Sun's heat on Earth in summer and winter

[psparky]

russ_watters, I must not be expressing myself well enough. Geometry has nothing to do with it, as far I can see. Forget that. I am talking of an instant in time, sun same place in the sky. The difference is in the time of year. NTW gets what I am saying. NTW, I particularly noticed this while inside a car, in which case the ambient conditions are WARMER in winter and colder in summer which would lend credibility to the likelihood it's a perception thing, although this would mean that the perception is based in a real physical effect...

So the sun is in the same place in the sky, but at different times of the year.

That is impossible. The Earth orbits around the sun. The sun is not in the "same place" from day to day and certainly changes drastically from season to season. The angle of the sun's light to the Earth has everything to do with it.

Please name an instance where the sun is in the same place in the sky at different times of the year.

Please, I beg of you.

 

[NTW]

So the sun is in the same place in the sky, but at different times of the year.

That is impossible. The Earth orbits around the sun. The sun is not in the "same place" from day to day and certainly changes drastically from season to season. The angle of the sun's light to the Earth has everything to do with it.

Please name an instance where the sun is in the same place in the sky at different times of the year.

Please, I beg of you.

Well, -for example, and forgetting refraction- for an observer at -say- 40º N latitude, the apparent altitude of the highest point of the celestial equator is 90º - 40º = 50º. Now, the declination of the sun varies, along the year, between +23,5º and -23,5º. Thus, at its lowest possible declination, on Winter's solstice, the sun reaches an maximum apparent altitude of 50º - 23,5º = 26,5º. It is obvious that -in that day- the sun can be observed, at a given moment, at any altitude between zero and 26,5º. Well, in fact, two times that day for any given altitude comprised between 0º and 26,5º

In the other 364 days of the year, the sun can also be observed, and twice a day, at any altitude reached by the sun in Winter's solstice.

 

[Baluncore]

I believe it comes down to the length of the path through the atmosphere. In winter and at dawn or dusk the sun is close to the horizon and more energy is absorbed over the longer path through the atmosphere. In summer at noon the path can be much shorter and so less scattering or absorption takes place. I notice that moving a few thousand feet above sea-level makes a noticeable difference to the intensity, also crops that mature in late summer often do better at higher altitudes. That is consistent with atmospheric path length determining intensity.

 

[DaveC426913]

...the ambient conditions are WARMER in winter and colder in summer...

You sure about that? What would you say is the actual temperature in your car, even with the heater on, in the winter? You think it's going to be 70F in your car in the winter? You'd be tearing your clothes off and cracking the window.

What about in the summer, with the AC on? You think it'll be colder than your car is in the winter?
BTW, note that there is more at-play than conduction with the ambient air. The windows of your car are radiating huge amounts of energy at you in the summer (yes, even with the AC on). In the winter, they are effectively radiating cold (actually, drawing heat off your body that is not being returned). Touch the window - or even get near it - in winter and again in summer. You will feel the diff.

Also, do not dismiss humidity. Dry air in the winter means insulation from cold. Humid air in the summer means more conduction of heat from air to skin.

So, my point is: it is virtually impossible to set "all other factors equal", as everyone keeps saying.

 

[NTW]

I believe it comes down to the length of the path through the atmosphere. In winter and at dawn or dusk the sun is close to the horizon and more energy is absorbed over the longer path through the atmosphere. In summer at noon the path can be much shorter and so less scattering or absorption takes place. I notice that moving a few thousand feet above sea-level makes a noticeable difference to the intensity, also crops that mature in late summer often do better at higher altitudes. That is consistent with atmospheric path length determining intensity.

But that path is the same with the same elevation, and any elevation lower than that of the sun at noon in the winter's solstice is reached twice a day in any other day of the year, if at different hours, of course... There are only two places on the Earth from which, within the year, you never observe the sun with the same elevation, and those places are the north and the south poles... Starting from a pole, and if you begin moving your observatory to the equator, you will have a growing number of days of the year with the sun reaching twice in that day the same altitude. And, from the artic circles to the equator, that number of days is 365...

