Update on the prediction for the relative positions of the binary star Xi Bootis

In my previous post I described a program to calculate the relative positions of a binary star for any given date. Using this program I can now verify the prediction that I had obtained previously using Roger Wesson's online calculator (or perhaps, more honestly, I can ensure that my code is performing the same as Roger's). Using the parameters for the Xi Bootis in that previous post Roger's utility predicted that separation and PA of Xi Boo on the 1st May 2024 (2024.333) was 4.81 arcseconds and 289.8 degrees, respectively. Running my code using the same input values I get that the PA is 289.8163749 degrees and the separation is 4.814400629 arcseconds which agrees exactly with Roger's results to the same number of decimal places as he quotes.

To make sure that my program works in a wide variety of cases I have also recomputed the predicted positions of these stars over the period 2024 to 2174 in gaps of 10 years (again refer to my earlier post) and, again, to the same number of decimal places I see no difference between the results of my code and those of Roger's. This is all good!

I have since realised that the BAA produces tables of ephemerides of visual binaries in the BAA Handbook (see page 111 for the 2025 book). I notice that the orbital elements they use come from the Sixth Catalog of Orbits of Visual Binary Stars at the US Naval Observatory. For Xi Bootis (ADS number 9413) they give the following parameters:-

Orbital Period P (years) : 152.9614

Date of periastron T : 1909.6213

Semi-major axis of orbit a (arc seconds): 4.93454

Eccentricity of orbit e : 0.51385

Inclination of orbit to plane of sky i (degrees) : 140.453

Argument of periastron ω (degrees) : 25.492

PA of ascending node Ω (degrees) :168.795

Using these parameters for Xi Bootis in my program for the 1st May 2024 (2024.333) I get that the PA is 290.9 degrees and the separation 4.97 arcseconds (as opposed to 289.8 and 4.81 found previously). This is slightly better in agreement with my observation for this date of 291 +/- 3 degrees and 5.5 +/- 0.4 arcseconds, respectively.

Finally, just to check my results against those of the BAA, using these parameters I get for 2025.0 that the PA is 289.8 degrees and the separation is 4.91 arcseconds. The BAA gets 289.7 and 4.91. For 2026.0 I get 288.2 and 4.82 and the BAA gets 288.1 and 4.82. So the BAA seems off by 0.1 in the PA each time.

All text and images © Duncan Hale-Sutton 2025

A program to calculate binary star orbits

Back in June last year I wrote a blog entry about predicting the relative positions of the binary star Xi Bootis. At that time I relied on an online calculator provided by the astronomer Roger Wesson. I thought it would be a good idea to have a means of computing these values for myself and towards that end I have now written some code for a programmable calculator (a Casio fx-4500P). This is not a particularly modern calculator (it dates from 1989/1990) but it is quite useful as it can store up to 1103 programmable steps and has 26 standard memory locations. The algorithm I have used to calculate the binary star positions comes from Practical Astronomy with your Calculator by Peter Duffet-Smith. I bought this book in 1981 but it has served me well over the years. The section (59) on binary star orbits can be found on page 130. This is my program:-

T"PERIOD"

E"PERIASTR"

Y"YEAR"

Z"E"

A"A"

L"LONG PERI"

I"I"

H"PA OF NODE"

F = (π/180)

X = (Y - E)/T

M = 2π(X - Int X)

X = M + Zsin M/(1 - Zcos M)

Lbl 1

D = X - Zsin X - M 

Abs D ≤ 1E-6 → Goto 2 ∆

W = D/(1 - Zcos X)

X = X - W

Goto 1

Lbl 2

N = 2tanˉ¹ (√((1 + Z)/(1 - Z))tan (X/2))

R = A(1 - Zcos X)

Q = sin (N + FL)cos FI

P = cos (N + FL)

X = tanˉ¹ (Q/P)

P < 0 → X = X + π: Goto 3 ∆

Q < 0 → X = X + 2π ∆

Lbl 3

O = (X/F) + H

O > 360 → O = O - 360 ∆

O < 0 → O = O + 360 ∆

O"PA" = O ▲

K"SEP" = RP/cos X ▲

Here is an explanation of some of the items in this code. The first 8 lines are the input parameters for the binary star as follows:-

 

T - the orbital period in years

E - the epoch or date of periastron

Y - the date of observation (in a decimal of a year)

