Start of Summer

Updated 1 May 26

The Spring equinox this year was on the 20th of March. With the Summer solstice on the 21st of June, I estimate the mid-point - and surely the proper start of Summer -  will be about the 5th of May. However, the first of May seems a good enough time to celebrate it.

Traditionally, May Day is associated with fertility, so we have the symbolism of the May Pole and the May Queen in many European countries. In Beltane, there is a tradition of lighting bonfires, no doubt to symbolically to drive away the dark and the evil spirits (rather than to defiantly face them as done at Halloween).

In Germany (and elsewhere in Europe), the celebration has been co-opted by the Church, to celebrate Saint Walpurga, an Anglo-Saxon nun tasked with converting the Germanic tribes in the eighth century. There are bonfires the night before (Walpurgisnacht in Germany), indicating a common root with Beltane. Given Walpurga is known for having fought against witchcraft, she was an inspired choice for re-branding the event!

As noted in an earlier post, I've read that having four seasons in the year is a fairly recent innovation and that traditionally there were just Winter and Summer. (Ireland seems to be the exception, but some scholars have said that the first references to Imbolc were by Christian writers, casting doubt on an ancient four seasons tradition). However, if that is so, why celebrate or note the mid-points between solstices and equinoxes?

In addition, it seems there was a Germanic and Nordic custom of Summer starting with May Day and ending with Halloween. Perhaps this was nothing to do with seasonal mid-points, but because of the time it takes for earth and sea to heat up and to cool down again, although that seems to be at odds with the fire festivals, which suggest they were about daylight. Either way, mid-Summer and mid-Winter celebrations would still have been held on the solstices and not early August (Lammsas) and early February (Imbolc).

Perhaps, in the Germanic tradition, Summer and Autumn (albeit the latter un-named) were combined to form one large season with Winter and Spring (also un-named) the other? 

Measuring the elongation of Venus with a watch?

Updated 3 May 26

As an alternative, I thought instead about just using the timings of the settings of the Sun and of Venus (I'm not a morning person!). Thinking that no doubt someone has already investigated this, after some digging, I came across this site with a method for such a measurement. However, there's no derivation of the method and it seems to assume that both will set at the same azimuth, which doesn't seem right. Anyway, I thought I'd check it with Stellarium and chose the greatest elongation East for Venus of 15 August 2026:

• Sun and Venus are about 45° apart, as we'd expect
• the Sun's declination (angular distance above the celestial equator) is around 14°
• that of Venus is around -5° (i.e. it's below the celestial equator), so very different from that of the Sun
• the Sun straddles the horizon, i.e. is half set at around 20.20 (BST?) at azimuth 294°
• Venus sets about 80 minutes later, at azimuth 261°, i.e. there's a whopping 33° difference!

If we use the formula, it says for a 45° separation, Venus sets 3 hours, not 1 hour, after the Sun, so the formula is indeed wrong - the Sun and Venus will set at the same point on the horizon only if they are at the same declination, which is not the case here nor in general. As a comparison, the star Denebola *is* at about the same declination as the Sun at this time, with an angular separation from it of 31°. The formula gives about 2 hours between the Sun and Denebola setting and Stellarium confirms that.

In addition to the time between the setting of the Sun and of Venus, then we'd need the difference in azimuths at setting or the declinations - and a new formula.

Verdict: What's the point?

Measuring the elongation of Venus

Updated 30 Apr 26

Given Mercury's proximity to the Sun, let's take it that measuring elongation will always be a challenge, so let's limit our investigation to Venus for now.

Measuring the elongation of Venus when it's distant from the Sun may seem easy enough in principle, if you have a clear view of the horizon -  just measure the direction on the horizon (azimuth) of the centre of the setting Sun and at the same time, record the azimuth of Venus and its altitude above the horizon and then determine the angular difference. However, given that the Sun will be bright and Venus much less so, perhaps it's going to be difficult in practice to observe both at the same time?

Wait then until the Sun has almost fully set, with just a remnant showing on the horizon, and Venus may be more obvious. Take measurements as before, but note the Sun's altitude is now one solar diameter (~ 0.5°) below the horizon, which we should probably take into account.

