Math Is Fun Forum

hi meanshape101

Welcome to the forum.

Bob

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

where x1 is a measurement on the circular dial.

I'll use phi for angles on the elliptical dial

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

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

I was, of course, hoping you'd ask that.

Bob

There's enough here https://en.wikipedia.org/wiki/Complex_number  to keep you busy for a few weeks.  The applications are about 3/4 of the way down the article with links if you want more details.

What I particularly like is the completeness of the complex numbers. Any algebraic equation has a solution in complex numbers.  It also leads to my favourite equation.

Bob

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

You can mark all real numbers on a horizontal line extending left and right to - ∞ and + ∞

Jean-Robert Argand extended this idea to include the imaginary numbers on an axis at right angles to the real axis.

I heard recently that Decartes ridiculed the idea of such numbers, calling them imaginary as an insult. That's not what I believe however. The imaginary axis is a perfect reflection of the real axis in the line y=x, so I prefer the name to be because it's an image of the real axis.

A complex number can be represented by a point somewhere in the plane of the diagram, having its real part as its x coordinate and its imaginary part as its y coordinate.

I've marked a typical point a + ib with coordinates [rcos(theta), rsin(theta)]

It's modulus (distance from the origin) is r where r^2 = x^2 + y^2 and its argument (angle the line makes with the + x axis) theta.

So far all that is true whatever the scales marked on the axes.

But now look at the scale I've chosen.  The picture now shows a circle radius 1, centred on the origin. The cube roots of 1 are three points at angle theta = 0°, 120° and 240°. 

Bob

Ok; using what I have just proved:

and if I multiply that result by another of the same:

and so on.  de Moivre's theorem is the general result:

It might not seem like much but it is really useful when dealing with complex number on the Argand Diagram.

eg. Let's say I want to find all three cube roots of 1.

Mod(1) is 1 so all three must have mod  1 too. So, on an Argand Diagram, draw a circle radius 1 around (0,0).

We know that a cube root is 1 but complex theory says there must be two more. They must lie on the circle or their mod will be wrong. And their arguments must add up to zero (or a multiple of 360.  So 120 and 240 will work.  That gives

I'll multiply out the first to prove it and leave the other as an exercise for you.

The i terms here cancel leaving

Wow! Getting all the LaTex right here was very tricky.

Bob

I've now tried it and it is very 'messy'.

a = 13i     b = 1     c = -101

Is that sufficient for an answer?  I checked with Wolfram alpha and that's the answer it gave.

If you've got another hour or so to spare, it is possible to find the complex square root of any number using De Moivre's theorem.

First find the modulus and argument of the complex number.

A square root will have a modulus that is the square root of the above mod. and an argument that is half the argument of the above.

That's where it gets messy. I had hoped it would come out to a 'nice, easy' number but it doesn't.

Bob

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

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.

Thanks.  Don't laugh ... I'm using Windows 11 on my laptop.

Bob

Every morning the Sun rises in the East and in the evening sets in the west. The Sun appears to move across the sky.  It's all relative.

If you set on a pendulum swinging to and fro you trace out a certain back and forth movement.  But if that pendulum is itself supported by another pendulum which is also swinging the resultant motion is surprising and I think you'll enjoy discovering what it is .

https://www.youtube.com/watch?v=uPbzhxYTioM

Bob