i.e. 15^o.

Angle XOP is the altitude of star X at the observer’s position (that is it is the angle between the visible horizon and star X

Angle ZOP = 90^o therefore, angle XOP = 90^o – 15^o = 75^o

So we can conclude that the approximate altitude of star X is 75^o

Summary of Data for star X:

Lat of observer: 50^oN

Dec = 65^oN,

LHA = 22^o

Approx. Alt = 75^o

Formulating The Method For Calculating The Approximate Altitude.

We can easily see that the approximate altitude is 90^o minus the difference between the observer’s latitude and the body’s declination.

We can formulate this as follows:

Approx. Alt. = 90^o – (Lat ∆ Dec)

We can apply this formula to the above example as follows:

Approx. Alt = 90^o – (50 ∆ 65) = 90^o – 15^o = 75^o

We can test the method by using it to calculate the approximate altitude of star Y.

Calculating approximate altitude of star Y. We can confirm the validity of the formula by calculating the approximate altitude of star Y by the same method as above.

Data previously determined for Star Y:

Latitude of observer’s DR position = 50^oN

Declination Star Y = 15^oN. Approximate Azimuth = 140^o

Y only approx altitude correction

In the diagram, Angle, ECZ is equal to 50^o, the latitude of the observer.

Angle ECY is equal to 15^o, the declination of star Y.

Angle BCZ is equal to 90^o and is the angle between the celestial horizon and the zenith of the observer.

Angle ZCY is 35^o and is the angle between the zenith of the observer and the zenith of star Y measured at the Earth’s centre.

Angle ZOY is equal to angle ZCY i.e. 35^o.

Angle YOQ is the altitude of star Y at the observer’s position (that is it is the angle between the visible horizon and star Y

Angle ZOQ= 90^o therefore, angle YOQ = 90^o – 35^o = 55^o

So we can conclude that the approximate altitude of star Y is 55^o and as previously calculated, the approximate azimuth is 150^o which puts star to the East of the observer.

Summary of Data for star Y:

Lat of observer: 50^oN

Dec = 15^oN,

LHA = 330^o (30^oE)

Approx. Alt = 55^o

Testing the Formula for star Y. The formula previously derived for calculating the approximate altitude is: Approx. Alt. = 90^o – (Lat ∆ Dec)

If we apply this formula to the data for star Y we have:

Approx. Alt. = 90^o – (Lat ∆ Dec) = Approx. Alt. = 90^o – (50 ∆ 15) = 90^o – 35^o = 55^o which is the same result that we got from the diagram and so we may conclude that this verifies the formula.

Does the formula work for stars in the opposite hemisphere? In each of the two cases above, the declination of the celestial bodies have been in the same as the latitude of the observer. We now need to see if the formula that we have derived works when the body’s declination is in the opposite hemisphere to the latitude of the observer, that is when the latitude and declination are contrary. To this end, in the example below, we repeat the procedure used in the last two examples to calculate the approximate altitude of star Z which has a declination of 10^oS while the observer’s latitude remains at 50^oN.

The latitude of the observer is 50^oN; therefore, in the diagram, angle, ECZ is equal to 50^o. The declination of star Z is 10^oS; therefore, measuring from the centre of the Earth, the angle between East and the direction of star Z is 10^o. The angle between star Z and the celestial horizon is 30^o and the angle between star Z and the visible horizon is also 30^o so we can conclude that the approximate altitude of star Z is 30^o(see note below) .

Note. Because of the vast distances of the stars from the Earth, we can assume that, in their cases, the celestial horizon and the visible horizon correspond with very little error.

Testing the Formula for star Z.

The formula previously derived for calculating the approximate altitude is Approx. Alt. = 90^o – (Lat ∆ Dec). If we apply this formula to the example of star Z we have:

Latitude of observer = 50^oN (+50^o)

Declination of star Z = 10^oS (-10^o)

Therefore, Approx Alt = 90^o – (+50^o ∆ -10^o) = 90^o– 60^o= 30^o

The answer agrees with our findings from the diagram so we can conclude that the formula works in cases where the latitude of the observer and the declination of the celestial body are contrary.

