The last fit for the Sun's declination wasn't a fair test for uniform circular motion. One can get a better fit to declination that results from using a Keplerian eliptical orbit for the Earth. To fit a sine function to the declination of Kepler's orbit one can choose it be symmetric about the Summer Solstice and let the fit parameters be a constant term and the coefficient of a cosine term. If one assumes the average error is zero then this gives one linear equation with two unknowns which can be solved for the constant term, ε_0. One can then look at the maximum of the absolute value of the errors for each value for the magnitude of the declination, ε_1, and plot this as a function of ε_1. This function has a clear minimum which is easily determined. The solution for the two unknowns is given above. This is a much better fit than before with the maximum error being less than half a degree. The green solid line is the declination using Kepler's method of solution and the dotted yellow line might be called that of Eudoxus. The presence of a constant term might add a little error to the determination of the equator and as a result add an error to the determination of one's latitude. Actual measurements would add additional errors which could also skew the results. One would have to accurately measure the declination to something on the order of a minute of arc to get a good measure of the errors in uniform circular motion. Ptolemy may have been the first to do this. Tycho Brahe's observations at the end of the 16th Century were intended to be accurate to 1 minute of arc but may have had errors of 3 minutes in some cases but they allowed Kepler to deduce his laws of motion. The telescope greatly improved the accuracy of measurements.

Edit (30 Mar 2011): A slightly better fit was found by doing a 2D search which gave the values 173.289, 0.379° & 23.360° and a maximum error of 0.468°.

## Saturday, March 26, 2011

## Friday, March 25, 2011

### The Uniform Circular Motion Approximation for the Declination of the Sun

The Gregorian Calendar was designed to keep the date of the Equinoxes fixed and this can be used to approximate the equinox year. With leap days there are 146097 days in every 400 years which is equivalent to an average of 365.2425 days per year. In 2009 we found a formula for converting month and day to day of year which can be used to simplify the formula for the uniform circular motion approximation of the declination of the Sun. The Spring Equinox occurs on March 20/21 which is doy 79/80. The uniform circular motion approximation is a simple sine function. (The floor function reduces a decimal number to just its integer part.)

A comparison of this approximation with that of a Keplerian orbit is shown below.

The deviation on the right side is due to a nonuniformity in the motion of the Sun along the Ecliptic. The Earth moves faster in its orbit about the Sun near perihelion on about Jan 3 than near aphelion on about Jul 4 (doy 185).

A comparison of this approximation with that of a Keplerian orbit is shown below.

The deviation on the right side is due to a nonuniformity in the motion of the Sun along the Ecliptic. The Earth moves faster in its orbit about the Sun near perihelion on about Jan 3 than near aphelion on about Jul 4 (doy 185).

### For Further Reading

Here are some links if you would like a little more information on the apparent motion of the Sun relative to the Earth and its effect on the climate. Assuming uniform circular motion one gets a first approximation to the motion of the Sun along the Ecliptic which is used in the link on the Declination of the Sun. This can be used with in the derived formula for an estimate of the amount of daylight for any day of the year.

Ecliptic

Plane of the Ecliptic

Sun Path

Effect of the Sun Angle on Climate

Day Length

Declination of the Sun

Uniform Circular Motion

Eudoxan Astronomy

Right Ascension

From Stargazers to Starships

The last link is to a website which gives a basic introduction to Astronomy.

Ecliptic

Plane of the Ecliptic

Sun Path

Effect of the Sun Angle on Climate

Day Length

Declination of the Sun

Uniform Circular Motion

Eudoxan Astronomy

Right Ascension

From Stargazers to Starships

The last link is to a website which gives a basic introduction to Astronomy.

## Thursday, March 24, 2011

### The Number of Hours of Daylight Throughout the Year by Latitude

I did a plot of the number of hours of daylight as a function of the number of days since the winter solstice with steps in latitude of 10° and obtained the following curves.

Above the arctic circle the number of hours of daylight is 24 hours a day for a period of time depending on the latitude. The derived formula indicates that this is indeed the land of the midnight sun. The formula also indicates that at the poles there is either one very long day or one very long night since the Sun either stays above the horizon for one half of the year or below it for the rest of the year.

Above the arctic circle the number of hours of daylight is 24 hours a day for a period of time depending on the latitude. The derived formula indicates that this is indeed the land of the midnight sun. The formula also indicates that at the poles there is either one very long day or one very long night since the Sun either stays above the horizon for one half of the year or below it for the rest of the year.