Z - the eccentricity of the orbit e

A - the semi-major axis of the orbit a (in arc seconds)

L - the longitude or argument of the periastron ω (in degrees)

I - the inclination of the orbit to the plane of the sky i (in degrees)

H - the position angle (PA) of the ascending node Ω (in degrees)

The last two lines are the output values for the stars:-

O - the position angle θ of star B relative to star A (in degrees)

K - the separation ρ of star B from star A (in arc seconds)

The calculator should be set to run in radians. The symbol → is an implication sign and it forms part of a logic test. If the statement before it is true the statement after it is actioned otherwise the next line of the code is run. The symbol ∆ terminates this logic test. The symbol ▲ stops the processing at this point and displays the contents of the memory variable before the equals sign. The section of code between Lbl 1 and Lbl 2 is an iterative solution to Kepler's Equation to obtain the true anomaly (more about this later).

All text and images © Duncan Hale-Sutton 2025

Observation of Venus on the 25th January 2025

Last Saturday we had a clear patch of weather and so late afternoon, just after sunset, I thought I would have a look at Venus again to see how things have changed since half phase (dichotomy) was reached. Here is the page out of my notebook:-

The angular distance between Venus and the Sun is now decreasing but it is still pretty much in the same part of the sky as before. At the time of the observation it stood at 23 degrees above the horizon in the SSW. There was a bit of thin cloud about which glowed an orange colour in the sunset. I used the same set up as previously but due to how Venus is orientated the terminator was much more tilted than before and so I ended up just drawing it as I saw it in the eyepiece.

As you can see from my drawing the phase is now looking much more crescent in appearance. It has noticeably gone from convex to concave. This time I paid much more attention to the terminator itself and it is clear that it isn't a well defined boundary but has some fuzziness to it. I have tried to represent this in my drawing. What's more I thought I could at times detect a bit of cloud shading near the terminator - this is where there are subtle differences in the brightnesses of the clouds on Venus. Again I have very tentatively put what I saw in my drawing. One other thing was that the cusps looked somewhat brighter than the other parts of the illuminated side.

I had drawn myself some new phase diagrams on the computer to show what the predicted phase would look like from 40 to 50% in steps of 1%. With these on my phone I could compare them directly with what I saw at the telescope and I estimated that the phase was 41%. The predicted phase was a bit more than this at 42.5%, a difference of 1.5%. Here is what 42% should look like:-

All text and images © Duncan Hale-Sutton 2025

The Schröter effect or phase anomaly of Venus

The Schröter effect was first coined by Sir Patrick Moore and was named after Johann Schröter who first noted it when he was observing Venus in the 1790's. In essence, the effect is an anomaly that is observed when determining the phase of Venus. When the planet is approaching its Greatest Elongation East (GEE) in the evening sky dichotomy is observed to take place a few days earlier than predicted whilst when the planet is approaching Greatest Elongation West (GEW) in the morning sky dichotomy is observed to take place a few days later than expected. I think it is now generally accepted that this is a genuine optical effect caused by the thick atmosphere of Venus and is not due to observer error. 

I don't think it is too hard to understand in a hand-waving way why this occurs. Most predictions about the phase of a planet or moon assume that the body is a sphere with a well-defined surface. This is largely to help make predictions about phase relatively easy to compute (see my previous blog entry). To aid with this description I include my first diagram from my earlier post:-

Venus is shown here in the gibbous phase and let us imagine that it is in the evening sky. For this to be the case Venus would have appeared from behind the Sun after a superior conjunction and then steadily increased its angular distance from the Sun as it approaches its GEE. Its phase initially would be almost full and then it would be less and less gibbous until dichotomy (half phase) is reached around GEE. In this picture we assume that Venus has a hard spherical surface and that light from the Sun can just reach the edge of the planet at the terminator (see, for example, point T).

Now imagine that Venus has an upper atmosphere that is like a thick haze. Light can penetrate this haze to a certain degree and be reflected from it, but the well-defined surface is now obscured by the haze and lower atmosphere cloud. Light destined for a point on the terminator T enters the atmosphere at a point ahead of T towards O. As it penetrates the atmosphere some of the light is scattered by the haze and so the amount of light that would have reached T is much reduced. The effect is to soften the terminator and pull its apparent edge towards O.

