Expressions for Earth Fields A and B , and difference. And Integral Curves. ETC. Under Construction.

• Field A has given us the following equation / expression: 288 2^(2/3) E^(8/3) Power[log(8), (3)^-1]
• Field B has given us: (216 E log^5(3))/log^7(2)
• Unfortunately these two expressions are not quite similar albeit they both have 288 and 216 which are both 36*n. They also both include the E constant. And log()) functions.
• Equation / Expression for field-difference is in that regard not so different: 144*E^(5/2)*Sqrt[log(27)/log(2)] .
• The respective values of 36 are: 288=36*8, 216=36*6, 144=36*4.

• With the 2**(2/3)*e**(8/3)*pow(log(8),(3)^-1) we get close to 2*sqrt(72000/192000).
• Here we have 144000/2=72000 and also 32000*6=192000.
• 72/192 then is 0.75/2.
• So this end product might be 36*x*(2*sqrt(72000/192000)) for Field A.
• Interesting also is that this product can be expressed as either sqrt(2/3) or with the same output csc(θₜ/2), later being cosecant of tetrahedral angle.
• Investigating Field B in a similar manner has not yet proven ..anything..but we might just look at what we got from Field A possibly.
• We believe the system is might be pretty mathematical "tight" or systematic. So in a generalizations Filed A divided Field B might be just about 1.5.
• Otherwise the difference could of course be around 1.44 which is just as close to 1.455500168495149 as 1.5
• Many expressions surface looking at 1.455500168495149. sqrt(162000/72000) fits well for 1.5 though.
• 162000 being a Harmonic of 162 which in 10f²+2 , f=4, the number of spheres in the 4th layer of closest packed spheres. (See Synergetics section 223 again.) etc.
• So Field B might be expressed as 36*x*(2*sqrt(72000/192000))*1.5 where derivative of x is 66.1362, close to 27*sqrt(6).
• Replacing 36*x with 6*x give the derive 11.0227, giving close to 9*sqrt(3/2).
• However generalizing a bit (2*sqrt(72000/192000)) = sqrt(6)/2.

• The two expressions seem to simplify to Field A=288*29 and Field B=432*29. We are going to to do various calculations eventually to see if these fit.
• The original estimate 2**(2/3)*e**(8/3)*pow(log(8),(3)**-1) equals 29.160000075669835. Thus 29 maybe implied.
• Though e*log(3,5)/log(2,7); 5.209092016164562 which should be multiplied by 216 to give 1125.1638754915455, though we double 216 to 432 and multiply by 29 here too. We trust the first expression we had to be the most correct. They were all only based on guesses from our software and anyway. Thus we get 12528 instead, a higher exponent but..so that it at least fit somewhat the original digits.
• So the original Field B is now larger than original field A? But otherwise it fits.
  
  
• So at least that leaves us with 8352+12528=20880 that looks like a castling of 28800..which might be correct.
• In 2nf²+2 here 432 would have it reduced to 216 by dividing by 2 which fit 2*3**2+2=218 and 218*2=436
• In 2nf²+2 here 288 would fit as 1*12²+2=290.
• The sum would then have to be 21054.
• And another possibly better explanation is 288=2â?¸*1.125 (s6), and 432 is 2â?¸*1.6875. (1/(2**7)*216)
• Anyway we had sum(1-144)*2 = 20880 as total strength. 4176 as difference which divided by 2 also is 2088.
• The conceptual of it all may be difficult to explain, with as few details as here.
• 12528 seems like a Harmonic of Schuman Resonance → 783*2*2*2*2=12528.
• Also 12528/6 is 2088 again. But 12528 → /2/2 we have 3132 to subtract 2**11..which leaves 1048 which is 24 away from 1024.
• Both field values squared and divided on each other, give multiplied with 36 exactly 16.
• 69755904/156950784*36=16. The reverse give 2.25 divided.

