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Why isn't the sun's rate-of-change in elevation constant?

  1. Jan 14, 2009 #1
    I have been experimenting with the 'solpos' library from NREL, which is a very nice library for computing solar position (and related) calculations given date/time and coordinates.

    One thing that surprised me is that the sun's rate-of-change in elevation (angle from the horizon) is not constant throughout the day. In fact, the elevation changes about 7-8 times faster at sunrise/sunset than it does at solar noon.

    Here's sample calculations for Toronto Canada for January 19, 2009:
    07:52:00: azim=118.788 elev=0.109222
    07:59:00: azim=120.007 elev=1.209
    ...
    12:28:00: azim=179.905 elev=25.9437
    13:18:00: azim=192.848 elev=24.9487

    I'm sure there's some perfectly good trigonometric reason why this should be the case, but at the moment I haven't been able to come up with a good explanation.

    Can anyone help unravel this mystery?

    Thanks,
    Adam
     
  2. jcsd
  3. Jan 14, 2009 #2

    mgb_phys

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    You are measuring elevation from the horizon - you are not at the centre of the Earth.
    Draw the picture - it will help ;-)
     
  4. Jan 14, 2009 #3

    russ_watters

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    The sun moves across the sky at a constant rate, but in two dimensions, not just one.
     
  5. Jan 14, 2009 #4
    Is there really a simple picture to clarify this observation? I've been scribbling diagrams all day and they all end up terribly complex and totally unenlightening. Any advice on how to draw this out?
     
  6. Jan 14, 2009 #5
    Yeah, I realized that a while ago, but I'd still like to be able to draw a simple picture to help illustrate the point. I might end up having to demonstrate this to higher-ups in my company so I want to make absolutely sure I've got a simple clear picture that they can intuitively understand.

    I'll continue to work on this, but any guidance is certainly appreciated.
     
  7. Jan 14, 2009 #6

    russ_watters

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    The diagram would be a sinusoidal arc, depicting the motion of the sun across the sky as viewed by a person standing on the earth, facing south. The sun travels along that arc at a constant rate and at any point in the arc, you can draw a triangle and do some trig to figure out how fast it is moving in the x and y directions.
     
    Last edited: Jan 14, 2009
  8. Jan 15, 2009 #7
  9. Jan 15, 2009 #8

    russ_watters

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    You got it!
     
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