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Yes, I understand what you mean. I will try to create an equation because I actually need the equation. It´s just part of a personal project. If I will not have a good equation during the following weeks, I will try to use Blender for this to get an approximation. In a worst case scenario, I might just have a chart with a list of different distances d and the corresponding values for the distance between the intersections. At least we know that a is -b/d and the distance from the vertex to the intersection with the sphere is a little more than 4 and it becomes less if the point p gets closer to the sphere because the angle would increase, not only because the distance would decrease.
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The problem is that all of these volumes still have an unknown value for "b" that still has to be calculated. This b does not only modify the slope of the negative equation, it also moves the points of the intersections on the x axis. I actually would prefer the formula, but if a computer program would help to get that formula, that would be great. Unfortunately, I am not a programmer to do that.
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Yes, it should be something like the double integral of the unknown equation minus the double integral of the sphere between -r and the second intersection minus the double integral of the sphere between the first intersection and +r is equal to 2/3 pi r^3, where the equation is -(b/d)x + b and actually what I am looking for of the equation is b which is the value of the equation at x0. d is the distance between x0 and the point P.
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Yes, I will have to study this during the weekend. How I can apply this to my problem. Thank you! I hope I can solve it (or at least get close to a solution).
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Yes, that makes sense. But how do I express it when the ground and roof of that cone is a dome? And the angle of that cone still depends on the position of point P. At first I thought that this would be easy using the tan functions, considering that the cone would always hit the y axis at the same height, but not even that is the case. Even the height at the y axis goes up very little as point P gets closer to the center of the sphere.
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Hello, I am currently dealing with a mathematical problem I am still not able to resolve: Imagine there is a sphere with radius 1 and the center is just in the middle of the coordinate system, so the equation is (x - 0)^2 + (y - 0)^2 + (z - 0)^2 = r^2 And there is a point with the coordinates (5, 0, 0) which is the tip of a cone that is directed directly towards the center of the sphere till the end of the sphere. What I now want to calculate is the exact angle this cone needs so that the volume it occupies within the sphere is exactly half of the sphere. And I need to figure out how long the line between the first intersection of the cone when hitting the sphere for the first time and the end of the cone on the other side of the sphere (second intersection) is. Here I have a little drawing to better explain what I need: The problem with this is that the remaining part of the sphere that is not within the cone is actually a ring going around the cone, making the calculation much more difficult. I know that the further away the point P is from the center, the closer the distance between both intersections gets to exactly 2r. And this distance decreases very little to values from 2 to 1,6 when the point P gets closer to the sphere. Another problem is that the area where the cone is touching the sphere is not even flat, so this would even include the calculation of two domes on both sides. Please, any help will be highly appreciated!
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I have lived in Argentina for many years and I can tell you of several very particular circumstances for Argentina: 1. During the 90s, president Carlos Menem established the famous "uno a uno" which means that one dollar was worth one peso and this exchange rate was maintained until December 2001. The problem is that at the beginning this was stable but during the year 2000 and 2001 with President De la Rua a lot of Argentinians converted their pesos into dollars and opened bank accounts in dollars instead of pesos. This created an excess of pesos and a shortness of dollars, creating a crisis that finally, in December 2001, forced the government to freeze all bank accounts in dollars for at least 6 months. Then they devaluated the peso from one dollar one peso to one dollar three pesos and they devaluated these bank accounts converting them first to pesos and than devaluating them. When people finally had access to their original dollars, their money was now worth only one third. This maneuver somehow fixed the debt of Argentina, but this created a huge distrust since then so Argentinians never again put their money in the bank, at least not more than for example 1000 dollars. Instead, they have all their money in cash dollars somewhere hidden at home or in a safe-deposit box. So this is some kind of cultural thing now that increases the demand for dollars and always makes it difficult for a new government to control inflation. Maybe there are more cash dollars in Argentina than in the U.S.! I would estimate some 500 billion US dollars in cash in Argentina. 2. The second problem is that people in Argentina on purpose do not have access to foreign investments, they cannot open a bank account and for example buy stocks at Wall Street. Instead, if you ask them, the best investment you can have is "to buy dollars". So, Argentinians are always expecting the value of the dollar to increase so their money is somehow "working". This is quite similar to the Bitcoin actually, where everybody is buying only because they believe that the value will increase, but there is actually no reason why the value should increase, but the government for many years was like forced to print more money to "increase" the value of the dollar, but what they actually did is only decreasing the value of the peso. But a lot of people do not realize that. 3. Another important point is that all properties in Argentina are for sale only in dollars. This has to do with the circumstance that sellers would never trust in the local currency and as they do not want to lose money, they all want to sell in dollars. You must understand that for example if you offer your home for sale for 100 million pesos (100k dollars) it´s possible that while you are in the middle of the transaction (which usually lasts 3 months) the peso can drop and at the end when you receive your 100 million, these might only be worth 60k dollars. Anyway, properties in Argentina are usually highly overvalued. But it is important to note that Javier Milei now ended this crazy inflation period and since last year the peso is quite stable. But instead there are now a lot of people suffering and even starving because they are no longer receiving "help" from the government. Please ask me further questions about Argentina. I have lived there between 1997 and 2017 and real experiences are sometimes much better than books or can complement them.
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I really need your help with some calculations
tmdarkmatter replied to tmdarkmatter's topic in Astronomy and Cosmology
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. -
I really need your help with some calculations
tmdarkmatter replied to tmdarkmatter's topic in Astronomy and Cosmology
You should give me some points for my reputation, because at least I did not offend anybody this time. -
I really need your help with some calculations
tmdarkmatter replied to tmdarkmatter's topic in Astronomy and Cosmology
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. -
I really need your help with some calculations
tmdarkmatter replied to tmdarkmatter's topic in Astronomy and Cosmology
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. -
I really need your help with some calculations
tmdarkmatter replied to tmdarkmatter's topic in Astronomy and Cosmology
But we are just saying things and I would need equations to calculate these shadows. -
I really need your help with some calculations
tmdarkmatter replied to tmdarkmatter's topic in Astronomy and Cosmology
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. -
I really need your help with some calculations
tmdarkmatter replied to tmdarkmatter's topic in Astronomy and Cosmology
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? -
I really need your help with some calculations
tmdarkmatter replied to tmdarkmatter's topic in Astronomy and Cosmology
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?