The Great Beyond

E=mc²/2
2009 is the end of Einstein's space-jail of time and Fraud symbol E=mc²
Joenahhas1958@yahoo.com
Time is not a structure like space to allow space-to time-back to space jumping claimed by Physicists regardless of what physicists have to say about it because Physics is a business and not necessarily science or scientific and like every business it comes with fraud and fraud is Einstein's space-time (x, y, z, it) continuum that led to fraud symbol E=mc² and yes I am saying that 109 years of Nobel prize winners physics and physicists are all wrong and space-time physics is based on scientific fraud. When "results" expected and "No" discovery, Physicists rigged Physics for grant money since the start of the industrial revolution. Physics today is at least 51 % fraud!
r ------------------>>Exp (ì w t) ---------->> S=r Exp (ì wt) Nahhas' Equation
Orbit-------->> Orbit light sensing------>> Visual Orbit; Exp = Exponential
Particle ---->> light sensing of moving objects------------ >> Wave
Newton--------->>light sensing---------->> Quantum
Quantum = Newton x Visual Effects
Quantum - Newton = Relativistic = Optical Illusions
E (Energy by definition) = mv²/2 = mc²/2; if v = c
m = mass; v= speed; c= light speed; w= angular velocity; t= time
S = r Exp (ì w t) = r [cos (wt) + ì sin (wt)] Visual effects
P = visual velocity = change of visual location
P = d S/d t = v Exp (ì w t) + ì w r Exp (ì w t)
= (v + ì w r) Exp (ì w t) = v (1 + ì) Exp (ì w t) = visual speed; v = wr
E (visual energy= what you see in lab) = m p²/2; replace v by p in E = mv²/2
= m p²/2 = m v²/2 (1 + ì) ² Exp (2ì wt)
= mv²/2 (2ì) [cosine (2wt) + ì sine (2wt)]
=ì mv² [1 - 2 sine² (wt) + 2 ì sine (wt) cosine (wt)];v = speed; c = light speed
wt = π/2
E (visual) = ìmv² (1 - 2 + 0)
E (visual) = -ì mc² ≡ mc² (absolute value;-ì = negative complex unit) If v = c
w t = π/4
E (visual) = imv² [1-1 +ỉ] =-mc²; v = c
wt =-π/4+ỉln2/2; 2ỉ wt=-ỉπ/2 - ln2
Exp (2i wt) = Exp [-ỉπ/2] Exp [ln(1/2)]=[-ỉ (1/2)]
E (visual) = imv² (-ỉ/2) =1/2mc² v = c
Conclusion: E = mc² is the visual Illusion of E = mc²/2 joenahhas1958@yahoo.com. All rights reserved.
PS: In case of E=mc² claims to be rest energy claims then
E=1/2m (m v + m' r) ² = (1/2m) (m' r) ²; v = 0
E = (1/2m) (mc) ²; m' r =mc
E=mc²/2