## Tuesday, March 22, 2011

### A Check on the Amount of Daylight

I live in an agricultural community and information on sunrise and sunset is published daily in the local newspaper on the weather page. The hours of daylight listed for the last few days differs from that computed using the formula for a local latitude of 37° and the declination of the Sun computed for midday in our time zone.

There are two main reasons for the discrepancy. The first is the astronomical definition of the time of sunrise which is taken to be the time at which the Sun first appears on the horizon. At this time the center of the Sun is half its diameter below the horizon. The second correction needed is for the refraction of the Sun's rays by the Earth's atmosphere usually taken to be 34 minutes of arc*. These two corrections combine to add an additional Δt equal to 6 minutes 40 seconds to the amount of daylight. The differences between the published lengths of time and corrected formula times is about one minute.

*see P. Kenneth Seidelmann, ed., Explanatory Supplement to the Astronomical Almanac, 1992, p.483

There are two main reasons for the discrepancy. The first is the astronomical definition of the time of sunrise which is taken to be the time at which the Sun first appears on the horizon. At this time the center of the Sun is half its diameter below the horizon. The second correction needed is for the refraction of the Sun's rays by the Earth's atmosphere usually taken to be 34 minutes of arc*. These two corrections combine to add an additional Δt equal to 6 minutes 40 seconds to the amount of daylight. The differences between the published lengths of time and corrected formula times is about one minute.

*see P. Kenneth Seidelmann, ed., Explanatory Supplement to the Astronomical Almanac, 1992, p.483

### Changes in the Amount of Daylight Throughout the Year

Yesterday was the Vernal Equinox at which time the amount of daylight and nighttime are equal to each other and this day marks the beginning of spring. The seasons are characterized by the amount of sunlight each day receives and the amount of warming that results. Ptolemy combined astronomy with geography noted the connection between latitude and the amount of the daylight. Like Aristotle before him who divided the world into five climate zones Ptolemy was concerned with the habitable parts of the world. But Ptolemy also noticed that longest and shortest periods of daylight could be used to determine a location's latitude. As an example he notes that for Rhodes the latitude is 36° and the longest period of daylight is about 14½ hours (at the summer solstice).

I thought it might be interesting to derive a formula to find the amount of daylight throughout the year. In what follows θ and φ indicate the latitude and longitude of the Sun and a location on the Earth's surface. The subscripts, S and L are used to distinguish the two sets of coordinates. Z indicates a location and its zenith direction. P is the pole of the Earth's rotation, M a point on the Equator through which the location's meridian passes and M' is another meridian which is 90° to the east of M.

Using these quantities we can specify both the position of some location and the relative position of the Sun. The Sun is above the horizon when the projection of the two directions onto each other is positive. This condition gives in an equation involving three angles.

The times of sunrise and sunset depend on the latitudes of Sun, θ_S, and the location, θ_L, and the difference between these times, Δt, is known if Δφ is known. For the day with longest daylight the Sun is over the Tropic of Cancer and its latitude is 23.44°. At the latitude of Rhodes, 36°, this gives the maximum amount of daylight as being 14.4 hours which is in good agreement with Ptolemy's value. One can also reverse the procedure and estimate the latitude if the maximum amount of daylight is known.

I thought it might be interesting to derive a formula to find the amount of daylight throughout the year. In what follows θ and φ indicate the latitude and longitude of the Sun and a location on the Earth's surface. The subscripts, S and L are used to distinguish the two sets of coordinates. Z indicates a location and its zenith direction. P is the pole of the Earth's rotation, M a point on the Equator through which the location's meridian passes and M' is another meridian which is 90° to the east of M.

Using these quantities we can specify both the position of some location and the relative position of the Sun. The Sun is above the horizon when the projection of the two directions onto each other is positive. This condition gives in an equation involving three angles.

The times of sunrise and sunset depend on the latitudes of Sun, θ_S, and the location, θ_L, and the difference between these times, Δt, is known if Δφ is known. For the day with longest daylight the Sun is over the Tropic of Cancer and its latitude is 23.44°. At the latitude of Rhodes, 36°, this gives the maximum amount of daylight as being 14.4 hours which is in good agreement with Ptolemy's value. One can also reverse the procedure and estimate the latitude if the maximum amount of daylight is known.

