darkoshi: (Default)
Follow-up to this post.

In the prior post, I pondered about angles. In particular, about the ~13 degrees per day that the moon orbits around the earth, and whether that angle would still look like 13 degrees to me, when measured from the surface of the earth.

The angle would *not* be exactly the same. However, because of how far away the moon is compared to the size of the earth, the difference in angle is very small. That difference can be calculated using trigonometry.

Here's a new diagram. All figures mentioned below are approximations or averages.



The angle measured from the center of the earth is 13 degrees.
"X" is the corresponding angle measured from the surface, which will be calculated.
"D" is the distance from the center of the earth to the moon: 384400 km
"R" is the radius of the earth : 6371 km

I've drawn 2 right triangles such that both have the same "opposite" side, with length "O".
The length of the adjacent side for the X-angled triangle is "A".
The length of the hypotenuse for the X-angled triangle is "B".
The length of the adjacent side for the 13-degree-angled triangle is A + R.
The length of the hypotenuse for the 13-degree angled triangle is D.

sin 13 degrees = O / D
O = D * sin 13 = 384400km * 0.2249511 = 86471 km

cos 13 degrees = (A + R) / D
A = (cos 13) * D - R = (0.9743701 * 384400) - 6371 = 368176.85

tan X = O / A
X = arctan(O / A) = arctan (86471 / 368176.85) = 13.217

So, the corresponding angle from the surface of the earth is ~13.2 degrees.

Date: 2017-08-28 08:23 am (UTC)From: [personal profile] marahmarie
marahmarie: (M In M Forever) (Default)
Saw this and thought of you: https://www.independent.co.uk/news/science/babylonians-trigonometry-develop-more-advanced-modern-mathematics-3700-years-ago-ancient-a7910936.html

(Boils down to: lost trig is better than modern trig because base60, not base10!)

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