Einstein's Physics+ MIT Harvard Cal-Tech Princeton
Stanford Perimeter institute + All space-time Dollar stores Physics/Astrophysics departments + NASA = 0 Physics
Visual Effects and the Confusions of "Modern" physics
r ------------------Light sensing of moving objects ------- S
Actual object--------------- Light ------------------ Visual object
r - -------cosine (wt) + i sine (wt) - S = r [cosine (wt) + i sine (wt)]
Line of Sight: r cosine wt : light aberrations
A moving object with velocity v will be visualized by
Light sensing through an angle (wt); w = constant and t= time
Also, sine wt = v/c; cosine wt = √ [1-sine² (wt)] = √ [1-(v/c) ²]
A visual object moving with velocity v will be seen as S
S = r [cosine (wt) + i sine (wt)] = r Exp [i wt]; Exp = Exponential
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 the visual effects caused by light aberrations of a moving object along the line of sight.
In a right angled velocity triangle A B C: Angle A = wt; angle B = 90°; Angle C = 90° -wt
AB = hypotenuse = c; BC = opposite = v; CA= adjacent = c √ [1-(v/c) ²]
And what is the visual effect for angular velocity of Perihelion of Mercury along the line of sight?
Areal velocity is constant: r² θ' =h Kepler's Law
h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity
r² θ'= h = S² w'; S = r Exp (ỉ wt) ----------------------------------------------------->
h = [r² Exp (2iwt)] w'=r²θ'; w' = (θ') exp [-2(i wt)]
w'= (h/r²) [cosine 2(wt) - ỉ sine 2(wt)] = (h/r²) [1- 2sine² (wt) - ỉ sin 2(wt)]
w' = w'(x) + ỉ w'(y) ; w'(x) = (h/r²) [ 1- 2sine² (wt)]
Δ w'= w'(x) – (h/r²) = - 2(h/r²) sine² (wt) = - 2(h/r²) (v/c) ² v/c=sine wt
(h/ r²)(Perihelion/Periastron)= [2πa.a√ (1-ε²)]/Ta² (1-ε) ²= [2π√ (1-ε²)]/T (1-ε) ²
Δ w' = [w'(x) – h/r²] = -4π {[√ (1-ε²)]/T (1-ε) ²} (v/c) ² radian per second
{x [180/π;degrees]x[100years=36526days;century]x[3600;seconds in degree]
Δ w" = (-720x36526x3600/T) {[√ (1-ε²]/(1-ε)²} (v/c)² seconds of arc per century
The circumference of an ellipse: 2πa (1 - ε²/4 + 3/16(ε²)²- --.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)
v=√ [G m M / (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m Advance of Perihelion of mercury.
G=6.673x10^-11; M=2x10^30kg; m=.32x10^24kg
ε = 0.206; T=88days; c = 299792.458 km/sec; a = 58.2km/sec
Calculations yields:
v =48.14km/sec; [√ (1- ε²)] (1-ε) ² = 1.552
Δ w”= (-720x36526x3600/88) x (1.552) (48.14/299792)²=43.0”/century
Conclusions: The 43" seconds of arc of advance of perihelion of Planet Mercury (General relativity) is given by Kepler's equation better than all of Published papers of Einstein. Kepler's Equation can solve Einstein's nemesis DI Her/AS Cam/V1143cgyni/V541cgyni Binary stars motion and all the other dozens of Binary stars motions Puzzles posted for past 40 years on NASA website SAO/NASA as unsolved by space-time physics or any published physics. Joenahhas1958@yahoo.com

2009 total collapse of Einstein's general relativity theory
Case of CD Draconis Binary stars apsidal motion puzzle solution

By Professor Joe Nahhas

Abstract: This is the solution to the 150 years apsidal motion puzzle solution that is not solvable by space-time physics or any said or published physics including 109 years of noble prize winner physics and 400 years of astronomy. Binary stars apsidal motion or "Apparent" rate of orbital axial rotation is visual effects along the line of sight of moving objects applied to the angular velocity at Apses. From the thousands of close binary stars astronomers picked a dozen sets of binary stars systems that would be a good test of relativity theory and collected data for all past century and relativity theory failed every one of them. This rate of "apparent" axial rotation is given by this new equation

W° (ob) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²]} [(v° + v*)/c] ² degrees/100 years
T = period; ε = eccentricity; v° = spin velocity effect; v*= orbital velocity effect
In general v* = v* (p) + v*(s) and v° = v° (p) + v° (s); p = primary; s= secondary
And v* + v° = 170.117 km/sec; ε = 0.0051; T = 1.238389985 days
W° (ob) = 6.731598944°/century = 0.06731598944°/year
U [years] = 360/[0.06731598944°/year]
U = 5348 years Nahhas
U (observed) = 5400+/-3200years
Einstein's and space-timers U = 360/ [0.00191x365.26] = 516 years

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 visual effects 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]

W (ob) = -4π { [√ (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) = (-720x36526/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° Then W° (ob) ≈ (-720x36526/Tdays) {[√ (1-ε²)]/ (1-ε) ²} x sine² Inverse tan [v°/c + v*/c] degrees/100 years

Approximations II

With v°

W° (ob) = (-720x36526/Tdays) {[√ (1-ε²)]/ (1-ε) ²} x [(v° + v*)/c] ² degrees/100 years
This is the equation that gives the correct apsidal motion rates -----------------------III

The circumference of an ellipse: 2πa (1 - ε²/4 + 3/16(ε²)²- --.) ≈ 2πa (1-ε²/4); R =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 [Mercury] = v*(p)
And v* (M) = √ [Gm² / (m + M) a (1-ε²/4)] = v* (s)

1- Planet Mercury 43" seconds of arc per century elliptical orbit axial rotation rate
[No spin factor]; data supplied does not include spin factor

W (obo) = (-720x36526x3600/T) {[√ (1-ε²]/ (1-ε) ²} (v/c) ² seconds of arc per century

The circumference of an ellipse: 2πa (1 - ε²/4 + 3/16(ε²)²- --.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)
v=√ [G m M / (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m

G =6.673x10^-11; M=2x10^30kg; m=.32x10^24kg
ε = 0.206; T=88days; c = 299792.458 km/sec; a = 58.2km/sec

Calculations yields:
v =48.14km/sec; [√ (1- ε²)] (1-ε) ² = 1.552

W (ob) = (-720x36526x3600/88) x (1.552) (48.14/299792)²=43.0”/century
This is the solution to Mercury's 43" seconds of arc per century without space-time fictional forces or space-time fiction

Stars apsidal motion measurements have spin factor in them; new rules have to be made different form planetary rules because binary stars are light emitters and their spin affects the reading of their motion.