However,  refraction by the Earth's atmosphere will also make a difference to the measurements of azimuth of both Venus and of the Sun. At the moment we don't know how to quantify this effect, but hope that it will be similar in both. Let's not worry too much about this and consider our measurement an estimate only. One or two degrees in measuring greatest elongation is unlikely to make a massive difference to our estimate of the radius of the orbit of Venus, although it will of course have more significance for estimating that of Mercury.

Determining the orbital radius for Saturn

Updated 28 Apr 26

To complete the orbital radii of the classical planets, let's now find Saturn's:

rₛ ~ (dₑ + sin θ)/ ((dₑ/pₛ) + sin θ)

where pₛ is its orbital period, for which we'll give the value of 29.45 that we found earlier.

This time, let's double Δt to 4 days, because we know θ may be too small to measure otherwise. As a result, distance dₑ will now be 8π/365.25 AUs. However, as this is an approximation (chord vs. arc of a circle), we will have increased the error compared with a time of 2 days.

Considering the opposition of Saturn on October 4 2026 and using Stellarium to determine its position 48 hours before and 48 hours after, and this tool, θ, the angular distance Saturn is seen to have moved in those four days is approximately 0.32° of arc, slightly more than for Jupiter over 2 days, again hopefully measurable.

Using these values, rₛ ~  9.4 AUs, while the accepted figure is around 9.6 AUs, which perhaps we could have achieved by averaging the results from many opposition events.

Again, checking Kepler's third law:

• the cube of its orbital radius is 9.4³ ~ 831
• the square of its orbital period is 29.45² ~ 867
• the former divided by the latter is ~ 0.96

Determining the orbital radius for Jupiter

Updated 27 April 26

Adapting the formula that we used for Mars, the radius of Jupiter's orbit, rⱼ, is given by:

rⱼ ~ (dₑ + sin θ)/ ((dₑ/pⱼ) + sin θ)

where pⱼ is its orbital period, for which we'll give the value of 11.85 that we found earlier.

Having said earlier that for Jupiter etc, it would be better to consider a larger Δt, for now, I'm going to use exactly the same value as for Mars, i.e. 2 days, and see how it goes, so as before, distance dₑ is 4π/365.25 AUs.

Considering the opposition of Jupiter on January 10, 2026 and using Stellarium to determine its position 24 hours before and 24 hours after, and this tool, θ, the angular distance Jupiter is seen to have moved in those two days, is approximately 0.27° of arc. This is smaller than was the case with Mars, now about half the angular size of the full moon, but perhaps still measurable with reasonable accuracy using the naked eye, a good star chart and a quadrant, say?

Using these values, rⱼ ~  5.1 AUs, which is very close to the accepted figure of about 5.2 AUs, which we would likely have arrived at by averaging the results from many opposition events.

Again, checking Kepler's third law:

• the cube of its orbital radius is 5.1³ ~ 132.7
• the square of its orbital period is 11.85² ~ 140.4
• the former divided by the latter is ~ 0.94

If we instead use the accepted value for the orbital radius, then this ratio becomes ~1, as we should expect.

For Saturn, the angle θ over just two days will be smaller again, so in that case it really will make sense to increase the time interval Δt and hence the value of dₑ.

Earth's gravitational pull vs. the Sun's

Updated 26 Apr 26

To compare the Earth's gravitational pull with the Sun's, let's look at the Moon's orbit about the Earth, but to keep it simple, we'll ignore its mass. (Assuming a similar density to the Earth's, its mass would be about 1/50th of it, and this approximation is fine for our purposes).

The Moon's orbital period is 27.3 days or 27.5/365.25 years.

Square that and it's about 5.6 x 10⁻³.

From memory, it's about 240 thousand miles away. Again, from memory, the Sun is about 93 million miles away, so the radius of the Moon's orbit about the Earth is about 2.6 x 10⁻³ AUs. 

(Actually, I've done this back-to-front and imagine that the size of the AU was first determined in relation to the Earth-Moon distance, something that had been measured fairly accurately by the Ancient Greeks).

Cubing that is about 17.1 x 10⁻⁹.

The ratio of this number to the square of the orbital period is about 3 x 10⁻⁶. This indicates the strength of the gravitational field of the Earth (from Newtonian Gravitational theory). Let's call it Γₑ.

The equivalent figure for the Sun (let's call it Γₛ)  is 1 (see my earlier post on Kepler's third law). So, the gravitational pull of the Sun is Γₛ/Γₑ, i.e. about 330,000, times that of the Earth. We know from Newton that this is also the ratio of the Sun's mass to that of the Earth.

