SIZE OF THE UNIVERSE AT BIG BANG
Let us find an Engineering method to calculate the minimum size of the universe at big bang. For this analysis we will use a method similar to the method used in Doppler Space Time except we will use the Plank length as the minimum size of the dot. Thus the dot-diameter is equal to the Plank length.
First need to find out the mass of the universe. Since we know the mass of a neutron, we can find out how many neutrons there are in the universe. Then we can calculate the number of dots in the universe. Finally we can calculate the size of a ball containing that many dots.
We know that the universe exists as a spherical shell 15.9 billion light years from the common center. The force equations operating upon a mass Mx within the spherical shell is:
G Mu Mx / Ru^2 = Mx C^2 / Ru (8-8)
In equation 8-8 the mass of the universe Mu stands at the common center of the universe at a distance Ru from the mass Mx upon the surface of the universe. This force is counterbalanced by the mass Mx that is moving away from the center of the universe at light speed C. This force is similar to an orbital force. Thus we have the mass times the velocity squared and divided by the distance.
The equation is independent of the mass Mx. It could have been a proton. It could have been a dot mass. The equation simply says that the gravitational force on any object depends upon the total mass of the universe and the gravitational constant and the distance from the universe to the common center. Solving for the mass of the Universe we get:
Mu = [C^2] Ru / G (8-9)
Since G = 6.67428E-11, Ru = 1.5052393E26, and C = 2.99792458E8, we get:
Mu = 2.0269477E53Kg. (8-10)
Since Mn = 1.67492721E-27Kg, the number of neutrons in the universe is:
# Neutrons = Mu/Mn = 1.21017E80 (8-11)
Since the # dots per neutron is 2.2813727E41, the dots per universe is:
# Dots per universe = 2.276085E121 dots (8-12)
The volume occupied by the dots will be the number of dots times the volume per dot. The radius of the package will be the radius of the dot times the cube root of the number of dots. Therefore:
R universe = (1.6162525E-35) 2.8339756E40 = 4.580418E5 (8-13)
Since I miles = 1609.334 meters, we get:
R universe = 284.4157 miles (8-14)
Equation 8-13 gives us the size of the universe at big bang. However this is the size of the sum of all the lower light speed universes. The light speed C universe sits above this size. It has the same volume but a different thickness.
If we assume that we have the series:
C, C/2, C/4, C/8,……………….C/2N (8-15)
We can assume that the volume of each higher lightspeed universe is equal to the total volume occupied by all the lower light speed universes. Therefore:
V' = Va + Va/2 + Va/4 + Va/8 + ………. Va/ 2N (8-16)
Therefore:
V (total) = 2Va (8-17)
We can now solve for the thickness of our universe at big bang. The total radius of the universe will be:
R = [2^0.333333] Ru (big bang) = 1.25992 Ru (big bang) (8-18)
R = 5.77095E5 meters (8-19)
The thickness of our universe is:
Thickness = 0.125992 Ru = 1.19054E5 meters (8-20)
Thickness = 73.9767 miles (8-21)
The universe at big bang is a layer 74 miles thick upon a radius of 284 miles from the common center.
It is worth noting that Planks length of 1.616E-25, the radius of the proton at 1.321E-15, and the radius of the universe at 1.505E26 form a series of measurements which are apart by approximately a power of 10 to the 20th. However one measurement is missing. That is:
L = 1.616E-35 x 1.505E26 / 1.321E-15 = 1.841E6 (8-22)
We see that the radius of the universe shown by Equation 8-19 fits into the series. Thus
R x pi = 1.8130 E6 (8-23)
Equation 8-22 is only 1.5- percent difference from equation 8-22. Therefore the radius of the universe at big bang fits perfectly into the series of constants.
R min = π [PL] x Ru / Rp (8-24)
R min = 5.78398E5 meters (8-25)
The percent difference with equation 8-19 is:
Percent difference = 0.225 percent (8-26)
We see that equation 8-24 gives us the minimum radius to within 0.225 percent of the original method. Therefore we have great confidence that we have the minimum radius of the universe at big bang.