 

[Graeme M]

I am surprised about the number of comments discussing the elevation of the sun in this question. It seems perfectly simple to me. Of course at noon in summer and noon in winter the sun is at a different elevation, and indeed even a different position relative to a fixed point at my location.

That wasn't what I meant, nor did I mean anything especially exact. I was simply saying that for an apparently similar elevation, say midday in winter vs maybe 9 AM in summer (again, don't take that as an exact statement) the sun feels weaker in winter than in summer. It even LOOKS weaker.

I am not saying that is actually the case, I was musing whether it is or is not, and what is the explanation for the perception if it is not.

I agree that there are many variables, even more than I'd have thought of.

Dave C, my point about conditions in the vehicle are again not to be taken so literally. In terms of my skin, in a car in summer with the A/C on strongly, my skin will be cool to the touch. In winter, with the heater on, it will be warm to the touch. I think then that similar amounts of insolation will 'feel' different - my colder skin will I assume absorb more energy and heat up more, than my already warm skin will in summer.

I do take your point about the windows though, I definitely hadn't thought of that.

 

[olivermsun]

The windows of your car are radiating huge amounts of energy at you in the summer (yes, even with the AC on). In the winter, they are effectively radiating cold (actually, drawing heat off your body that is not being returned).

I do take your point about the windows though, I definitely hadn't thought of that.

;) See some of the early replies in the thread:

If the you're driving along on a bright sunny day in 85°F weather, then you're feeling the radiation from the sun, the warmth from the surrounded air, IR coming from the windows, etc., etc.

You are probably feeling radiant heat from other sources--especially the windscreen.

 

[klimatos]

So the sun is in the same place in the sky, but at different times of the year.

That is impossible. The Earth orbits around the sun. The sun is not in the "same place" from day to day and certainly changes drastically from season to season. The angle of the sun's light to the Earth has everything to do with it.

Please name an instance where the sun is in the same place in the sky at different times of the year.

Please, I beg of you.

On the equator, the Sun rises directly in the East and sets directly in the West every day of the year.

 

[NTW]

On the equator, the Sun rises directly in the East and sets directly in the West every day of the year.

Well, no... Not every day... Only two days...

For an observer on the equator, the sun rises exactly in the East and sets exactly in the West only in two days of the year, precisely in the equinoxes of spring and fall. Why? Because, in the equinoxes, the declination of the sun is precisely zero, and its apparent trajectory coincides with the celestial equator. In the rest of the year, the sun rises and sets a bit to the North or to the South, but not much, as the magnitude of that 'deviation from the east/west' reaches a maximum of 23,5º on the days of the solstices. In one of them (summer solstice for the northern hemisphere) the sun rises at a point 23,5º to the north (measured from the east), and sets at a point 23,5 to the north (measured from the west). On the other solstice, a similar thing same happens, but with the sun rising at a point 23,5º to the south (measured from the east) and setting at a point 23,5º to the south (measured from the west).

It sounds complicated in words, but is really very easy: the sun wanders in the south celestial hemisphere during one-half of the year (when its declination is negative) and in the north celestial hemisphere during the other half, when its declination is positive...

 

[Graeme M]

Fair point Olivermsun, I did tend to just scan some of those earlier responses and missed those particular statements. :)

 

[psparky]

I believe it comes down to the length of the path through the atmosphere. In winter and at dawn or dusk the sun is close to the horizon and more energy is absorbed over the longer path through the atmosphere. In summer at noon the path can be much shorter and so less scattering or absorption takes place. I notice that moving a few thousand feet above sea-level makes a noticeable difference to the intensity, also crops that mature in late summer often do better at higher altitudes. That is consistent with atmospheric path length determining intensity.