Determining the orbital radius for Mars

As mentioned in an earlier post, the (average) orbital radius for an inner planet can be readily determined from its (average) maximum elongations observed from the Earth. However, another way is needed for the outer planets and the sketched diagram below illustrates one such method.

As an approximation, we assume orbits are circular.

In a short interval of time, Δt, centred on the opposition of Mars (corresponding to position 2 in the diagram) seen from the Earth, the Earth will move a short distance, dₑ, in green, approximated as a short straight line segment of length rₑ.ѡₑ.Δt, where rₑ is the radius of the Earth's orbit and ѡₑ is its angular velocity.

Similarly, in the same interval of time, Mars will move a short distance, dₘ, in red, approximately equal to rₘ.ѡₘ.Δt, where rₘ is the radius of Mars' orbit and ѡₘ is its angular velocity, which we know (see earlier post) is around ѡₑ/1.9. Hence dₘ is about dₑ.rₘ/1.9 AUs.

Let's use the Mars opposition of January 16, 2025. It happened at 00:48 UT (position 2 in the diagram). Using Stellarium to determine its position 24 hours before (position 1) and 24 hours after this time (position 3), and this tool, from the Earth, Mars is seen to have moved approximately 0.8° of arc (0.014 radians) in those 2 days.

In this case, distance dₑ is 4π/365.25 AUs.

We see that the angular distance, let's call it θ, is the angle between the line joining the Earth and Mars 24 hours before opposition and the line joining the planets 24 hours after opposition.

These lines meet at a distant point, which is at distance l₁ from Mars and distance l₂ from the Earth. The distance between the Earth and Mars at opposition is therefore l₂ - l₁.

We can see that the distance travelled by the Earth in that time, dₑ, is approximately equal to l₂.sin θ.

Also that the distance travelled by Mars in that time, dₘ, is approximately equal to l₁.sin θ.

Putting this all together:

rₘ ~ (dₑ + sin θ)/ ((dₑ/1.9) + sin θ) ~ 1.5 AUs.

Being based around a single astronomical event (unlike the results determined for the inner planets), this result is fortuitously close to the the average of Mars' perihelion (1.38 AUs) and aphelion (1.66 AUs) and also to the semi-major axis of its orbit, according to wikipedia.

Let's check Kepler's third law for Mars as we did for the inner planets and the Earth:

• the cube of its orbital radius is 1.5³ = 3.375
• the square of its orbital period is 1.9² = 3.61
• the former divided by the latter is ~ 0.93

Actually, the accepted figure for the orbital period is 1.88 (years). Squaring this gives 3.53 and the ratio is now 0.95.

We could use the same method for the other outer planets, but as the angle, θ, would be smaller, and less easy to measure, it would be better to increase Δt, perhaps even extending it to the full period of retrograde motion. If we did so, however, the distance dₑ, at least, would need to be calculated as being the chord of a circle centred on the Sun.

Analogue Step Sequencer

 

Updated 4 Apr 26

(clideo was used to rotate video)

In this video from maybe 10 years' ago, here I am twiddling the cutoff and resonance controls of a Gakken synthesizer hooked up to a breadboard-based Baby 8 analogue step sequencer while it plays a sequence. Interesting, and to me, somewhat unexpected results, but I mustn't give up the day job! I later made a permanent version of the sequencer on stripboard.

Checking Kepler's Third Law

(You'll need to view this in landscape mode if using a small screen).

For Earth and the inner planets:

Planet	Orbit radius (r) in AUs	r^3	Period (t) in years	t^2	r^3/t^2
Mercury	sin 22.5° *	~0.06	7/29	0.06	0.96
Venus	sin 46° *	~0.37	8/13	0.38	0.98
Earth	1	1	1	1	1

(*approximate average greatest elongation from data here)

So far, so good, but to extend this to the outer planets, I'll need a method of determining the orbital radii for those. 🤔

Determining the oribital periods of the inner planets

Updated 30 Mar 26

Greatest elongation etc. Source: wikimedia

Using a different approach, let's consider again two planets moving around the Sun in concentric circular orbits in the same plane.

Let the outer planet orbit the Sun at rate a (number of revolutions per time period).

Let the inner planet orbit the Sun at rate b (number of revolutions per time period).

