brianv » November 25th, 2017, 6:41 am wrote:Great job debunking. Even I nearly understood.

So

what is the calculated distance to the Sun using trig with these values? Which I think was the object of the exercise.

You can't really calculate the sun's distance accurately with those values due to the

sun position calculator having an accuracy of 0.5 arcminutes (30 arcseconds or

0.0083°), and calculating the sun's distance from those values would require the angles be accurate within 1 or 2 arcseconds.

With that limitation in mind, we can come up with a possible range of values for the sun's distance using the data for locations identified in the video as "B" and "F" because they are very close to the same distance apart from "D". Be warned, maths ahead.

At B the sun's elevation is 47.0832°. At F it is at 47.8468°.

We adjust by 0.3817 to account for the sun's altitude at D (89.6183°) being offset by 0.3817 from 90°, so add 0.3817 to B and subtract it from F (because these locations are in opposite directions from D):

B = 47.0832 + 0.3817 =

47.4649°F = 47.8468 - 0.3817 =

47.4651°Now all we have to do is multiply the B-F chord length (straight-line distance) by the cosecant of the difference between the two angles. If we use those angles as-is

without accounting for the

± 0.0083 uncertainty, the difference is 0.0002°. Given that the B-F chord length is 8613 km, that equals:

8613*csc(0.0002°) =

2,467,442,744 kmBut we know that distance isn't accurate because it's based on an angle far smaller than the stated accuracy of the data.

If we account for the

± 0.0083 uncertainty, the difference between angles B and F could be as high as 0.0168°, which corresponds to a distance of:

8613*csc(0.0168°) =

29,374,318 km. This would be the lower bound for distance based on the uncertainty of the angles - the sun can be further away than that, but it cannot be closer (closer would require a larger angular difference, and we can't get that even with the uncertainty).