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Plotting azimuth and sun angle on a spherical(ish) building

 
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Hello Community! I'm designing my own off-grid home and I'm stuck on how to plot the travel of the sun across my building.

Since the structure is more or less spherical I could theorectically plot the sun's full arc across the sky across my home. The windows on the winter solstice could receive direct light from sun up to sun down. Where I am stuck is on how to find the actual height, on the structure, of where the sun will land, and how to change that calculation should the diameter of the home change, or the amount of the home above/below grade change.

the pic attached is an 18' diameter sphere, first 3' below grade, next 8' below grade. how would I find where the sun lands on the actual structure? how would I change my calculations if the sphere expanded to 24' diameter or sank below grade, or was elevated above grade on some type of foundation?

I appreciate all the genius minds at work here on this forum and I thank you for any advice you can give
Photo-on-3-26-21-at-10.44-AM.jpg
18' diameter sphere- top is 3' below grade, bottom is 8' below grade
18' diameter sphere- top is 3' below grade, bottom is 8' below grade
 
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I've done similar graphs for determining the angles I should mount my solar panels at.  You can do this with a protractor and your paper chart.  First, we need to know your latitude?  Where are you located?  For the sake of example, let's say you live in north Texas at a latitude of 35degrees North.

At the spring and fall equinoxes, the position of the sun will be 35degrees lower than directly overhead, or 90 degrees - 35 degrees = 55 degrees above horizontal at noon.
At the summer solstice, the sun will be 23 degrees higher in the sky, or 90 degrees -35 degrees + 23 degrees = 78 degrees above horizontal at noon.
At the winter solstice, the sun will be 23 degrees lower in the sky, or 90 degrees -35 degrees -23 degrees = 32 degrees above horizontal at noon.

Now you can just draw three lines on your graph that all pass through the center of your structure at GROUND level, one at 78degrees, one at 55degrees, and one at 32degrees.  Where they intersect the curved wall of the building will be were the sun will shine at noon.  You just substitute your latitude for the one I used in the example to get YOUR values.

Keep in mind that these numbers will be the MAXIMAL heights at noon, and for the remainder of the day the sun will be lower in the sky.

For the azimuth of the sun, assume that at the equinoxes the sun will rise and set due East, and due West respectively.  At the summer solstice, the sun will rise 23degrees north of East and set 23degrees north of West.  In winter, it will rise 23degrees south of East, and set 23 degrees south of West.

Good luck!

 
Daniel Burnam
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This is great I really appreciate that advice. I've been playing around with similar techniques using sun tracker apps and the like.

I live at roughly 40 degrees north latitude so my max winter sun angle is 26 degrees, 49 degrees on the equinox and 72 degrees at summer solstice.

What I understand you to mean by

Now you can just draw three lines on your graph that all pass through the center of your structure at GROUND level, one at 78degrees, one at 55degrees, and one at 32degrees.  Where they intersect the curved wall of the building will be were the sun will shine at noon.  You just substitute your latitude for the one I used in the example to get YOUR values.



is something like this first picture where every part of the sphere above grade is convex and that all seems to make sense where the winter sun would fall.

but if I were to raise the whole structure above grade and apply that same process it would look like the second picture, which looks funny. haha

my suspicion is that there would be an angle at which the sun would reach the equator of the structure and then start to track predictably. and that leaves a short portion of the day where the sun is tracking on the "underside" of the sphere and then it would begin tracking over the top and the underside would be shaded

any thoughts?

E5FD9CF8-3257-4FE3-B9A2-9EBF8241560B.jpeg
sphere 1/2 below grade blue=winter sun, green=spring/fall, red=summer sun angles
sphere 1/2 below grade blue=winter sun, green=spring/fall, red=summer sun angles
99147EB0-8148-4C81-A610-4013A7C299EE.jpeg
sphere fully above grade
sphere fully above grade
 
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Michael Qulek wrote:Where I am stuck is on how to find the actual height, on the structure, of where the sun will land, and how to change that calculation should the diameter of the home change, or the amount of the home above/below grade change.



As the great Carpenter once said:  "He causes his sun to rise on the evil and the good, and sends rain on the righteous and the unrighteous."  So when drawing for solar design, remember that the sun, being so very far away, is directly shining on everything that is facing it that isn't shaded out by a tree or structure or dirt or what have you.  Just as the rain from the sky will hit the whole roof, the solar radiation travelling in many parallel lines, hits the whole half of the sphere minus the front part below ground. (See attachment)

Draw some parallel lines, and from there you can make your design decisions more easily based off what is inside the structure that you want to have hit by the sun or shaded out.  Most people want the winter sun reaching far back into the structure as practical, not just at the center of the room.  Then, as far as calculations go, you can use trigonometry, or graphing paper or digital drawings with your structure to scale to figure out heights above ground if that is of for construction purposes.  Fortunately, the math will become more apparent based on the sun angles and radius of the sphere once you see some more lines.
SolarSphere.png
Solar effects during solstices and equinoxes.
Solar effects during solstices and equinoxes.
 
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Excellent description of how to approach solar design.
Another aspect to keep in mind (which solar tracker programs would do) is that in winter, the sun direction will fall between southeastish to southwestish, and never hit east or west windows straight on. In summer, the sun will hit the walls all the way from northeastish around the south to northwestish.Any windows on east or west, if you don't want the sun shining deep into them in summer, will need some form of shading. If you are looking at solar energy considerations, most of the windows would be best on the south face with only a few for view and cross-ventilation on other sides.