• Curiously total field strength 20880*6 is 125280 a Harmonic from Field B.
• Next subtract Field A: 125280-8352=116928 and divide by 216 = 541.3333333333334.
• x=541.3333333333334 and x=1/x then x=x*6*6*6*6*6 = 14.364532019704434. Subtract this from 1.44 and do inverse = 28.194444444444.
• x=28.194444444444*6 = x=169.16666666666**2 = 28617.361110885555..
• Divide by 2 = x=14308.680555442777 speed of light and inverse = 6.988764590314494e-05 gravity factor.
• The problem with the above calculation is as usual to create a "concept" out of it.
• Here we have a value for Field B inside both Field A and B..which subtracted Field A and manipulated..
• Though 169.16666666666 which let the Wadsworth constant turn up again in it's equation:
• 169.16666666666 = (5075*Wwad)/9 from which we get 5075.
• To 1/(12528/5075)*216 == 87.5 → *2*2*2 = 700 Harmonic?
• 700*2**11=1433600 Harmonic and have 1440000-x = 6400..
• Then 1/(12528/5075)*216*2048 = Harmonic 1433600. Which in Harmonic equation 3 give: 69698.22108069912 as a result.
• Curiously 20880/16 is composed of 783+522, but in hertz 1305 is equal to about 78300 RPM.
• 1305/522 is exactly 2.5.
• 20880 RPM is about 348 hertz which is 2*6*29.
• 522 hz is equal to 31320 RPM which divided by 4 is 7830.
• 522 *2 *2 is 2088 again.
• Ayers Rock, or "Uluru" is standing 348 meters up into the air?
• 29*1.125*16 = 522.0.
• 3393 (Pulse of the universe p. 43) divide by 8352 * 32 = 13.
• 3392 / 12528 = 0.2708333333333333. Multiply by 6 = 1.625 divide by 8 = 13.
• 13*216=2808.
• Base of the Great Pyramid is 5832 is a castling of 8352.
• 1/(8352/5832)=1.4320987654320987 into 1.44-x into 1/x = 126.562499999999 or 126.5625 to x/2/2. Then to x*8*8/(9*9) == 25 divide by 6000 and divide by 60 == 6.944444444444444e-05.
  
  
• Then 1/x == 14400.0. The Great Pyramid and Field A share the Harmonic 14400.

• 12528/5832*216=464. 464=29*2*2*2*2. → 2880-(464*6)=96. → 96/2 → 144/48=3. → 12528/3 → 5832/4176 = → 1/1.396551724137931
  
  
• → *2 → 1.44-1.4320987654320987 = → → 0.7901234567901234*1296 = 1024 == 2*2*2*2*2*2*2*2*2*2...
  
  
• We get 4176 from Field B and combine with pyramid digit to get a light-speed eventually. Light speed is then subtracted from max speed and ..., and multiplied by 6â?´ to get difference as harmonic 1024 or 2**n. Roughly. The whole operation is rather complicated....but shows that also Field B can be associated with The Great Pyramid.
  
  
  
• 12528/1024 → 12.234375*64 = 783. Of course → *16 = 12528..
  
  
• This leaves out that we usually think of Schuman Resonance as 1/128=0.0078125. Harmonic different with 7.83 hz would be harmonic 175.
  
  
• 175*2*2*2*2*2*2*2*2*2*2*2*2*2=143360 as stated earlier. Difference from max light speed harmonic, is 6400.
  
  
• Working backwards from 781.25 → *16 we get 12500, 28 less than Field B. → 12500/432 → 29-x → x*216 = 14?
  
  
  
• Boolean: 12528 and 8352 is 8352. 12528 or 8352 is 12528. 12528 xor 8352 is 4176 (field difference). Thus the values, if it's important, seems compatible in boolean algebra..
  
  
  
• Maybe some of these things above show that the values 8523 and 12528 might work. We suppose much more is needed to prove anything.