What Baluncore has posted here is the remaining piece of the puzzle. The angle of refraction is signifigant, but the "length of path through the atmosphere" is the biggest contributor. The sun just loses UV rays when it takes a long time to penetrate atmoshere. In summer, "the length of the atmosphere is much shorter due to the almost perpendicular penetration".

 

[DaveC426913]

Yes, I had intended to bring that up, but that is all part of the "sun's angle" issue - which has been addressed and factored out.

If the OP is claiming that he's comparing two events (at different times of the year), when the sun is at the same position in the sky, then it will be (ipso facto) traversing the same path length through the atmo

 

[psparky]

Yes, I had intended to bring that up, but that is all part of the "sun's angle" issue - which has been addressed and factored out.

If the OP is claiming that he's comparing two events (at different times of the year), when the sun is at the same position in the sky, then it will be (ipso facto) traversing the same path length through the atmo

So are you saying that there is some truth to the OP's claim...original question?

Give me the dates of the possible scenario you have above? The sun is at the same angle and same atmoshpere traveling length at two different times of the year?

What are the exact dates please?

 

[DaveC426913]

So are you saying that there is some truth to the OP's claim...original question?

Give me the dates of the possible scenario you have above? The sun is at the same angle and same atmoshpere traveling length at two different times of the year?

What are the exact dates please?

You seem to have misread what I wrote. I said "if same angle (irrespective of time of year) then same path length". So I am in agreement with you.

The longer version:
Several members suggested that a lower incident angle would make for less insolation.
The OP said 'thanks I get that, and I am ruling out sun angle as a factor because reasons'.
So, if the OP is ruling out incident angle as a factor in his hypothesis (for whatever reasons), then he is implicitly ruling out any loss in isolation due to length of path through the atmo. (because same angle means same path length).

Early on, I was going to suggest 'differing path length due to differing angle of Sun' - but that was before the OP said 'same angle'. So my suggestion became moot.

 

[psparky]

No, you misunderstand.

Several members suggested that a lower incident angle would make for less insolation.
The OP said 'thanks I get that, and I am ruling it out as a factor because reasons'.
So, if the OP is ruling out incident angle as a factor in his hypothesis (for whatever reasons), then he is implicitly ruling out any loss in isolation due to length of path through the atmo.


You also seem to have misread what I wrote. I said "if same angle then same path length". So I am in agreement with you.

Ok, but I am confused. If we neglect the incident angle and path thru atmosphere...what was the point of the original question?

That's like playing a football game without a football...it just doesn't make any sense.

 

[DaveC426913]

Ok, but I am confused. If we neglect the incident angle and path thru atmosphere...what was the point of the original question?

The OP is looking for alternate causes for a discrepancy in the feeling of warmth on his face.

At first analysis, we all agree that sun angle would be the most obvious cause for discrepancy, but the OP is ruling that out and asking for other causes. (He's ruling it out based on his experiments that suggest that, even when the Sun is in the same position in the sky (yet different times of day) between winter and summer, he feels this discrepancy).

Some of us have proposed factors - such as ambient air temp, radiation off windows, humidity, etc. - that would throw a wrench in his perception of the sun's insolation.

 

[klimatos]

Well, no... Not every day... Only two days...

For an observer on the equator, the sun rises exactly in the East and sets exactly in the West only in two days of the year, precisely in the equinoxes of spring and fall. Why? Because, in the equinoxes, the declination of the sun is precisely zero, and its apparent trajectory coincides with the celestial equator. In the rest of the year, the sun rises and sets a bit to the North or to the South, but not much, as the magnitude of that 'deviation from the east/west' reaches a maximum of 23,5º on the days of the solstices. In one of them (summer solstice for the northern hemisphere) the sun rises at a point 23,5º to the north (measured from the east), and sets at a point 23,5 to the north (measured from the west). On the other solstice, a similar thing same happens, but with the sun rising at a point 23,5º to the south (measured from the east) and setting at a point 23,5º to the south (measured from the west).

