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Astonishingly, the solution to the problem is similar to that of calculating the angular sizes. After focusing on the problem I realized that: Center = Real Center / Depth. The real center is the center of the object at distance = 1. The same way: Angular Size = Real Size / Depth The real size is the size of the object at distance = 1. Center2 = ...

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This of course has been solved many years ago in computer graphics. Look up something called perpsective projection. The simplest and most common way to do this is a point to plane projection. All parts of the scene are projected to a single point, usually called the view point or the eye point. You conceptually suspend the image plane at a fixed ...

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Observer, if you assume as someone watching the experiment or activity, is plainly wrong. Anything that can be detected and measured and thus, in principle, from infinitely hard calculation can tell us about the past or previous states, can be said to be information, and thus, entropy increased while the process was being carried out. Take for example, a box ...

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Where's the flaw? Here's one: Let γ be the squareroot of 1−v²/c² but, in fact, $$\gamma_v = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$ But, there is a flaw in your reasoning too so correcting the error in the formula for $\gamma$ will not get you to the correct answer. Let's work it out with coordinates to see explicitly what's going on. Let the ...

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I don't have the reputation to comment, so I'll comment about David Hammen's (accepted) answer here. His conclusion is correct, but mentions "Those formulae do imply a singularity for the clock that is closest to you" in reference to the forumale he thought you were referring to. He also mentions "In between, you'll get a nice continuous change from faster ...

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I am assuming we are talking about the one dimensional case in which we can "move through" clocks. Assuming you synchronise the clocks in your own frame, those further away from you will show an older time than those closer because it takes longer for light to reach you from them. On top of that, all clocks will be equally time dilated since they are all ...

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I assume you used the formulae $f_o = fs\sqrt{\frac{1+v/c}{1-v/c}}$ for the clocks ahead of you and $f_o = fs\sqrt{\frac{1-v/c}{1+v/c}}$ for the clocks behind you. Those formulae do imply a singularity for the clock that is closest to you. Which equation to use? The answer is neither. Those expressions assume the travel is along the line of sight to the ...

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If you consider a straight-on trajectory, there really will be a discontinuity. That is the same as with the audible Doppler effect. There is a smooth drop in frequency of a fire truck's siren passing you on the street. The reason is, that there is a certain distance between you and the truck at the closest point. If that were not the case, i.e. the siren ...

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The signal from the clock moving towards you is the Doppler shifted version of the value you "know" it to be - that is, first slow it down by gamma (clock moving relative to your frame of reference), then speed it up by Doppler shift. Ditto, with sign reversed, for clocks behind you. Now the clock moving at right angles shows what you expect and there is no ...

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