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Oct 29

Time Machine 2: Space Time Machine

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I grabbed a new 1TB drive to put Time Machine backups and my media on. I can’t believe that you can get 1TB for just a couple of hundred bucks. Wow.

I really like what Time Machine is trying to be. Running backups has been a royal pain for the longest time. I know that there are other packages out there, and tar can even be your friend, but I have seen constant pain from non-technical family trying to get backups sorted out.

Being able to open up Time Machine and see the Star Wars time dimension is great. Jumping around in time is going to be a real saver. However, Time Machine still has problems:

a) California Fires

It is great to backup your system. If something messes up on your machine, rebuilding will be a breeze. But what if something happens to your location. Seeing your computer and backup drive go up in fumes is a nightmare. There are two solutions here:

  • Off site: Have multiple drives and slave copy the main drives and take them offsite. Make sure that you don’t forget for a few months at a time (gulp)
  • On line: Have the data stored in the sky. Maybe you use a hack on top of S3 (make sure to diff so you don’t send huge amounts all the time). .Mac is another choice for Apple folk, but the limits are poor. The future is surely in the “data in the sky” approach though.

b) Space Time

Backing up the data on your main machine is simple. The problem that I have is that the data that I have can not fit on my machine anymore. If I take my video, music, source code, and all of the other assets I need a lot of space. Too much for my laptop hard drive.

I need a solution that maps to space as well as time. I want to be able to selectively tag data as “this goes on my laptop” and “this can go on my mac pro” etc. I need the system to understand that I have a lot of data, and a subset is sync’d for me. My use case seems to break a lot of the tools, and means that I have custom scripts to manage it all. And, it is a pain.

So, I am looking forward to Space Time Machine, a solution that takes this all into account.

4 Responses to “Time Machine 2: Space Time Machine”

  1. Sam Says:

    It already is a space-time machine. Just delete files you no longer want on your machine after they have been backed up on your 1TB disk. Then later when you want them, go and restore them…

    Though thinking this through, it would be cool to have iTunes like sync for laptops connected to desktop machines where you select the folders and items you want synced each time you connect it.

  2. Scott Morgan Says:

    Carbonite is an online backup service I hear about on the radio all the time. I have read a lot about them. They have unlimited space and it is $50 per year. Unfortunately they do not have a mac client yet but if you are using Windows you can give it a try for free for 15 days. They say the mac version is coming soon, i’ve heard that before.

    http://www.carbonite.com/

  3. joe nahhas Says:

    Einstein’s Physics Dollar Store on Campus
    MIT Harvard Cal-Tech
    Sponsored by NASA
    Why Relativity theory is not Physics and why Einstein’s “thought” = 0
    Walking the walk and talking the talk taking on all space-time confusion of physics by
    MIT Harvard and Cal-Tech and all other Physics dollar stores departments

    Visual Effects and the confusions of “Modern” physics

    r ————————- Light sensing of moving objects ——————————- S
    Actual ———————————– Light ———————————————- Visual
    Object at r ———- Light aberration = cosine (wt) + i sine (wt) —– S = r [cosine (wt) + i sine(wt)]
    Newton space particle ——— Kepler’s time dependent visual effects —- Time dependent Newton
    Particle ——————————– Visual effects —————————– Wave

    Line of Sight: rcosine wt

    r ————————————————————————- r cosine (wt) due light aberrations

    B
    A moving object with velocity v [C-B] will ^
    meet light at B hypotenuse . ^ Object
    sine wt = v/c light . ^ Velocity
    cosine wt = √ [1-(v/c)²] Velocity . ^ v
    c . Angle ^
    .A= wt ^
    . . . . . . .
    A c √ [1-(v/c) ²] C

    S = r [cosine (wt) + i sine (wt)] = r Exp [i wt]; Exp = Exponential

    Also, sine ω(r) t= v/c; cosine ω t = √ [1-sin²ω(r) t] = √ [1-(v/c) ²]

    S = r [√ [1-(v/c) ²] + ỉ (v/c)] = S x + i S y

    S x = Visual along the line of sight = r [√ [1-(v/c) ²]
    This Equation is special relativity length contraction formula and it is just visual effects along the line of sight.