Reliability of the Formula when for bodies close to the horizon.

The formula is not reliable for calculating the approximate altitude of celestial bodies whose LHA’s are between 65^o and 90^o east or west, that is when they are less than 25^o above the horizon. There is no great problem here because sextant measurements of bodies which are close to the horizon are not very reliable anyway because pronounced error due to refraction is likely to occur when the altitude is below 20^o. particularly during twilight.

Overall Conclusion. The results of the tests above, show that the Where To Look Method could aid the navigator in locating a chosen celestial body’s position in the sky so that accurate measurements of altitude and azimuth can be measured. All the navigator needs to employ the WTL method is to apply the simple formula Approx. Alt. = 90^o – (Lat ∆ Dec) and to visualize the simple diagram for calculating approximate azimuth.

Proof of The WTL Method. There is not sufficient space to show the method of testing and proving the ‘Where To Look’ Method in this post; however, for those who may be interested, it is shown in detail in my book ‘Astronomy For Astro Navigation’.

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The Importance Of Morning And Evening Star Sights

Posted on February 19, 2019 by Jack Case

Part 1 – Checking to see which stars and planets will be above the observer’s horizon during Nautical Twilight. By Jack Case

Introduction.

The Astronomical Position Line. A position line is a line drawn on a nautical chart along which a vessel’s position is known to lie. When a position line is derived from an observation of a celestial body, it is known as an astronomical position line.

The Observed Position. The position of a vessel can be established by the intersection of two or more position lines; such a position is known as a position fix. A position fix derived from astronomical position lines is known as an observed position and is marked on the chart as ‘Obs’.

Inaccuracy in Astronomical Position Lines Inaccuracy may be introduced into an astronomical position line by a number of contributory errors which are described here. For this reason, the accuracy of an observed position derived from only two astronomical position lines cannot be relied upon and therefore, it is generally accepted that at least three position lines are required in order to obtain an accurate fix.

Daylight Fixes. During the hours of daylight, we are mainly restricted to obtaining fixes from just one celestial body, the Sun. Sometimes, we can achieve a two point fix from the Sun and the Moon but mostly, all we can do is obtain two time-separated astronomical position lines using the Marcq St. Hilaire method to give us a two point fix.

Stars and Planets. There are 59 navigational stars and 4 navigational planets which we can use to achieve position fixes derived from three or more position lines but there are only two short periods during the day in which we can do this because we need it to be dark enough to see the bodies in the sky yet light enough to see the horizon.

Twilight Time. In general, the optimum conditions for taking observations of stars and planets occur during the times of civil twilight and nautical twilight when it is likely to be light enough for the horizon to be seen yet dark enough for those celestial bodies to be visible.

Civil Twilight. Morning civil twilight begins when the geometric centre of the Sun is 6^o below the horizon and ends at sunrise. Evening Civil Twilight begins at sunset and ends when the geometric centre of the Sun reaches 6^o below the horizon. During civil twilight, the horizon is clearly visible and the brightest stars as well as Venus, the brightest of the navigational planets, can be seen as long as they are above the horizon at that time.

Nautical twilight is the time when the centre of the Sun is between 6° and 12° below the horizon. Thus, Morning Nautical Twilight begins when the Sun rises to 12° below the horizon and ends when Morning Civil Twilight begins, that is when the Sun has risen to 6° below the horizon. Evening Nautical Twilight begins when Evening Civil Twilight ends, that is when the Sun has sunk to 6° below the horizon and ends when the Sun is 12° below the horizon . Nautical twilight is so named because it is when navigators are able to take reliable sights of stars and planets using a visible horizon for reference. During Nautical Twilight, the horizon is visible and most of the navigational stars and all of the navigational planets can be seen as long as they are above the horizon at that time.

How Do We Establish Which Stars/Planets Will Be Visible During Twilight? Unlike the Sun and the Moon which are easily identified, the approximate positions of stars and planets need to be established before observations can be made. The times of rising and setting of the Sun and Moon can be found in the daily pages of the Nautical Almanac so we know when they will be visible above the horizon. However, the risings and settings for stars and planets are not listed so we have to calculate these for ourselves.

