Solar Questions to Bob

KerimF · 2025-03-01 01:38:21

Hi Bob,

[1] Let us assume that on the 21st June at noon (12:00 AM) the sun was at its highest elevation relative to an observer on the ground. Will the sun be also at its highest elevation on the 21st December at noon to the same observer?
I mean, is it true that the sun will be at its highest elevation at noon (relative to the same observer), no matter the day of the year is?

[2] Let us assume the angle A is of the arc on which the sun moved from noon to 1:00 PM. Will this angle change on the 21st December at noon?
I mean, is it true that the sun will move the same arc angle (from its highest elevation) from noon to 1:00 PM (relative to the same observer), no matter the day of the year is?

I try to design a simple one-dimension solar positioner (much like a satellite dish positioner). The controller will have an IC for time and date. It is supposed to move the solar panel, based on the time and date, read from that IC. The vertical movement of the positioner is supposed to be made manually (to simplify its mechanical structure), say 4 positions; one for each 3-month season.

Although there are various formulas to find out the azimuth and elevation of the sun in function of time and date, the accuracy is not critical in this application.

Thank you.
Kerim

Bob · 2025-03-01 11:21:50

hi KerimF

No to both. I've made a simple diagram that will start to explain this. It's definitely NOT to scale. It shows the Earth in orbit around the Sun going anticlockwise. The parallel lines mark the polar axis (ie the axis that the Earth rotates around) with North at the top. (If you were looking down on the North Pole that rotation is anticlockwise too.)

This means that at mid-summer the northern hemisphere of the Earth is 'leaning' towards the Sun which is why it's warmer then for folks that live there. Sunlight strikes the southern hemisphere at a lower angle so it gets less hot there.

6 months later the situation is reversed which is why it's summer then in the Southern hemisphere.

The angle between the polar axis and the plane of the Earth's orbit around the Sun (called the plane of the ecliptic) is about 67°. This year the 20th Of March is the Spring equinox. The exact date varies a bit because of leap years. On that date and at noon the Sun will be exactly overhead the equator and the day length is 12 hours. After that date the Sun is overhead of places north of the equator, gradually 'moving' north to the summer solstice when it is overhead points on the line of latitude called the tropic of Cancer. Then it gradually 'moves' back towards the equator at the Autumnal equinox. In winter the Sun at noon is lowest in the sky reaching its lowest at the Winter solstice .

So the angle between the Sun and the horizon depends on the observer's latitude and the date in the year.

If you're hoping to point a solar panel directly at the Sun for different dates then I need to know your latitude.

Bob

ps. I don't like the common practice of using the term 12 AM. AM is an abbreviation of the Latin phrase ante meridiem which translates as 'before noon'. So how can 12 before noon make any sense. We already have a perfectly good term for this moment in time. It's 12 noon. PM stand for post meridiem which means after noon. So for the next 11 hours and 59 minutes it's after noon and at exactly 12 hours later it's 12 midnight and definitely not 12 PM. In some parts of the world 12 AM and 12 PM have the opposite meanings which also shows why neither term should be used.

KerimF · 2025-03-01 20:27:55

Thank you, Bob.

But I am sorry because I forgot to add in my first question that 'its highest elevation' is also relative to the actual arc of the sun during the daylight.
I assumed that a 'sun watch' works in summer and winter as well (blue sky), only the length of the shadow varies.

Bob · 2025-03-01 20:37:12

I've been thinking some more about this.

Keen astronomers get an 'equatorial mount' for their telescope. The mount can rotate in two ways. One of the axes is set to point at the pole star. The other allows the astronomer to point at a star and then follow it as the Earth rotates. The polar axis rotation can automatically rotate at the right speed using a suitable motor.

Following the Sun would need a similar arrangement. The reason the Sun moves across the sky is totally due to the rotation of the Earth so, once you are pointing at the Sun, you'll need to rotate your device around the polar axis. The angle between the polar axis and the horizontal is the latitude.

