View Full Version : calculating winter sun
16-02-2008, 07:18 AM
Is there any reasonably low tech way to calculate where the sun will rise, set and be at its zenith for a given location at winter solstice? Other than waiting until the solstice and observing ;-)
I'm trying to figure out if the winter sun will be above the trees on the north side of a house (southern hemisphere).
16-02-2008, 11:46 AM
You can use this link http://solardat.uoregon.edu/SunChartProgram.html enter your latitude and longitude(latitude negative in Oz) and then get the sun angle throughout the year.
Is that low tech enough? :)
16-02-2008, 02:44 PM
Just about but not quite :wink:
I've done the chart for the location. Now I need to know how to tell where 23 degrees is above the horizon, and where 52 degrees is on the horizon from true north.
I've got a compass, but need to know if true north is to the left or the right of magnetic north. I'm in NZ and the difference between them is around 22 degrees east.
I also apparently need a clinometer to get the degree above horizon, but I've never heard of one of those.
17-02-2008, 08:01 AM
OK, we'll get there, but folks should chime in if I'm not doing it right
If your declination is 22degrees East, then True North is 22 degrees West of where your needle is pointing. So its simple subtraction to find True North
My compass has a cheap clinometer built in, so that's how I'd do it. You can also set the compass rings to your sun altitudes (23 and 52 degrees), turn the compass on its side, level it, and look down the ring arrow to your appropriate altitude.
It'll be rough, but way cheaper than a clinometer. (which you could rent, if that kind of accuracy was needed)
how about this for a clinometer..
1)get a protractor, drill a little hole in it or find some other way of connecting a string to the center of the protractor.
2) Connect a heavy object such as a stone, bolt etc to the other end of the string.
3)Sit, stand or lay in the location which you want to find where the sun will be.
4) You can measure an angle by looking along the axis of the protractor and reading the angle off the string.
?? what do you think? I'm fairly sure it will work, but haven't actually made one since I have a clinometer from work.
19-02-2008, 08:09 AM
That would do the trick.......
19-02-2008, 02:31 PM
PPP, I didn't quite follow that. Should I have gotten a 360 protractor instead of a 180?
No, a 180 degree protractor is what you want.
Let me consider an example and hopefully this will help me to explain myself.
.. for simplicity let's try to measure 0 degrees (ie horizontal).
With the protractor held such that the flat edge is on the top, and the curve underneath, held exactly horizontal, you should be able to read 0 degress.. aahhh no, I understand the problem..
You read 90 degrees with it horizontal right?
OK, well to measure 22 degrees you need to have the string reading 68 degrees (ie 90 - 22).
So, now to find out where 22 degrees is, look along the straight edge of the protractor up towards your eve or tree or whatever. Vary the angle untill it shows 68 degrees (can you get someone to look at the protractor while you look along it?)
Does this help? Sorry, ask again if it still isn't clear... (there's another way you can do it with a measuring tape..)
19-02-2008, 06:04 PM
Thanks! I get that now. I'll have a look in the morning.
Part of the difficulty is that even though I can get the angle, I can't seem to 'see' the line that will extend from my eye ahead say 50 metres, especially where it's sky. I guess I can only do this for an object right? I'm trying to figure out how high the zenith is above the nearest object, in this case a roof.
I got the sunrise/sunset positions from Rich's tip but I'm also going to have to figure out the angle at sunset as there are hills there so I can't see the horizon.
instead of trying to measure something off into the sky, can you keep the angle fixed and move toward the tree untill the top of the tree is at the angle. You will then know where the sun will hit the ground.
20-02-2008, 05:34 PM
Good idea. For some reason I was trying to do all the measurements standing in the same spot, but I don't really need to do that.
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