At time t=0, let the inner plant be at greatest elongation (its greatest angular distance) from the Sun, in a specified direction, when viewed from the outer planet. While in principle, we could have chosen inferior conjunction (when the inner planet is directly between the outer one and the Sun), in practice this is (near) impossible to observe other than in the rare instance when the planet transits the Sun.

We can say that this configuration is repeated whenever:

(b - a).t(n) = n, where n is an integer > 0.

The time interval between the event at t(n) and the next, t(n+1), by definition is:

t(n+1)-t(n) = 1/(b - a)

This is independent of n. Let's call it 𝛥t.

Hence:

𝛥t = 1/(b - a)

So:

b/a = 1 + 1/(a.𝛥t)

Now, let's make this more concrete:

• the outer planet is Earth, so a is equal to 1 revolution per year
• the inner planet is Venus
• we'll choose the repeated event as the greatest elongation West, i.e. as observed in the morning sky - see table here.

From this table, we can see that an instance of greatest elongation West for Venus, observed from the Earth, occurred on the morning of March 22 2014 and then for a fifth time on almost the same day 8 years later on the morning of March 20 2022, so, 𝛥t is on average almost exactly 8/5 years.

Hence:

b/a = b ~ 1 + 5/8 = 13/8 (revolutions per year)

So, one orbit of Venus about the Sun takes 1/b, i.e. around 8/13 years or 224.8 days. The correct figure is 224.7 days, so we have achieved an accuracy of closer than 1 part in 1000.

Now consider Mercury in place of Venus. We can see there was an instance of greatest elongation West on March 14 2014 and then 22 instances later, another on March 6 2021. In this case:

b/a = b ~ 1 + 22/7 = 29/7 (revolutions per year)

Hence, one orbit of Mercury about the Sun takes 1/b, i.e. around 7/29 years or 88.2 days, which is close to the correct figure of 88.0 days, although with less accuracy than was the case for Venus. No doubt the relatively high inclination of its orbit to the ecliptic is responsible.

(We can adapt this method to the outer planets, with the inner planet now representing Earth and its rate of rotation, b, equal to 1. The rate of rotation of the outer planet, a, is given by:

a/b = 1 - 1/(b.𝛥t)

The time interval being measured is once more (the average) between one opposition and the next of the outer planet as seen from Earth.

)

Determining the orbital period of outer planets

 Updated 22 Apr 26

This sketch represents a view from above of the Earth in its orbit (green), around the Sun (bright orange in the original sketch!), and an outer planet, Jupiter say, (originally red!).

The orbits are circular and are in the same plane.

At time 1, for an observer on the Earth, Jupiter is said to be at opposition, when at its closest to the Earth.

At time 2, 6 months later, the Earth has completed half an orbit. Being much further away from the Sun than is the Earth, Jupiter's angular velocity is much lower and it has completed only a small part of its orbit in that time.

At time 3, Jupiter is at opposition again and since it has moved along in its orbit, after completing one orbit the Earth also has to travel an extra distance (solid green line), in a time interval represented by the delta symbol, 𝛥, in green, between the points marked 1 and 3. Jupiter has now completed 𝛥/365 of an orbit in 365 + 𝛥 days.

In 2026, opposition was January 10th and in 2027 it will be February 11th. In that time, Jupiter will have completed 32/365 of an orbit in 397 days. 

From this, its orbital period in years is therefore 397/32 or 12.4.        (Method 1)

The accepted period is 11.86, so what happened? Mainly it's because I've assumed the orbits are circular and they are not - they are elliptical. Now, I'm going to try to eliminate that error, by considering a longer period of time when the eccentricities of the orbits of Jupiter and Earth will have been smoothed out, somewhat.

Let's look ahead to 2038, when opposition is on January 14th and Jupiter will have completed 369/365 ((365+4)/365) of an orbit in 12 and 4/365 years. That means it will have completed a full orbit in 365/369 of those 12 and 4/365 years, i.e. its orbital period is:

((12x365) + 4)/369 or 11.88 years.        (Method 2)

That's very close to the accepted figure. I've not taken into account the exact times of day of the consecutive oppositions (available here), nor that the Earth takes 365 and 1/4 days to orbit the Sun. If I do that, then I make it 11.85 years.        (Method 3)

A similar analysis using method 2 above for Saturn over July 15th 1990 to July 21st 2020 gives an orbital period of 29.45 years (actual=29.46) and for Mars over March 4th 2012 to April 9th 2014 gives 1.91 years (actual=1.88). Interestingly, the deviation from the accepted period is greater in the latter case (closer to us, shorter time interval) and smaller in the former (further from us, bigger time interval).

