Not until I looked it up. OK Does that have something to do with why you copy others work and then do some minimal algebra and pretend it is your work? I don't think that is what is happening here. Since you and Reiku are the same person, I think the answer has to be yes. Glad your life is excellent. You don't seem to ever discuss it on here in any of your different sock puppets. Oh gee, lots. But none are a big deal - nobody will give a shit about them in 100 years.
Please Register or Log in to view the hidden image! Such honest answers, they have to be true. Quick, send for the cavalry and have my account deleted at once! Please Register or Log in to view the hidden image!
So add this to the list of things that I'm the first to do. Did you work out the areas of the different parts in my diagram yet? Do you know the length of the various lines? Does it all add up perfect using 3.14?
The complexity of the lines, will involve very difficult mathematics... Certainly it doesn't add up to the quantity given in the previous diagram since it wasn't extended past a two-dimensional image.
They each have 3.14 square units of area. The circle has a radius of 1 unit, and the square has sides of \(sqrt{3.14}\) units.
There are more components than that... I can't even fathom them at the moment. The circle for instance has a simplicity of a radius equal to 1,the square of the sides may then be \(\sqrt{3.14} + \epsilon\) in which \(\epsilon\) is a continuous function, if it isn't, it quickly becomes the famous approximation to \(3.14159\)... either way fits really. You can't apply trigonometry to a spherical system without perturbations not accounted for, like affine angles? Extend a two dimensional sphere, extended by it's symmetry on a three dimensional hypersphere, you have something much more bizarre than your ''static background lines.'' You have to start thinking Tessaract- style. ... ... and time is one of those things, which should be like politics.... should never be discussed in public, but often is.
The area of the parts must add to 3.14 square units of area. MUST! Pythagorean theorem MUST hold true at ALL times!
MD, Please Register or Log in to view the hidden image! Assuming these circles have a radius of 1.000, please tell me the lengths of these blue lines. Thanks.
Which axis? I can do the 3d thing, remember? Remember how component velocities don't equate to the right coordinate in 3d? You taught me well, Neddy!! Thanks!!
I sound like the borg queen when I say this (lol) but you are not thinking four dimensionally. Am I right in thinking you want to impose ''time'' on this image? If so, you do realize the image presented is an incomplete version of 3-dimensions, the background lines don't curve into the pi-lines according to symmetry.
I can tell you when and where an expanding light sphere hits a moving point that's traveling at any speed and direction. In other words, I can tell you the exact time light will travel from any point in space to another point in space, 3 dimensionally, and tell you the coordinate that it impacts that point at!! Right Neddy?
If time exists then yes, there will be an symptotic time relative to ''large bubbles in space.'' These bubbles represent the boundaries at which GR breaks down, a subject I spoke about not long ago. But there is a catch... that is that maybe time doesn't exist and therefore neither can these relative bubbles. Instead, the universe acts on us according to the rules of Machian relativity, where all systems relative to us in the coordinate frame of the universe, are static to each moment in it's relative changes. It's only static because the changes can be symmetrical, a dilation ''here'' is accounted by a dilation ''there'' even in the quantum scale.
Along the axis that is parallel to the blue line. I want to know the length of the blue line itself, not any other measurement.
Let's see... we have... \(1.... 2... 4 ^2 = 8\) dimensions on his axis which makes the ''innards'' of the circle. What frame did we have in mind to mix?
The reason I want to know the length of the blue line is because I am going to apply the Pythagorean Theorem to this right triangle... Please Register or Log in to view the hidden image! ...which was obtained by bisecting the equilateral triangle in this image: Please Register or Log in to view the hidden image! Get it now?