## Monday, March 21, 2011

### Using a Scientific Calculator to Compute Chords

One can use a scientific calculator to compute the chords and convert them into sexigesimal notation. I used an old TI-34 Solar Scientific Calculator but an updated version of this is the TI-36X Solar. One needs a scientific calculator which can convert between decimal degrees (DD) and degree-minute-second (DMS).

The image above displays the length of the chord of 90° which is 60 √2. To calculate this one enters 2, performs the √ operation, multiplies by 60 (the radius of the circle), and then converts the DD value to DMS. Similarly, the chord of 72° is 60 √((5-√5)/2) or 70;32,03.

The image above displays the length of the chord of 90° which is 60 √2. To calculate this one enters 2, performs the √ operation, multiplies by 60 (the radius of the circle), and then converts the DD value to DMS. Similarly, the chord of 72° is 60 √((5-√5)/2) or 70;32,03.

## Sunday, March 20, 2011

### Ptolemy's Chord for 1°

One cannot trisect a given angle using just a compass and straightedge but Ptolemy needs to find the chord of 1° so he has to find a way of numerically trisecting a 3° angle. Using the procedure for computing the chord of half an angle he can find the chords of 3/2° and 3/4°. He then makes two estimates of the chord of 1° by projecting these chords onto the vertical 1° line. The effect of these projections is exaggerated in the figure below.

Ptolemy shows that both estimates give, approximately, the same value of 1;02,50 for the chord of 1° and and he therefore concludes that this its value. One can also arrive at this value by considering a trapezoid whose sides are the heights of the chords at the known angles and finding its height at 1°.

Ptolemy shows that both estimates give, approximately, the same value of 1;02,50 for the chord of 1° and and he therefore concludes that this its value. One can also arrive at this value by considering a trapezoid whose sides are the heights of the chords at the known angles and finding its height at 1°.

## Saturday, March 19, 2011

### Ptolemy's Chord Theorems

One can extract a number of procedures for computing other chords from known chords by studying his example of the use of the Pythagorean Theorem to find the chord of a supplementary angle and the proofs concerning the chord of half an angle and those of the difference and sum of two angles. The proofs are in the style of Euclid's Data but through doing the proofs he also shows how to find unknown chords from known chords.

Menelaus of Alexandria wrote six books on chords which unfortunately have been lost over time and like Hipparchus he may have used a half angle theorem. The procedure found in Ptolemy is as follows.

The proofs for the chords of the difference and sum of two angle make use of Ptolemy's Theorem. This theorem may be due to Ptolemy since it is not found prior to its appearance in the Almagest.

The procedure for chord of the sum of two angles is similar to that of the difference but involves the use of an intermediate angle.

By using these procedures we can find the chords of other angles in Ptolemy's chord table down to the chord of 3° and halves of these angles. The last three procedures involve multiplication and division by 60 which for sexagesimals only involves a shift in the sexagesimal place. Ptolemy's choice of 120 for the diameter of his circle is seen as one of convenience.

EDIT (20 Mar 2011): Division by 60 corrected to division by 120 for chords of difference and sum.

Menelaus of Alexandria wrote six books on chords which unfortunately have been lost over time and like Hipparchus he may have used a half angle theorem. The procedure found in Ptolemy is as follows.

The proofs for the chords of the difference and sum of two angle make use of Ptolemy's Theorem. This theorem may be due to Ptolemy since it is not found prior to its appearance in the Almagest.

The procedure for chord of the sum of two angles is similar to that of the difference but involves the use of an intermediate angle.

By using these procedures we can find the chords of other angles in Ptolemy's chord table down to the chord of 3° and halves of these angles. The last three procedures involve multiplication and division by 60 which for sexagesimals only involves a shift in the sexagesimal place. Ptolemy's choice of 120 for the diameter of his circle is seen as one of convenience.

EDIT (20 Mar 2011): Division by 60 corrected to division by 120 for chords of difference and sum.

## Thursday, March 17, 2011

### Ptolemy's Circle

So it would seem that the price of admission to Ptolemy's Circle is a thorough knowledge of the Works of Euclid and perhaps some acquaintance with the tradition of the Pythagoreans. There are some other things that we need to know about his circle. He uses the sexagesimal number system for measurement of both the central angle and the length of its chord. A circle is divided into 360 degrees (°) which are further subdivided into minutes, seconds, etc. The diameter of the circle is divided into 120 parts (p) which are also subdivided into minutes, seconds, etc. Note that the ratio of the circumference to the diameter is 3 (°/p) instead of π but with these mixed "units" the irrational is avoided.