The Coherer - the simplest detector of radio waves?

 

Here's a simple circuit consisting of a power supply, an LED and series resistor, in series with a coherer made of some scrunched up aluminium foil balls inside a plastic beaker in contact with each other and a couple of aluminium foil strips wired to complete the circuit. 

Except that the circuit is only completed once there is a nearby source of radio waves, which somehow creates a low resistance pathway for the electrons in the aluminium balls. In demonstrations, this is usually achieved with a piezo-electric gas igniter. However, in this video, made perhaps 18 months ago, I instead touch one of the foil strips with a short length of wire which acts as an aerial to pick up radio waves from, IIRC, the electric motor of an adjacent freezer. The LED (top left) then lights up. If I tap the side of the beaker sharply, the pathway is disrupted and the LED turns off again.

An even better calendar?

Here's a modified version of that new calendar. There are now fewer changes (6),  March and September having the same number of days as they do now:

January	31	30	-1	-1
February	28	29	+1	0
March	31	31	0	0
April	30	31	+1	+1
May	31	31	0	+1
June	30	31	+1	+2
July	31	31	0	+2
August	31	31	0	+2
September	30	30	0	+2
October	31	30	-1	+1
November	30	30	0	+1
December	31	30	-1	0

Let's see what difference this makes to the dates for equinoxes and solstices: 

• first, we'll fix the 2025 Autumn Equinox at Sep 22 (i.e. unchanged)
• the 2025 Winter Solstice is now on Dec 22, one day later than before, as there are now 90, not 91, days in total in the three months of September, October, November
• the 2026 Spring Equinox, almost exactly 89 days later, i.e. 3 full months, is on Mar 22
• the 2026 Summer Solstice, nearly 93 days later, again 3 full months, is on Jun 22
• the 2026 Autumn Equinox, again nearly 93 days later, and again 3 full months, is now on Sep 22.

Wow - all these events, at least for 2025/6, are now on the 22nd of the month, so this mk2 version is even better! I find that quite surprising given the somewhat lopsided look with 2 months with fewer than 31 days at the start of the year and 4 such months at the end.

An improved calendar?

Updated 28 Feb 26

Following my last post, here is one suggestion for a new calendar, having a different number of days for 8 of the months in the year (highlighted). The current number of days per month is in the second column, the proposed number is in the third, the difference is in the fourth and the accumulated difference is in the fifth, which we'd expect to vary over the year, but to be 0 by the end, as indeed it is!

January	31	30	-1	-1
February	28	29	+1	0
March	31	30	-1	-1
April	30	31	+1	0
May	31	31	0	0
June	30	31	+1	+1
July	31	31	0	+1
August	31	31	0	+1
September	30	31	+1	+2
October	31	30	-1	+1
November	30	30	0	+1
December	31	30	-1	0

Let's see what difference this makes to the dates for equinoxes and solstices: 

• first, we'll fix the 2025 Autumn Equinox at Sep 22 (i.e. unchanged)
• the 2025 Winter Solstice is also unchanged on Dec 21, as there are still 91 days in total in the three months of September, October, November
• the 2026 Spring Equinox, almost exactly 89 days later is now on Mar 21, one day later in the calendar than now, as there are now 89 days in the three months December, January and February, as opposed to the actual 90
• the 2026 Summer Solstice, nearly 93 days later is also unchanged on Jun 21, as the number of days in total in March, April and May is still 92
• the 2026 Autumn Equinox, again nearly 93 days later is now on Sep 22, one day earlier in the calendar than now, as the number of days in total in June, July and August is now 93, rather than 92.

So, we can see that with the new equinox dates there would be less variation in these day numbers than before, i.e. on the 21st or 22nd of the month, as opposed to the 20th to 23rd (at least for the years quoted).

If there is a leap year, then we could give the extra day to February, as now, to make it a "proper" month with 30 days, so the number of days in any month would now only ever be 29, 30 or 31.

One final point - why not make December the shortest month, at 29 days and 30 in a leap year? Well, think of the potential for confusion, e.g. do we celebrate New Years Eve on the 29th or is there a 30th this year? January then? Well, it's so close to February, so let's stick with tradition.