Tuesday, August 14, 2012

The Last Math Solution, by Ezekial Davis

The Last Math Solution, by Ezekial Davis
   
Math to solve energy, time, speed of light, and the interaction between different Electromagnetic Fields and Alternating Energy.

1. Understanding the Math Solution,
    1. Pressure X Orbits per cycle = The length a strand travels to compete a unit of Time.
    2. Heat Intensity = The speed a strand travels
    3. Energy = The resistance of the Strand. It`s strength to change and power to resist, Heat Transfer to and from other strands.
    4. Orbits per cycle = The number of times a strand completes 360° turns in tell it reach its starting point.
    5. Cycle = A unit of Time. the point a strand repeats its path, crossing over its starting point.
   
2. Math Solution.
       
    1. ( Pressure X Orbits per cycle / Heat Intensity= Time Intensity ), or ( Length / speed = Time ).
        * a equal change in Heat intensity and pressure (as a %) changes the Energy Level, not the time Intensity.
    2. ( Pressure X Orbits per cycle X Heat Intensity = Energy ), or ( Length X Speed = Energy ).

3. Scales.
   
    A. Scales must solve from theoretical absolute zero.
    B. the use of the scales on a object will calculate for it`s mass in the solutions.
            * I`ll use BTU`s and Psi a in the example solution.
        1. Absolute zero, −459.67° on the Fahrenheit scale.
        2. British thermal unit, It is approximately the amount of energy needed to heat 1 pound (0.454 kg) of water, its a traditional unit of energy equal to about 1.055 K Joules.
            A. Sensible Heat as a solid 491.67 BTU`s, (-459.67° to 32°), (~ - 491.67)
            B. Latent Heat of Fusion for Ice which is 144 BTU`s, (32°change in state), (491.67 - 635.67)
            C. Sensible Heat as a liquid 180 BTU`s, (32°to 212°), (635.67 - 815.67)
            D. Latent Heat of vapor is 970 BTU`s, (212°change in state), (815.67 - ~)
                *There is 673.67BTUs on 1 pound of water at 70°F
        3. Pounds per Squire Inch Absolute (Psi a) about 14.7 at sea level.
            A. 2.2 pound-mass of water fills a container whose volume is 4.2ft³
            B. 2.2 pound-mass of water has a volume of 15.12 Squire inches
            C. 2.2 pound-mass of water at 14.7 Psi a has 222.264 pounds of pressure on it.

4. Example Solution.

    A. Solve for 2.2pounds of 70°F water at 14.7psia.
       
        1. Pressure = 222.264 pounds of pressure
        2. Orbits per cycle = 3
            *(100% 666.801)(90% 600.1209)
        Heat Intensity = 1482.074 BTU`s

    B. Example 1, 2.2pounds of 70°F water at 14.7psia.
       
        1. Time Intensity
            222.267 X 3 / 1482.074 = 0.44991073320225575780966402487325
        2. Energy
            222.267 X 3 X 1482.074 = 988248.425274
   
    C. Example 2, Solve for 10% less Heat Intensity and pressure,
           
        1. Time Intensity
            2000.403 X 3 / 1333.8666 = 4.4991073320225575780966402487325
        2. Energy
            2000.403 X 3 X 1333.8666 = 8004812.2447194
   
    D. Compering the difference, 10% less Heat Intensity and pressure.
       
        1. Time Intensity    = 100%, with a  0% change.
        2. Energy        =  81%, with a 19% change.