1-These are the rules and situations encountered in binary stars

Looking from top or bottom at two stars they either approach each other coming from the top (↑) or from the bottom (↓)

Knowing this we can construct a table and see how these two stars are formed. There are many combinations of velocity additions and subtractions and one combination will give the right answer.

CD Draconis apsidal motion table:

Primary →
Secondary ↓ v°(p) ↑ v* (p)↑ v° (p) ↑v* (p)↓ v° (p) ↓ v* (p) ↑ v° (p) ↓V* (p) ↓
v°(s) ↑ v* (s)↑ Spin=[↑,↑][↑,↑]=orbit [↑,↑][↓,↑] [↓,↑][↑,↑] [↓,↑][↓,↑]
Spin results v°(p) + v°(s) v°(p) + v°(s) - v°(p) + v°(s) -v°(p) + v°(s)
Orbit results v*(p) + v*(s) -v*(p) + v*(s) v* (p) + v*(s) -v* (p) + v* (s)
Examples CD Draconis
v° (s) ↑v* (s)↓ [↑,↑][↑,↓] [↑,↑][↓,↓] [↓,↑][↑,↓] [↓,↑][↓,↓]
Spin results v°(p) + v°(s) v°(p) + v°(s) -v°(p) + v°(s) -v°(p) + v°(s)
Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)
Examples
v° (p) ↓ v*(s) ↑ [↑,↓][↑,↑] [↑,↓][↓,↑] [↓,↓][↑,↑] [↓,↓][↓,↑]
Spin results v°(p) - v°(s) v°(p) - v°(s) -v°(p) - v°(s) -v°(p) - v°(s)
Orbit results v*(p) + v*(s) -v*(p) + v*(s) v*(p) + v*(s) -v* (p) + v* (s)
Examples
v° (s) ↓V*(s) ↓ [↑,↓][↑,↓] [↑,↓][↓,↓] [↓,↓][↑,↓] [↓,↓][↓,↓]
Spin results v°(p) - v°(s) v°(p) - v°(s) -v°(p) - v°(s) -v°(p) - v°(s)
Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)
Examples CD Draconis

Data: T=1.268389985days; m = 0.231 M (0); M = 0.2141 M (0); a = 3.7634 R (0); ε = 0.0051
And [v° (p); v° (s)] = [9.5 +/- 1; 10.0 +/- 1]; W° = 1.91x10^-3/day

Calculations
m + M = 0.4451 M (0)
1-ε = 0.9949; (1-ε²/4) = 0.99993498; [√ (1-ε²)] / (1-ε) ² = 1.01
G=6.673x10^-11; M (0) = 1.98892x19^30kg; R (0) = 0.696x10^9m

Calculations
With v* (p) = √ [GM²/ (m + M) a (1-ε²/4)] = 72.436 km/sec
And v* (s) = √ [Gm²/ (m + M) a (1-ε²/4)] = 78.153 km/sec

And v° (p) = 9.5 km/sec; v° (s) = 10 km/sec
Then v* (p) + v* (s) + v° (p) + v° (s) = 170.117 km/sec

Apsidal motion is given by this formula:

W° (ob) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²]} [(v° + v*)/c] ² degrees/100 years

W° (ob) = (-720x36526/1.238389985) (1.01) [170.09/300,000] ² degrees/100 years
W° (ob) = 6.731598944°/century = 0.06731598944°/year

U [years] = 360/[0.06731598944°/year]

U = 5348 years Nahhas
U (observed) = 5400+/-3200years
Einstein's and space-timers U = 360/ [0.00191x365.26] = 516 years
Can it get any better?

References: Absolute properties of the low-mass eclipsing binary CD Draconis; 2009
By : Juan Carlos Morales; Ignasi ribas; carme jordi; Guillermo Toress; Jose Gallardo; Edward F. Guinan; David Chardonneau; Marek wolf; David w.latham; Guillem Angalada Escude; David H.Bradstreet; Mark E.Everett; Francis T. O, Donavan; Georgi Mandushev; Robert D. Mathieu; 9 other co-authors.

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