From the diameter of the circle and the sides of the first three regular polygons Ptolemy finds the set of initial chords for his table.

Note the chords are given in sexagesimals. The number before the semicolon is the number of parts in the same units of length as the length of the diameter of the circle. The following three numbers are the minutes, seconds and "thirds" of the parts. I calculated these chords to one more sexagesimal place than Ptolemy does so that one can see his accuracy.

From the diameter of the circle and the sides of the first three regular polygons Ptolemy finds the set of initial chords for his table.

Note the chords are given in sexagesimals. The number before the semicolon is the number of parts in the same units of length as the length of the diameter of the circle. The following three numbers are the minutes, seconds and "thirds" of the parts. I calculated these chords to one more sexagesimal place than Ptolemy does so that one can see his accuracy.

### Some Greek Math

I have been concerned about some points of Ptolemy's solution to the side of the pentagon lately and the book, Science Awakening I by B. L. van der Waerden, has provided some answers. Ptolemy considers it sufficient that the lengths in his construction are in "extreme and mean ratio" in order to show that they yeild the chords of the decagon and pentagon. On page 101 of van der Waerden's book the author states that the Pythagoreans knew that lines of the five pointed star divided themselves in extreme and mean ratio and that they knew how to solve the resulting quadratic equation. Van der Waerden later goes further into the procedures found in the works of Euclid, in the Elements and the Data, and shows that they provide solutions to standard problems (see pages 118-24). The methods are primarily geometrical. He also states that there seems to have been a work in ancient times known as "The Tradition of Pythagoras" which contained their knowledge of geometry and possibly drew on Babylonian sources.

But Ptolemy is rather emphatic with his, "I say that ZD is the side of the [regular] decagon, and BZ the side of the [regular] pentagon." on p. 48 of Toomer's Ptolemy's Almagest.

But Ptolemy is rather emphatic with his, "I say that ZD is the side of the [regular] decagon, and BZ the side of the [regular] pentagon." on p. 48 of Toomer's Ptolemy's Almagest.

## Thursday, March 3, 2011

### Ptolemy's Construction for the Side of a Pentagon as a Formula

Ptolemy shows how to construct the length of the side of a pentagon in Fig. 1-1 on page 48 of Toomer's

One possibility is that he used two procedures relating the chords of a pentagon to solve the equivalent of two equations with two unknowns. Ptolemy's theorem can be used to relate the sides of an inscribed quadrilateral and for the pentagon there are only two unknown chords. Applying Ptolemy's theorem to three neighboring sides of a pentagon and the larger chords needed to complete the quadrilateral results in a quadratic equation involving the two unknown chord lengths. The quadratic equation can be solved for the ratio of the lengths of the two sides which turns out to be the golden ratio.

One can then use the formula for finding the chord of half an angle to obtain another relation between the two chords. Substituting the golden ratio into this formula simplifies it and reduces it to an equation involving one unknown which can be easily solved for the length of the side of a pentagon.

The figure above is a copy of the one in

Following the construction through one step at a time and converting the lengths involved into numbers we end up with the same formula obtained by the algebraic solution above. Ptolemy follows the steps of the construction computing the lengths and sides of the triangles in the figure and thus obtains the chord of 72°.

Ptolemy's method for finding the chord of half the angle given the chord of an angle is algebraically equivalent to the formula used above. He also uses the proportion of the golden ratio in his proof. So it is not unlikely that the solution of these equations was involved in finding the formula for the side of the pentagon.

*Ptolemy's Almagest*. It is BE in the figure below but Ptolemy does not explain how this was arrived at. There seems to be another missing key needed to understand his methods.One possibility is that he used two procedures relating the chords of a pentagon to solve the equivalent of two equations with two unknowns. Ptolemy's theorem can be used to relate the sides of an inscribed quadrilateral and for the pentagon there are only two unknown chords. Applying Ptolemy's theorem to three neighboring sides of a pentagon and the larger chords needed to complete the quadrilateral results in a quadratic equation involving the two unknown chord lengths. The quadratic equation can be solved for the ratio of the lengths of the two sides which turns out to be the golden ratio.

One can then use the formula for finding the chord of half an angle to obtain another relation between the two chords. Substituting the golden ratio into this formula simplifies it and reduces it to an equation involving one unknown which can be easily solved for the length of the side of a pentagon.

