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the spiral are called a wormhole, or "time tunnel"), at center of the mind part.

Tác giả: Mr.Smile
We will calculate the cross-sectional area of ​​the spiral as follows:

The diameter of the circle is Φ = 1.618 and its reverse is 1/Φ = 0.618.

The circumference of the circle is: π × Φ or π × 1/Φ

The cross-sectional area of ​​the circle is: π × (Φ/2)^2 or π × (1/2Φ)^2

At each point in time, a length of a constant e = 2.71828 is created, which is a time interval t, where t is a Fibonacci sequence arranged on the time axis of an infinite number of different time intervals:

Thus, the volume of the spiral with one time interval with all the mathematical constants (π, Φ, and e) related to each other is:

V1 = π × (Φ/2)^2 × e (are called a wormhole, or "time tunnel"), at center of the Mind part.

Or:

V2 = π × (1/2Φ)^2 × e (are called a wormhole, or "time tunnel"), at center of the Buddhism Precepts part.

* For example, at the time If a particle moves in a spiral with a cross-section of t = 89, the total volume of the spiral with a cross-section of t = 89 will be:

V1 = π × (Φ/2)^2 × e × 89

Or:

V2 = π × (1/2Φ)^2 × e × 89

(With t from 0 to 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377....etc).

If a particle moves in a spiral with a cross-section time of t = 89, with the speed of light c = 3 × 10^8, and a distance of S = e × 89, then the time it takes to travel that distance is:

t1 = (e × 89)/(3 × 10^8) = 8.1 × 10^-7 (s).

We have: V1 × A1 = V2 × A2 ( A is cross-sectional area).

<=> c × (π × (1/2Φ)^2) = V2 × (π × (Φ/2)^2)

<=> c × (1/4Φ^2) = V2 × (Φ^2/4)

If V1 is the speed of light, then V2 will be 1/Φ^4 times V1 and vice versa.

=> V2 = c × (1/4Φ^2) × (4/Φ^2) = 1/Φ^4 × c

=> V2 = 43.770.061,3 (m/s).

=> t2 = 8.1 × 10^-7 / 1/Φ^4 = 55,5174 × 10^-7 (s)

=> If the Top Head part moves at the speed of light, the Buddhism Precepts part will be 1/Φ^4 times slower. Conversely, if the Buddhism Precepts part moves at the speed of light, the Top Head part will be 1/Φ^4 times slower.
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