The Stars. The stars are at such an immense distance from the Earth that the movement of their relative positions in the sky, which is so slow and so small, can be discounted without any great loss of accuracy and we can assume therefore, that they are in fixed positions in the celestial sphere. However, for the reasons explained in my articles ‘Stars for all Seasons’, the stars that we see in the night sky change from season to season and the ones that we do see, rise earlier each night.

The Planets. The planets in the solar system orbit the Sun in an anti-clockwise direction when viewed from the north pole of the celestial sphere. The apparent motions of the planets, when viewed from the Earth, are complicated by the facts that they are at varying distances from the Sun, have different orbital patterns, retrograde motions and orbital speeds. Only those planets that are sufficiently prominent to be observed with an ordinary sextant are considered to be ‘navigational planets’. These are: Venus, Mars, Jupiter and Saturn.

So what we need is a method that will enable us to calculate not only which of the navigational stars and planets will be above the celestial horizon during civil and nautical twilight but also what their approximate position in the sky will be during those times. There are various methods and devices that can be used for this purpose such as celestial navigation star-globes, star Identifiers, ABC tables and navigation software systems. However, my book ‘Astronomy For Astro Navigation’ describes my ‘Where To Look’ Method which is a simple ‘rule of thumb’ method of predicting which stars and planets will be visible during during civil and nautical twilight without the need to clutter the chart table with cumbersome globes, computers and large books of tables, some of which are quite expensive and complicated. The Where To Look Method, of which I claim ownership, is particularly advantageous in the close confines of a small-boat chart table and is cost-free.

The ‘Where To Look’ Method. The ‘Where To Look’ Method enables us to quickly and easily establish whether or not a particular star or planet will be above the celestial horizon during civil and nautical twilight and if it is, what it’s approximate position in the sky will be. It is emphasised that this is not intended to be a method of accurately pin-pointing the position of a celestial body; it is just a method of locating it so that we can begin the process of using it to calculate an astronomical position line.

Before proceeding with the explanation, I would like to say that Although it will take many words to explain this method thoroughly, it is something which, once fully understood, can be put into practice quickly and easily with just a little mental arithmetic and logic.

Stage 1. Planning which stars and planets will be above the celestial horizon during civil and nautical twilight . We should remember that unlike the navigational planets which will be in the sky at some point during a 24 hour period, the stars that we can see will change from season to season because of revolution as explained in ‘Stars For All Seasons Part 1’. We should also remember that even those stars that are ‘in season’ will not necessarily be visible during Nautical Twilight because of reasons of rotation which is also explain in ‘Stars For All Seasons Part 1’. By studying all seven of ‘Stars For All Seasons’ articles, you will come to know which stars will be available for your star sights throughout the year.

Method. Firstly, in the Nautical Almanac, look up the time of nautical twilight for your latitude. To find out if a star or planet will be above the celestial horizon at your estimated position during nautical twilight, you need to take two things into account, its local hour angle and its declination.

Local Hour Angle (LHA). For a star or planet to be visible, its geographical position must be within 90^o east or west of our estimated longitude at the time of the planned observations. If a celestial body’s LHA is greater than 0^o but less than 90^o , then the body will be above the celestial horizon to the west of the observer’s longitude.

If a body’s LHA is greater than 270^o but less than 360^o then the body will be above the celestial horizon to the east of the observer’s longitude. We can explain this in the following way: if LHA is greater than 270^o but less than 360^o, then 360^o – LHA will be less than 90^o and so the body will be above the celestial horizon.

We can formulate the above statements as follows:

Body is above horizon to the west if: LHA = 0^o to 90^o.

Body is above horizon to the east if: LHA = 270^o to 360^o

or 360^o – LHA = 0^o to 90^o

We can demonstrate what has been discussed above with the aid of the following diagram.

In the diagram, Point O is the position of an observer on the Earth’s surface at latitude 50^oN.

NS is the meridian of the observer and in terms of LHA is 0^o.

WE is the celestial equator.