There must be a reasonably simple formula for fixing the angle between the polar axis and a line pointing towards the Sun. Don't know what it is yet but I'm hopeful I can work it out or look it up if the former fails. It will depend on the date.

Bob

Bob · 2025-03-01 22:03:57

I looked it up. I'll think about why it works later.

First calculate the Sun's declination:

The declination of the sun is:

δ = 23.45° * sin( (360° / 365) * (n + 284) )

where "δ" represents the declination angle, and "n" is the day of the year with January 1st being day 1 (where we are in the leap year cycle has a negligible effect on this).

Next calculate the Sun's maximum altitude:

A=90−(φ−δ)

where

A is the altitude of the sun (degrees)
φ is the latitude of the observer (degrees)
δ is the declination of the sun (degrees)

Use a clock set to your local time rather than the Country time across the time zone. You can calculate this using your longitude. The Earth rotates 15° per hour. I live East of Greenwich so, for me, the Sun is 'ahead' of GMT by about 3 minutes. The altitude value occurs when the Sun is due South of the observer (for the Northern hemisphere).

There is another factor that will alter the local noon value because of small inaccuracies due to the elliptical nature and tilt of the Earth's orbit. This shifts local noon by some minutes but is only about 15 minutes at worst so you can probably skip this factor. (ref: the equation of time)

I think that's all you need.

Bob

KerimF · 2025-03-02 02:30:28

You did help a lot. Thank you, Bob.

Therefore, the declination of the sun is now:
δ = 23.45° * sin( (360° / 365) * (n + 284) )
where:
n= 31+28+2 = 61 days

δ = 23.45° * sin( (360° / 365) * (61 + 284) ) = -7.914911995°

================

the Sun's maximum altitude at Aleppo city is:
A=90−(φ−δ)
where:
φ = the latitude of the observer (degrees) = 36° 12' 7'' = 36.20194444°
δ = -7.914911995°

A= 90 - (36.20194444 + 7.914911995) ≈ 45.9°

Here, it is around 5.30 PM. So, tomorrow noon, I will try to measure the angle A (though for n=62, A≈46.3°)

================
================

As I mentioned earlier, to simplify the positioner installation, I will try to move automatically the solar panel, horizontally only (The vertical movement will have to be done manually, every 3 months, for example).
In this case, if the controller's clock is set to 12 noon when the sun is at mid sky (at its maximum altitude), is it possible to calculate the sun deviation angle, θ, at 1 PM? Edited: Sorry again, you gave me the answer already which is 15° / hour (=360/24)

Last edited by KerimF (2025-03-02 17:10:21)

Bob · 2025-03-03 22:41:12

hi KerimF

I've begun to research this with more understanding. The formula for declination that I gave in post 5 is not correct! My apologies.

I've looked elsewhere and found a formula that does give sensible values so I'm happy with it.

I couldn't understand where the number 284 came from. That date is 11th October which has no significance over any other date. If Jan 1st is 1 then adding 10 takes us back to the Winter solstice. I've changed to radians (hence the 2pi rather than 360) . Here are the results of that formula for the 4 key dates in the year.

Spring day 79 delta = -0.90804
Summer day 172 delta = 23.44913
Autumn day 265 delta = -0.50455
Winter day 355 delta = - 23.45

Strictly those equinox values should be zero but there will always be a small difference unless you take account of leap years. The solstice values are very close to the correct values.

More later.

Bob

KerimF · 2025-03-04 02:13:27

Hi, Bob

For instance, I remember you attached once a photo of your solar watch, I couldn't find its post.

Kerim

Bob · 2025-03-04 05:50:39

Did you mean this:

https://www.mathisfunforum.com/viewtopi … 62#p442362

I'm glad to get onto sundial theory because I think it is going to be needed here.

The Sun appears to track along a great circle at 15° /hour. But the shadow on my sundial does not track at that rate. Only the horizontal movement affects the shadow so it is necessary to calculate each hour angle. The angle between 12 and 1 is about 12° and between 6 and 7 is about 16°. I've got the theory for this written down somewhere ... I'll dig it out.

