GodlyCharmander
He/Him- 6,735
- 6,135
Here:Wow, nevermind, quite the clever way to get the Sun Disk's size, it has basically no contradictions, and all the logic is consistent. The amount of assumptions made were reasonable enough.
I would include at least some corrections because the screenshot used has the disk at an angle, not facing directly the camera, it matters because it's a disk, and not a sphere like the star.
It's a minor adjustment, but we can use the cosine law to compensate for the angle, assuming a low angle of 15° since they are so close together.
I could try it.
Law of Cosines states that c2 = a2 + b2 - 2ab·cosC
Point A: Disk
Point B: Screen
Point C: Sun
c = Distance from the Disk to the screen (AB)
b = Distance between the Disk and the Sun (AC)
a = Distance between the Sun and the Screen. (BC)
CosC = angle between the lines from the sun and the two other points. Assumed 15°
c^2 = a^2 + b^2 - 2ab·cosC
c^2 = 9967693387.91^2 + 1350993153.4^2 - 2*9967693387.91*1350993153.4^2*cos15deg = 7.5165228e+19
c = sqrt7.5165228e+19 = 8669788232.71m
Slightly higher. Since it's not facing directly at us.
The angle could be completely wrong, a higher angle would yield a higher result. (Around 9 bil). I can see argument for 20-30°, I am eyeballing quite a bit.
For reference, that makes for a 1020171653.72m diameter, and a
(1020171653.72/1013924803.23)^5*9.6800782e+37J = 9.981973e+37J or 24.23 Ronnatons. Barely a difference. Lmao. I just wasted my time.