Hi all, I have these compass bearings for the suns position at sunrise and sunset at my location. sunrise 118 degrees ESE and sunset 242 degrees WSW.
How do you work out the degrees of movement the sun has made during the day? I need this for a sketch I am doing. It is a problem I have not been able to solve for some time now. There must be a formula for this?
If the sunrise bearing to the Sun is 242 degrees and the sunset bearing is 242 degrees then surely it moved through 242 - 118 degrees. ***
So what is your problem in calculating the degree of movement ?
***Actually, it did not move at all, it was the Earth that rotated
The earth rotates 360° in 24 hours, or 15° per hour.
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What's your problem exactly
And when you need the sol. Needed
That's only going to be true at a particular time of year. In the summer the sun will rise further East and set further West than it does in the winter (assuming northern hemisphere)
For me the sun was around 150° at 10:00.
I'm expecting it to be close to 180° at 12:00.
Information above from timeanddate.com/sun/uk/lincoln
12:00 Update:
In 2 hours the sun has moved 27.68°, slightly less than the 30° I was expecting from my comment in post #3.
I assume the discrepancy is due to the tilt of the earth's axis, and my latitude.
When you project the position of the sun to the horizon, i.e. a compass bearing, the rate is not exactly 15 deg/hour. I think if you track the sun via an equatorial mount, you would see the polar axis change by exactly 15 deg/h.
If only we knew what problem @petercl14 was trying to solve ...
Thanks for that explanation bobcousins.
I got confused trying to do 3 dimensional geometry in my head.
I am guessing he has fixed numbers from an example, but wants to know how to calculate it dynamically. I have used the formula to calculate time of sunrise/sunset, but I never needed bearings. Solar power installers like to know that sort of thing.
Eh?
Data lacks your lat/lon.
This calculation/measurement depends upon your season and location. Only on the two equinox does the sun "rise" and "set" at the equator. For example, if you are in the northern hemisphere and you are in summer, the sun will "rise" south of the equator and "set" south of the equator, so your days would be longer than 12 hours. However if you are in the southern hemisphere and you are in winter, the sun will still "rise" south of the equator and "set" south of the equator, but your days would be shorter than 12 hours. The first example will have a largeer degree movement, the second example would have a smaller degree movement.
For time measured fractional hours, 24 h clock,
= (set_time - rise_time)*15
I don't believe it is that simple Bob. Where I am the daylight time is 14hrs 6 min 25s. Lets say 14. In 24 hours the sun can be assumed to have moved 360 degrees. So 14/24x 360=210 degrees. That sounds like the more correct figure for the suns movement(earth actually).
How can you get this figure from the magnetic compass direction of the sun I have given?
refer John Lincolns reply. Earth rotates 360 degrees in 24 hours or 15 degrees per hour. Then 14 X 15=210. The same result I got before. 14/24 x360=210.
This must be the correct degree of movement.
Noted in post #13, reference (at this moment in this season) the southern hemisphere.
If you are two meters tall, and the horizon is 20 miles if you live on a calm body of water, sunrise to you is slightly earlier than shorter individuals, and sunset is slightly later.
What is you reference plane? The horizon, the celestial equator or ...?
There are many solar position calculators on the web, here's one:
https://planetcalc.com/4270/
The problem Bob is the degree of movement of the sun at my location at the time of the year it is now. That is the problem I am trying to solve. Your solution is way out.
There is approximately 14 hours of daylight now. Then at 15 degree movement per hour this is a 210 degree movement in 14 hours. Your solution gives 124 which is way out.