# tmdarkmatter

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1. ## I really need your help with some calculations

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.
2. ## I really need your help with some calculations

You should give me some points for my reputation, because at least I did not offend anybody this time.
3. ## I really need your help with some calculations

Ok, I have done these simple calculations and the results are as follows: If the observer is 1 km over the surface of earth, according to the tangent ecuation, the object on the surface of earth must have the size of 4,8481 mm (the size of a very small coin) to block Proxima Centauri. So if you are staring at a rooftop that is 1 km away and Proxima Centuari is just showing up behind the rooftop, you might estimate that a tiny coin would be enough to block the light coming from Proxima Centauri. But what is interesting is that the blocking object on the other side of earth at a distance of 12,742 km would also increase in size by 4,8481 mm, so the equation is totally linear. At a distance of 100 km above earth, the object on the surface would have a size of 0,48481 meters to block the light coming from Proxima Centauri. But what is interesting is that if I insert the total distance to Proxima Centauri in that equation, the result is an object 907 times the size of Proxima Centauri, meaning that the light we see coming from Proxima Centauri is 907 times bigger than the star itself, because we do not only see the star but also its bright surroundings. (Correct me if I am wrong) But the ratio between blocking object on the surface and blocking object on the opposite side of earth would begin at zero when the observer is on the surface to a value of almost 1 when the observer is light years away. And the size of the blocking object would in both cases become bigger than earth itself. So if we are standing on earth and would fit in this area of 0,1844 square meters, the object to block the light from Alpha Centauri from us on the opposite of earth (maybe I can call it antishadow) would have to be 16.630,78 times bigger than ourselves. This seems to be a constant for all objects of all sizes on the surface in the case of the angle of Alpha Centauri.
4. ## I really need your help with some calculations

Yes, this is getting closer to what I need. We can also write it as theta = s/r indicating that if r increases with an increasing distance, s also has to increase for theta to be the same. Later today I will do some calculations moving the observer away from the surface of earth. This would provide me a certain surface on earth as a blocker which can be compared to the blocker surface on the opposite side of earth. This way I can compare them and figure out how both values increase if the observer is moving away from the surface.
5. ## I really need your help with some calculations

But we are just saying things and I would need equations to calculate these shadows.
6. ## I really need your help with some calculations

If I want to block the whole star from all of earth half way to Proxima Centauri, I might need a star about half the size of Proxima Centauri.
7. ## I really need your help with some calculations

But the size of the object needed depends on the distance where the object is located. How do you calculate the size according to the distance?
8. ## I really need your help with some calculations

Yes, of course, this is obvious. But if I am standing on earth, I only see a dot in the sky. But this dot means a certain circle on the opposite side of earth, a circle of a certain size that is necessary to completely block that beam. Imagine if I am standing on a mountain and asking somebody to climb on another mountain about 100 kilometres away, I can see at the horizon and that Alpha Centauri will be visible just a few meters above that mountain from my perspective. So I am asking him to block Alpha Centauri for me. What is he going to use? A coin? A big sign? A house? But even worse, I want him to create a shadow to Alpha Centauri that has the size of 1 square meter in my position. How do I calculate that?
9. ## I really need your help with some calculations

I think you misunderstood my questions. I am not questioning how eclipses work. I am just trying to find a solution to my problem with my calculations. But as we can see in your images, the sun is much bigger than the moon and the moon is much bigger than its shadow on earth. Therefore, the size of an object at a distance of the diameter of earth should be much bigger than our thumb to cover Proxima Centauri, but how big should it be?
10. ## I really need your help with some calculations

In that case how do you explain the existence of solar eclipses if the diameter of the moon is 3,474.8 kilometres and the diameter of the sun 1,392,700 kilometres? And even worse, how do you explain that the solar eclipe is only visible on a line of just a few hundred kilometres of width on Earth considering the size of the moon?
11. ## I really need your help with some calculations

If you can cover the moon with your thumb, how is this image possible?
12. ## I really need your help with some calculations

But if the light coming from the star was perpendicular, the star would have the size of just a few millimetres and that is not the case. The further away we are, the smaller we see the star. At a further distance, the angle of the visible light coming from the star becomes smaller.
13. ## I really need your help with some calculations

That´s the problem. I need to go from a point to an area or distance. Maybe I can also ask you the question: How big must an object be so at a distance of the diameter of earth it is able to block the light coming from Proxima Centauri so it creates a shadow that is just big enough to cover an object of the size of 1 metre, 2 metres, 10 metres? If you consider the trajectory of the light from Proxima Centauri as parallel, that would mean that a coin would block the light and that is not the case.
14. ## I really need your help with some calculations

I am sorry, what I mean is that I need help and that I am not interested in "defending my point". I just need a solution to this problem.