It sounds complicated in words, but is really very easy: the sun wanders in the south celestial hemisphere during one-half of the year (when its declination is negative) and in the north celestial hemisphere during the other half, when its declination is positive...

Sorry, NTW, but you are in error. The Equator is a "great circle" . The Sun's path through the sky (including the path below the horizon) is also a great circle. Two great circles that intersect always cut one another in half. This is evidenced by the fact that every day on the equator has exactly the same period of daylight (sunrise to sunset) as every other day. It is slightly more than twelve hours (between twelve and thirteen minutes) because the sun appears as a disc rather than a point, but it does not vary from day to day. If the sun were a point rather than a disc, this period would be exactly twelve hours--half a day. In order for this to be true, the sun must rise directly in the east and set directly in the west. From the June Solstice to the December solstice, its path through the sky will be in the northern half of the sky as seen from the Equator. In the other half of the year, its path will be in the southern half of the sky. Twice a year, on the equinoxes, people on the equator will see the sun directly overhead at local noon (cloud cover permitting).

I speak as a professional climatologist, but I do not want you to take my word for it. Check with any professional astronomer who is conversant with Earth-Sun relationships.

 

[NTW]

Sorry, NTW, but you are in error. The Equator is a "great circle" . The Sun's path through the sky (including the path below the horizon) is also a great circle. Two great circles that intersect always cut one another in half. This is evidenced by the fact that every day on the equator has exactly the same period of daylight (sunrise to sunset) as every other day. It is slightly more than twelve hours (between twelve and thirteen minutes) because the sun appears as a disc rather than a point, but it does not vary from day to day. If the sun were a point rather than a disc, this period would be exactly twelve hours--half a day. In order for this to be true, the sun must rise directly in the east and set directly in the west. From the June Solstice to the December solstice, its path through the sky will be in the northern half of the sky as seen from the Equator. In the other half of the year, its path will be in the southern half of the sky. Twice a year, on the equinoxes, people on the equator will see the sun directly overhead at local noon (cloud cover permitting).

I speak as a professional climatologist, but I do not want you to take my word for it. Check with any professional astronomer who is conversant with Earth-Sun relationships.

No. I'm not in error. I ratify myself in everything that I have written in this thread.

Just to comment a single point of what you say, please note that the apparent path of the sun through the sky is not a great circle. It's easier to see that if one positions himself precisely on one of the poles and forget refraction. The path of the sun will be seen as an spiral, starting at the summer solstice with an elevation of 23,5º, and that elevation diminishing with every passing day, till, at the equinox day, with an elevation of zero, the sun disk is exactly parted in two by the horizon. (That day, and only that day, the 360º path of the sun comes close to a great circle. The rest of the days of the year, the path, always tracing a 360º arc of spiral every day, will approximate a small circle, smallest at the solstices). After that equinox, with each passing day, the sun, invisible for the observer, continues its spiral path below the horizon, till -at the winter solstice- it reaches a depression (or negative elevation, if you wish) of -23,5º.

The path of the sun, with its continuously varying declination, is always an spiral, independently of the latitude of the observer. If I have put the example of an observer sited at the pole, it's just because it's easier to visualize.

Yes, twice a year, people sited on the equator can observe the sun at the zenith. But that's not a privilege of 'equatorials'. Any observer sited in the tropical zone -between the tropics- can observe the sun twice a year at the zenith, at local noon... The separation of those two dates, minimum for an observer based precisely on one of the tropics, (where they become just one day), grows gradually, as the latitude of the observer diminishes, reaching a maximum of six months for the equator-based observer.

I'm editing to add a comment: just think, klimatos, in the apparent path of a star with any declination different from zero. It's the case of the sun at any time except in the equinoxes. You'll easily see that that stellar path is not a great circle. Take Polaris, for example, that is a little off the north celestial pole. Its apparent path is a very small circle... Isn't it..?

 

[klimatos]

No. I'm not in error. I ratify myself in everything that I have written in this thread.

