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Effective angular diameter of the sun for this experiment

by omoplata
Tags: angular, diameter, effective, experiment
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omoplata
#1
Mar7-12, 06:08 AM
P: 315
Hi,

We are using a radio interferometer to measure the angular diameter of the Sun.

An in depth description of how this is done is not relevant to solve the immediate problem I have. But briefly, we rotate the interferometer in azimuth so that the direction it's looking at traverses across the center of the Sun. We take readings of an interference pattern from two mirrors off to the sides during this traverse. Then we use this pattern to find out the angular diameter of the Sun that we saw.

My problem is that the interferometer rotates in azimuth around an axis that is perpendicular to the surface of the Earth, but the Sun is up at some latitude. So the interferometer doesn't trace a straight line diameter across the Sun. It traces an arc. The length of this arc is going to be the effective angular diameter that we finally obtain.

This is described in the attached diagram. The real angular diameter of the Sun is the length of the straight line. But the value for the angular diameter that we actually obtain is the length of the curved line. That curved line is the path the interferometer actually traverses.

I want to find a relationship between the length of the curved line, the length of the straight line, and the latitude of the Sun, so I can find the real angular diameter of the Sun from the value that I obtained.

Could someone please suggest a way I can do this? I think it has to do with spherical trigonometry, but I don't know where to start. I don't want to go through a whole textbook of Spherical Trigonometry just to solve this. What I need is a relevant equation.

Thanks.
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sun_line.jpg  
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omoplata
#2
Mar7-12, 06:12 AM
P: 315
Sorry, that diagram is wrong. Attached is a better diagram.
Attached Thumbnails
sun_line.jpg  
omoplata
#3
Mar11-12, 04:55 AM
P: 315
OK, I found an expression for the effective angular diameter of the first diagram (but not for the second one yet) using the spherical law of cosines.

If the latitude of the sun is [itex]\theta[/itex], and the actual angular diameter of the sun ( straight line ) is [itex]d[/itex], the effective angular diameter ( curved line ) as shown in the first diagram ( when the direction the interferometer is pointing towards doesn't go through the center of the sun ) is,
[tex]d \phi = \cos (\theta) \arccos \left( \frac{\cos( d ) - \sin^2 ( \theta )}{\cos^2 ( \theta )} \right) [/tex]


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