Article 9, No Big Whoop Part IV B. THE THEORY OF EVERYTHING...

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⚠️WARNING ⚠️HARD HAT AREA: THE INITIAL CONSTRUCTION OF THIS ARTICLE IS DONE, BUT UNTIL I SEE THE SUN—the macro realm—THERE MAY BE SUBSTANTIAL MODIFICATIONS NEEDED IN THE MICRO REALM OF ARTICLE 9 THAT I COULD NOT FORESEE DURING THE CONSTRUCTION PHASE.

Topic 12, continued

Q&A 7 through Q&A 12

We can’t go on like this, pretending that everything modern-science says with numb3rs is sacred, when all we are actually getting from modern-science is what I call FRUIT BOWL MATH, by which I mean MATHEMATICAL EQUATIONS THAT REPLICATE OBSERVATIONS.

Math that merely REPLICATES WHAT OUR EYES SEE is like a fruit bowl painting—it is a work of ART, not a LAW of the Universe—and before we can ELEVATE the FRUIT BOWL MATH to the STATUS OF LAW, we must APPLY LOGIC to ensure that the fruit bowl math is TELLING THE WHOLE STORY.

Logic is what is missing—and self-interested magical thinking is abundant—in the FRUIT BOWL MATH we get from modern-science.

Now Imma go Crazy On You and show: the cause of ACCELERATION is NOT unbalanced, aka NET, CONSTANT FORCE!

And we already know that RELATIVE SPACETIME IS NEVER GOING TO BE THE RIGHT ANSWER TO ANY QUESTION.

“By their fruit you will recognize them. Do people pick grapes from thornbushes, or figs from thistles?” —Jesus

No, but that doesn’t stop modern-science from mocking intelligent life, as evidenced by the two fruitless pickings on the topic of ACCELERATION—“CONSTANT FORCE did it” and/or “SPACETIME did it” 😂—that modern-science expects us “believe in,” so therefore I respectfully submit that the only option any rational individual REALLY has is to figure-out how Newton’s Second Law of Motion, F = ma, REALLY works.

No one can credibly argue that once you eliminate the impossible, what remains is multiple alternatives.

“Once you eliminate the impossible, whatever remains, however improbable, must be the truth.” —Sherlock Holmes

There’s undoubtedly only ONE way to explain Newton’s Second Law of Motion, F = ma, and before now, no other theory of motion has even TRIED TO EXPLAIN Newton’s First Law of Motion (an object staying in motion without external force application); but today, we can EXPLAIN BOTH PHENOMENA SIMPLY by applying previously-derived principles of the TOEWC.

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Q&A 7

Q: Continuing to refer-to the ATM Accelerating Down a 3DADL Diagram, what would happen if, IN THE ON-TRACK TIME-SLOT OF INSTANT 2, Father Time FORCELESSLY UN-STEERED the two BTPs directly across from each other—i.e., on opposite sides of the MACOM—in the Lowest Circle of MACOM SheLL, causing their 423DCTCCs to tilt toward the MACOM again?

A: It’s still unclear.

We know that it’s safe to assume that all BTPs in the same ATM have the SAME NORMAL WPD, and present-day science’s interpretation of Newton’s Second Law of Motion (F = ma) tells us that CONSTANT FORCE causes acceleration, so therefore we could invoke Newton’s First Law of Motion on that basis alone.

Let’s do it: we would conclude that if the BTPs in the Highest Circle of MACOM SheLL have the SAME WPD in instant 3 as they did in instant 1, then the MSs they are providing to the MACOM have effectively been BALANCED (by the equal-magnitude, opposite-direction MSs provided by the two UN-STEERED BTPs in the Lowest Circle of MACOM SheLL), and have therefore CEASED TO ACT-ON THE MACOM, and as such, the ATM should follow Newton’s First Law of Motion, continuing to move at CONSTANT-VELOCITY in the 3-d Down direction on the ATM COMPASS unless the ATM experiences a new unbalanced MS^^.

^^that’s indeed what present-day science tells us that Newton’s First Law of Motion says.

It’s not *exactly* 😂 clear how to determine *what is persevering* and *why*.

And that’s important to clarify.

Fortunately, the question itself—how to determine *what is persevering* and *why* in Newton’s First Law of Motion?—affords us THE OPPORTUNITY to investigate.

And here’s what we need to INVESTIGATE: we’ve ENCOUNTERED A CONUNDRUM that demands reconciliation, to wit: how is it physically-possible for our ATM to satisfy Newton’s First Law of Motion if the UN-STEERING of the two BTPs in the Lowest Circle of MACOM SheLL caused the MACOM to experience BALANCED NET FORCE, i.e., if SETFx = 0 at 0 on the MACOM SheLL Compass?

In other words, are we prepared to say that BALANCED NET FORCE CAN CAUSE CONSTANT-VELOCITY MOTION, INSTEAD OF REST??

Spoiler Alert: NO! But that’s the conundrum we SEEM to be facing if we assume that the MS FORCE provided by the two MSP BTPs in the Highest Circle of MACOM SheLL remained CONSTANT!

So before we go crazy on Newton’s First Law of Motion, let’s start with that assumption and VERIFY OUR CONCLUSION that SETFx = 0 at 0 after the UN-STEERING of the two BTPs in the Lowest Circle of MACOM SheLL, by re-doing the MOTION ANALYSIS (between the 🐢 icons, below):

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Regarding MAGNITUDE SETFx in instant 3—after the UN-STEERING of the two BTPs in the Lowest Circle of MACOM SheLL—both MSP BTPs in the Highest Circle of MACOM SheLL are continuing to do what they were doing previously (they are tipped UD, so we assign a negative value to the MSs they are providing.)

And now we also have to account-for the UN-STEERED BTPs in the Lowest Circle of MACOM SheLL (they are tipped UU, so we assign a positive value to the MSs they are providing).

