Thanks for your comment. However, I don’t know whether your name is Ashli or Mathew. I find it kind of odd that the web page you include is to a profile of someone named Mathew, but the email contains Ashli. Anyway, on to your question.

I haven’t specifically set up a Twitter account yet for Randosity because 160 character limit isn’t really enough to write an article of any value. I also haven’t set up a Facebook page because that’s like setting up a second blog site and I already have plenty of work around here to manage WordPress. I’m also not thrilled by all of the social and privacy issues surrounding Facebook. In fact, I might write a Randosity article on this very topic.

You can always follow my blog right here by subscribing. Once subscribed, you will receive a notification each time I write a new article. The subscription area is to the right of this article in the side panel.

Thanks for commenting.

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]]>Could it be simply me or do a few of these responses come across like they are written by brain dead folks?

:-P And, if you are writing on other online social sites, I’d like to keep up with you. Could you post a list of the complete urls of your shared sites like your twitter feed, Facebook page or linkedin profile?

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]]>Regarding time travel CERN seems to have the answers and its only a matter of time and of course travel . The answer lies within

Mick

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]]>If I’m understanding the DI Herculis precession issues correctly, apparently the visual observations can be more simply explained as that neither Einstein’s nor Newton’s corrected equations took into account negative precession. That is, their equations assumed that the precession would continue in a positive numeric amount. As This Article states:

“To the rescue may come, they say the “nonsymmetric gravity theory” of John W. Moffat of the University of Toronto (SN: 9/3/83, p. 152). Unlike Einstein’s theory, Moffat’s permits backward precession. The Moffat correction for DI Herculis is -1.40[deg.] per century. Combining that with the classical value yields 0.53[deg.] per century, very close to the observation.”

Whether or not this binary star system is actually exhibiting negative precession cannot be known. So, these are obviously just theories. Note, I am not a mathematician, so I have no idea if your formulas describe this situation correctly or not. I am also not exactly sure how this binary star discrepancy relates to John Titor or time travel… or how that (your?) formula will break open ‘Einstein’s space-jail of time’ … but, it’s definitely a random thought and fits well with this blog.

In the spirit of randomness, thanks for your post.

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]]>The problem that the 100,000 PHD Physicists could not solve

This is the solution to DI Her “Quarter of a century” Smithsonian-NASA Posted motion puzzle that Einstein and the 100,000 space-time physicists including 109 years of Nobel prize winner physics and physicists and 400 years of astronomy and Astrophysicists could not solve and solved here and dedicated to Drs Edward Guinan and Frank Maloney

Of Villanova University Pennsylvania who posted this motion puzzle and started the search collections of stars with motion that can not be explained by any published physics

For 350 years Physicists Astrophysicists and Mathematicians and all others including Newton and Kepler themselves missed the time-dependent Newton’s equation and time dependent Kepler’s equation that accounts for Quantum – relativistic effects and it explains these effects as visual effects. Here it is

Universal- Mechanics

All there is in the Universe is objects of mass m moving in space (x, y, z) at a location

r = r (x, y, z). The state of any object in the Universe can be expressed as the product

S = m r; State = mass x location

P = d S/d t = m (d r/dt) + (dm/dt) r = Total moment

= change of location + change of mass

= m v + m’ r; v = velocity = d r/d t; m’ = mass change rate

F = d P/d t = d²S/dt² = Force = m (d²r/dt²) +2(dm/d t) (d r/d t) + (d²m/dt²) r

= m γ + 2m’v +m”r; γ = acceleration; m” = mass acceleration rate

In polar coordinates system

r = r r(1) ;v = r’ r(1) + r θ’ θ(1) ; γ = (r” – rθ’²)r(1) + (2r’θ’ + rθ”)θ(1)

F = m[(r”-rθ’²)r(1) + (2r’θ’ + rθ”)θ(1)] + 2m'[r’r(1) + rθ’θ(1)] + (m”r) r(1)