At the 'poles' of the planet at P and P' something else happens. Light enters the atmosphere ahead of point P and P' as it does at T. The same scattering and light reduction occurs here too due to the haze. Light that would have terminated at P or P' on a hard surface continues diffuse into the haze beyond P and P' into the dark. However, unlike at T the effect is amplified because we are looking at the atmosphere edge on - we are seeing this forward scattering over a wider depth. The net effect is that we don't see the terminator move inwards at P and P' as we did at T.

I hope you can see from my description that the overall effect is that the terminator is pushed by this scattering process to be a little ahead (sunwards) of its predicted position. This means that as we approach EGE the observed phase is always less than its predicted phase and it is this that causes dichotomy to be observed a few days early.

Now if Venus is in the morning sky then just after inferior conjunction it is a very thin crescent. On the following days its angular separation from the Sun increases until GEW is reached. As this happens the crescent phase of Venus gradually thickens until dichotomy occurs. Again because Venus has a thick atmosphere the terminator of Venus will be a little bit sunwards of where it is predicted to be. This means that the observed phase of Venus will be less than the predicted value. This will cause dichotomy to be seen a few days later than expected.

For a full description of this model of how the atmosphere affects the phase of Venus you can read the paper in the BAA journal by Anthony Mallama published in 1996. I think the idea that the observed phase is altered by scattering is compelling and it is supported by the fact that the effect is more pronounced in blue and ultraviolet light rather than yellow (see, for example, this nice observation by David Basey in 2017). It is well-known that blue and ultraviolet is more scattered than yellow light. Also, the idea that light scattering in the upper atmosphere can diffuse into the dark portion of the visible disc is supported by the observation of the extension of the cusps of the terminator into areas which are predicted to be dark when Venus is a very thin crescent (see, for example, this image by John Sussenbach in 2017).

For further discussions of the Schröter effect have a look at this article by William Sheehan in 2017.

All text and images © Duncan Hale-Sutton 2025

Measuring the phase of a planet or moon

In a previous blog I had promised that I would prove something about measuring the phase of a planet (or moon) and here I will try to do this. The phase of a planet or moon is defined as the proportion of the sunlit side that is visible from Earth. In more detailed terms it is the ratio of the illuminated area to the total area of the visible surface. This diagram will help illustrate what is going on:-

I will assume that the body in question is a sphere. From Earth the body appears to be illuminated from the left by the Sun and the shaded area represents the dark side. The centre of the observed body is O and the points T and S lie on a diameter that passes through O. T lies on the terminator - the boundary between the light and dark sides of the body. P and P' are the points on the terminator where they intercept the apparent edge of the body. They also lie on a diameter and the line POP' is perpendicular to the line TOS.

To make things more clear we should also view this arrangement by looking down on the body from directly above P:-

This makes it clear how the gibbous phase in the first diagram is seen. Note that it is assumed that the observer and the Sun are along way from the body so that rays from the Sun arriving at the body are parallel as are the rays of reflected light arriving at the observer.

We now wish to calculate the apparent area of the illuminated part of the body (refer to the first diagram). Let the apparent radius of the body (the distance OP or OS) be a. Let the distance TO be b. We now note from both diagrams that the terminator is a great circle that passes through T, P and P'. This great circle is viewed obliquely from Earth and as a consequence is the shape of an ellipse (you can see the hidden remainder of the terminator as a dashed line in the first diagram). This ellipse has semi-major axis a and semi-minor axis b. The area of the ellipse is πab. So it follows that the illuminated area is made up of half this area plus half the full area of the body as seen from Earth. That is (π/2)(ab + a²). As a ratio of the total area this is 

(π/2)(ab + a²)/πa² = (1/2)(1 + b/a). 

This is the phase. If we let b vary from -a to a, then we can also account for phases which are crescent as well as gibbous. Note that when b = -a, then the phase is 0, when b = 0 the phase 1/2 and when b = a the phase is 1 as expected.

We can get a slightly better relationship if we note that the distance TS = a + b and as 1 + b/a = (a + b)/a then

Phase = TS / 2a = TS / D

where D is the diameter of the apparent body. This is a simple and useful measure of phase assuming that the body behaves like an illuminated sphere.