And if you want a dial on a vertical face that too is possible but needs a different set of angles. Your tracker will need both the altitude (vertical) and azimuth (horizontal) calculations I think.

Bob

KerimF · 2025-03-04 06:57:47

Bob wrote:

Did you mean this:
https://www.mathisfunforum.com/viewtopic.php?pid=442362#p442362

Yes, Thank you. It inspired me to think of this project (tracking the sun horizontally).

Bob wrote:

I'm glad to get onto sundial theory because I think it is going to be needed here.
The Sun appears to track along a great circle at 15° /hour. But the shadow on my sundial does not track at that rate. Only the horizontal movement affects the shadow so it is necessary to calculate each hour angle. The angle between 12 and 1 is about 12° and between 6 and 7 is about 16°. I've got the theory for this written down somewhere ... I'll dig it out.
And if you want a dial on a vertical face that too is possible but needs a different set of angles. Your tracker will need both the altitude (vertical) and azimuth (horizontal) calculations I think.
Bob

You are right. It is not simple as we like it to be.
I hope the formulas (the set of formulas) will give the azimuth and altitude (maximum, I guess) in function of date and time only, besides the laltitude and longitude of the observer

I guess, the horizontal limits will be made around ±60° for practical reasons (mechanical).

Last edited by KerimF (2025-03-05 23:54:56)

KerimF · 2025-03-05 23:48:01

Hi Bob,

By the way, I had the impression that your solar watch works (± small θ) on all days of the year when the sunlight is clear. Could it be that simple?
If it is, I guess the function of the sun azimuth versus o'clock could be found empirically.

Kerim

Bob · 2025-03-05 23:50:57

It could be found empirally. But I think I've got two formulas, one for the alttiude and one for the azimuth.

I'm helping a neighbour with their garden at the moment.

I've post what I have later today.

Bob

Bob · 2025-03-06 22:02:05

Sundial theory:

Imagine you are at the North Pole. Make a disc with 24 sectors, 15° apart. As the Earth rotates, the Sun appears to revolve around the polar axis at a rate of 15° per hour. As you're at the pole it's not clear which time zone you're in so you'd have to decide which sector line represents zero hour but that need not concern us here. A vertical stick along the polar axis will make a shadow and we can use that shadow to tell the time.

Now take your stick and disc to latitude alpha. If you move so that the stick is still parallel to the polar axis you can continue to use the dial to tell the time. But sometimes the Sun is lower than the disc so no shadow. What lots of sundial makers do is convert the disc to a horizontal one. You have to keep the stick pointing along the polar axis so only the dial is changed.

This diagram is a side view of the situation. The polar axis is the line FA at angle alpha to the horizontal. The side view of the circular sundial is the line BAC with A at the centre and radius r. The dial is projected down onto a horizontal at EFD. The circle becomes an ellipse with major axis ED, half length a. The minor axis will retain the width of the circular dial so half width r. The equation of this ellipse is

You could have an elliptical sundial face but it's more common to make it circular once more. But, what needs to change are the hour angles. Because the face is stretched along the ED axis all the hour angles change. What follows shows how to work out the new angles.

I'll use theta for the angle between the 12 noon sector line and the secor line for any other hour (eg one o'clock theta = 15, two o'clock theta = 30 and so on)

So on the circular dial

........(1)

where x1 is a measurement on the circular dial.

I'll use phi for angles on the elliptical dial

..........(2)

y is unchanged by the projection but x2 is stretched by the equation

.........(3)

Dividing (1) by (2) and using (3)

So we get the equation

I used that formula to make the hours angles for my dial. I also checked it by an on-line source. And my dial gives the right time so I'm fairly happy that's the correct equation and I think you can use it for azimuth.

If FG is a vertical dial up to the middle (G) of the dial then a similar calculation will give the new hours angles. This time the x coordinates stay constant and the y's are stretched so the equations are

so

I haven't got independent confirmation of that formula but I think it works. You can use it for the altitude angles.

Bob

KerimF · 2025-03-07 02:57:02

Thank you, Bob. You gave me homework to think of

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Solar Questions to Bob

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