F = [d²(m r)/dt² – (m r)θ’²]r(1) + (1/mr)[d(m²r²θ’)/d t]θ(1) = [-GmM/r²]r(1)

d² (m r)/dt² – (m r) θ’² = -GmM/r²; d (m²r²θ’)/d t = 0

Let m =constant: M=constant

d²r/dt² – r θ’²=-GM/r² —— I

d(r²θ’)/d t = 0 —————–II

r²θ’=h = constant ————– II

r = 1/u; r’ = -u’/u² = – r²u’ = – r²θ'(d u/d θ) = -h (d u/d θ)

d (r²θ’)/d t = 2rr’θ’ + r²θ” = 0 r” = – h d/d t (du/d θ) = – h θ'(d²u/d θ²) = – (h²/r²)(d²u/dθ²)

[- (h²/r²) (d²u/dθ²)] – r [(h/r²)²] = -GM/r²

2(r’/r) = – (θ”/θ’) = 2[λ + ỉ ω (t)] – h²u² (d²u/dθ²) – h²u³ = -GMu²

d²u/dθ² + u = GM/h²

r(θ, t) = r (θ, 0) Exp [λ + ỉ ω (t)] u(θ,0) = GM/h² + Acosθ; r (θ, 0) = 1/(GM/h² + Acosθ)

r ( θ, 0) = h²/GM/[1 + (Ah²/Gm)cosθ]

r(θ,0) = a(1-ε²)/(1+εcosθ) ; h²/GM = a(1-ε²); ε = Ah²/GM

r(0,t)= Exp[λ(r) + ỉ ω (r)]t; Exp = Exponential

r = r(θ , t)=r(θ,0)r(0,t)=[a(1-ε²)/(1+εcosθ)]{Exp[λ(r) + ì ω(r)]t} Nahhas’ Solution

If λ(r) ≈ 0; then:

r (θ, t) = [(1-ε²)/(1+εcosθ)]{Exp[ỉ ω(r)t]

θ'(r, t) = θ'[r(θ,0), 0] Exp{-2ỉ[ω(r)t]}

h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity

h = 2πa²√ (1-ε²); r (0, 0) = a (1-ε)

θ’ (0,0) = h/r²(0,0) = 2π[√(1-ε²)]/T(1-ε)²

θ’ (0,t) = θ'(0,0)Exp(-2ỉwt)={2π[√(1-ε²)]/T(1-ε)²} Exp (-2iwt)

θ'(0,t) = θ'(0,0) [cosine 2(wt) – ỉ sine 2(wt)] = θ'(0,0) [1- 2sine² (wt) – ỉ sin 2(wt)]

θ'(0,t) = θ'(0,t)(x) + θ'(0,t)(y); θ'(0,t)(x) = θ'(0,0)[ 1- 2sine² (wt)]

θ'(0,t)(x) – θ'(0,0) = – 2θ'(0,0)sine²(wt) = – 2θ'(0,0)(v/c)² v/c=sine wt; c=light speed

Δ θ’ = [θ'(0, t) – θ'(0, 0)] = -4π {[√ (1-ε) ²]/T (1-ε) ²} (v/c) ²} radians/second

{(180/π=degrees) x (36526=century)

Δ θ’ = [-720×36526/ T (days)] {[√ (1-ε) ²]/ (1-ε) ²}(v/c) = 1.04°/century

This is the T-Rex equation that is going to demolished Einstein’s space-jail of time

The circumference of an ellipse: 2πa (1 – ε²/4 + 3/16(ε²)²—) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)

v (m) = √ [GM²/ (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<<M; Solar system

v = v (center of mass); v is the sum of orbital/rotational velocities = v(cm) for DI Her

Let m = mass of primary; M = mass of secondary

v (m) = primary speed; v(M) = secondary speed = √[Gm²/(m+M)a(1-ε²/4)]

v (cm) = [m v(m) + M v(M)]/(m + M) All rights reserved. joenahhas1958@yahoo.com

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