  4. joe nahhas Says:

    Modern Physicists: Legacy of incompetence

    Mercury’s “apparent” advance of perihelion of 43″ seconds of an arc per century

    By Professor Joe Nahhas

    Abstract: This is the solution to the 150 years old “apparent” Advance of Perihelion motion puzzle solution using Newtonian mechanics and not space-time confusions of physics. Planetary advance of perihelion or rate of orbital axial rotation is visual effects along the line of sight of moving objects applied to the angular velocity at perihelion. This rate of “apparent” axial rotation is given by this new equation
    W° (ob) = (-720×36526x3600/T) {[√ (1-ε²)]/ (1-ε) ²]} [(v° + v*)/c] ² seconds/100 years
    T = period; ε = eccentricity; v° = spin velocity effect; v*= orbital velocity effect
    For Mercury: v* = 48.14km/s
    And v° = 0.002km/s = spin velocity of Planet

    Universal Mechanics Solution:

    For 350 years Physicists Astronomers and Mathematicians and philosophers missed Kepler’s time dependent Areal velocity wave equation solution that changed Newton’s classical planetary motion equation to a Newton’s time dependent wave orbital equation solution and these two equations put together combines particle mechanics of Newton’s with wave mechanics of Kepler’s into one time dependent universal mechanics equation that explain “relativistic” as the difference between time dependent measurements and time independent measurements of moving objects and in practice it amounts to light aberrations along the line of sight of moving objects

    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/d t) + (dm/d t) 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 = Total 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’ + r θ”)θ(1)
    r = location; v = velocity; γ = acceleration
    F = m γ + 2m’v +m” r
    F = m [(r"-rθ'²) r (1) + (2r'θ' + r θ") θ (1)] + 2m’[r' r (1) + r θ' θ (1)] + (m” r) r (1)
    = [d² (m r)/dt² - (m r) θ'²] r (1) + (1/mr) [d (m²r²θ')/d t] θ (1)
    = [-GmM/r²] r (1) ——————————- Newton’s Gravitational Law
    Proof:
    First r = r [cosine θ î + sine θ Ĵ] = r r (1)
    Define r (1) = cosine θ î + sine θ Ĵ
    Define v = d r/d t = r’ r (1) + r d[r (1)]/d t
    = r’ r (1) + r θ’[- sine θ î + cosine θĴ]
    = r’ r (1) + r θ’ θ (1)

    Define θ (1) = -sine θ î +cosine θ Ĵ;
    And with r (1) = cosine θ î + sine θ Ĵ

    Then d [θ (1)]/d t= θ’ [- cosine θ î - sine θ Ĵ= - θ' r (1)
    And d [r (1)]/d t = θ’ [-sine θ î + cosine θ Ĵ] = θ’ θ (1)

    Define γ = d [r' r (1) + r θ' θ (1)] /d t
    = r” r (1) + r’d [r (1)]/d t + r’ θ’ r (1) + r θ” r (1) +r θ’d [θ (1)]/d t
    γ = (r” – rθ’²) r (1) + (2r’θ’ + r θ”) θ (1)

    With d² (m r)/dt² – (m r) θ’² = -GmM/r² Newton’s Gravitational Equation (1)
    And d (m²r²θ’)/d t = 0 Central force law (2)

    (2): d (m²r²θ’)/d t = 0
    Then m²r²θ’ = constant
    = H (0, 0)
    = m² (0, 0) h (0, 0); h (0, 0) = r² (0, 0) θ’(0, 0)
    = m² (0, 0) r² (0, 0) θ’(0, 0); h (θ, 0) = [r² (θ, 0)] [θ'(θ, 0)]
    = [m² (θ, 0)] h (θ, 0); h (θ, 0) = [r² (θ, 0)] [θ'(θ, 0)]
    = [m² (θ, 0)] [r² (θ, 0)] [θ'(θ, 0)]
    = [m² (θ, t)] [r² (θ, t)] [θ' (θ, t)]
    = [m²(θ, 0) m²(0,t)][ r²(θ,0)r²(0,t)][θ'(θ, t)]
    = [m²(θ, 0) m²(0,t)][ r²(θ,0)r²(0,t)][θ'(θ, 0) θ' (0, t)]
    With m²r²θ’ = constant
    Differentiate with respect to time
    Then 2mm’r²θ’ + 2m²rr’θ’ + m²r²θ” = 0
    Divide by m²r²θ’
    Then 2 (m’/m) + 2(r’/r) + θ”/θ’ = 0
    This equation will have a solution 2 (m’/m) = 2[λ (m) + ì ω (m)]
    And 2(r’/r) = 2[λ (r) + ì ω (r)]
    And θ”/θ’ = -2{λ (m) + λ (r) + ỉ [ω (m) + ω (r)]}

