### Algebraic Equation for Time

Posted:

**March 3rd, 2018, 2:57 pm**
by **Eodnhoj7**

Presented argument:

Time, as movement, is inseperable from a dualism of actual movement (locality as the observable direction of space) and potential movement (non-local movement where the direction of space is not observed) where the potential movement acts as a “barrier” through which actual movement exists.

T= Aω/(Bϕ + (1/(1 ≤ x) ≤ BΔ ≤ (A-B))ϕ) → 1ω/(1ϕ + (1/(1≪(n →∞)) = BΔ)ϕ)

### Re: Algebraic Equation for Time

Posted:

**June 12th, 2018, 12:29 pm**
by **Eodnhoj7**

What we observe as time is strictly relations between movement. For simplicity we will use a 24 hour clock as an example. Aω is equivalent 86,400 seconds being the potential relations of one cycle as “day”. Bϕ, is the actual relations at 3:00 a.m or 10,800 seconds.

As actual relations, Bϕ is a grade of Aω as:

T=________Aω________ → T= ________86,400ω______

.................Bϕ.............................10,800ϕ

Bϕ has a constant state of change added to it, considering it is in a constant state of movement. This span of change relative to Aω, maintains a window of movement through A – B equivalent to 75,600.

T=________Aω________ → T= ________86,400ω______

.............Bϕ + (A-B)ϕ.............10,800ϕ+ 75,600ϕ

Adding (A – B) to Bϕ, or 10,800+ 75,600, does not take into account the change as progression from B → A as 1 cycle of movement, or 10,800 → 86,400 as 1 cycle.

This change begins with 10,800+1/(1 ≤ x) . This is considering all measurements of unity begin with 1 or a fraction of 1 as potential unity, with this unity itself equivalent to the second.

It ranges to 10,800 + 75,600 therefore is equal to 1/(1 ≤ x) ≤ BΔ ≤ (A-B) where BΔ is equivalent to a constant change.

T=___________Aω___________ → T= ____________86,400ω_____________

......Bϕ +(1/(1 ≤ x) ≤ BΔ ≤ (A-B).........10,800ϕ + (1/(1 ≤ x) ≤BΔ ≤75,600)ϕ

This constant change ranges from 1/(1 ≤ x) to 75,600 and is indefinite as pinpointing one movement causes a change in the measurement. Take for example observing three seconds later at 10,803 causes a change in the measurements as:

T= ____________ 86,400ω_____________

.....10,803ϕ + (1/(1 ≤ x) ≤ BΔ ≤ 75,597)ϕ

BΔ is equivalent to a constant state of change as relation. This change acts as linear relation between 1/(1 ≤ x) and (A – B). In these respects BΔ, as change observes an approximation between 1/(1 ≤ x) and 75,597.

Using the example above and observing a measurement where the cycle is complete the equation can be observed as:

T= __________86,400ω__________ → T= _________1ω___________

.........86,400ϕ+ (1/(1 ≤ x)≤ BΔ )ϕ...............1ϕ+ (1/(1 ≤ x) ≤ BΔ )ϕ

Considering 1ϕ+ (1/(1 ≤ x)≤ BΔ ) would require 1ω to exist as a fraction of:

_____1ω_____ → ____1ω____

.(1ϕ+ 1ϕ = BΔ)........(2ϕ = BΔ)

if 1/( 1 )= BΔ then; 1ϕ+ (1/(1 ≤ x)≤ BΔ ) must change to:

T= __________1ω___________

....1ϕ + (1/(1≪(n →∞)) = BΔ )ϕ

Where x is equivalent to a number that tends towards infinity. In these respects Time is always approximate as it is always divided by a continuous change at the peak of its cycle as the perpetual relation of particulate. In these respects what we understand of time is merely approximation of movement.

In summary Time is equivalent to Potential Particulate relations divided by Actual Particulate Relations plus a fraction less than or equal to one that is less than or equal to Actual particulate change which is less than or equal to A minus B.

T= ____________Aω____________ → ___________1ω___________

.... Bϕ + (1/(1 ≤ x) ≤ BΔ ≤ (A-B))ϕ .......(1ϕ + (1/(1≪(n →∞)) = BΔ )ϕ