Plenty of room

Just a quick note before I drift off to study for my exams.

I re-read the famous ‘there’s plenty of room at the bottom‘ speech made by Richard Feynman recently. Aside from being inspired by his genius and foresight (as usual) I think I hit on an interesting idea.

At the end of the speech Feynman half-jokingly proposes a contest for high school students with the goal of writing smaller than anyone else. I think we have enough industrial infrastructure and technical expertise to make that contest come true, albeit with possibly different goal than simply ‘writing small’ and perhaps geared towards undergraduate students.

Those of you who have been following this blog or any other one of my web presence knows that I am deeply interested in synthetic biology, to the extent that I ventured into the recent Synthetic Biology conference 4.0 in Hong Kong armed with my meager knowledge of genetics and molecular biology. In fact, I’ve been so interested in the discipline that I’ve been driving my professors crazy with questions, delving deep into molecular biology texts and courses outside my proclaimed field of expertise (which is plasma physics), even touching up with a bit of crude wet work.

The reason why I became aware of the field of synthetic biology and began taking its possibilities and my involvement with it seriously, was the International Genetically Engineered Machine competition or iGEM. It is an international competition for high school-to-undergraduate students to build the best synthetic organism (or genetically engineered machine) using opensource biological parts termed BioBricks, which can be pieced together like puzzle to form a working genetic system complete with chassis (usually E.Coli or Yeast). The quality of the competition entries have been phenomenal so far. The winning entry in this year’s iGEM competition actually prototyped a whole new vaccine against gastritis. It took undergraduates six months to come up with that stuff (with help of graduate level faculty). Just imagine what people will be able to do once we streamline the whole process and work out some kinks inherent in dealing with biological systems!

Now, let’s imagine something similar with nanotechnology. I believe that it is possible to put together some minimal nanotech components/chassis in the fashion of the BioBricks, opensource them, and apply it toward high school-undergraduate level competition. Of course, the things we can come up with using today’s technology won’t be as vibrant as the projects pursued by those of iGEM teams, but I still believe that we have enough room for ingenuity and improvisation in constructing minimal nanotechnological systems and parts. With suitable industrial support the international nanoengineered machine competition (iNEM?) might lend the field of nanotechnology accessibility and interest the field rightly deserves.

One thought on “Plenty of room

  1. Einstein’s Nemesis: DI Her Eclipsing Binary Stars Solution
    The problem that the 100,000 PHD Physicists could not solve

    This is the solution to the “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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