    Then (m’/m) = [λ (m) + ì ω (m)]
    Or d m/m d t = [λ (m) + ì ω (m)]
    And dm/m = [λ (m) + ì ω (m)] d t
    Then m = m (0) Exp [λ (m) + ì ω (m)] t
    m = m (0) m (0, t); m (0, t) Exp [λ (m) + ì ω (m)] t
    With initial spatial condition that can be taken at t = 0 anywhere then m (0) = m (θ, 0)
    And m = m (θ, 0) m (0, t) = m (θ, 0) Exp [λ (m) + ì ω (m)] t; Exp = Exponential
    And m (0, t) = Exp [λ (m) + ỉ ω (m)] t
    Similarly we can get
    Also, r = r (θ, 0) r (0, t) = r (θ, 0) Exp [λ (r) + ì ω (r)] t
    With r (0, t) = Exp [λ (r) + ỉ ω (r)] t

    Then θ’(θ, t) = {H(0, 0)/[m²(θ,0) r(θ,0)]}Exp{-2{[λ(m) + λ(r)]t + ì [ω(m) + ω(r)]t}} —–I
    And θ’(θ, t) = θ’ (θ, 0)]} Exp {-2{[λ (m) + λ (r)] t + ì [ω (m) + ω (r)] t}} ——————–I
    And, θ’(θ, t) = θ’ (θ, 0) θ’ (0, t)
    And θ’ (0, t) = Exp {-2{[λ (m) + λ(r)] t + ì [ω (m) + ω(r)] t}
    Also θ’(θ, 0) = H (0, 0)/ m² (θ, 0) r² (θ, 0)
    And θ’(0, 0) = {H (0, 0)/ [m² (0, 0) r (0, 0)]}

    With (1): d² (m r)/dt² – (m r) θ’² = -GmM/r² = -Gm³M/m²r²
    And d² (m r)/dt² – (m r) θ’² = -Gm³ (θ, 0) m³ (0, t) M/ (m²r²)
    Let m r =1/u
    Then d (m r)/d t = -u’/u² = – (1/u²) (θ’) d u/d θ = (- θ’/u²) d u/d θ = -H d u/d θ
    And d² (m r)/dt² = -Hθ’d²u/dθ² = – Hu² [d²u/dθ²]

    -Hu² [d²u/dθ²] – (1/u) (Hu²)² = -Gm³ (θ, 0) m³ (0, t) Mu²
    [d²u/ dθ²] + u = Gm³ (θ, 0) m³ (0, t) M/ H²

    t = 0; m³ (0, 0) = 1
    u = Gm³ (θ, 0) M/ H² + A cosine θ =Gm (θ, 0) M (θ, 0)/ h² (θ, 0)

    And m r = 1/u = 1/ [Gm (θ, 0) M (θ, 0)/ h (θ, 0) + A cosine θ]
    = [h²/ Gm (θ, 0) M (θ, 0)]/ {1 + [Ah²/ Gm (θ, 0) M (θ, 0)] [cosine θ]}
    = [h²/Gm (θ, 0) M (θ, 0)]/ (1 + ε cosine θ)

    Then m (θ, 0) r (θ, 0) = [a (1-ε²)/ (1+εcosθ)] m (θ, 0)
    Dividing by m (θ, 0)
    Then r (θ, 0) = a (1-ε²)/ (1+εcosθ)
    This is Newton’s Classical Equation solution of two body problem which is the equation of an ellipse of semi-major axis of length a and semi minor axis b = a √ (1 – ε²) and focus length c = ε a
    And m r = m (θ, t) r (θ, t) = m (θ, 0) m (0, t) r (θ, 0) r (0, t)
    Then, r (θ, t) = [a (1-ε²)/ (1+εcosθ)] {Exp [λ(r) + ỉ ω (r)] t} ———————————- II
    This is Newton’s time dependent equation that is missed for 350 years
    If λ (m) ≈ 0 fixed mass and λ(r) ≈ 0 fixed orbit; then
    Then r (θ, t) = r (θ, 0) r (0, t) = [a (1-ε²)/ (1+ε cosine θ)] Exp i ω (r) t
    And m = m (θ, 0) Exp [i ω (m) t] = m (θ, 0) Exp ỉ ω (m) t

