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couldabin
16-Aug-2017, 13:09
I plan to photograph the solar eclipse (weather-permitting) with both my 4x5 (210mm lens) and 8x10 (12" lens). From my viewing spot in Nebraska, the sun's azimuth will change nearly 80 degrees from start to finish. Does anyone know how to convert that to angle of view when the elevation is 60 degrees above the horizon? (It's not a simple calculation, since the vertical lines converge.) Thanks.

Jim Noel
16-Aug-2017, 18:36
First, I would put the 12" lens on the 4x5. Even then you image is going to be quite small. The Sun and Moon each intercept an angle of only about 0.5 degrees.

tgtaylor
16-Aug-2017, 19:43
I'm thinking about observing and photographing the eclipse from San Francisco with a 10" Meade SCT at prime focus (2500mm equivalent) equipped with a solar filter. The Meade has dual axis electronic drive (tracks in both declination and right ascension) which eliminates problems due to the earth's rotation making a several second exposure doable.

Thomas

Pfsor
17-Aug-2017, 02:32
I plan to photograph the solar eclipse (weather-permitting) with both my 4x5 (210mm lens) and 8x10 (12" lens). From my viewing spot in Nebraska, the sun's azimuth will change nearly 80 degrees from start to finish.

Are you mad? In Nebraska the sun's azimuth changes in several minutes nearly 80°?

Nodda Duma
17-Aug-2017, 03:44
The total event time (moon just touching the sun to moon just leaving) in Nebraska will be ~3 hours, with the sun traversing (3*360/24) 45 degrees of sky during that time frame. (15 degrees per hour). Totality time will be about 1 min, 30 s

An extended fist (back of the hand towards you) covers approximately 10 degrees.

This should give you an idea of the amount of sky the sun will traverse.

To satisfactorily fill the frame of a 35mm camera with an image of the eclipsed sun (or of the moon if doing astrophotography), you need a focal length of about 1250mm. To capture the entire corona during full eclipse would call for about 200-300mm.

You should be able to scale to required 4x5 focal length from there.

Pfsor
17-Aug-2017, 05:50
Does anyone know how to convert that to angle of view when the elevation is 60 degrees above the horizon? (It's not a simple calculation, since the vertical lines converge.) Thanks.

You don't need any calculation at all. Make yourself a viewing frame for whatever film format and FL you want to use, and see the sun through it at whatever time lapse you want to use. Up to you how to keep the frame and you eye in the position for the time needed.

Pfsor
17-Aug-2017, 05:52
Totality time will be about 1 min, 30 s

According to this source https://www.space.com/33797-total-solar-eclipse-2017-guide.html the totality time will last about 2min 40s.

Nodda Duma
17-Aug-2017, 07:56
According to this source https://www.space.com/33797-total-solar-eclipse-2017-guide.html the totality time will last about 2min 40s.

"At most". That, is, at the center of the eclipse path.

You need to look up the information for your specific location.

http://www.eclipse2017.org/2017/communities/states/NE/Lincoln_1125.htm

Pfsor
17-Aug-2017, 08:17
"At most". That, is, at the center of the eclipse path.

You need to look up the information for your specific location.

For Lincoln, Nebraska, the totality will last 2 min. According to this source - https://www.space.com/33797-total-solar-eclipse-2017-guide.html

Ted R
17-Aug-2017, 09:01
I plan to photograph the solar eclipse (weather-permitting) with both my 4x5 (210mm lens) and 8x10 (12" lens). From my viewing spot in Nebraska, the sun's azimuth will change nearly 80 degrees from start to finish. Does anyone know how to convert that to angle of view when the elevation is 60 degrees above the horizon? (It's not a simple calculation, since the vertical lines converge.) Thanks.