This is the expression that W.M. Smart uses in his Foundations of Astronomy (first published 1942) but even he doesn't explain how it relates to the ratio of the areas. He just says (in Section 79, page 97), "This fraction measures the phase." Paul Abel, director of the Mercury and Venus section, suggests to observers that they measure the phase using this fraction TS / D as seen in this article (look for the heading 'Measuring the phase'). One of the problems with this measure is the assumption that the object being looked at is a solid sphere. In the case of Venus the issue is complicated by its thick atmosphere and the phase anomaly (and other visual anomalies) are an indictor that this assumption might not be a valid one.

All text and images © Duncan Hale-Sutton 2025

Phase of Venus on the 9th January 2025

Three days ago we had some more cold clear weather late in the afternoon. Venus is now quite a prominent object in the sky and seems very bright. It will get brighter still even though the illuminated area of the planet will be shrinking. This is because Venus is getting closer to us as it heads towards inferior conjunction. Maximum brightness will be around the 20th February when it reaches magnitude -4.9. On the 9th it was 1 day away from greatest elongation east and at about 26 degrees high in the SSW.

I was again out the front of the house with my 102mm Celestron:-

The most important thing about this observation was that in my opinion dichotomy (or half phase) has been reached! This was the feature that I had been waiting to see these last few weeks. I was again using an eyepiece and barlow lens combination that gave a magnification of 147x. When the image had settled enough (the breeze and seeing playing its part) I almost felt like I could detect some concavity in the shape of the illuminated side (which would mean that the phase was less than 50%). It definitely looked like half phase at the very least. This is against what would be expected for a solid sphere at Venus's current position. The predicted phase on this date is still 51.6% and it indicates to me that the phase anomaly (that dichotomy is observed about 4 days ahead of its prediction) is a real effect. Why this is so may be to do with the atmosphere of Venus and I will try and explain this another time.

One effect that I haven't been able to understand is that, to my eyes, the northern cusp still looked slightly rounded with the indentation of dark eating into the light side. This has been apparent all through this string of recent observations (the southern cusp on the other hand went straight to the edge with no deviation). I will be interested to see what happens a few days after the predicted date of dichotomy.

All text and images © Duncan Hale-Sutton 2025

Drawing of the crater Atlas on the Moon (3rd January 2025)

After I had completed my phase observation of Venus on the 3rd, I thought I would have a go at drawing some feature on the nearby Moon. The Moon at this stage was only 4 days old and a nice thin crescent. Near the terminator were a number of nicely lit craters and I settled on this one to draw:-

Atlas is an 89 km diameter impact crater that can be found at latitude 47 degrees North and longitude 44 degrees East (lunar coordinates). I was using my Celestron NexStar 102 SLT with a 9mm orthoscopic eyepiece and a star diagonal. This gave a magnification of 73x. I started observing at 16:25 UT and finished at 16:48. The seeing was pretty good and AII on the Antoniadi scale.

The first thing to point out was that I didn't know I was drawing Atlas before I started. I had to work this out after I had finished (I did draw the positions of craters Franklin and Cepheus as a guide to recognition). What amazed me was my lack of understanding of the orientation of what I was looking at! Fortunately, I have a very good Atlas (the Collins Atlas of the Night Sky by Storm Dunlop) which has lunar maps for each type of telescope combination. So with a refractor with a star diagonal in the position I had it in, lunar West and East were in the right position but North and South were flipped (see my diagram). This was a surprise to me and, as I had been using the same set-up for my Venus observations, it meant that I had been assuming North and South were the wrong way round (not so easy to tell this for a featureless surface like Venus). Ho hum, as they say. Rookie error! So if you want to view my drawing in the usual manner you would have to reflect in a line along the east-west axis.

Note also another very confusing thing that I didn't appreciate. When you look at the moon directly what is labelled as the East side of the Moon is not on the east side of the sky - it is on the west! It was decided to have this type of labelling because then the terminator moves from lunar East to West, a bit like on Earth. Very confusing if you are trying to orientate yourself.

Anyway, I was able to identify that it was Atlas that I had been drawing. Note that Atlas was just beyond the terminator and in the illuminated part of the Moon. West of Atlas the Moon was in shadow. You can see there is quite a dark wall on the North side of the crater and this because there is another crater next to it (Atlas E) which causes this feature (have a look at this Sky at Night article and the image of this area). I think another smaller crater (Atlas A) is to the South.

My drawing may look a bit small but in my defense I would say that drawing at the telescope in freezing conditions is not easy! This is the first time I had tried this in a serious way and it was quite taxing to try and get the details right.

All text and images © Duncan Hale-Sutton 2025