The limits of the observer’s western and eastern horizons are at LHA 90^o and LHA 270^o which are both 90^o from the observer’s meridian.

Suppose that point A in the diagram represents the position of a celestial body whose LHA is 40^o and that point B represents the position of another celestial body whose LHA is 295^o. Since the LHAs of these bodies are either less than 90^o or between 270^o and 360^o, they will be visible above the celestial horizon to the west and the east respectively.

Bodies whose LHAs are greater than 90^o but less than 270^o would be below the celestial horizon and therefore would not be visible.

The articles ‘Stars For All Seasons’ on this site will give help in establishing which navigational stars will be visible above the horizon during Nautical Twilight at certain times of the year. Chapter 5 of my book ‘Astronomy For Astro Navigation’ will give more detailed help of the same topic.

How To Calculate The LHA Of A Star. The LHA of stars is not listed in the Nautical Almanac so we must calculate this ourselves. The following method can be used.

From the nautical almanac daily pages, find the Greenwich Hour Angle (GHA) of Aries (to the nearest degree) at the planned time of the observation.
From the ‘Index to Selected Stars’ in the Nautical Almanac, find the Sidereal Hour Angle (SHA) of the chosen star to the nearest degree.
Calculate your estimated longitude (to the nearest degree) at the planned time of the observation.
Combine the SHA, GHA Aries and the estimated longitude to find the approximate LHA.

Example. Using the following scenario, calculate LHA of Arcturus.

Scenario:

Date and Time: 30 May 2017, 21^h 00^m(+3)

Estimated Latitude and Longitude: 45^oN, 47^o30’W

Sunset: 19^h 37^{m (from the Nautical Almanac).}

Civil twilight: 20^h 13^{m (from the Nautical Almanac).}

Naut. twilight: 21^h 00^{m (from the Nautical Almanac).}

Chosen Star: Arcturus (‘Stars For All Seasons Part 7’ tells us that this star will be visible during nautical twilight on 30 May).

SHA of Arcturus: 146^o, Declination: N19^o (from the Nautical Almanac).

To Calculate LHA of Star Arcturus

SHA Arcturus: 146^o

GHA Aries : 203^o

SHA + GHA: 349^o

Estimated Long: -47^o (subtract if long is west, add if long is east)

LHA Arcturus: 302^o

Since LHA is greater than 270^o but less than 360^o, Arcturus will be above the horizon to the east of the observer’s position. (i.e. 360^o– 302^o03’ = 57^o57’ so we can conclude that the meridian of the GP of Arcturus will be 57^o57’ to the east of the observer’s longitude during nautical twilight).

Declination. We have discussed how we ascertain, from a celestial body’s LHA, whether or not it will be above the celestial horizon to the west or the east of the observer’s position. We also need to discuss how we can ascertain from the body’s declination whether or not it will be above the horizon to the north or south of the observer’s position. To be visible above the horizon to the north or south, a celestial body’s declination must be within 90^o of the latitude of the observer’s position. If the latitude is north, then the declination must be within the range 90^o north to (90^o – latitude) south. If the latitude is south, then the declination must be within the range 90^o south to (90^o – latitude) north. We can formulate the above statements as follows:

Latitude North: visible range = 90^oN to (90^o – Lat)S.

Latitude South: visible range = 90^oS to (90^o – Lat)N.

Returning to the example above, given that the declination of Arcturus is 19^oN we can calculate whether or not it will be above the horizon to either the north or south of the observer’s position as follows:

The latitude of the observer = 45^oN so by the rules stated above, the visible range of celestial bodies from the observer’s position will be 90^oN to 45^oS. Furthermore, since the declination of Arcturus is 19^oN, we can conclude that the latitude of its GP will be 26^o to the south of the observer’s latitude during nautical twilight.

Conclusion. Arcturus will be visible above the observer’s horizon to east and south of his position during nautical twilight.