I wonder about the "spherical" aspect of the planned house... I think it obvious that there needs to be a mostly flat floor; what height in the sphere will the floor be at? If the sphere were set only three feet into the ground, would the floor be small, or above ground? An 18' diameter sphere, even with a floor several feet above its bottom, would have a very high ceiling. If buried (and floored) 8' deep, I suppose a 10' maximum is not excessive given that much of the rest will be significantly lower.

What sort of material and construction method do you have in mind for this sphere? How thick will the walls/ceiling be?
 
Glenn Herbert
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Something to consider if you want to maximize solar input in winter and minimize it in summer is that a sphere is a very ineffective shape for doing that. It has little direct south exposure compared to east/west exposure, so it will be difficult to maximize south-facing windows, and also difficult to exclude direct summer solar gain without added overhangs beyond the spherical profile. Effective overhangs for solar control would conceal the spherical shape, especially if practical leak-resistant roof configuration is included in overhang projections.
 
Daniel Burnam
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YES thank you. That makes a lot of sense now and a good bit easier to wrap my head around. The parallel lines will be a huge help. The sun will be shining on everything not shaded, from it’s given angle, given the time of day and year. And yet there is a focal point on any given structure? And that focal point will coordinate to angle of the sun, measured from center of structure at ground level

For example, on the winter solstice, at 8:15 am, an hour after sunrise where I am, all of the structure facing 144 degrees SE will be irradiated by a sun shining at an angle of +8.5 degrees from the horizon. And the focal point of that irradiation will be concentrated on the structure as measured from the center at ground level at an angle of 8.5 degrees?

Did I get that right?

Sounds like I would be greatly aided by some computer modeling. I have so far remained ignorant of this wonderful modern tool, but I imagine it’s time for me to take the time to learn it. Any recommendations for easy CAD 3-D modeling type software that I could start to model without having to do a ton of self-education?

Thanks again for all the help
 
Daniel Burnam
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And yes, Glenn, I think I understand what you’re saying about the shortcomings of the sphere re: maximizing winter sun and minimizing summer sun. In the end, I will almost certainly wind up with an octagonal shape, but the sphere seemed like the easiest way to get my head around plotting out the sun path and also to play with my own ideas of having some part of the structure be above grade and also shaded by the larger diameter part of the building, for the sake of extra edge and potential microclimates.
 
George Yacus
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Daniel Burnam wrote:And yet there is a focal point on any given structure?



Negative.

I'm betting you may be a victim of someone's bad sector analysis teachings.  Some permies can easily get confused because many permaculture design sector analysis drawings do solar azimuth a big disservice zoomed in on a house or Zone 0.  Remember...parallel rays.  It's a Zone 0, not a "point 0".  The sun hits the whole  house or structure, and doesn't go through a single tiny "point" at the center of a house.  Really, any area, spot or element in a design essentially has its own sector analysis at play.  And then you do good design by blocking or channeling in the energies you don't or do want.  Let's talk focal points in a permaculture context.

Focal points are unique to specific structural shapes and materials.  Focal points will only naturally happen with certain shapes, namely parabolas.  A flat wall won't usually have a focal point, unless it is intentionally made with certain engineered glass or mirror-like materials.  Focal points are indeed a valid permaculture solar technique, but are uncommon though.

Focal points deal with either:
1) Concentrating energy via reflection of rays.
  • A beneficial permaculture-related example of this is a parabolic solar cooker.
  • Planting trees in a parabola shape with apex poleward of a homesite...aka a "sun trap" (very weak reflection but can help block winds, too, creating an even better microclimate).
  • A long linear solar concentrator can be used to heat water for household use or cooking.  Commercial example is GoSun grill and solar cooker.
  • Parabolic microphones.
  • Satellite antennas.
  • A detrimental example of it in architecture is this Las Vegas 'death ray hotel'

  • 2) Concentrating energy with refraction (bending) of rays.
  • A beneficial permaculture-related example of refraction is the Fresnel lens.  Such a lens bends the originally parallel radiation of energy that hits it, and it puts it all on one single spot at a fixed distance.  You can cook with a large Fresnel lens.
  • Mollison's Permaculture, A Designers' Manual has an example of lenses made of ice in a linear concentrator in Chapter 12.
  • Another detrimental example is a kid mercilessly burning ants with a magnifying lens, or starting a forest fire while playing unsupervised.


  • It will help to think of focal points as a very separate thing beyond the scope of basic passive solar design for any home.  That said, there could be some interesting acoustics and maybe some light bending effects if you intentionally shape it that way, but you wouldn't gain any "extra energy" benefits doing this.
     
    Daniel Burnam
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    Roger that.

    So if I had a flat wall facing due south at high noon in the winter, the intensity of the sun at 5’ up the wall is no different than the intensity of the sun 10’ up the wall, assuming there’s no shading.

    Window placement has to do with how far back into the space the sun will shine and so is about the relationship between the window and the floor, whether that floor is at ground level or 50’ off the ground. period...question mark...explanation point

    To be sure I misunderstood that. Thank you for the clarity.

    And for the links.
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