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How is the sun's equatorial circumference measured by scientists?

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Suggestion to the question formulation (v1): Replace the word calculated with the word measured. –  Qmechanic Jun 15 '13 at 16:07
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Of interest: physics.stackexchange.com/q/29363 –  dmckee Jun 17 '13 at 20:02
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2 Answers

It is quite simple to get a reasonable measure of the sun's diameter, using a pinhole, a screen and the formula for experiment 2 on this At Home Astronomy article: "Finding the size of the sun and moon".

From the 'At Home Astronomy' article, the sun's diameter can be calculated as:

$$\frac{\text{Diameter of the sun's image}}{\text{Distance from the pinhole to the paper}}\times\text{Earth-sun distance} = \text{Diameter of the sun}$$

Edited to add:

The first recorded attempts to determine the distances and sizes of the Earth and Moon were performed by Aristarchus (310 - 230 B.C.), these were not very accurate (this information is also in the first link I gave originally). Once the distance was known (or estimated), a measure for the size could be determined.

His method was to determine the angle to the sun when the moon was half full, hence at a right angle to the observer (a potential source of error) and measure the angle to the sun (as in the diagram below)

Aristarchus' observational geometry

(Source)

I hope this clears up the inadequacies of my original post.

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@DrStrangeLove the formula they use is a derivation of relative size ratios - a bit more explanation of the maths involved is here en.wikipedia.org/wiki/Pinhole_camera_model –  user24901 Jun 15 '13 at 12:08
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@DamienIgoe While links to more info are good, we discourage link-only answers, since they lose all value when the link inevitably dies. It would help if you provided a brief summary of the technique. –  Chris White Jun 15 '13 at 15:28
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Okay, added a relevant image. –  user24901 Jun 15 '13 at 23:41
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Doesn't answer the question because we don't know the object distance. This equation correlates size and distance, and this leaves us needing one more equation. –  AlanSE Jun 16 '13 at 0:36
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No one is going to delete your question on you (and indeed no one has even downvoted it), I just suspected you didn't understand Alan's objection because it was rather minimally stated. –  dmckee Jun 17 '13 at 19:53
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It should be simply as below: $C = \pi d$

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Great, now what is $d$? –  santa claus Jun 16 '13 at 0:00
    
d = diameter, can be determined using the 'At home astronomy article' in my reply. –  user24901 Jun 16 '13 at 0:45
    
The sun area is assumed circle! and cicumference of a circle is school math and $\pi d$. However glad to help $d$ is the diameter. –  al-Hwarizmi Jun 16 '13 at 8:00
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