MAGNITUDE SETFx = -(Fmack + FfSet) - (Fmack + FfSet) + (Fmack + FfSet) + (Fmack + FfSet) = 0

We verify that the MACOM is experiencing NO UNBALANCED FORCE in instant 3, after the UN-STEERING of the BTPs in the Lowest Circle of MACOM SheLL.

Regarding DIRECTION SETFx in instant 3, the two MSP BTPs in the Highest Circle of MACOM SheLL have a midpoint of 180-degrees on the MACOM SheLL Compass {(120 +240)/2 = 180}, and the two MSP BTPs in the Lowest Circle of MACOM SheLL have a midpoint of 0-degrees on the MACOM SheLL Compass {(330 +30)/2 = 180, but we add 180 because those BTPs are tipped UU}, so there are ZERO degrees between those midpoints, which means that they are forces in opposite directions along the same line.

RECALL from the MOTION ANALYSIS How-To (between the 🐢 icons in Topic 4):

And that’s the case we’ve seemingly got now with the UN-STEERED BTPs!

Ergo, DIRECTION SETFx = 0

SETFx = ZERO at ZERO on the MACOM SheLL Compass.

This means the MACOM is NOT EXPERIENCING UNBALANCED FORCE, it is experiencing REST.

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And therefore *if* we assume CONSTANT WPD, *then* Newton’s First Law of Motion cannot be satisfied, because MOVEMENT CANNOT HAPPEN UNLESS THERE IS AN EXPERIENCE OF UNBALANCED NET FORCE to cause it.

That conclusion cannot be contradicted; it’s a simple TRUISM!

I am NOT going to argue otherwise, and say that BALANCED FORCE can cause MOTION; nope! BALANCED FORCE CAUSES REST.

And we can’t argue with THE OBSERVATION of Newton’s First Law of Motion; it’s too simple to find any fault with!

Ergo, we know that SETFx is *NOT REALLY* ZERO at ZERO on the MACOM SheLL Compass—the MACOM is *not really* experiencing NO UNBALANCED FORCE, aka REST—after the UN-STEERING of the two BTPs in the Lowest Circle of MACOM SheLL!

And that’s FINE, because it leads us inexorably to our SECOND conundrum—the ($64 BAZILLION MONEY)² QUESTION—which we have mentioned but not emphasized yet: WHY/HOW DOES ACCELERATION OCCUR?

It’s DEFINITELY NOT because AN UNBALANCED *CONSTANT FORCE* is causing acceleration!

I’m 100 entitled to make that statement and let CRITICS hoist themselves on their own petards.

Here I cite proof that IT IS COMMON-SENSE KNOWLEDGE that it’s IMPOSSIBLE for present-day science to EXPLAIN why/how Newton’s Second Law of Motion, F = ma, works by making the assumption of CONSTANT FORCE.

“What is this power being represented by, if the force is constant?”

How does an ordinary Moving Service Provider INCREASE POWER TEMPORARILY IN A SINGLE DIRECTION TO ACCELERATE A BOX LIKE A BOSS, then to top it all off, turn around in a different direction and GO BACK TO NORMAL POWER to move the same box in the new direction as if the SUPERPOWER was TURNED-OFF?

This is the question.

In case it’s difficult to read my notes, I wrote:

“This highlights the issue about the source of power for F = ma. The reason why the box-pushing example is important is because it shows how non-sensical it is for the box’s speed to be CAUSED by the force F, but yet the force F cannot ‘keep up’ with the box!! We can solve that problem by putting the box on top of the Force-provider. And that’s when the OG question is seen clearly: what is the source of acceleration of the CONSTANT Force-provider???”

I am writing in June, 2026.

Gary Allen and Not_Einstein were writing in November, 2020.

Isaac Newton published F = ma in 1687.

“Once you eliminate the impossible, whatever remains, however improbable, must be the truth.” —Sherlock Holmes

Begin SOLVING F = ma MYSTERY with THREE (3) CLUES (between the 🔍 icons)

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Clue 1 🔍. There’s nothing physically-impossible about what our motion theory is saying so far.

So far, our motion theory is telling us this:

*if* THE MOVING SERVICE (in this case, the MOVING SERVICE is the WPD of the two MSP BTPs in the Highest Circle of MACOM SheLL) REMAINS CONSTANT, *then* the UN-STEERING of the two BTPs in the Lowest Circle of MACOM SheLL will cause the MACOM to experience BALANCED NET FORCE, with SETFx = 0 at 0 on the MACOM SheLL Compass.

That’s PERFECT.

Clue 2 🔍 . Newton’s First and Second Laws of Motion are telling us WHAT IS *APPARENTLY* HAPPENING, i.e., what our eyes and other detecting instruments SEE HAPPENING.

That’s also PERFECT, because we are SELF-AWARE, which means that WE KNOW that WHAT WE SEE does not always reveal to us WHY WE ARE SEEING what we see; as adults, WE ARE AWARE there are MANY REASONS WHY what we are seeing MIGHT BE MISLEADING US, at worst, or SIMPLY MIGHT NOT BE THE FULL PICTURE, at best.

Ergo, we don’t have to SLAVISHY WRITE AND FOLLOW MATH EQUATIONS THAT REPLICATE WHAT OUR EYES SEE to “do science right”!

In fact, math equations that merely REPLICATE WHAT OUR EYES SEE are like fruit bowl paintings—they are ART, they are NOT LAWS—which means that before we can ELEVATE “FRUIT BOWL MATH” TO THE STATUS OF LAW, we must APPLY LOGIC to ensure that the fruit bowl math is TELLING THE WHOLE STORY.

Sometimes, ADDING ASSUMPTIONS will complete the picture, and if that’s the case, then the fruit bowl math becomes a priceless work of art that illustrates A NATURAL LAW.

Other times, the picture is poisoned ☠️ by self-centered dreams, in which case the fruit bowl math is a worthless piece of science fiction that illustrates magical thinking that might nevertheless be well preserved, like a fossilized turd, as a reminder that every creature on God’s green earth sh*ts the bed in their youth, but only humans know enough to deprioritize mental hygiene in favor of a sweaty wad of cash and zero—or worse: *feigned*—respect from every ordinary mortal “doubting Thomas,” Dick and Harry who would gather together to “celebrate a life of love” after declaring a death of love.

The point is that adding more fruit bowl math on top of poison fruit bowl math would be FRUITLESS, so therefore we can’t HONOR Newton’s Laws of Motion again until we understand whether we’re looking at fruitful fruit bowl math or poison fruit bowl math.

Logically, we know that it’s PHYSICALLY-IMPOSSIBLE for ACCELERATION per Newton’s Second Law of Motion (F ‎ = ma) to be caused by a CONSTANT FORCE.

We IMAGINE that’s what we’re seeing, but when we get into the nitty of the gritty of an acceleration thought-project, WE REALIZE that cannot be the whole story.

“Something has been worked on us,” to quote insurance fraud investigator Barton Keyes in the 1944 movie “Double Indemnity.”

So now at least we have found the THREE PROBLEMS.

THE FIRST PROBLEM is that Newton’s First Law of Motion (constant-velocity motion according to F = mv) cannot even kick-in until the WINNING MS ceases to act on the object (and no new unbalanced NET FORCE in a DIFFERENT DIRECTION acts-on the object); but as we have seen, whenever there is motion, THERE IS ALWAYS GOING TO BE A WINNING MS (a MSP BTP *continuing to act* on the MACOM) INSIDE OF AN ATM, and so the problem is that Isaac Newton did not account-for that fact (he was only OBSERVING objects EXTERNALLY), and therefore Newton’s First Law of Motion doesn’t actually describe any REAL PHYSICAL SITUATION! To repeat: it’s physically-impossible for the WINNING MS to cease to act on the object UNLESS the object experiences a new unbalanced NET FORCE—a new WINNING MS—in a DIFFERENT DIRECTION! And again (as we will show, below), the reason why is also the explanation for acceleration, to wit: if the MACOM of an ATM has experienced SETFx > 0, then unless DIRECTION SETFx changes, the WINNING MSP BTP(s) inside of the ATM will always have INCREASED WPD MAGNITUDE as compared to NORMAL WPD, so therefore the WINNING MSP BTP(s) can’t technically *cease* acting-on the MACOM until DIRECTION SETFx changes.

THE SECOND PROBLEM IS THE F = ma MATH: Newton’s Second Law of Motion, F = ma, is PHYSICALLY-IMPOSSIBLE TO BE TRUE for a force that remains *constant* over time.

THE THIRD PROBLEM IS PROVING HOW IT WOULD BE POSSIBLE for THE MAGNITUDE of a force with a constant value at one time TO INCREASE TO ANOTHER CONSTANT VALUE AT A SUBSEQUENT TIME, and thereby cause OBJECT ACCELERATION per F = ma, for so long as no NEW unbalanced NET FORCE began acting on the same object in a NEW DIRECTION.

All three problems can be SOLVED by EXPLAINING WHAT IS REALLY HAPPENING TO CAUSE ACCELERATION per Newton’s Second Law of Motion.

Before we do that we should ACKNOWLEDGE that Isaac Newton is not necessarily THE CAUSE of the problems!

Isaac Newton could argue: #1, F = mv is an OBSERVATION only, NOT a THEORY OF MOTION!; and #2, F = ma IS NOT WRONG—AND WAS NEVER WRONG—per se, because IT IS TRUE THAT ALL FORCES ARE CONSTANT FORCES IN THE SAME INSTANT when they are acting!

Isaac Newton could also argue that he was making the assumption of Absolute Time, so by no means was he CONCLUDING that a force had to remain CONSTANT FROM TIME TO TIME, but rather, Isaac Newton was LEAVING THE DOOR WIDE OPEN FOR OTHER SCIENTISTS TO COME IN AND *EXPLAIN* THE “INNER-WORKINGS” of Newton’s Second Law of Motion!

In particular, Isaac Newton could credibly argue that if—instead of scientists coming-in and explaining the “inner-workings” of Newton’s Second Law of Motion—the scientists went-ahead and DENIED ABSOLUTE SPACE AND TIME, and thereby COMPLETELY LOST ALL HOPE OF *PROPERLY* APPLYING NEWTON’S SECOND LAW OF MOTION (in other words, if the scientists-in-charge were CORRUPT and got all wrapped-up in Einstein’s invisible fabric of spacetime in a Faustian bargain with Satan to CALL THE CREATOR NON-ESSENTIAL TO SUCCESS IN THE SCIENTIFIC STUDY OF CREATION), then that was not Isaac Newton’s fault.

Begin pause for Mini-Q&A

Mini-Q: As long as a MSP continues to act-on an object, why should the MSP have to GIVE BACK *ANYTHING* AFTER WINNING A NET FORCE COMPETITION FOR CONTROL OF THE OBJECT in a certain direction (the “Controlling MS Direction”) unless and until THERE IS A NEW WINNER OF THE NET FORCE COMPETITION THAT IS COMING FROM ANOTHER DIRECTION???

Mini-A: An object can’t ARGUE AGAINST THE NET FORCE it is experiencing; our theory of motion must ACKNOWLEDGE that it’s POSSIBLE for the DIRECTION of the NET FORCE to remain the same, but yet the MAGNITUDE of the NET FORCE could change (either INCREASING or DECREASING.)

In the limited situation we’re studying—lone BTPs and BTPs inside of ATMs—that’s only going to happen *if and only if* a new unbalanced force (a “NUF”) acting-in THE SAME DIRECTION OR IN THE OPPOSITE DIRECTION ALONG THE SAME LINE AS THE CONTROLLING MS DIRECTION crops-up; this is because THE APPLICATION OF FfSet WILL ALWAYS CHANGE THE CONTROLLING MS DIRECTION, and FfSet is applied whenever one lone BTP acts-on another in a different direction than the CONTROLLING MS DIRECTION, and whenever a BTP acts-on the MACOM inside of an ATM.

The limited situation we’re studying is actually the SIMPLEST CASE, but don’t worry, it’s not actually the most INTUITIVE case, so therefore MOTION will actually get EASIER to understand as we enter “the real world” of massive objects, where FfSet is not noticeable (outside of an ATM, FfSet is not seen or heard-from again), and mass is highly variable.

In “the real-world” of massive objects, there can be many and varied SMALLER new unbalanced forces (“SNUF”s) ACTING ALONG DIFFERENT LINES THAN THE CONTROLLING MS DIRECTION that come and go and CHANGE THE MAGNITUDE OF THE NET FORCE that an object is experiencing WITHOUT CHANGING THE CONTROLLING MS DIRECTION.

But in every situation—in our limited situation in the micro-realm (with the DIRECTION-DICTATING Big Bang Force FfSet ever-present), *and* in the more nuanced universe of massive objects with which we are intuitively familiar—if the MAGNITUDE of the NET FORCE *increases* in the Controlling MS Direction (which would only be the case if a NUF was acting-in THE SAME DIRECTION along the same line), then the MSP would actually have to GAIN MORE POWER per P = Fv, and the object would have to ACCELERATE FASTER THAN BEFORE per F = ma (with the amount of POWER/ACCELERATION GAIN being equal to THE MAGNITUDE OF THE NET FORCE INCREASE in the Controlling MS Direction.)

But OTOH, if the MAGNITUDE of the NET FORCE *decreases* in the Controlling MS Direction (which would always be the case if a NUF was acting-in THE OPPOSITE DIRECTION along the same line *OR* if a SNUF was acting along a different line), then the MSP would have to GAIN LESS POWER per P = Fv, and the object would have to REDUCE THE RATE OF ACCELERATION per F = ma (with the amount of POWER/ACCELERATION LOSS being equal to THE MAGNITUDE OF THE NET FORCE REDUCTION in the Controlling MS Direction.)

For example, imagine that you are in a frictionless environment and you are riding on top of an ACCELERATING MOTORIZED VEHICLE.

Assume that whatever external physical force is applied to get the vehicle’s engine working, the effect of that external physical force on all *engine components* is LINEAR MOTION, NOT ROTATION (because we haven’t contemplated ROTATION yet.)

The Moving Service Provider (“MSP”) is the engine, and the Moving Service (“MS”) is the engine power, and the Moving Service Recipient (the “vehicle”) is the object with mass “m” that is being acted-on by the MS, which is accelerating the vehicle in a certain direction (the “Controlling MS Direction”) per Newton’s Second Law of Motion, F = ma.

Imagine that YOU CANNOT STEER THE VEHICLE, but YOU CAN APPLY THE BRAKES TO THE VEHICLE, so you go-ahead and do that—assume that you apply a constant braking force (“BF”) along the same line but in the *opposite direction* as the Controlling MS Direction (let’s call that the “Braking Direction”)—and assume that magnitude of the BF is less than the magnitude of the MS.

NOTE that the BF is a NUF, not a SNUF; we have not studied SNUFs yet.

NOTICE that whether or not you stopped “pushing the gas” when you applied the BF is RELEVANT.

If you stopped “pushing the gas,” then this is the situation covered by Newton’s First Law of Motion—an *external* unbalanced force stops acting-on an object and no other *external* unbalanced force acting along *a different line* begins acting-on the object—and in this situation, ACCELERATION CEASES.

Also, if you stopped “pushing the gas” when you applied the BF, then WHEN YOU STOPPED APPLYING THE BF YOU WOULD NOT AUTOMATICALLY *RESUME THE OG RATE OF ACCELERATION*; you would only resume the OG rate of acceleration if you “pushed the gas” again.

But as we will show, below, the fact that ACCELERATION CEASED does not mean that the vehicle ceased being FORCEFULLY DRIVEN!

The vehicle didn’t just start COASTING ALONG when the acceleration ceased!

OMG! NO!!

And that is what Newton’s First Law of Motion is MISSING, to wit: THE PRESENCE OF POWER after an external acceleration-causing force ceases; THERE IS POWER CAUSING THE MOTION, there is NOT “COASTING”!!!

Once the mechanism of action of ACCELERATION is known (and we’re getting there soon), then it will become very clear why IT’S IMPOSSIBLE TO STOP THE MS FROM ACTING-ON THE VEHICLE *INSIDE OF THE ATM* AT *THE COM OF THE VEHICLE* unless and until THE DIRECTION OF THE NET FORCE experienced by the vehicle CHANGES. And on this set of facts, that didn’t happen!

To repeat: whether or not you were still “pushing the gas” when you applied the brakes, THE PEDAL WAS STILL TO THE METAL *INSIDE* OF THE ATM AT THE COM OF THE VEHICLE.

So if STOPPING THE VEHICLE WAS THE GOAL, then the strategy was a FAIL; you either needed BIGGER BRAKES, or a way to CHANGE THE ENGINE DIRECTION (e.g., put it in reverse), because unless the vehicle experiences NET FORCE in a NEW DIRECTION (NOT MERELY A FORCE IN THE *OPPOSITE DIRECTION* ON THE *SAME LINE*), the vehicle is going to keep moving in the Controlling MS Direction.

And if you didn’t stop “pushing the gas” when you applied the lesser-magnitude NUF BF, then you are forked; now the problem isn’t figuring out how much THE VEHICLE VELOCITY was reduced, as was the problem when you stopped “pushing the gas” during the BF application, but rather, now the problem is figuring out how much THE BASE RATE OF ACCELERATION was reduced; in other words, the velocity is going to keep increasing—merely at a slower rate of increasing than before lol—if you didn’t stop “pushing the gas” when you applied the brakes.

“A train that size going that fast will vaporize anything in front of it.” —Frank, “Unstoppable”

The Unstoppable movie presents a contrived situation, but it highlights the physics issues that aren’t easy to see when all you’re looking at is the unexplained-since-1687 fruit bowl math F = ma.

Think about it some more: the train in the Unstoppable movie was UNMANNED, meaning that the train’s acceleration was happening NATURALLY, per F = ma, simply because it was “in gear.” SURELY THE TRAIN HAD BIG BIG BRAKES (and I assume that they do make BRAKES THAT ACCELERATE), but the brakes were not accessible, and nothing that could be moved into the path of the train could generate enough FORCE to COMPETE with the engine power (sh*te just sitting there on the tracks can’t compete with a missle the size of the Chrysler Building *accelerating* against the friction of rusty metal.) What they needed to stop the train was a bigger force in a different direction (a bigger force than AN ACCELERATING MISSLE THE SIZE OF THE CHRYSLER BUILDING), e.g., a derailment event with a city-sized event radius, OR a second train accelerating in the opposite direction, and they chose the second train option.

In theory, a force-substitute in the form of an electromagnetic smack 👋 in the caboose with an Emack TFG—blasting the output of the motor with a burning hot radiation gun—could have interfered destructively ☠️ with enough of the train engine’s WPD to stop the train, but there isn’t any such a thing as an industrial-sized portable radiation machine that’s effective for *preventing* widespread destruction.

End of pause for Mini-Q&A

NOW WE’RE READY TO PROVE HOW IT WOULD BE POSSIBLE for THE MAGNITUDE of a force with a constant value at one time TO INCREASE TO ANOTHER CONSTANT VALUE AT A SUBSEQUENT TIME, and thereby cause OBJECT ACCELERATION per F = ma, for so long as no NEW unbalanced NET FORCE began acting on the same object in a NEW DIRECTION.

Clue 3 🔍 . RECALL from Topic 3:

ALSO RECALL from Topic 9:

NOTE that it is the FIRST WAY in which a NEW WOPR Stream Spectra 🐳 md becomes an OLD WOPR Stream Spectra 🐳 md that is relevant to accelerated motion per Newton’s Second Law of Motion, F = ma.

Now CONTINUE RECALLING from Topic 3:

Next, we need to explain how, *exactly*, a BTP’s NORMAL WPD could be INCREASED TO CAUSE ACCELERATION (and subsequently DECREASED TO CAUSE DECELERATION!) per Newton’s Second Law of Motion, F = ma.

In every ATM, the Universal O/S installed in the 6-d COM Segment of the MACOM must be “clocking” SETFx (MAGNITUDE SETFx and DIRECTION SETFx) at the end of the On-Track Time-Slot (which is the beginning of the Pit-Stop Time-Slot), by monitoring the ENERGY-MAIL BOXES at all the doors of the MACOM SheLL Compass.

Knowing SETFx, the Universal O/S can determine whether or not to put the ATM into a new “ACCELERATION CYCLE.”

Universal O/S:

Begin ACCELERATION CYCLE—STAY OR GO—DECISION-TREE 🎄example (between the 🎄 icons), which the Universal O/S could implement at the beginning of every Pit-Stop Time-Slot.

Also see “F1 The Movie”.

No pressure.

NOTE that the following Decision-Tree is only applicable to LINEAR MOTION; after we study ROTATION, we can easily update the Decision-Tree.

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Decision-Tree 🎄. At the beginning of the Pit-Stop Time-Slot, the Universal O/S would check DIRECTION SETFx, aka the Controlling MS Direction, to determine SHOULD I STAY OR SHOULD I GO NOW? in the On-Track Time-Slot:

Option 1, STAY in the active acceleration cycle if DIRECTION SETFx is unchanged, and keep doing Acceleration Cycle Math (between the 🧮 icons), then proceed to evaluate BRANCHES 🕊️ 1 and 2 (between the 🕊️ icons) to determine what MOTION OR ATTEMPTED MOTION to instruct BTP WPD Management to ENABLE for the WINNING MSP BTP(s) during the On-Track Time-Slot, and relax during the Pit-Stop Time-Slot while BTP WPD Management re-sets the WPD of all of the BTPs in the ATM to ENABLE THAT MOTION OR ATTEMPTED MOTION to be done during the On-Track Time-Slot;

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Branch 1 🕊️. If DIRECTION SETFx is the same, and if no NUF or SNUF is acting-on the MACOM (RECALL that a NUF is a new unbalanced force—in the same direction or in the opposite direction along the same line as DIRECTION SETFx, aka the Controlling MS Direction—that wasn’t present in the same measure in instant 1, when the acceleration cycle began; ALSO RECALL that a SNUF is a SMALLER new unbalanced force along a different line than DIRECTION SETFx), then at the beginning of the Pit-Stop Time-Slot, the Universal O/S uses the base rate of acceleration determined in instant 1 (see Acceleration Cycle Math, MATH 1 🧮) to calculate the present-instant velocity (see MATH 2 🧮), which becomes the new present-instant WPD MAGNITUDE Number (see MATH 3 🧮) that is sent to BTP WPD Management, and BTP WPD Management re-sets the WPD of the WINNING MSP BTP(s) in the active acceleration cycle to the new WPD MAGNITUDE Number (and simultaneously, BTP WPD Management re-sets the WPD of the other BTPs in the ATM to NORMAL WPD), and that causes the ATM to display ACCELERATED MOTION in the On-Track Time-Slot, according to Newton’s Second Law of Motion.

Branch 2 🕊️. If DIRECTION SETFx is the same, but if a NUF or a SNUF is acting-on the MACOM, then at the beginning of the Pit-Stop Time-Slot, the Universal O/S ADJUSTS THE BASE RATE OF ACCELERATION (see MATH 2 🧮), and uses the adjusted base rate of acceleration to calculate the present-instant velocity (also see MATH 2 🧮), which becomes the new present-instant WPD MAGNITUDE Number (see MATH 3 🧮) that is sent to BTP WPD Management, and BTP WPD Management re-sets the WPD of the WINNING MSP BTP(s) in the active acceleration cycle to the new WPD MAGNITUDE Number (and simultaneously, BTP WPD Management re-sets the WPD of the other BTPs in the ATM to NORMAL WPD), and that causes the ATM to display ADJUSTED ACCELERATED MOTION in the On-Track Time-Slot, according to Newton’s Second Law of Motion.

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Option 2, STAY at rest, if no acceleration cycle is active and if DIRECTION SETFx is unchanged, then instruct BTP WPD Management to ENABLE BUPKIS during the On-Track Time-Slot, and relax during the Pit-Stop Time-Slot while BTP WPD Management re-sets the WPD of all the BTPs in the ATM to NORMAL WPD to ENABLE THE BUPKIS to be done during the On-Track Time-Slot;

Option 3, GO chuck-out the active acceleration cycle or prior rest and begin a new acceleration cycle if DIRECTION SETFx has changed and MAGNITUDE SETFx > 0, doing new Acceleration Cycle Math between the 🧮 icons, then proceed to evaluate BRANCH 🕊️ 3 to determine what MOTION OR ATTEMPTED MOTION *and* WINNING MSP BTP(s) to instruct BTP WPD Management to ENABLE for the On-Track Time-Slot, and relax during the Pit-Stop Time-Slot while BTP WPD Management re-sets the WPD of all of the BTPs in the ATM to ENABLE THAT MOTION OR ATTEMPTED MOTION to be done during the On-Track Time-Slot;

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Branch 3 🕊️. If DIRECTION SETFx is different, then that would mean that there was a NEW NET FORCE (and also a fresh MAGNITUDE SETFx caused by a new set of WINNING MSP BTPs acting-on the MACOM), so then at the beginning of the Pit-Stop Time-Slot, the Universal O/S GOes and begins a new acceleration cycle by doing new first-instant Acceleration Cycle Math (see MATH 1 🧮), then the Universal O/S sends the first-instant WPD MAGNITUDE Number (see MATH 3 🧮)— and identifies the WINNING MSP BTP(s)—to BTP WPD Management, which uses the first-instant WPD MAGNITUDE Number to re-set the WPD of the WINNING MSP BTP(s) in the new acceleration cycle (and simultaneously, BTP WPD Management re-sets the WPD of the other BTPs in the ATM to NORMAL WPD), which causes the ATM to CHANGE DIRECTION AND SPEED in the On-Track Time-Slot.

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Option 4, GO to rest, if DIRECTION SETFx has changed and MAGNITUDE SETFx = 0, then instruct BTP WPD Management to ENABLE BUPKIS for the On-Track Time-Slot, and relax during the Pit-Stop Time-Slot while BTP WPD Management re-sets the WPD of all the BTPs in the ATM to NORMAL WPD to ENABLE THE BUPKIS to be done during the On-Track Time-Slot.

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End of ACCELERATION CYCLE—STAY OR GO—DECISION-TREE 🎄example

Begin Acceleration Cycle Math (between the 🧮 icons); NOTE that all force-values are in units of *wave-length force per instant*, and all velocity values are in units of *m/s*.

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MATH 1 🧮.

In an acceleration cycle, the ATM’s velocity in the first On-Track Time-Slot (the first instant, aka “instant 1,” of the acceleration cycle)—which is also the base rate of acceleration in DIRECTION SET Fx, aka the Controlling MS Direction—is calculated according to: {MAGNITUDE SETFx = mv},

so therefore {v1 = base rate of acceleration = MAGNITUDE SETFx/m}

MATH 2 🧮.

*if* no new NUF or SNUF is acting-on the MACOM (a NUF is a new unbalanced force—*in the same direction or the opposite direction* along the same line as DIRECTION SETFx, aka the Controlling MS Direction—that *wasn’t present in the same measure* in instant 1, when the acceleration cycle began; a SNUF is a smaller new unbalanced force along a different line than DIRECTION SETFx),

*then* the base rate of acceleration remains the same, and subsequent-instant ATM velocities are calculated according to:

{Newton’s Second Law of Motion, F = ma},

where a = v²,

and where v² = {present-instant ATM velocity = (base rate of acceleration, or if no acceleration, then 1) MULTIPLIED BY (prior-instant ATM velocity)};

*if* a NUF or a SNUF is acting-on the MACOM,

*then* there is an ADJUSTMENT CAUSED BY THE NUF OR THE SNUF, calculated according to:

{adjustment caused by NUF = NUF/m}

{adjustment caused by SNUF = SNUF/m},

and if the NUF was in *the same direction* as SETFx, {adjusted base rate of acceleration = base rate of acceleration + adjustment caused by NUF},

and if the NUF was in *the opposite direction* as DIRECTION SETFx, {adjusted base rate of acceleration = base rate of acceleration - adjustment caused by NUF},

and in the case of a SNUF, {adjusted base rate of acceleration = base rate of acceleration - adjustment caused by SNUF},

then subsequent-instant ATM velocities are calculated according to:

{Newton’s Second Law of Motion, F = ma},

where a = v²,

and where v² = {present-instant ATM velocity = (adjusted base rate of acceleration, or if no acceleration, then 1) MULTIPLIED BY (prior-instant ATM velocity PLUS OR MINUS adjustment caused by NUF or SNUF)}

NOTE that MATH 2 assumes that if an unbalanced *EXTERNAL* source of NET FORCE caused the active acceleration cycle, then that EXTERNAL source of force is continuing to act; if such an EXTERNAL source of force CEASED TO ACT, then acceleration would have to cease—the adjusted base rate of acceleration would be NO ACCELERATION—and the adjustment caused by NUF or SNUF would only be used to determine present-instant VELOCITY.

MATH 3 🧮.

Establish {present-instant WPD MAGNITUDE Number = present-instant ATM velocity}

Begin Mini-explanation of present-instant WPD MAGNITUDE Number (MATH 3 🧮)

RECALL that in general: {POWER = (Force) MULTIPLIED BY (velocity)}

This tells us that if an ATM is in an acceleration cycle, then the present-instant ATM POWER (the POWER of the ATM as-a-whole) is calculated according to:

{ATMPower = (the total number of WINNING MSP BTPs contributing to MAGNITUDE SETFx in the acceleration cycle) MULTIPLED BY (present-instant velocity)}

This also tells us that if an ATM is at-rest, then the present-instant ATM Power is calculated according to:

{ATM Power = “internal force of mass” only, aka ATM Relative Mass}

This further tells us that in the case of an individual WINNING MSP BTP, which is going to receive a certain WPD MAGNITUDE Number, aka WPD POWER Number, the present-instant WPD MAGNITUDE Number is calculated according to:

{WPD MAGNITUDE Number = (1 BTP) MULTIPLIED BY (present-instant velocity)}

NOTE that the WPD MAGNITUDE Number has to be implemented by the specific Wave-Length Logs in the Atomic Spectra of the ATM (i.e., the SPECIFIC WAVE-LENGTH LOGS in the NORMAL WPD of the ATM.)

Now we can envision the WPD MAGNITUDE Number as the means of BUYING WINNING MSP BTP VELOCITY with different “denominations” of “currency”—different Wave-Length Logs with different force-values—in an Atomic Spectra: ANY VELOCITY UP TO THE TERMINAL VELOCITY for a particular MSP BTP can be BOUGHT WITH THE DIFFERENT FORCE-VALUES (DENOMINATIONS) OF WAVE-LENGTH LOGS IN THE BTP’s ATOMIC SPECTRA, simply by increasing the WPD Magnitude Number, and adding or subtracting individual Wave-Length Logs in ONE COPY OF THE WPD to “make change” to “pay for” the EXACT VELOCITY required!

Then the Universal O/S could send the increased WPD Magnitude Number to BTP WPD Management at the beginning of the Pit-Stop Time-Slot.

RECALL from Topic 3:

During every Pit-Stop Time-Slot, BTP WPD Management is responsible-for GLOWING-UP EVERY PP/3-d SheLL WITH AN ARTIFICIAL *INTERNAL* FORCE OF MASS, aka WPD, AND THEREBY CREATING AN ARTIFICIAL BABY TURTLE PARTLE (BTP) by instructing the 6DCT in the same Snowman of God to skim from the 4-d Space River Spectrum 🔥a specific collection of Wave-Length Logs, which we’re calling a 4-d Spectra Assembly 🪵.

And again: it is possible to SKIM MULTIPLE COPIES OF THE 4-d SPECTRA ASSEMBLY 🪵—adding or subtracting individual Wave-Length Logs in ONE COPY to “make change” to “pay for” the EXACT VELOCITY required—so it’s like GOING TO THE LOCAL MONEY-MINTING FACILITY AND PRINTING COPIES OF DENOMINATIONS OF MONEY, because the force of the one Eternal unit of mass in the 4-d Space River Spectrum 🔥is “c,” so there are plenty of COPIES of any given Wave-Length Log in the 4-d Space River Spectrum to buy any force/velocity < c.

BTP WPD Management: *receiving the new WPD MAGNITUDE Number from the Universal O/S*

BTP WPD Management UP-CYCLES the 4-d Spectra Assembly 🪵 to the AIP inside of the 3-d SheLL, where it becomes the 3-d WOPR Stream Spectra 🐳 md, which is heading in the same direction that the 423DCTCC has been STEERED (either FORCEFULLY or FORCELESSLY), so that when the 3-d SheLL makes the WPD Tax Payment 🔫 (by reversing the 🐳 md and THRUSTING IT OUT to satisfy Newton’s Third Law of Motion), the total THRUST, Emack, causes the 3-d SheLL to apply an equivalent physical force, Fmack, at the AIP, with Emack and Fmack together comprising artificial WPD—the ARTIFICIAL *INTERNAL* FORCE OF MASS—that the 3-d SheLL can GO! and work with during the On-Track Time-Slot.

BTP WPD Management: *cheerfully concluding the tour of the Eternally-Burning Radioactive Waste Recycling Center*

And that’s how the WPD MAGNITUDE Number becomes the new WPD of the WINNING MSP BTPs in a particular ATM (i.e., the BTPs that contributed to MAGNITUDE SETFx in the active acceleration cycle), which is what enables the ATM-as-a-whole to achieve its velocity.

End of Mini-explanation of present-instant WPD MAGNITUDE Number (MATH 3 🧮)

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End of Acceleration Cycle Math

Resuming the Clue 3 🔎 discussion, the significance hits us: that is a VERY BIG WHOOP!

CONSTANT FORCE DOES NOT CAUSE ACCELERATION!

In fact, AN ATM’S INTERNAL FORCE OF MASS—which is the physical force an ATM applies along its line of movement or rest, so in the case of an ATM with velocity, we could call it “FmackATMPower”—IS NOT CONSTANT FROM TIME TO TIME DURING ATM ACCELERATION!