The figure above is a copy of the one in

*Ptolemy's Almagest*. Ptolemy assumed that the diameter of the circle had 120 parts but in what follows it is assumed that the radius of the circle is 1.Following the construction through one step at a time and converting the lengths involved into numbers we end up with the same formula obtained by the algebraic solution above. Ptolemy follows the steps of the construction computing the lengths and sides of the triangles in the figure and thus obtains the chord of 72°.

Ptolemy's method for finding the chord of half the angle given the chord of an angle is algebraically equivalent to the formula used above. He also uses the proportion of the golden ratio in his proof. So it is not unlikely that the solution of these equations was involved in finding the formula for the side of the pentagon.

### Half-Angle Chord Formula

Why is Ptolemy's chord table more precise than Hipparchus'? The 7 1/2° steps suggest that he arrived at the angles by successive division 60° by 2 (7 1/2° = 15°/2). Subdivisions of 72° do not exist so he appears not to have know the chord of 72° which is the side of the pentagon.

As mentioned before, one can use the golden ratio and a method for the chord of a half angle to find the side of a pentagon. In the figure above c_1 is the chord of the angle and c_2 that of half the angle. The sides, x and y, are the required sides of the right triangles. Here the circle is assumed to have a unit radius.

One can then use the Pythagorean Theorem to find a formula for the chord of the half angle in terms of the chord of the original angle. This along with the golden ratio will allow us to find an expression for the side of a pentagon.

As mentioned before, one can use the golden ratio and a method for the chord of a half angle to find the side of a pentagon. In the figure above c_1 is the chord of the angle and c_2 that of half the angle. The sides, x and y, are the required sides of the right triangles. Here the circle is assumed to have a unit radius.

One can then use the Pythagorean Theorem to find a formula for the chord of the half angle in terms of the chord of the original angle. This along with the golden ratio will allow us to find an expression for the side of a pentagon.

### Knowledge of Euclid's Elements is Assumed by Ptolemy

One can find the theorems concerning inscribed angles in Euclid's Elements which would explain why Ptolemy assumes that they are known. They are Propositions XX through XXII of Book IV and roughly state the following.

Prop. XX The central angle of a circle is double the angle at the circumference.

Prop. XXI Inscribed angles in same segment of a circle are equal.

Prop. XXII The sum of opposite angles of a quadrilateral inscribed in a circle is two right angles.

One also wonders how he arrived at a value for the chord of 120° which is the side of a pentagon. The Pythagoreans knew how to construct a pentagon from a isosceles triangle with base angles of 72° and a peak angle of 36° (see Elements, Book IV, Prop. XI). But that doesn't tell how Ptolemy arrived at a magnitude for the side of the pentagon. One possibility is the use of the golden ratio which is found in Euclid's Elements (Book VI, Prop. XXX) and what is known as Ptolemy's theorem or some other method to find the chord of half the angle if the chord of an angle is known.

Ptolemy was not the first to construct a table of chords. The credit for that goes to Hipparchus who constructed a 7 1/2° chord table which Ptolemy later improved on with a 1/2° table. Hipparchus anticipated much of what Ptolemy did concerning astronomy. He is believed to have traveled to Alexandria. Hipparchus made astronomical observations at Rhodes where the Antikythera mechanism is believed to have come from.

Prop. XX The central angle of a circle is double the angle at the circumference.

Prop. XXI Inscribed angles in same segment of a circle are equal.

Prop. XXII The sum of opposite angles of a quadrilateral inscribed in a circle is two right angles.

One also wonders how he arrived at a value for the chord of 120° which is the side of a pentagon. The Pythagoreans knew how to construct a pentagon from a isosceles triangle with base angles of 72° and a peak angle of 36° (see Elements, Book IV, Prop. XI). But that doesn't tell how Ptolemy arrived at a magnitude for the side of the pentagon. One possibility is the use of the golden ratio which is found in Euclid's Elements (Book VI, Prop. XXX) and what is known as Ptolemy's theorem or some other method to find the chord of half the angle if the chord of an angle is known.

Ptolemy was not the first to construct a table of chords. The credit for that goes to Hipparchus who constructed a 7 1/2° chord table which Ptolemy later improved on with a 1/2° table. Hipparchus anticipated much of what Ptolemy did concerning astronomy. He is believed to have traveled to Alexandria. Hipparchus made astronomical observations at Rhodes where the Antikythera mechanism is believed to have come from.

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