I don't have an answer to your request for a calculation, however if I understand correctly you intend to capture the complete eclipse period from first contact to last. Since you have already determined your lens focal lengths, and this determines the angle of view, the primary task that remains is the aiming of the camera to set the point of totality in the center of the frame. To do this the azimuth of totality at your location is needed. Here is a website that may help https://www.timeanddate.com there is a comprehensive database that allows the location to be entered and precise values obtained. Good Luck.

tgtaylor
17-Aug-2017, 09:50
According to Schneider, the 360mm Symmar-S will provide 70 degree coverage at f22; the Rodenstock Apo-Sironar-S gives about 75 degree coverage at f11; except for the 120-SW, Nikon's lens provide for a 60 (f5.6) to 70 (f/22) degree coverage with the 120-SW gives 80 degree coverage at f/8 and 105 degree at F/22. I forget how it goes but you can guesstimate the arc in the sky by pointing one arm directly at the sun and keeping the other at 90 degrees (?) turn in the direction of solar travel.

Thomas

Pfsor
17-Aug-2017, 13:39
I plan to photograph the solar eclipse (weather-permitting) with both my 4x5 (210mm lens) and 8x10 (12" lens). From my viewing spot in Nebraska, the sun's azimuth will change nearly 80 degrees from start to finish. Does anyone know how to convert that to angle of view when the elevation is 60 degrees above the horizon? (It's not a simple calculation, since the vertical lines converge.) Thanks.

This on-line calculator http://keisan.casio.com/exec/system/1224682277 will give you exactly the sun's azimuth and elevation (altitude)changes for your position, the time of the year and the length of your observation period. From these values you get the horizontal and vertical angles you need to cover by the angle of view of your lens (you have to calculate that from the FL of your lens and the film format you want to use).
More problems wait for you on the way if you want to make a composite picture of the different eclipse phases on one single film frame, though.

couldabin
19-Aug-2017, 10:53
No, the 12" goes on the 8x10. Yes, the sun's disc is roughly 0.5 degrees.

First, I would put the 12" lens on the 4x5. Even then you image is going to be quite small. The Sun and Moon each intercept an angle of only about 0.5 degrees.

couldabin
19-Aug-2017, 10:54
No, I'm not mad. I live 2 hours from the center of totality, so I thought that might be a convenient viewing spot.

couldabin
19-Aug-2017, 11:03
Thanks, but I think I didn't make my question sufficiently clear. Yes, the sun "moves" 0.25 degrees/minute. Totality at my viewing location will be 2:30. The duration of start of eclipse to end of eclipse is ~3 hours. Earth will rotate roughly 45 minutes during that period. But the azimuth changes nearly 80 degrees. I was asking how to convert that information to one that relates to the camera's angle of view. This is a calculation that would take into account the fact that azimuth is expressed at the horizon, but that all on the compass converge overhead.

couldabin
19-Aug-2017, 11:05
Yes, I ended up doing this empirically -- the sun traverses the 4x5 view in ~2:15. I should be able to capture nearly 2/3rds of the event on the 4x5, but all of it (with about 8 min. to spare) with the 8x10. But I would like to know the trig involved.

couldabin
19-Aug-2017, 11:07
Won't be in Lincoln, which will have 1:25 of totality.

couldabin
19-Aug-2017, 11:12
Thanks, Ted. I was interested in finding out how to account for the fact that we have the azimuth for all relevant points in the event, but the event takes place at 52-61 degrees above the horizon. When you point your camera upwards, the angle of view covers something greater than the values implied by the azimuth -- point it directly overhead, for example, and the narrowest angle of view will capture the entire circle. I ended up simply timing it, which is effective for this but doesn't advance my ability to calculate such things in the future.

couldabin
19-Aug-2017, 11:16
I think the relevant angles of view are determined by the focal length and film size. It's conventional to express the angle of view using the diagonal distance of the film, but that's of course the greatest angle the film can capture. The horizontal angle of view is less, and the vertical less still. My question related to the fact that the visual distance between two celestial objects (one with azimuth 100, one with azimuth 150), depends on their elevation, since all points converge directly overhead.

Pfsor
19-Aug-2017, 13:47
Earth will rotate roughly 45 minutes during that period. But the azimuth changes nearly 80 degrees.

Couldabin, I'm sorry you don't want to use the on-line calculator. To think that the Sun will change its azimuth with nearly 80 degrees in 45 minutes is nothing less than madness. I think you don't know at all what you're talking about.

couldabin
20-Aug-2017, 13:53
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.

Ted R
20-Aug-2017, 18:28
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.

couldabin
22-Aug-2017, 07:56
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.

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.

couldabin
23-Aug-2017, 08:19
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 ...

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.