    We Have θ’(0, 0) = h (0, 0)/r² (0, 0) = 2πab/ Ta² (1-ε) ²
    = 2πa² [√ (1-ε²)]/T a² (1-ε) ²
    = 2π [√ (1-ε²)]/T (1-ε) ²

    Then θ’(0, t) = {2π [√ (1-ε²)]/ T (1-ε) ²} Exp {-2[ω (m) + ω (r)] t
    = {2π [√ (1-ε²)]/ (1-ε) ²} {cosine 2[ω (m) + ω (r)] t – ỉ sin 2[ω (m) + ω (r)] t}
    And θ’(0, t) = θ’(0, 0) {1- 2sin² [ω (m) + ω (r)] t}
    – ỉ 2i θ’(0, 0) sin [ω (m) + ω (r)] t cosine [ω (m) + ω (r)] t

    Then θ’(0, t) = θ’(0, 0) {1 – 2sine² [ω (m) t + ω (r) t]}
    – 2ỉ θ’(0, 0) sin [ω (m) + ω(r)] t cosine [ω (m) + ω(r)] t

    Δ θ’ (0, t) = Real Δ θ’ (0, t) + Imaginary Δ θ (0, t)
    Real Δ θ (0, t) = θ’(0, 0) {1 – 2 sine² [ω (m) t ω(r) t]}

    Let W (ob) = Δ θ’ (0, t) (observed) = Real Δ θ (0, t) – θ’(0, 0)
    = -2θ’(0, 0) sine² [ω (m) t + ω(r) t]
    = -2[2π [√ (1-ε²)]/T (1-ε) ²] sine² [ω (m) t + ω(r) t]

    If this apsidal motion is to be found as visual effects, then
    With, v ° = spin velocity; v* = orbital velocity; v°/c = tan ω (m) T°; v*/c = tan ω (r) T*
    Where T° = spin period; T* = orbital period

    And ω (m) T° = Inverse tan v°/c; ω (r) T*= Inverse tan v*/c
    W (ob) = -4 π [√ (1-ε²)]/T (1-ε) ²] sine² [Inverse tan v°/c + Inverse tan v*/c] radians
    Multiplication by 180/π

    W (ob) = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} sine² {Inverse tan [v°/c + v*/c]/ [1 - v° v*/c²]} degrees and multiplication by 1 century = 36526 days and using T in days

    W° (ob) = (-720×36526/Tdays) {[√ (1-ε²)]/ (1-ε) ²} x
    sine² {Inverse tan [v°/c + v*/c]/ [1 - v° v*/c²]} degrees/100 years

    Approximations I

    With v° << c and v* << c, then v° v* <<< c² and [1 - v° v*/c²] ≈ 1
    Then W° (ob) ≈ (-720×36526/Tdays) {[√ (1-ε²)]/ (1-ε) ²} x sine² Inverse tan [v°/c + v*/c] degrees/100 years

    Approximations II

    With v° << c and v* << c, then sine Inverse tan [v°/c + v*/c] ≈ (v° + v*)/c

    W° (ob) = (-720×36526/Tdays) {[√ (1-ε²)]/ (1-ε) ²} x [(v° + v*)/c] ² degrees/100 years
    This is the equation that gives the advance of perihelion rates ———————–III

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

    Where v (m) = √ [GM²/ (m + M) a (1-ε²/4)]
    And v (M) = √ [Gm² / (m + M) a (1-ε²/4)]
    From Newton’s laws for a circular orbit: m v²/ r (cm) = GmM/r²; r (cm) = [M/m + M] r
    Then v² = [GM r (cm)/ r²] = GM²/ (m + M) r
    And v = √ [GM²/ (m + M) r = a (1-ε²/4)]
    And v* = v (m) = √ [GM²/ (m + M) a (1-ε²/4)] = 48.14 km

    Advance of Perihelion of mercury.

    G=6.673×10^-11; M=2×10^30kg; m=.32×10^24kg
    ε = 0.206; T=88days; c = 299792.458 km/sec; a = 58.2km/sec

    Calculations yields:
    With v* =48.14km/sec; v° = spin = 2 meters per second
    Then v* + v° = 48.14 km/sec
    And [√ (1- ε²)] (1-ε) ² = 1.552

    W” (ob) = (-720×36526x3600/Tdays){[√(1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ²seconds /100 years

    = (-720×36526x3600/88) x (1.552) [(48.14/299792)]²=43.0”/century

    Joenahhas1958@yahoo.com all right reserved

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