Planets. The procedure for calculating whether or not a planet will be above the horizon during civil and nautical twilight is the same as that for stars except for one thing. Whereas the Nautical Almanac does not list the GHA for stars, it does for planets, so for this reason, the procedure is made simpler as shown below:

GHA Mars : 117^o

Estimated Long: -47^o(w) (subtract if long is west, add if long is east)

LHA Mars : 70^o (i.e. visible above the horizon to the west)

Stage 2. Calculating Approximate Altitude And Azimuth. It is all very well knowing whether or not a particular star or planet will be above the horizon during nautical and civil twilight but if we intend to use it to calculate an astronomical position line, we also need to know where to look for it in the sky; that is, we need to know its approximate azimuth and altitude. We will discuss stage 2 in part 2 of this series.

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Stars For All Seasons – Part 1

email: astrodemystified@outlook.com

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Stars For All Seasons Part 7 – Spring Stars in the Northern Hemisphere.

Posted on January 22, 2019 by Jack Case

Spring Stars in the Northern Hemisphere.

Spring is just around the corner and when the winter constellations begin to make their exits to the west, we will find Boötes, Leo, Cancer, Hydra and Virgo lurking in the wings ready to take their places. They will soon join the ever-present circumpolar constellations of Ursa Major and Minor, Cassiopeia, Perseus and Auriga. in the northern hemisphere’s night sky.

Leo, The Lion. Leo is one of the largest constellations in the sky and is visible throughout the Northern Hemisphere during the spring months and in the northern regions of the southern hemisphere during the summer and autumn months.

Leo is home to two navigational stars, Denobola and Regulus which are shown in the diagram above. Regulus, the brightest star in the constellation, is said to mark the lion’s heart with Denobola marking the tip of its tail. From a navigation perspective, these stars are best seen for star sights during evening nautical twilight during the month of April.

The name Leo means Lion in Latin and the constellation, which is depicted as a crouching lion, is associated in Greek mythology with the lion of Nemea which was killed by Heracles as one of his twelve labours.

How to find Leo. When the line of pointers in Ursa Major is produced in the opposite direction to the Pole Star, that is from Dubhe to Merak, it will point to Leo as shown in the diagram below.

Boötes The Herdsman. If we take a line from Alioth to Alkaid in the Great Bear and extend that line in an imaginary curve for about roughly three hand-spans as shown in the diagram below, it will point to Arcturus, the brightest star in the constellation Boötes.

Boötes, the 13th largest constellation in the sky, is located in the northern hemisphere and can be seen from 90^oN to 50^oS. The ancient Greeks visualized it as a herdsman chasing Ursa Major round the North Pole and its name is derived from the Greek for “Herdsman”.

Arcturus, the brightest star in Boötes, is the fourth brightest star in the sky and is a navigational star which, for navigators, is best seen during morning and evening nautical twilight during April and May. The ancient Greeks named Arcturus the “Bear Watcher” because it seems to be looking at the Great Bear (Ursa Major).

Corona Borealis The Northern Crown. If we next take a line from Nekkar to Princepes in Boötes, and extend that line by about one and a half hand-spans, it will point to Nusakan in the close by Corona Borealis constellation.

Corona Borealis, whose name in Latin means ‘northern crown’, is a small constellation in the northern hemisphere and can be seen between latitudes 90^oN and 50^oS. Alphecca, the brightest star in the group, is a navigational star and is best seen for star sights during morning and evening nautical twilight in April and May.

The main stars in Corona Borealis form a semi-circle which is associated with the crown of Ariadne in Greek mythology. It is said that the crown was given by Dionysus to Ariadne on their wedding day and after the wedding, he threw it into the sky where the jewels became stars which were formed into a constellation in the shape of a crown.

Virgo, the Virgin. Virgo lies over the southern hemisphere and is one of the largest constellations in the sky; it is visible between latitudes 80^oN and 80^oS.

The brightest star in Virgo is Spica, the 15th brightest star in the sky and a very important navigational star which can be seen during morning nautical twilight from December through to May and during evening nautical twilight from April through to September.

In ancient Greek mythology, Virgo is associated with the goddess Dike, the goddess of justice and the constellation Virgo takes its name from the Latin for virgin or young maiden.

