Angle Of The Sun

Jack3.142

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The position of the sun overhead moves back and forth about 47 degrees from the tropic of Capricorn to the tropic of Cancer every 180 days, and then back.
The earth rotates 360 degrees every 365 days.

What is the trigonometry calculation that can give you the sunrise at 20 degrees north on the 13th May, assuming the earth is a perfect sphere?

Thoughts?
 

Cubist

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To me it would seem sensible to use 3d vectors, planes, and matrices (rotation/ translation matrices) for this type of calculation.

The sunrise time could be found by considering the motion of a plane that touches a point on the surface of the Earth and is tangential (touching a point at 20 degrees latitude in your example!). If this is confusing to you (I don't know your background) then think of a very large piece of thin, stiff, card glued to a spinning ball.

If you consider the sun as a point in distant space then the spinning plane will intersect it twice a day, at the times when the sun appears/ disappears over the horizon. For better accuracy, considering the sun as a sphere would not make the necessary calculation a lot harder.

An exception exists if the point is near the axis of spin then there may be a situation of constant daylight/ night - so the plane may not intersect the sun during a 24 hour period

As input you'd need to know the coords of the point on the earth's surface, the relative position of the sun to the earth, the rotation of the earth, and the tilt of the earth's axis (at, say, midnight on the particular date).
 

Jack3.142

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Thanks cubist.

What's interesting as I currently understand it is that the earth does not wobble into and away from the sun as each year progresses to get winter and summer. It's tilt in respect of everything in the universe is constant.

What causes winter and summer is the orbit of the earth around the sun, because on one "side" of the orbit the tilt puts the northern hemisphere more in the sun, and on the other side of the orbit that same tilt puts the southern hemisphere more in the sun.

I figure that this is important.

Because clearly if you are going to be able to calculate the position of the sun relative to a place on earth on any particular day and time, that position will include this Declination, as you say the tilt of the earths axis.

We observe that this declination changes back and forth over a year in pattern that is regular year in year out.

Now, you don't need to be alive long to know that when you see anything that clusters around a central point, waves, sine waves seem to pop up, the idea that the amount of something is less the further away you are from that central point.

So, of course, when you see the angle of the sun in the sky at midday rising and falling and then rising again over a year its is not unexpected that you would think it would be because the earth wobbles. Then, having turned to the mathematics, you would start thinking I bet the rate each day at which the position the sun changes throughout the year is fastest when it is smack bang in the middle of the "same" direction daily change, and slows to a stop at an ever increasing rate all the way to the change in direction in the middle of winter and summer.

I mean, if it was a wobbling earth with one wobble back and forth taking a year that was causing all this, and that slowing rate of change to zero was not the case, we would all have "Earth Direction Change Chairs" that we strap into twice a year and hear an announcement an hour before world wide saying "Warning. Warning. Strap in. Its about to flick the switch to the other direction!"

The question then becomes:

Is the rate of change of the angle that the sun hits the earth a constant each day, or is it a sin wave.

If it is a constant per day, and we do not have strap in chairs, than that is a truly wondrous thing!

Would that be Saw Tooth wave?

Thanks very much for the pointers. I'll lookup the mathematics and definitions of 3d vectors, planes, and matrixes. I am thinking 3 d computer games code and so on must build up from all this, there is a lot of activity in that area, so I assume there might be plenty around down that path.

XYZ direction and magnitude, as resolving to any location in 3d space, and a plane as a great illumination of the sun rays touching at the edge of the day time range, I think I conceptually get.

When I think of Matrixes I think of ordered numbers, ordered in one or more dimensions, each number related in some way to each number next in order in any dimension in the matrix. Is that what you are thinking? Is there anything in particular you have in mind as to what might be in that matrix in this instance?
 

Cubist

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You obviously have a passion for this subject. That's great! Maybe you should also consider posting to an astronomy forum if you're interested in "higher level" discussions of this nature.

But if you are also interested in the learning the maths behind it then maybe there are others on this forum who can recommend a modern website/ book for a beginner in maths/mechanics? - 3d vectors, lines, planes, intersections, matrices, equations of motion etc. (I myself learnt it from handwritten notes on a blackboard many years ago !). Stick at it and you'll get there. Do plenty of example questions as you learn.

Your intuition is right, the sin and cos trig functions are used in the construction of a rotation matrix. Then this matrix can be multiplied with a vector, and you'll get another vector as output which has been rotated.

- To see why we don't need seatbelts for a sudden shift of the Earth's tilt - type the words "earth tilt orbit around sun" into a google and click on the "images" button. Many of the images will show the Earth, with its axis highlighted, in 4 positions around the sun. When you see this you'll probably realise what is happening. The axis itself doesn't actually move much (not in a year anyway)
 

j-astron

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Hi Jack3.142
I'm an astronomer and wanted to comment a little on some of things you said:
Thanks cubist.

What's interesting as I currently understand it is that the earth does not wobble into and away from the sun as each year progresses to get winter and summer. It's tilt in respect of everything in the universe is constant.
The tilt of the Earth's rotation axis with respect to its orbital plane (the ecliptic plane) is fairly constant. But the axis does change direction (with constant tilt, so that it sweeps out a cone every 26,000 years). This motion is called precession. Because of precession, the direction of the axis isn't constant with respect to "everything [else] in the universe." The axis changes direction with respect to the "fixed" (very distant and nearly unmoving) stars. There is also a tiny bit of wobble called 'nutation' that changes the angle of the axis a little bit. See the diagram at the top of this page:



What causes winter and summer is the orbit of the earth around the sun, because on one "side" of the orbit the tilt puts the northern hemisphere more in the sun, and on the other side of the orbit that same tilt puts the southern hemisphere more in the sun.
What causes the seasons is the tilt, in the sense that the tilt leads to longer (or shorter) hours of sunlight, and the sunlight that is received being more direct (or more oblique). Yes it is true that orbital motion is required in order for the tilt of a given hemisphere to change from being "away from" to being "towards" the sun. But it's not like it's possible for orbital motion not to happen, and orbital motion with no tilt would result in no seasonal changes.

Yes, the transition from a hemisphere being tilted away from the sun to being tilted towards it is gradual and continuous, not abrupt.

From our perspective, on Earth, what really changes over the course of a year is the Sun's coordinates on the Celestial sphere. As you seem to already know, these coordinates are measured in terms of a celestial longitude (Right Ascension) and a celestial latitude (Declination). The exercise of figuring out where the sun will be on a given date in local "horizon" coordinates (azimuth and elevation) is then an exercise in transforming between two different spherical coordinate systems: from RA and Dec to Az and El. To do this, you also need to know your latitude, because this tells you the angular offset between the poles of the two coordinate systems. Finally, you must know the hour angle of the object (how far away it is in RA from your meridian), which depends not only on the object's RA, but also on the local sidereal time. Once you have all of these inputs, the calculation is an exercise in spherical trigonometry. Here is a link to website that explains the math:


I hope that this is somewhat helpful. :)
 

Jack3.142

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That's great thanks!

I am struggling to understand how it can be that the rate of change of the declination angle per day during in a year, it's magnitude, can be anything other than a constant ((4 * 23.5)/265), given that all that decides it with respect to earth is of uniform change.

By with respect to earth I mean that if one stands some distance away from the edge of a circle on the circles plane, puts a mark on the circle, and rotates the circle, the mark will appear to move up and down with at a rate that that is a sine wave. Where are we not in the centre of the "circle" in this instance?

The Tilt does not change (constant with respect to time)
The orbit around the sun sees the earth dropping towards the sun at a constant rate as it moves perpendicular to the sun at a constant rate, and the definition of the angle that is declination has no difference whether it is summer winter spring or fall.

I am expecting a Saw Tooth graph of sun location per day expressed as an angle to the equator not a curve.

What am I missing?

Cheers.
 

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Jack3.142

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Typo: ((4 * 23.5)/265) in the above should be ((4 * 23.5)/365)

That is, up and back then down and back from the equator north and south in one year, ignoring leap years for moment.
 

Cubist

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I am expecting a Saw Tooth graph of sun location per day expressed as an angle to the equator not a curve.

What am I missing?
Do you have access to a globe and set square? If so then you could do a physical experiment and plot a graph to convince yourself that it's not a saw tooth wave. Move the set square around the globe at 15 degree intervals and measure the line of latitude where the set square touches as you go. If you don't have a globe then maybe you have a cheap ball that you could mark with some latitude lines and then balance it, tilted, on a cup/ ramekin?
 

Jack3.142

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Great idea I’ll have a go at that!

I am thinking a plumb bob.

In terms of what changes with respect to what that could cause a differing rate of change, I am thinking all I have left is the variance between the celestial equator and the ecliptic of the suns orbit during the year.

I have been focusing on the angles at the earths centre.

Maybe I need to head back the the surface.
 

Jack3.142

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No luck with the Plumb Bob. I set the globe up with a tilt and moved the plumb around it and saw where it just touched the globe, and there was no variation in the amount of degrees latitude it it moved per degrees longitude.

The other source of cyclical difference of something is the elliptical nature of the earths orbit. I’ve regarded it as a circular orbit so far.

I’m going to turn my attention to that.
 

Jack3.142

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Tested with a built model the tilt effect on the rate of change in declination after I realised the plumb bob method just confirmed the location of the ecliptic circle, which was not much point.

I definitely now say that the rate of change of declination is driven either fully or partly by the geometry of the relationship of the Equatorial and Ecliptic circles.

Good idea the physical model check Cubist.
 

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Cubist

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I was surprised to see your globe, it was more like a large slice of edam :). But it did the trick!

It is more usual to let the "change" figures go negative when the numbers start to decrease.

I extrapolated your results to a full circle by copying them down and making them -ve, and then re-plotted them alongside results caclulated from sin function...

graphs2.png
 

Jack3.142

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That's so good Cubist thanks!

And thanks j-Astron - I am working through your post it will take a me a little longer to comprehend, but I really appreciate you taking the time to put me in the most productive direction.
 
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