### Derivation of s / t = 1

Posted:

**Sat Apr 27, 2019 1:45 pm**This is not meant to be a formal definition.

That is beyond me at this point in my development.

There exists the distinct possibility this derivation is wrong and/or false (hope not!).

Please check my work.

Simple integration over time yields this Newtonian classic:

v (s/t) = a (s/t

Given:

acceleration (a) = 10 m/s

distance (x) = 250 m

time (t) = 5 s

Solve:

velocity (normalized to time): v

velocity (normalized to distance): v

These, of course, are averages as we are dealing with acceleration, or

If we multiply an

If we divide an

Yes,

In the case we are most familiar (Case 1), we create that ratio by normalizing to time and not to space as this is the ordinary bias that forms our perception of 3D coordinate space -- three dimension of motion in space with one dimension of scalar (clock) time.

Therefore:

acceleration (spacial): a

acceleration (temporal): a

Sidebar: temporal acceleration would be the conjugate of spacial acceleration and therefore would be our "causal" "force" (in actuality just another aspect of motion and neither causal nor a-causal in its own right)

Assert:

a = a

Ergo:

v

but v

so 1 × x = 1 ÷ t

or

put otherwise, F = ma

Why mass though? Why is mass the fulcrum?

Is this an inherent bias from our spacial viewpoint? It must be as mass ceases to have traditional meaning outside of a larger gravitational field (i.e. 3D space with clock time). Would its conjugate be the Aether?

More questions than answers.

"Thanks to Einstein and others, we know that time and distance are equivalent and interchangeable, in many ways." - Miles Mathis

That is beyond me at this point in my development.

There exists the distinct possibility this derivation is wrong and/or false (hope not!).

Please check my work.

Simple integration over time yields this Newtonian classic:

v (s/t) = a (s/t

^{2}) × tGiven:

acceleration (a) = 10 m/s

^{2}distance (x) = 250 m

time (t) = 5 s

Solve:

velocity (normalized to time): v

_{t}= a × t = 10 m/s^{2}× 5 s = 50 m/svelocity (normalized to distance): v

_{x}= a ÷ x = 10 m/s^{2}÷ 250 m = 0.04 s^{-2}These, of course, are averages as we are dealing with acceleration, or

*change*in velocity over time (or conjugate velocity as*change*in energy, our "forcing function" [Ed: lulz], over distance).If we multiply an

*acceleration*by a time, we get a*velocity*(s/t^{2}× t = s/t → v)... magnitude, no direction.If we divide an

*acceleration*by a distance, we*also*get a*velocity*... but how can this be? The units don't work? Or do they...Yes,

**inverse seconds squared**is an**equivalent velocity**. We define it directly as 1/t^{2}being the equivalent motion in unit time and so being an angular velocity is a velocity nonetheless. This is the basis for the second power relationship in moving from the first to second*equivalent*dimension of motion (linear velocity, v → orbital velocity, v^{2}).- In Case 1 (velocity normalized to time) -- MOTION IN
**SPACE**or SPEED -- we do this all the time and is our standard practice. 50 m/s is 50 meters per 1 second. We have normalized time (to unity) by setting 1 = 5 seconds.

- Conversely, in Case 2 (velocity normalized to distance) -- MOTION IN
**COUNTERSPACE**or ENERGY -- we invert our operation (as we*also*invert our operand, i.e. space for time) where we then normalize distance (to unity) by setting 1 = 250 meters. This is only valid if what is stated above is true. i.e. we take inverse time to the second power to indeed be an equivalent velocity.

*change*).In the case we are most familiar (Case 1), we create that ratio by normalizing to time and not to space as this is the ordinary bias that forms our perception of 3D coordinate space -- three dimension of motion in space with one dimension of scalar (clock) time.

Therefore:

acceleration (spacial): a

_{x}= v_{x}× x = 0.04 s^{-2}× 250 m = 10 m/s^{2}acceleration (temporal): a

_{t}= v_{t}÷ t = 50 m/s ÷ 5 s = 10 m/s^{2}Sidebar: temporal acceleration would be the conjugate of spacial acceleration and therefore would be our "causal" "force" (in actuality just another aspect of motion and neither causal nor a-causal in its own right)

Assert:

a = a

_{t}= a_{x}= 10 m/s^{2}Ergo:

v

_{x}× x = v_{t}÷ tbut v

_{x}= v_{t}as the kinetics described remain the same, we only change our perspective of motion (and have shown in both cases to have created an equivalent motion, either temporal or spacial)so 1 × x = 1 ÷ t

or

**x ÷ t = 1**put otherwise, F = ma

Why mass though? Why is mass the fulcrum?

Is this an inherent bias from our spacial viewpoint? It must be as mass ceases to have traditional meaning outside of a larger gravitational field (i.e. 3D space with clock time). Would its conjugate be the Aether?

More questions than answers.

"Thanks to Einstein and others, we know that time and distance are equivalent and interchangeable, in many ways." - Miles Mathis