I really need your help with some calculations

[tmdarkmatter]

You should give me some points for my reputation, because at least I did not offend anybody this time.

[Bufofrog]

1 hour ago, tmdarkmatter said:

You should give me some points for my reputation, because at least I did not offend anybody this time.

Gee, that's a pretty low bar... but what the hell. +1.

[KJW]

On 3/18/2024 at 1:02 AM, tmdarkmatter said:

the star Proxima Centauri which has a size of about 1.02 ± 0.08 milliarcsec in the sky. This is about 0,00027778 degrees (correct me if I am wrong).

I see no one thought to check this... but I did. This is what I got:

[math]1.02 \text{ milliarcsec} = 1.02 \times \dfrac{1}{1000} = 0.00102 \text{ arcsec}[/math]

[math]0.00102 \text{ arcsec} = 0.00102 \times \dfrac{1}{60} = 0.000017 \text{ arcmin}[/math]

[math]0.000017 \text{ arcmin} = 0.000017 \times \dfrac{1}{60} = 2.83 \times 10^{-7} \text{ degree}[/math]

Continuing further:

[math]2.83 \times 10^{-7} \text{ degree} = 2.83 \times 10^{-7} \times \dfrac{2 \pi}{360} = 4.95 \times 10^{-9} \text{ radian}[/math]

Calculating in radians allows you to calculate without the explicit use of trigonometric functions.

 

Edited March 22 by KJW

[tmdarkmatter]

22 hours ago, KJW said:

Calculating in radians allows you to calculate without the explicit use of trigonometric functions.

Thank you for your assistance! I can see the mistake I made above. But now that we have this angle in radians, how can I convert the observer who is just a dot into a surface or at least a two-dimensional line? Because I am not looking for the object needed at a distance of the diameter of Earth that would be big enough to block Proxima Centauri for one point, I want to calculate the new angle that is created having in mind that the "observer" is for example an object of a diameter of 1 meter and how big the object creating the shadow on the other side of Earth should be. The main issue is to convert the observer from a point into a surface. You might say that the size of the observer can be disregarded, but it is clearly not the same to block the light of Proxima Centauri for just one point than to block it for a certain surface. And it does not seem that it is just an addition of the surface of the observer to the object calculated with this amount of radians, because the angle of the light coming from Proxima Centauri in total changes. We would see the star bigger if our eyes would be more separated.

I have been analyzing this and it is as if the point of the observer would be closer to the object creating the shadow, so an inverted shadow is arriving at the surface. It´s like moving the apex towards the opposite side and creating a cross with a second side where the observer is located, converting the observer into a surface.

This would mean that the bigger the observer becomes, the more the cross would displace towards the opposite side and the bigger the angle would become.

Edited March 23 by tmdarkmatter