Finding Virgo. As we learned when studying the constellation Boötes, an imaginary curved line from Alkaid in the Great Bear leads to the bright orange star Arcturus, in the constellation Boötes. If, as shown in the following diagram, we continue that curved line by another hand span from Arcturus we will come to the bright bluish-white star Spica in the constellation Virgo.

Hydra, the sea serpent. The constellation Hydra, the sea serpent, is the largest constellation in the night sky and it is also one of the longest constellations. Hydra’s head is located south of the constellation Cancer and its tail lies between Centaurus and Libra. It is best seen from the southern hemisphere, but can be observed in the northern hemisphere between January and May. Hydra contains one navigational star and that is Alphard which, for star sights, is best observed during evening nautical twilight during February, March and May.

In Greek mythology, Hydra represents the water snake brought to the god Apollo by the crow Corvus as an excuse for being late from his errand to fetch water. It may also represent the hydra from the myth of Hercules and his twelve labours. The Hydra was a giant beast with the body of a dog and 100 snake-like heads. It was slain by Hercules on the second of his twelve labours for the king of Mycenae. As each head was cut off, two more would grow in its place. Hercules burned the roots of the heads to prevent them from growing back.

Finding Hydra. Hydra is such a large constellation that it really depends on which part of it you want to see. If you want to see head, then, as the diagram below shows, you should look between the constellations Canis Minor and Leo (look for the bright stars Procyon in Canis Minor and Regulus in Leo). If you the tail, then you should look to the south of Virgo (look for the bright star Spica).

Cancer The Crab. Cancer is a relatively small constellation in the northern hemisphere and is visible between latitudes 90^oN and 60^oS.

Cancer consists of mainly faint stars, none of which is a navigational star and for this reason, it is not a very useful constellation for astro navigation. However, it does help us in one way: Although astrology has no place in astro navigation, the signs of the zodiac can be very useful to navigators because the order in which they follow one another can tell us the position of one zodiac constellation in the sky with respect to another. For example, we know that Cancer’s position on the ecliptic falls between Leo and Gemini and as the following diagram demonstrates, Cancer can easily be found nestling between those constellations with Gemini lying to the west and Leo to the east (remember that in star maps, east and west are reversed with respect to conventional maps).

From this, it can be seen that Cancer can be an aid to locating both Gemini and Leo.

The Tropic of Cancer. These days, the Sun passes through Cancer in late July; however, in the time of Ptolemy, around 2000 years ago, this occurred during the summer solstice when the Sun reached 23.4^o N, the northern limit of the ecliptic. The latitude 23.4^o N is still called the Tropic of Cancer even though the Sun now resides in Taurus at the summer solstice.

In Greek mythology, Cancer is associated with the crab in the story of the Twelve Labours of Heracles. The goddess Hera sent the crab to attack Heracles while he was fighting the Lernaean Hydra but Heracles kicked it all the way to the stars where it formed the constellation Cancer. In another version of the story, Hera placed the crab in the sky in gratitude for its efforts even though it was killed by Heracles. (Heracles is the Roman name for the Greek god Hercules).

Stars For All Seasons Part 6

Posted on December 12, 2018 by Jack Case

Winter Stars in the Northern Hemisphere

“Know The Stars And You Will Always Have A Compass”. (Michael Punk. 2002. The Revenant)

As we move into winter, new constellations take their place in the night sky of the northern hemisphere including Taurus, Orion, Canis Major and Minor, Perseus, Gemini and Aries.

Taurus, The Bull.

Taurus is one of the most prominent constellations in the northern winter sky. It passes through the night sky from November to March and is visible at latitudes from 90^oN to 65^oS. Taurus is most visible in January and for the benefit of navigators, it is always on hand for star sights during morning and evening nautical twilight throughout that month.

Taurus is most famous for its red giant star, Aldebaran, which is known as Taurus’ Eye, is the 14th brightest star in the sky and is an important navigational star. The star at the tip of the northern horn of Taurus, Al Nath (sometimes spelled El Nath) is the second brightest star in the constellation and is also a navigational star. In earlier times this star was considered to be shared with the constellation Auriga, forming the right foot of the Charioteer as well as the Northern horn of the bull.

