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Thread: Angle of view as it relates to the solar eclipse

  1. #21

    Join Date
    Aug 2017
    Posts
    11

    Re: Angle of view as it relates to the solar eclipse

    I see I mistyped -- I mean 45 degrees, not 45 minutes. The problem you alluded to -- multiple exposures on a single frame -- is just what I wanted to resolve. I have been using NOAA's online solar position calculator to ascertain the azimuth and elevation for that location. I don't think I have expressed the issue very clearly, so I'll give it one more shot. The landscape angle of view for a 4x5 with a 210mm lens is 33.6 degrees. NASA tells us that the eclipse event, at my location, will last for 3 hours, with a 80 degree change in the sun's azimuth. On its face, that suggests I can capture less than half the event on my 4x5 (33.6 versus 80). This would be true if everything occurred at the horizon. However, the reality is that I can capture roughly 2/3rds of it. The reason for the apparent contradiction is that if the starting and stopping azimuths are extended vertically, they get closer together (eventually meeting directly overhead). What I wanted to know is if anyone can succinctly describe the calculation for determining what the apparent angle of view is when tilted for the eclipse (elevation will range from 52 to 63 degrees). I ended up simply timing it but would like to know how to make the calculation as well.

  2. #22

    Join Date
    May 2007
    Location
    Greenwood Lake NY USA
    Posts
    148

    Re: Angle of view as it relates to the solar eclipse

    The calculation is a piece of geometry probably falling under the heading spherical triangles.

    Here is another way to obtain an approximation from the basic facts.

    Using round numbers the eclipse begins at about 11.30 and ends at about 2.30, the duration is therefore about three hours. The apparent motion of the sun is 15 degrees per hour, therefore in three hours the sun will appear to move 45 degrees across the sky. In order to put the entire eclipse sequence on a single piece of film a field of view of at least 45 degrees is needed.

    Hope your skies are clear!

  3. #23

    Join Date
    Aug 2017
    Posts
    11

    Re: Angle of view as it relates to the solar eclipse

    Yes, that back-of-the-envelope calculation is correct. The skies were cloudier than we would have liked (we had to put plastic bags over the cameras when it rained!) but it was clear leading up to and following total eclipse, which was a delight. It would have helped to have remembered to take the filter off for the shot of the corona during total eclipse, but the experience was well worth it. I have ordered a book on spherical geometry.


    Quote Originally Posted by Ted R View Post
    The calculation is a piece of geometry probably falling under the heading spherical triangles.

    Here is another way to obtain an approximation from the basic facts.

    Using round numbers the eclipse begins at about 11.30 and ends at about 2.30, the duration is therefore about three hours. The apparent motion of the sun is 15 degrees per hour, therefore in three hours the sun will appear to move 45 degrees across the sky. In order to put the entire eclipse sequence on a single piece of film a field of view of at least 45 degrees is needed.

    Hope your skies are clear!

  4. #24

    Join Date
    Aug 2017
    Posts
    11

    Re: Angle of view as it relates to the solar eclipse

    OK, FWIW I think this is the quick-and-dirty way to convert a known change in azimuth to apparent angle of view, given a specific elevation.

    For elevation Ev, and change in azimuth of Az, then:

    a= COT(Ev)
    a'= a * SIN(Az/2)
    Apparent angle of view = 2 * ATAN(a' / SQRT(1+(a*a)))

    For the solar eclipse that just occurred, if it started at an elevation of 52 degrees, and incurred a change in azimuth of 80 degrees, the apparent angle of view for the event would be 43.2 degrees.

    I think that's the correct calculation. Let me know if I have this wrong ...



    Quote Originally Posted by Ted R View Post
    The calculation is a piece of geometry probably falling under the heading spherical triangles.

    Here is another way to obtain an approximation from the basic facts.

    Using round numbers the eclipse begins at about 11.30 and ends at about 2.30, the duration is therefore about three hours. The apparent motion of the sun is 15 degrees per hour, therefore in three hours the sun will appear to move 45 degrees across the sky. In order to put the entire eclipse sequence on a single piece of film a field of view of at least 45 degrees is needed.

    Hope your skies are clear!

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