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ralfcis

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Everything posted by ralfcis

  1. It seems that Willo is using a math method I have never seen before and am wondering if anyone else has. @Willo I've decoded what you tried to tell me into this Md. photos.app.goo.gl/JA8iwFX7yboAvYfdA. I've never seen this format and it's not used in popsci relativity. This is the format used photos.app.goo.gl/Cp7FjpQ3Rmhp9VYB6 t'=t/Y and t''=t'/Y=t/YY. The ' represents how many times gamma is used. t(E)=t′(E)=t′′(E)=0. t(F)=5, t′(F)=4, t′′(F)=3.2 from Alice's perspective outbound to t''(F)= 6.8 from Alice inbound. t(G)=10, t′(G)=8, t′′(G)=10. We don't use the same language so you'll never
  2. Arithmetic is not my forte but did you make one of the velocities negative? It doesn't seem like it could apply to anything besides 90 degrees but why would anyone develop a method that only solves one rare example? I think I just don't fully understand how to use his method. Here's what I'm trying to do. All perspective realities are illusions of perspective. Einstein says they're all equally real. I base relativity on a single proper time and space reality that is accessible through the Loedel perspective which is the half speed perspective. So for .6c, it's 1/3c. Then I show how relativi
  3. So, Popeye, you won't take me on because I'm a denier? My math is granite, no, diamond, it's that rock solid, not soft like BS but no one wants to take the trouble to understand it.
  4. Let's check if it works for a 45 degree vB . I assume the vector of w will be 45 degrees to vA. I see no justification for this assumption but let's check it out against the longform equation. So w2 = .92 -.3922 = .6563 so w=.81c Oops should have seen the angle can't be 45 if one of the sides is .392 and the other is .9. The longform is w = sqrt (u2+v2+2uv [email protected] - (vu [email protected]/c)2) / (1+vu [email protected]/c2) w= sqrt(.92+.92 + 2*.9*.9 cos45 - (.9*.9 sin45)2) /(1+.9*.9cos45) cos45=sin45=.707 w=sqrt( 1.62 + 1.145 - .328)/ 1.5727 =.9926c which is definitely the wrong answer. The longform answer is to
  5. So it looks like this example occupies 2 space dimensions and its projection on the linear dimension cuts off some of vB due to a trick of dividing by YA. I've never needed to consider more than 1 space dimension before. Maybe that guy Janus can explain further.
  6. If by "double" value you mean an extra Y that is not expected, I've noticed this in the formula (ct')2 = (ct)2 - x2. This formula is valid from the stationary perspective but not from the moving perspective. From the moving perspective the formula is (ct'')2 = (ct')2 - (x/Y)2. Use v=.6, t=5, t'=4, t''=3.2, Y=1.25 to check it out. The morons on the PSX would have no idea what I just said here.
  7. Ok good that's an honest answer. I've definitely learned something new, velocity dilation and velocities have perspective. I'm going to work that into my own theory which is totally based on the magic properties of Yv, not v like Einy's theory.
  8. But Victor, you don't understand. Victor, you're one of the cranks. Anyway here's something I don't understand. Instead of that big formula I used to solve the problem. I looked at that thread Tony provided on the sciforums and Janus has a different way of solving it. Third scene We will look at it from the rest frame for A where you get: 0.9c <--Earth-----------------A | | | \/ B Since the Earth-B pair share a 0.9c to the left velocity, this pair is time dilated by a factor of 0.436 as measured by A. This includes the vertical speed of B with respect to the Earth. That means that B ve
  9. You seem to be perplexed by how a minus sign is related to direction and how it can give such a different final value than a plus sign for the same velocity. When you get into the Doppler Shift Ratio, you'll see a minus sign acts like a division sign. This craziness doesn't exist in Newtonian physics so you believe relativity has made up crazy math that gives fake values to support its fake theory. Give up Tony. You just don't have the talent for this. No amount of explanation will ever convince you to abandon your ignorant beliefs.
  10. What are you talking about? What game are you playing over and over opening multiple threads on the same subject. Someone gave you a formula and it's like you discovered that if you put different values into it you get different answers. Are you then concluding relativity does not consistently produce the same answer every time?
  11. "On the contrary, you are looking for excuses to NOT understand, in order to keep your idee fixe alive." Yeah like all relativists who blindly believe in Einstein's theory and do not possess the intelligence to truly understand it. The vast majority of people on physics forums are compulsive liars, the mentally ill and total morons incapable of reason. I've only met 2.5 who weren't but I always get banned before I can mine them for true knowledge.