Let us find an Engineering method to calculate the minimum size of the universe at big bang. For this analysis we will use a method similar to the method used in Doppler Space Time except we will use the Plank length as the minimum size of the dot. Thus the dot-diameter is equal to the Plank length.
First need to find out the mass of the universe. Since we know the mass of a neutron, we can find out how many neutrons there are in the universe. Then we can calculate the number of dots in the universe. Finally we can calculate the size of a ball containing that many dots.
We know that the universe exists as a spherical shell 15.9 billion light years from the common center. The force equations operating upon a mass Mx within the spherical shell is:
G Mu Mx / Ru^2 = Mx C^2 / Ru (8-8)
In equation 8-8 the mass of the universe Mu stands at the common center of the universe at a distance Ru from the mass Mx upon the surface of the universe. This force is counterbalanced by the mass Mx that is moving away from the center of the universe at light speed C. This force is similar to an orbital force. Thus we have the mass times the velocity squared and divided by the distance.
The equation is independent of the mass Mx. It could have been a proton. It could have been a dot mass. The equation simply says that the gravitational force on any object depends upon the total mass of the universe and the gravitational constant and the distance from the universe to the common center. Solving for the mass of the Universe we get:
Mu = [C^2] Ru / G (8-9)
Since G = 6.67428E-11, Ru = 1.5052393E26, and C = 2.99792458E8, we get:
Mu = 2.0269477E53Kg. (8-10)
Since Mn = 1.67492721E-27Kg, the number of neutrons in the universe is:
# Neutrons = Mu/Mn = 1.21017E80 (8-11)
Since the # dots per neutron is 2.2813727E41, the dots per universe is:
# Dots per universe = 2.276085E121 dots (8-12)
The volume occupied by the dots will be the number of dots times the volume per dot. The radius of the package will be the radius of the dot times the cube root of the number of dots. Therefore:
R universe = (1.6162525E-35) 2.8339756E40 = 4.580418E5 (8-13)
Since I miles = 1609.334 meters, we get:
R universe = 284.4157 miles (8-14)
Equation 8-13 gives us the size of the universe at big bang. However this is the size of the sum of all the lower light speed universes. The light speed C universe sits above this size. It has the same volume but a different thickness.
If we assume that we have the series:
C, C/2, C/4, C/8,……………….C/2N (8-15)
We can assume that the volume of each higher lightspeed universe is equal to the total volume occupied by all the lower light speed universes. Therefore:
V' = Va + Va/2 + Va/4 + Va/8 + ………. Va/ 2N (8-16)
Therefore:
V (total) = 2Va (8-17)
We can now solve for the thickness of our universe at big bang. The total radius of the universe will be:
R = [2^0.333333] Ru (big bang) = 1.25992 Ru (big bang) (8-18)
R = 5.77095E5 meters (8-19)
The thickness of our universe is:
Thickness = 0.125992 Ru = 1.19054E5 meters (8-20)
Thickness = 73.9767 miles (8-21)
The universe at big bang is a layer 74 miles thick upon a radius of 284 miles from the common center.
It is worth noting that Planks length of 1.616E-25, the radius of the proton at 1.321E-15, and the radius of the universe at 1.505E26 form a series of measurements which are apart by approximately a power of 10 to the 20th. However one measurement is missing. That is:
L = 1.616E-35 x 1.505E26 / 1.321E-15 = 1.841E6 (8-22)
We see that the radius of the universe shown by Equation 8-19 fits into the series. Thus
R x pi = 1.8130 E6 (8-23)
Equation 8-22 is only 1.5- percent difference from equation 8-22. Therefore the radius of the universe at big bang fits perfectly into the series of constants.
R min = π [PL] x Ru / Rp (8-24)
R min = 5.78398E5 meters (8-25)
The percent difference with equation 8-19 is:
Percent difference = 0.225 percent (8-26)
We see that equation 8-24 gives us the minimum radius to within 0.225 percent of the original method. Therefore we have great confidence that we have the minimum radius of the universe at big bang.