DA Morgan offers:
Anyone that thinks they have a solution to the nature of the universe, and doesn't have a formal background in math and physics should start with proving the following:
e (raised to the power (pi * i)) + 1 = 0
The constant e is the base of the natural logarithm and has the value 2.718281828459045235360287471352662497757....
i is the square root of -1
pi, of course, is 3.141592653589793238462643383279502884197....
The proof is actually rather straight-forward. So if you can't handle this ... please don't try to tackle the rest of the universe.
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DA Morgan
Well DA, you get a quick victory with that proposition. Luckily for me I never got beyond the Solar System. It is possible for people to get useful and meaningful results even though they are not completely up on the physics of the universe. My efforts were based on simple mathematics because that was what I knew best and that was what I could work on a calculator and a computer in BASIC. I will offer you a simple test. My number for the minimal radius of this Solar System is 31,830,914,183.8. Let?s call it Sr. The Earth equatorial radius is published as 3963.3 for a diameter of 7926.6 miles. We will designate Er as the Earths radius in miles. Lastly the published orbital radius for the Earth is 92,961,440. We can call it Or. The published data is in conflict by authors but I have no control over that. This is an excerpt from my book. I am sharing an important solar radius determination with you all.
Sn/(Or-Er) = 342.424464537 with a square root of 18.504714657
Sn/(Or+Er)= 342.395268066 with a square root of -18.503925747
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.00078891
For ease of use I round this to .000789, let?s call it our significant number, sn.
What I have done is deduct the tracts of the Earths closest edge from the track of the Earths farthest edge, otherwise represented in orbital velocities of miles per second. This number is very useful and is offered and used for each planet in my book.
Example: Earths circumference divided by sn provides the seconds in one complete revolution around the sun. Earths years days of 365.25 times .000789 provides the equatorial surface velocity of the Earth in miles per second, or .2882 mps.
Your test is to use your standard physics to prove to me that the surface velocity of the Earth?s equator is .2882 miles per second without my Sn.
Jim Wood