  12. I said that, what's your point? Can you convert the formula into parallel form or not?
  13. Stop lying Victor. Your different way was not even close. It didn't even make the slightest bit of sense. You didn't even know an angle was required. Of course I had to look it up, it was on the link you gave me. Do you think I know everything. The difference is I don't pretend to. I'll bet you still have no clue how to use the formula. Can you even convert it into normal form? Do you know what the normal form is?
  14. Victor, baby, the formula was in your Wiki link. Please don't compliment me any more on my progress through relativity as you clearly have no grasp of even your own wiki articles. It's embarrassing for me to be promoted by people who know nothing. I'll just give you and the rest of the effort impaired and intellectually challenged on here the formula: w = sqrt (u2+v2+2uv [email protected] - (vu [email protected]/c)2) / (1+vu [email protected]/c2) @= 90 degrees so [email protected]=0 and [email protected]=1 u=v=.9c answer w=.9817c which is the relativistically combined u=v=.9c leaving at right angles from earth. See that answer is less than .9945c
  15. What's your numerical answer then Victor. Changing Yx to 1/X and the rest of your superficial lipstick isn't going to get you a correct answer. Show me.
  16. Are you just taking wild guesses now? How is this older form: γx = 1/(1- Vx2/C2)1/2 γy = 1/(1- Vy2/C2)1/2 then γx2 + γy2 = γf2 Then reverse the equation γf = 1/(1- Vf2/C2)1/2 Any different from your new form: X = (1- Vx2/C2)1/2 Y = (1- Vy2/C2)1/2 then x2 + y2 = r2 , r = (1- Vf2/C2)1/2 x2 + y2 +z2 = r2 You've never done math before have you. You can't read the link I gave you, can you. It's time to give up Victor. Maybe one of the other math frauds can step up. Take your time. Just translate the link I gave you into the much simpler problem of two ships leaving earth at
  17. I don't know this notation but I'm pretty sure u is a velocity and you're not using the formula correctly because your answer is wrong. Wiki is useless if you don't understand what you're flashing. Show me a formula that gets you the right answer and I'll translate it into a simple algebraic notation I understand. I'm not into equation porn. I've given you a link to an example, don't make me use your crank stamp on you.
  18. So I make a lot of mistakes with arithmetic but the answer I get is .9955c. This number happens to be greater than the .9945c of them separating at 180 degrees. Your answer can't be correct as the speed addition at 180 degrees must be less than at 90 degrees because the vector is smaller. I don't understand how you can be using gammas to calculate anything about velocity combination. γx2 + γy2 = γf2 means nothing. Even if you used velocities instead of gammas, it would still mean nothing. Try googling this question, I found an example. See if you can decipher it and apply it to this problem.
  19. Ok so you're just as much of a math fraud as all the other math frauds on this forum. Just man up to your ignorance. I don't think relativity is wrong, I think Einstein's theory is wrong but my answers don't disagree with his, only my math method does. Two .9c ships leaving earth at right angles to each other, what is their relative velocity to each other? How hard could this be for a non-math fraud? Get 006, P8ly and exchemist in on this but not Popeye because he would just give you guys the answer.
  20. A) Still don't see the gamma multiplication version of the velocity combo law. First you say addition and now it's a relativistic multiplication of gammas. I'm not familiar with that form. Show it please. Here's a form I like: DSRw =DSRv *DSRu where w is the velocity combination of u and v velocities using multiplication of Doppler Shift Ratio formulas. :cool: This is your go-to smoke bomb. There is no need to go to 3 spatial dimensions for the examples given so don't try to convince me there's a special answer there. C) I'm just asking a question. Do you have an answer or are you just
  21. It's called math salad. People do that to hide they don't know what they're talking about and count on the fact no one is math literate enough to check it.
  22. Victor I initially thought you were using some kind of new equations but your math makes no sense. You use gammas and not velocities? Ya = Yb =2.3 At first you say "The answer is your add the gamma's to find the speed " then you multiply the gammas contradicting what you said Ya * Yb = 5.26. Neither is the correct answer .9945. Tony is right. Then you try to hide you don't know how to calculate the third scenario by giving the same answer of .9945 and finally admit trig is necessary but don't show how to use it. From what you're inferring, the answer would be sqrt(.92 + .92) = .9
  23. No way to check out this math because it's in some kind of weird format and there's almost no verbal explanation of what it's doing.
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