Taurus is associated with several mythological beliefs. In Greek mythology, Zeus was said to have disguised himself as a bull to abduct Europa, the daughter of king Agenor. The ancient Egyptians believed that the constellation represented the sacred bull associated with spring and Babylonian astronomers called it the ‘Heavenly Bull’.

Finding Taurus. If we imagine a line from Phad to Meral in Ursa Major and extend it for a distance of 80^o or roughly 4 hand-spans it will point directly to the star Aldebaran in the constellation Taurus. Once we have located Aldebaran the remaining stars of Taurus can easily be identified.

The Pleiades, the ‘Sailing Stars’ or the ‘Seven Sisters’.

The Pleiades form a small star cluster close to Taurus; they are visible between latitudes 90^oN and 65^oS and are best seen during the month of January. Merope is the brightest of these stars but even so, it not considered to be a navigational star. In ancient times, the Pleiades were called the Sailing Stars because Greek sailors would not put to sea unless they could be seen in the sky.

Another name for the Pleiades is the ‘Seven Sisters’ after Alcyone, Maia, Anterope, Taygeta, Celaena, Electra and Merope from Greek Mythology. Although Pleione was not one of the seven sisters, she along with her consort Atlas is included in the Pleiades Cluster.

Finding The Pleiades The next diagram shows that if, in the constellation Taurus, we draw a line from Aldebaran to Tau Tauri (the two eyes of the Bull) this will point roughly in the direction of the Merope in the star cluster Pleiades which lies about 10^o (a palm-width) from Taurus.

Auriga the Charioteer.

The constellation Auriga is in the northern hemisphere and is visible between latitudes 90^oN to 40^oS. For navigator’s, it is best seen during nautical twilight in February and March.

Capella is one of the navigational stars; it is the brightest star in the constellation Auriga and the 6th brightest in the northern hemisphere.

The diagram below shows the star Al Nath (sometimes spelled El Nath) forming the right foot of the Charioteer and as has already been explained, it was once considered to be shared with the constellation Taurus where it forms the northern horn of the bull. When the constellation boundaries were changed in 1930, Al Nath was assigned solely to Taurus. However, if we still think of Al Nath (the second brightest star in Taurus and also the second brightest in Auriga) as the foot of Auriga, we will have an easy method of finding both Auriga and Taurus as we can see from the diagram below.

In mythology, Auriga is associated with Myrtilus the charioteer because the shape of the constellation was said to resemble a pointed helmet of a charioteer. It is also identified with Hephaestus, the god of the blacksmiths who invented the chariot.

Orion, Canis Major and Canis Minor, The Winter Triangle.

The Winter Triangle is an astronomical asterism formed from three of the brightest stars in the winter sky; Sirius, Betelgeuse, and Procyon, the primary stars in the three constellations of Canis Major, Orion, and Canis Minor which are discussed below. As the following diagram shows, imaginary lines drawn between these stars form an imaginary equilateral triangle drawn on the celestial sphere.

Orion, The Hunter.

Orion is one of the brightest and best known constellations in the night sky it lies straddles the celestial equator and is visible between latitudes 95^oN and 75^oS. This easily recognized constellation contains 4 navigational stars, Rigel, Belatrix, Anilam and Betelgeuse which, for navigation purposes are best seen during nautical twilight in the month of January.

In Greek mythology, this constellaion represents the mythical hunter Orion, who is often depicted in star maps as either facing the charge of Taurus, the bull, pursuing the Pleiades sisters with his two hunting dogs, represented by the nearby constellations Canis Major and Canis Minor.

Meissa marks the position of the Hunter’s head while Betelgeuse, and Belatrix are his shoulders. Alnitak, Alnilam and Mintaka form his belt and from this hangs his sword which is marked by the Orion Nebula. His right thigh is marked by Saiph and Rigel marks his left foot.

Finding Orion If we take a line from Tau Tauri to Aldebaran in the constellation Taurus and extend this line for roughly three hand-spans we will come to the star Belatrix in Orion as the diagram below shows.

Canis Major, The Greater Dog.