Just to comment a single point of what you say, please note that the apparent path of the sun through the sky is not a great circle. It's easier to see that if one positions himself precisely on one of the poles and forget refraction. The path of the sun will be seen as an spiral, starting at the summer solstice with an elevation of 23,5º, and that elevation diminishing with every passing day, till, at the equinox day, with an elevation of zero, the sun disk is exactly parted in two by the horizon. (That day, and only that day, the 360º path of the sun comes close to a great circle. The rest of the days of the year, the path, always tracing a 360º arc of spiral every day, will approximate a small circle, smallest at the solstices). After that equinox, with each passing day, the sun, invisible for the observer, continues its spiral path below the horizon, till -at the winter solstice- it reaches a depression (or negative elevation, if you wish) of -23,5º.

The path of the sun, with its continuously varying declination, is always an spiral, independently of the latitude of the observer. If I have put the example of an observer sited at the pole, it's just because it's easier to visualize.

Yes, twice a year, people sited on the equator can observe the sun at the zenith. But that's not a privilege of 'equatorials'. Any observer sited in the tropical zone -between the tropics- can observe the sun twice a year at the zenith, at local noon... The separation of those two dates, minimum for an observer based precisely on one of the tropics, (where they become just one day), grows gradually, as the latitude of the observer diminishes, reaching a maximum of six months for the equator-based observer.

I'm editing to add a comment: just think, klimatos, in the apparent path of a star with any declination different from zero. It's the case of the sun at any time except in the equinoxes. You'll easily see that that stellar path is not a great circle. Take Polaris, for example, that is a little off the north celestial pole. Its apparent path is a very small circle... Isn't it..?

Virtually every textbook on astronomy speaks of the Sun's apparent path through the sky as a circle--360 degrees. Part of this circle is below the horizon and part is above. How much of each depends upon your latitude and the time of year. You are absolutely correct in that this circle is not a great circle. Properly speaking, it is not a circle (nor a spiral) but is a double-helix.

The circle of illumination, however, is a great circle. It is this great circle that always cuts the Equator in half.

From the March equinox to the September equinox, every place in the northern hemisphere (outside of the Arctic) will see the Sun rise to the south of directly east and set to the south of directly west. How much south depends upon your latitude and the time of year. Every place in the southern hemisphere (outside of the Antarctic) will see the Sun rise to the north of directly east and set to the north of directly west. The same thing is true (in reverse) for the other half of the year. A moment's thought will show that, halfway between the two hemispheres (that is, on the Equator) the Sun will always rise directly east and set directly west.

This phenomenon, along with the brevity of twilight, has been commented on by generations of Equatorial travelers.

I believe that we have digressed substantially from the OP.

 

[NTW]

Facts are stubborn, klimatos...

Think it again... In the equator (and elsewhere outside the artic zones) the sun rises exactly in the east, and sets exactly in the west ONLY in the equinoxes... Twice a year... In the rest of the days of the year, the sun rises and sets somewhat due north or south, depending on its declination, that varies from a maximum of +23,5 degrees to a minimum of -23,5 degrees. The rising and setting point of the sun, for an observer on the equator, will reach a maximum deviation of 23,5 degrees, either to the north or to the south, on the days of the solstices.

That deviation of the rising and setting points from the east and west grows if the observer increases his latitude, growing through the temperate zone, and reaching a maximum of 90 degrees at any point of the arctic circle, where in the summer solstice the sun does not set, but is parted in half by the horizon exactly due north, and in the day of the winter solstice the sun does not rise, but is parted in half by the horizon exactly due south...

 

[NTW]

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...

 

[NTW]

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:

 

[klimatos]

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:

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.

 

[NTW]

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.

 

[NTW]

An extraordinary webpage on the position of the sun anywhere and at any time of the year:

http://drajmarsh.com/scripts/educational/solar-position-and-sun-path

 

[D H]

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.

 

[Graeme M]

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?

 

[klimatos]

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!

 

[NTW]

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...!

 

[Irol]

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.