Sun's heat on Earth in summer and winter

  • Thread starter Graeme M
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  • #51
NTW
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Another attempt to convince klimatos that, for an observer on the equator, the sun does not rise always exactly in the east and sets exactly in the west...

Let's imagine that the observer is on the equator, and that he somehow sees the intersection of the celestial equator with the horizon. One intersection point will lie exactly east, and the other exactly west of the observer. That' easy to visualize.

Now, let' see what happens a day when the sun's declination is -for example- +15 degrees. The sun's path will be (approximately) a small circle to the north of the celestial equator, keeping with it, at every point, a distance of 15 degrees measured along any great circle perpendicular to the celestial equator. In the case of the 'equatorial observer' he might -magically- see the circles of right ascension, and -because of the privilege of being sited in the equator- the circle of his horizon will coincide with one of those circles of RA. Hence, at sunrise or sunset, he will see that the point where the sun crosses the horizon deviates 15 degrees to the north, measured from the exact east and west points...
 
  • #52
NTW
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This picture, taken from a meteorological web page, helps to understand why the sun, as seen by an observer at the equator, has apparent paths of equal duration through the year, but rises and sets in the east/west only in the equinoxes (blue line). In the solstices (red lines), it rises and sets at points not exactly east/west, but deviated 23,5º to the north or to the south, depending on the sign of the extreme declination of the sun:


15411794259_f825b63303.jpg
 
  • #53
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This picture, taken from a meteorological web page, helps to understand why the sun, as seen by an observer at the equator, has apparent paths of equal duration through the year, but rises and sets in the east/west only in the equinoxes (blue line). In the solstices (red lines), it rises and sets at points not exactly east/west, but deviated 23,5º to the north or to the south, depending on the sign of the extreme declination of the sun:


15411794259_f825b63303.jpg

The above diagram, and many others like it on the internet is factually incorrect. Just look at those idiotic 90° angles at the two solstices. The only time that the Sun's path ever intersects the horizon at right angles is during the equinoxes. Again, I do not want you to take my word for it. Ask any astronomer! During these equinoxes, every place that gets twelve hours of daylight will see the Sun rise directly east and set directly west. The laws of spherical trigonometry require it!

The circle of illumination is a great circle. Agreed? The Equator is a great circle. Agreed? The principles of spherical trig mandate that any intersecting great circles will cut one another in half. That is why the period of daylight is always the same on the Equator at any time of year. Agreed? This means that when you are on the Equator, the sun will always make an 180° path through the sky. Agreed?

Look at the diagram. Do the three paths seem the same length to you? Of course not! Therefore, since the Sun's movement through the sky is always 15° per hour, the red path and the violet path are too short to be accurate. Spherical trig will tell you that there is no way for the Sun to trace an 180° path through the sky unless it starts at due east and ends at due west.

Why are you unwilling to accept the testimony of hundreds of millions of people who live on or near the Equator and will readily tell you that the Sun rises in the East and sets in the West (give or take a fraction of a degree) every day of the year? Ask any of the more than a million people who live in Quito, only 22 km from the Equator.
 
  • #54
NTW
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Try to see the problem in another way, so that you can free yourself from your error. You say that, for an observer at the equator, the sun rises and sets always at the same place, exactly in the east and in the west. I say that it doesn't, of course, and that it depends on the decination of the day. Only when that declination iz zero (in the equinoxes) does the sun rise and set exactly (for an observer at any place on the Earth, save for the exact poles) in the east and west.

Well, to see the problem in a new way, consider that, for an observer at the equator, ANY star in the heaven, whatever its declination, stays six hours above the horizon for that observer, but -except in the case when its declination is zero- they NEVER rise and set in the equator.

All this that I'm telling you is basic astronomy, so basic that it was taught at school in my time, at first in elementary terms, in primary education, and later as an application of spherical trigonometry. Thus, I'm not going to insist any more times on the subject. It's a pity that someone who says he's a 'professional meteorologist' is so poorly trained in elementary concepts of astronomy.
 
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  • #56
D H
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The above diagram, and many others like it on the internet is factually incorrect.
You are way off base here, klimatos. The diagram posted by NTW is factually correct.