The constellation Canis Major is in the Southern Hemisphere; it is visible from 90^oS to 60^oN and is best visible from November to March. Canis Major contains Sirius, which is otherwise known as the Dog Star. Sirius is the brightest star in the sky and is a navigational star which for navigation purposes is best seen during nautical twilight during the month of February.

Canis Major is Latin for ‘The Greater Dog’ and is so named because it was said to represent one of the two hunting dogs of Orion the Hunter; the other dog being Canis Minor. In Greek mythology, Zeus sent Laelaps, an alternative name for Canis Major, into the sky when it failed to outrun a fox.

Canis Minor The Lesser Dog.

Canis Minor is a small constellation in the northern hemisphere and is visible at latitudes from 85^oN to 75^oS. Its name means ‘the lesser dog’ in Latin and it is said to represent one of the dogs that follow Orion. In another mythological tale, Canis Minor represents Maera the dog which was sent to the sky by Zeus after it died of grief when its master Icarius was killed. As shown in the diagram, there are only two bright stars in Canis Minor, Gomeisa and Procyon which is a navigational star and which, for navigational purposes is best seen during nautical twilight in March.

Perseus.

Although Perseus is circumpolar at latitudes north of 40^oN, it can be seen from anywhere north of 35^oS during the northern hemisphere’s winter months. This constellation is associated with the Greek mythological hero Perseus who, on the orders of King Polydectes, slayed the Gorgon Medusa who had the power to turn people to stone. Polydectes had hoped that Perseus would not return and when he did, he became hostile. Perseus was so angered by this that he took out the head of Medusa and turned Polydectes to stone. The wife of Perseus was called Andromeda and the constellation that represents her lies side by side with the one that represents him. Even though Perseus’ stars are bright relative to other constellations, Mirfak is its only navigational star and this is best seen for navigation purposes during nautical twilight in December.

Finding Perseus. If a line is drawn from Navi to Ruchbah in Cassiopeia, it will point almost directly towards the star Mirfak of the constellation Perseus at about one hand-span as shown in the diagram below.

Gemini,The Heavenly Twins.

The name Gemini means Twins in Latin and for this reason, it has the alternative name the ‘Heavenly Twins’. The constellation is associated in, Greek mythology, with the twins Castor and Polydeuces. Pollux and Castor, the brightest stars in the constellation are said to form the eyes of the twins. (We use the Latin name Pollux instead of the Greek name Polydeuces).

Gemini is in the northern hemisphere and is visible between 90^oN and 60^oS. Pollux is a navigational star and is best seen for navigation purposes during nautical twilight in February.

Finding Gemini. As shown in the diagram below, a line running from Megrez to Merak in Ursa Major leads to the constellation Gemini.

Aries, The Ram.

The constellation Aries is located in the Northern Hemisphere and is visible between latitudes 90^oN and 60^oS. The name Aries, which means Ram in Latin, is associated with Jason and the Golden Fleece in Greek mythology.

Aries is a difficult constellation to see with the naked eye but it can be found about mid way between the Pleiades and the constellation Pegasus which we discussed in part 5 of this series. Pegasus lies to its west, Pleiades to its east as the following diagram shows. (Remember that, in star maps, east and west are reversed). Hamal is the brightest star in Aries and it is a navigational star which, for navigation purposes, is best seen during nautical twilight in the month of December.

The First Point of Aries. Just as the Greenwich meridian has been arbitrarily chosen as the zero point for measuring longitude on the surface of the Earth, ‘the first point of Aries’ has been chosen as the zero point in the celestial sphere. It is the point at which the Sun crosses the celestial equator moving from south to north (at the vernal Equinox in other words). The confusing thing is that, although this point lay in the constellation of Aries when it was chosen by the ancient astronomers, due to precession, it now lies in Pisces. The First Point of Aries is usually represented by the ‘ram’s horn’ symbol shown below:

Links:

Stars For All Seasons – Part 2

Stars For All Seasons – Part 3

Stars For All Seasons – Part 4

Stars For All Seasons – Part 5

Books of the Astro Navigation Demystified Series: