LoFi version for PDAs  Help Search Members Calendar 
Welcome Guest ( Log In  Register )  Resend Validation Email 
Pages: (35) « First ... 32 33 [34] 35 ( Go to first unread post ) 
Add reply · Start new topic · Start new poll 
norgeboy 
Posted: Apr 3 2012, 11:08 PM

Member Group: Power Member Posts: 198 Joined: 8January 12 Positive Feedback: 0% Feedback Score: 0 
Corrected Appendix G = density of matter and black holes:
mass = K x radius in fact. _____________________ Density of Matter and Black Holes As the intersection of one dimensional space (a line) with two dimensional space (a surface) is a single point with zero dimension, the intersection of two and three dimensional space is a line with one dimension, and the intersection of three and five dimensional space should be a surface with two dimensions, then the intersection between five and eight dimensional space should be threedimensional (observable in 3dimensions) and is suggested by the spherical volume of a black hole. From Appendix E, the maximum allowed energyevent is 2.700E+25 J. Then EsubB at a black hole surface should be bounded by the maximum allowed c^3 J kg^1. For the hole surface: EBmax = GmHrH / rH^2 or EsubBmax = G x m / r for the hole, and c^3 = G x m / r relating to the hole, or we can write mH / rH ≤ c3 / G or mH ≤ rH c^3 / G where G = 6.673E11 met^3 kg^1 sec^2, rsubH has units meters, and c (3E+8 numerical) has units J^1/3. Then c^3 / G ≤ 2.700E+25 / 6.673E11, and mH / rH ≤ 4.046E+35 kg met^1 for any black hole. If we let EsubB = ρ EsubBmax = ρ c^3 where 0 < ρ ≤ 1, and ρ = nb where 1 ≤ n ≤ 1/b = B, then mH = rH ρ c^3 / G or mH = KG rH where KG = ρ c^3 / G. 
Send PM · Send email ·

norgeboy 
Posted: Apr 3 2012, 11:21 PM

Member Group: Power Member Posts: 198 Joined: 8January 12 Positive Feedback: 0% Feedback Score: 0 
I also left one sentence out of E, the corrected E is attached.
0 <= energy event <= c/b = c^3 numerically FYI. ______________________ Appendix E The Nature and Meaning of Quantum Mechanics We postulate that any allowed energy quanta has a wavelength λ = nb where n is an integer and b = 1.111… E17 meters per the main text. For example, the 13.6eV H ground state transition is λ = 91.2nm And n = λ / b = 8208820882 + Δn where Δn is the integer adjustment for the repeating decimal 1.111…. Similarly, the H state 1 to state 2 transition is λ = 121.6nm And n = 10945094509 + Δn. An H state 3 to state 1 transition is λ = 486.1nm and n = 43753375338 + Δn and so on. Then 0 ≤ one energyevent ≤ c / b ( = 2.700E25 J) and quantum energy = hc/λ is defined as an integral operation of 1/b. Then the base of all quantum mechanics is 1/b = B where t = cB. Q.E.D. 
Send PM · Send email ·

norgeboy 
Posted: Apr 6 2012, 04:45 AM

Member Group: Power Member Posts: 198 Joined: 8January 12 Positive Feedback: 0% Feedback Score: 0 
Adding this here also regarding black holes.
______________________ Appendix H The Spatial Nature of Black Holes Per Appendix G, mH / rH = KG or msubH = KsubG x rsubH where KG = λ c^3 / G. The following surface density boundary conditions should apply for any black hole: The hole mass msubH / (surface“area”) of the 3dimensional 2D surface = msubH / (surface“area”) of the 5dimensional 3D surface. Similarly, mH / (4/3 x π x rH^3) = LT x rH^5 where LsubT defines the 5dimensional “surfacearea” for 8dimensional space. Then boundary conditions require: πrH^2 = LT38 x rH^5 and 4/3 πrH^3 = LT58 x rH^5 where LT = LsubT = fn(LT38, LT58) = fn( ∫∫∫∫∫∫∫∫dV8, ∫∫∫∫∫dV5, ∫∫∫dV3, ∫∫dV2, ∫dx) and rsubH = rH = 3LT58 / 4LT38 from boundary conditions. Then λ = (4 x LsubT38) / (3 x LsubT58) or λ = 4LT38 / 3LT58. Then λ represents the state of transition from 0 to 1 from 3dimensions to 8dimensions for the spatial intersection known as a black hole. 
Send PM · Send email ·

StayOpen42 
Posted: Apr 7 2012, 08:06 PM

Newbie Group: Power Member Posts: 23 Joined: 6April 12 Positive Feedback: 0% Feedback Score: 0 
I just wanna say guys this is awesome i've theorized like everything you've talked about i've just never had the place to learn the math and work it all out and i havn't had time to research so thanks for posting all this so i don't feel crazy lol! i'm not the only one

Send PM · Send email ·

Confused1 
Posted: Apr 8 2012, 12:12 AM

Member Group: Power Member Posts: 2091 Joined: 8August 10 Positive Feedback: 69.23% Feedback Score: 6 
I like to think I'm not naturally vicious BUT am I the only one thinking
Ego + ((1/2)wit)^4 = 0 ?  The only thing that distinguishes a troll from any other idiot is that the troll is intentionally malicious.

Send PM ·

brucep 
Posted: Apr 8 2012, 12:24 AM


Advanced Member Group: Power Member Posts: 4027 Joined: 3October 09 Positive Feedback: 88.37% Feedback Score: 146 
You referring to the numerologist? 

Send PM ·

norgeboy 
Posted: Apr 8 2012, 12:38 AM

Member Group: Power Member Posts: 198 Joined: 8January 12 Positive Feedback: 0% Feedback Score: 0 
Very funny. I love you guys.
Let's face it, I have now accurately: 1. Derived Plank's constant h. 2. Derived the linear massradius relationship for black holes. 3. Defined the base of quantum mechanics. I know it sounds weird, but it is mathematically indisputable. Newest and attached: deltalambda = csub3 / csub5 is the ratio of curvatures around mass. Why do you continue to disbelieve? _______________________ Appendix H The Spatial Nature of Black Holes Per Appendix G, mH / rH = KG or msubH = KsubG x rsubH where KG = Δλ c^3 / G. The following surface density boundary conditions should apply for any black hole: 1. The hole mass msubH / (“surfacearea”) of the 3dimensional 2D surface = msubH / (“surfacearea”) of the 5dimensional 3D surface (volume.) 2. Similarly, mH / (4/3 x π x rH^3) = CR x rH^5 where CsubR defines the 5dimensional “surfacearea” for 8dimensional space. Then boundary conditions require: 4 /3 (πrH^3) / rH = CR38 x rH^5 and 4/3 πrH^3 = CR58 x rH^5 where CR nm = CsubR for dimension n curving through dimension m and rsubH = rH = CR58 / CR38 from boundary conditions. Then Δλ = CsubR38 / CsubR58 or Δλ = CR3 / CR5 where CsubR3 and CsubR5 represent the curvature rates of 3 and 5 dimensional space respectively through 8dimensional space, where the ratio mH / rH is proportional to Δλ, and where we assume CsubR3 ≤ CsubR5. Then there is only an effective zerodensity “black hole” for CR3 ~ 0 while the highest density black hole occurs where CR3 = CR5, Then Δλ represents the ratio of curvatures of 3 and 5 dimensional space through 8 dimensions for the spatial intersections known as black holes. The higher the mass density in a spatial location, the more the effective radius of curvature should change. With dense enough matter, then curvatures among dimensions become more closely equivalent as density becomes large. To visualize in two dimensions, πr^2 and 4/3 πr^3 / r = 4/3 πr^2 are both two dimensional surfaces that curve in 3dimensions. The curvature (lack of) for a flat circle is 0 while the curvature for the closed spherical surface is 1. This post has been edited by norgeboy on Apr 8 2012, 01:27 AM 
Send PM · Send email ·

norgeboy 
Posted: Apr 8 2012, 03:20 AM

Member Group: Power Member Posts: 198 Joined: 8January 12 Positive Feedback: 0% Feedback Score: 0 
and I think,
Appendix H (cont.) If we set CRn = Cn x esubn where n = Fibonacci dimension n, then per Appendix C λ ~ 10^5 for common black holes. my guess is many black holes should have the "density" m/r ~ 10^+30 to be corrected if wrong, thanks. This post has been edited by norgeboy on Apr 8 2012, 03:23 AM 
Send PM · Send email ·

norgeboy 
Posted: Apr 9 2012, 12:25 AM


Member Group: Power Member Posts: 198 Joined: 8January 12 Positive Feedback: 0% Feedback Score: 0 
Please ignore the last post. I have now calculated the minimum lambda to be ~ 10^7 so that the minimum mass/radius for black holes (in 3dimensions) should be ~ 10^28 or so. Also, the surface area of a sphere is 4pir^2... I meant to write 4/3 pi r^3 /(r/3)... always watch my math(!) __________________________________ Appendix H The Spatial Nature of Black Holes Per Appendix G, mH / rH = KG or msubH = KsubG x rsubH where KG = Δλ c^3 / G. The following surface density boundary conditions should apply for any black hole: 1. The hole mass msubH / (“surfacearea”) of the 3dimensional 2D surface = msubH / (“surfacearea”) of the 5dimensional 3D surface (volume.) 2. Similarly, mH / (4/3 x π x rH^3) = CR x rH^5 where CsubR defines the 5dimensional “surfacearea” for 8dimensional space. Then boundary conditions require: 4πrH^2 = CR38 x rH^5 and 4/3 πrH^3 = CR58 x rH^5 where CR nm = CsubR for dimension n curving through dimension m and rsubH = rH = CR58 / CR38 from boundary conditions. Then Δλ = CsubR38 / CsubR58 or Δλ = CR3 / CR5 where CsubR3 and CsubR5 represent the curvature rates of 3 and 5 dimensional space respectively through 8dimensional space, where the ratio mH / rH is proportional to Δλ, and where we assume CsubR3 ≤ CsubR5. Then there is only an effective zerodensity “black hole” for CR3 ~ 0 while the highest density black hole occurs where CR3 = CR5, Then Δλ represents the ratio of curvatures of 3 and 5 dimensional space through 8 dimensions for the spatial intersections known as black holes. The higher the mass density in a spatial location, the more the effective radius of curvature should change. With dense enough matter, then curvatures among dimensions become more closely equivalent as density becomes large. To visualize in two dimensions, πr^2 and 4πr^2 are both two dimensional surfaces that curve in 3dimensions. The curvature (lack of) for a flat circle is 0 while the curvature for the closed spherical surface is 1. Appendix H (cont.) If we assume CsubR5 is closed (curvature 1) in 8dimensions, then CsubR3 has the possible range 0 1 in 8dimensions where 0 represents no intersection at all and 1 represents infinite intersections. To see/observe the intersection (black hole) it must be at least a 5 and 8 dimensional intersection. Then the “smallest” black hole is the “least dense mH / rH” black hole having Δλ ~ 0 but still large enough to represent an intersection of 5 and 8 dimensional space. Among other conditions, the boundary condition “inside” the 3dimensional hole: MH / V5 = MH / V8 where Vsub5 and Vsub8 are defined by Appendix C. Then the least dense black holes are functions of e8 and e5 and should conform to Δλ = Vsub5 / Vsub8, or Δλ ≥ ~ 10^7 for any observable (3dimensional) black hole. 

Send PM · Send email ·

norgeboy 
Posted: Apr 9 2012, 09:26 PM

Member Group: Power Member Posts: 198 Joined: 8January 12 Positive Feedback: 0% Feedback Score: 0 
Here is the anticipated correction for lambda (min) and should agree with the socalled "event horizons of known black holes.
_______________________________ Appendix H The Spatial Nature of Black Holes Per Appendix G, mH / rH = KG or msubH = KsubG x rsubH where KG = Δλ c^3 / G. The following surface density boundary conditions should apply for any black hole: 1. The hole mass msubH / (“surfacearea”) of the 3dimensional 2D surface = msubH / (“surfacearea”) of the 5dimensional 3D surface (volume.) 2. Similarly, mH / (4/3 x π x rH^3) = CR x rH^5 where CsubR defines the 5dimensional “surfacearea” for 8dimensional space. Then boundary conditions require: 4πrH^2 = CR38 x rH^5 and 4/3 πrH^3 = CR58 x rH^5 where CR nm = CsubR for dimension n curving through dimension m and rsubH = rH = CR58 / CR38 from boundary conditions. Then Δλ = CsubR38 / CsubR58 or Δλ = CR3 / CR5 where CsubR3 and CsubR5 represent the curvature rates of 3 and 5 dimensional space respectively through 8dimensional space, where the ratio mH / rH is proportional to Δλ, and where we assume CsubR3 ≤ CsubR5. Then there is only an effective zerodensity “black hole” for CR3 ~ 0 while the highest density black hole occurs where CR3 = CR5, Then Δλ represents the ratio of curvatures of 3 and 5 dimensional space through 8 dimensions for the spatial intersections known as black holes. The higher the mass density in a spatial location, the more the effective radius of curvature should change. With dense enough matter, then curvatures among dimensions become more closely equivalent as density becomes large. To visualize in two dimensions, πr^2 and 4πr^2 are both two dimensional surfaces that curve in 3dimensions. The curvature (lack of) for a flat circle is 0 while the curvature for the closed spherical surface is 1. Appendix H (cont.) If we assume CsubR5 is closed (curvature 1) in 8dimensions, then CsubR3 has the possible range 0 1 in 8dimensions where 0 represents no intersection at all and 1 represents infinite intersections. To see/observe the intersection (black hole) it must be at least a 5 and 8 dimensional intersection. Then the “smallest” black hole is the “least dense mH / rH” black hole having Δλ ~ 0 but still large enough to represent an intersection of 5 and 8 dimensional space. Then the boundary condition is a single event: CR3min = 1 / cB where B = 1 and Δλmin = 1 meter / c meters = 3.333E9. This post has been edited by norgeboy on Apr 9 2012, 09:44 PM 
Send PM · Send email ·

norgeboy 
Posted: Apr 9 2012, 11:14 PM

Member Group: Power Member Posts: 198 Joined: 8January 12 Positive Feedback: 0% Feedback Score: 0 
Or here is a more accepted way to say it:
... Appendix H (cont.) If we assume CsubR5 is closed (curvature 1) in 8dimensions, then CsubR3 has the possible range 0 1 in 8dimensions where 0 represents no intersection at all and 1 represents infinite intersections. To see/observe the intersection (black hole) it must be at least a 5 and 8 dimensional intersection. Then the “smallest” black hole is the “least dense mH / rH” black hole having Δλ ~ 0 but still large enough to represent an intersection of 5 and 8 dimensional space. Then the boundary condition is a single event: CR3min = 1 / cB where B = 1 and Δλmin = 1 meter / c meters = 3.333E9. Then c2 / G ≤ mH / rH ≤ c3 / G or c^2 ≤ mH / rH ≤ c^3 kg met^1 or we can write the expression rH = K(λ) mH G / c^2 meters This post has been edited by norgeboy on Apr 9 2012, 11:18 PM 
Send PM · Send email ·

norgeboy 
Posted: Apr 10 2012, 10:44 PM

Member Group: Power Member Posts: 198 Joined: 8January 12 Positive Feedback: 0% Feedback Score: 0 
This is about curvatures as in the black hole discussion:
Appendix K The Various Sizes of Black Holes and Curvatures of ThreeDimensional Space Appendix H defines the massradius relationship as observed in threedimensional space: mH / rH = KG(λ) where λ = Δλ = CR3 / CR5 and represents the ratio of curvatures from 3dimensional space and 5dimensional space through 8dimensional space respectively. We assume, for the three dimensional intersections, that CR5 = 1. The minimum CR3 = 1 / c and the maximum CR3 = c / c = 1. Allowed quantum are then n / c for n = 1 to c. The minimum (least dense) intersection is an intersection among 3, 5 and 8 dimensional space where CR5 = 1 and CR3 = CR3(min) = 1 / c. The next “largest” (more dense) intersection should occur for CR3 = 2 / c and so on. The most dense intersection occurs where CR3 = c / c = 1 and represent a closed third dimension in both eight dimensional and five dimensional space. To visualize curvatures, the diameter of a circle = d is a straight line with curvature CR13 = 0 while the circumference (length πd) closes upon itself (runs into the back of itself) and has the curvature CR13 = 1. The curvature CR23 is closed in 3dimensions visualized as a spherical (or elliptical, not reviewed in this scope) surface area that has closed itself around a centerofmass cM. The two dimensional surface does not alter or “grow” in three dimensions, but the one dimensional line, e.g. the straight path of a distant comet or ray of light (CR13 = 0) or the line of a planetary satellite CR13 = 1) both curve (or bend) around mass in three dimensions to the two extreme degrees of curvature. Then the ratio mH / rH = KG should represent a curvature of threedimensional space through eightdimensional space. 
Send PM · Send email ·

Mekigal 
Posted: Apr 10 2012, 10:59 PM


Advanced Member Group: Power Member Posts: 3213 Joined: 22March 12 Positive Feedback: 0% Feedback Score: 0 
thats funny 

Send PM · Send email ·

norgeboy 
Posted: Apr 12 2012, 01:17 AM

Member Group: Power Member Posts: 198 Joined: 8January 12 Positive Feedback: 0% Feedback Score: 0 
What would happen if we took a GE turbine (like at the Hoover dam) and placed it into geosync or any other orbit around the Earth. It would be weightless.
If it got spinning from a solar panel energy, would it provide more energy than from the waterfall at the Hoover dam? 
Send PM · Send email ·

Robittybob1 
Posted: Apr 12 2012, 01:35 AM


Advanced Member Group: Power Member Posts: 6593 Joined: 15October 11 Positive Feedback: 0% Feedback Score: 0 
Are you being serious? If you could get power from it, say from the Solar Wind it would soon get a very elliptical orbit and shortly thereafter crash back to the Earth. 

Send PM · Send email ·

Pages: (35) « First ... 32 33 [34] 35 
Add reply · Start new topic · Start new poll 