This is the source of your error:
The Sun's path through the sky (including the path below the horizon) is also a great circle.
That is not true. The path of an object on the celestial equator through the sky is a great circle. For an object that isn't on the celestial equator, the path of that object through the sky is a small circle rather than a great circle. For example, the path of Polaris (Alpha Ursae Minoris, aka the North Star) through the sky obviously is a tiny, tiny little circle about true north.

At the equator, the Sun is always in the southern half of the sky at this time of year, even when it's underfoot. The Sun does not make a great circle. The Sun is about 11.5 degrees south of the celestial plane at this time of year (late October). People on the equator will see the Sun rise 11.5 degrees south of east and set 11.5 degrees south of west.

Ask any of the more than a million people who live in Quito, only 22 km from the Equator.
Let's do just that. From http://www.timeanddate.com/astronomy/ecuador/quito, the Sun rises today (October 23, 2014) at 5:55 AM at an azimuth of 101° (11° south of due east) and sets at 6:02 PM at an azimuth of 258° (12° south of due west). Note the 11° (sunrise) and 12° (sunset) offsets. Those are due to the current 11.5° declination of the Sun.
 
  • #57
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Well, this thread has gone astray but it's very interesting. I can't at all visualise what is being discussed nor how it relates to my original question, and I definitely can't tell whether NTW or klimatos is correct. But one small question re DH's comment above. I think I dimly see what you are saying re the apparent path of distant objects such as the north star. But I can't see it's a valid comparison. In effect, those distant objects appear as points on a large 'background sphere' around the earth and hence their paths will depend on both the rotation of the earth and the axis of that rotation in relation to the background sphere. But the sun is a different kettle of fish. The earth revolves around the sun, the sun is not in effect part of the background sphere. Doesn't that make the relationship between earth/sun rather different from that of earth/distant object?
 
  • #58
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You are way off base here, klimatos. The diagram posted by NTW is factually correct.

This is the source of your error:

That is not true. The path of an object on the celestial equator through the sky is a great circle. For an object that isn't on the celestial equator, the path of that object through the sky is a small circle rather than a great circle. For example, the path of Polaris (Alpha Ursae Minoris, aka the North Star) through the sky obviously is a tiny, tiny little circle about true north.

At the equator, the Sun is always in the southern half of the sky at this time of year, even when it's underfoot. The Sun does not make a great circle. The Sun is about 11.5 degrees south of the celestial plane at this time of year (late October). People on the equator will see the Sun rise 11.5 degrees south of east and set 11.5 degrees south of west.


Let's do just that. From http://www.timeanddate.com/astronomy/ecuador/quito, the Sun rises today (October 23, 2014) at 5:55 AM at an azimuth of 101° (11° south of due east) and sets at 6:02 PM at an azimuth of 258° (12° south of due west). Note the 11° (sunrise) and 12° (sunset) offsets. Those are due to the current 11.5° declination of the Sun.

I stand corrected, and I apologize for my arrogance, especially to NTW. Damn! I was so cocksure that my logic was impeccable, despite NTW's best efforts to show me my error. There must be something wrong with my mental three-dimensional imaging. But that's no excuse. The evidence from the Quito Observatory is irrefutable. Mea culpa!
 
  • #59
NTW
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Well, this thread has gone astray but it's very interesting. I can't at all visualise what is being discussed nor how it relates to my original question, and I definitely can't tell whether NTW or klimatos is correct. But one small question re DH's comment above. I think I dimly see what you are saying re the apparent path of distant objects such as the north star. But I can't see it's a valid comparison. In effect, those distant objects appear as points on a large 'background sphere' around the earth and hence their paths will depend on both the rotation of the earth and the axis of that rotation in relation to the background sphere. But the sun is a different kettle of fish. The earth revolves around the sun, the sun is not in effect part of the background sphere. Doesn't that make the relationship between earth/sun rather different from that of earth/distant object?

The sun is a very bright celestial body, and with a large disk, perceptible to the unaided eye. But it's just -with the Moon and the planets- another celestial body that 'wanders through the skies'. That's the only difference with the stars, that have -save for a small 'proper motion' measurable in some- fixed positions in the celestial sphere. On the other hand, the sun, the moon and the planets have variable positions, as their equatorial coordinates change constantly.

Of course, we don't live in a geocentric universe, but it is a useful way of visualizing those things.

Having said that, I feel happy that klimatos has finally understood the problem of the sun's apparent path... Thanks, klimatos...!
 
  • #60
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Not buying this. I mean, yes I understand the angle, but the sun feels week in late fall, winter and early spring.
What evidence do I have to support this? Well, the average UV index in the summer is around 8 or 9 during peak times....
And in the winter the average UV index is around 1 or 2 during peak times of the day as well.

Also, one can lay on a blanket in summer and get "flat on" diret rays.
Are you saying one can get more of a sun tan in the winter by standing outside for a half hour in 32 degree weather with a 1 UV index?
Or would I get more of a suntan laying on a blanket on a 90 degree day with full 9 or so on UV index?

Hope this helps,

From October to May, here in Canada at least, there is actually less uvb in the sun's radiation. UVB is the "burning ray" .
as for your tanning question, uvb kick starts the tanning process by causing melanocytes to start producing melanin. as it absorbs into surrounding skin cells ,uva oxidizes the melanin turning your skin brown. So "tanning" in the winter would be a much longer process, as you would not be producing melanin, just browning the melanin already in your skin. Although UVA can burn you in large quantities, ski burn is more likely from elements such as cold and wind. along with the dryer winter air.
a uv index of 1 or 2 is ideal exposure times,( about a 95% uva and 5%uvb) ... A Uv index of 9???!! cover up, burning would be likely.
 
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  • #61
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That's an interesting comment Irol. I haven't gone back to review the other comments on this thread but it does raise a good point in relation to my original post. What part of the spectrum carries 'heat'? I am pretty fuzzy on the physics of this but does sensible light necessarily have the highest effect in terms of heating an object? Is it likely that other less visible frequencies (eg UV) are more efficient at heating and are more intense during summer than winter? Why should there be less UV in winter than summer?

Which leads me to wonder. The earth's atmosphere must attenuate the sun's rays a LOT. I know that the moons surface temp in full sun is very high (120C) yet even in desert conditions ambient temperature on earth are never that high. However that is not an apples and apples comparison, because the moon's surface is the actual ground. What temperature does the desert sand in say the Gobi desert reach? Obviously the sand radiates off heat but then again, air temps have never exceeded what? 60C?

I wonder what a Stevenson Screen would measure on the moon?
 
  • #62
Baluncore
Science Advisor
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Why should there be less UV in winter than summer?
Fundamentally, because the path through the atmosphere is longer and so there is more absorption. You also expect less UV early in the morning and late in the afternoon for the same path length reason. Likewise UV is also higher in the mountains and at lower temperatures.

There are some counter effects and complexities. The ozone hole over the poles breaks up and moves towards the tropics over summer. That permits more UV to pass in summer. In the tropics, UV levels may be more dependent on atmospheric temperature, that is because more water vapour can be dissolved in warmer air. Below –20°C there is very little water vapour in air.
 
  • #63
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sorry for resurrecting an old thread but i think it is closely related to what i want to ask. Imagine the situation (i have so expertly drawn) below in real dimensions. 2 people hold a plate exactly perpendicular to the sun rays, the plates are the same size. Which one catches more energy and why? From experience I think A should receive radically less energy (otherwise it wouldnt matter if solar power plants are built on equator or on arctic circle if the PV cells track the sun), but i dont know why. Is it just because of the longer path through the atmosphere? A is also further from sun but i think the ~6400km are negligible.
hokus.png
 
  • #64
Bystander
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longer path through the atmosphere?
Yes.
 
  • #65
Baluncore
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Welcome to PF.
Yes, atmospheric path length and cloud is by far the most important.

fayn said:
A is also further from sun but i think the ~6400km are negligible.
That may be true for the inverse square law applicable to radiation, but the tides are driven by the inverse cube. Neglect that difference at your cost, and you may be washed away by the rising tide, driven by that very small difference.
 
  • #66
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Yes.
Welcome to PF.
Yes, atmospheric path length and cloud is by far the most important.
thanks for the answers, now that i look at it it really seems logical.
 
  • #67
russ_watters
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thanks for the answers, now that i look at it it really seems logical.
Well, there's another important issue here too: if you want to build a solar array of any decent size in the arctic circle, to arrange it as in your picture would require mounting it vertically on a large tower! (or use a very long, skinny piece of land).
 
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