End of Clue 3 🔎

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End of SOLVING F = ma MYSTERY with THREE (3) CLUES

Resuming the discussion of Q&A 7 where we left off, we RECALL that we didn’t actually get anywhere further than ASKING THE QUESTION and NOTICING that present-day science’s interpretation of Newton’s Second Law of Motion, F = ma, cannot be correct; so NOW, having found and fixed-up the problem with present-day science’s interpretation of Newton’s Second Law of Motion, we are left with the task of applying our own solution to answer our original question.

RECALL Q&A 7 :

Returning to the MOTION ANALYSIS of the scene illustrated in the ATM Accelerating Down a 3DADL Diagram, what we’ve determined so far is that in instant 2, when the UN-STEERING occurs, the ATM will already be moving in the 3-d Down direction along the 3DADL with Terminal Velocity, according to Newton’s Second Law of Motion.

RECALL the exact MOTION ANALYSIS we did in Q&A 6 to arrive-at the ATM’s Terminal Velocity:

Whew!

Now all we have to do is step into the shoes of the Universal O/S at instant 3, then go through the Decision-Tree 🎄:

We NOTICE that we’ve got to do two tasks before we can rule-in or rule-out Option 1:

task (i), figure-out where we’re at in the Acceleration Cycle Math;

and

task (ii), check DIRECTION SETFx.

Regarding task (i), we already did most of the Acceleration Cycle Math when we completed the MOTION ANALYSIS in Q&A 6; we notice that MATH 3 🧮 at instant 1 and instant 2 is the only Acceleration Cycle Math we didn’t already do.

If we had done MATH 3 🧮 at instant 1 and instant 2, then we would have known that the WPD MAGNITUDE Number of the two MSP BTPs in the Highest Circle of MACOM SheLL increased in instant 2.

Instant 1 WPD MAGNITUDE Number = v1 = 2,326 wave-length force per instant

Instant 2 WPD MAGNITUDE Number = v2 = 5,410,276 wave-length force per instant

Whoop there it is! THAT’S NOT CONSTANT FORCE causing object acceleration! That’s ACCELERATING FORCE causing object acceleration! Now FINALLY, Newton’s Second Law of Motion, F = ma, MAKES SENSE!!!

Regarding task (ii), now when we check DIRECTION SETFx in instant 3, we no longer find that DIRECTION SETFx = 0, as it was when we falsely assumed that the FORCE, aka WPD MAGNITUDE Number (the magnitude of Fmack and Emack), of the two MSP BTPs in the Highest Circle of MACOM SheLL REMAINED CONSTANT AND EQUAL-TO the FORCE, aka NORMAL WPD (the magnitude of Fmack and Emack), of the two UN-STEERED MSP BTPs in the Lowest Circle of MACOM SheLL!

Here I repeat from above regarding HOW TO FIND DIRECTION SETFx:

RECALL from the MOTION ANALYSIS How-To (between the 🐢 icons in Topic 4):

Yes! This is the situation we see in instant 3, with EACH of the two MSP BTPs in the Highest Circle of MACOM SheLL having WPD MAGNITUDE Number = 5,410,276 at 180-degrees, and with EACH of the two UN-STEERED BTPs in the Lowest Circle of MACOM SheLL having WPD MAGNITUDE Number = 2,326 at 0-degrees.

Ergo, in instant 3, DIRECTION SETFx = 180, which is THE SAME as it was in instant 2 and instant 1.

So the Decision-Tree 🎄 is done, and what happens is Option 1: the ATM must STAY in the active acceleration cycle in instant 3.

Now we proceed to evaluate BRANCHES 🕊️ 1 and 2 to determine what MOTION OR ATTEMPTED MOTION the Universal O/S will instruct BTP WPD Management to ENABLE during the On-Track Time-Slot, and we see that Branch 2 🕊️ is applicable, because DIRECTION SETFx is the same, and now a NUF—the two UN-STEERED BTPs in the Lowest Circle of MACOM SheLL—is also acting-on the MACOM in the opposite direction along the same line as the two WINNING BTP MSPs in the Highest Circle of MACOM SheLL.

Yes!

We’re ready to run this b*tch down!

We don’t have an EXTERNAL force acting-on the ATM, so we can’t eliminate the *possibility* of acceleration ourselves, we have to let the MATH tell us that answer; all we’ve got to do is consult MATH 2 🧮 to calculate the ADJUSTED BASE RATE OF ACCELERATION, if any, with the NUF present, then we can figure-out the ATM’s INSTANT-3 VELOCITY, v3, which will become the instant-3 WPD MAGNITUDE Number for the WINNING MSP BTPs in the Highest Circle of MACOM SheLL.

MATH 2 🧮:

We NOTICE that the MAGNITUDE of the NUF must be the same as MAGNITUDE SETFx was in INSTANT 1 (which means that the MAGNITUDE of the NUF = v1), since all of the BTPs have the same NORMAL WPD.

Ergo, adjustment caused by NUF = v1.

Next, let’s calculate the adjusted base rate of acceleration.

adjusted base rate of acceleration = base rate of acceleration - adjustment caused by NUF

Ergo, the adjusted base rate of acceleration = v1 - v1 = 0 (meaning that there will be NO ACCELERATION in instant 3!)

Finally, let’s calculate the velocity, “v3,” in instant 3.

v3 = (adjusted base rate of acceleration, or if no acceleration, then 1) MULTIPLIED BY (prior-instant ATM velocity - adjustment caused by NUF)

v3 = (1) MULTIPLIED BY (v2 - v1)

v3 = v2 - v1 = 5,410,276 - 2,326 = 5,407,950 m/s

The ATM should hold steady at CONSTANT-VELOCITY, v3, which is LESS THAN TERMINAL VELOCITY.

Ta-da, ta-done!

v3 = the new WPD MAGNITUDE Number that the Universal O/S sends to BTP WPD Management, which uses it to re-set the WPD of the two MSP BTPs in the Highest Circle of MACOM SheLL (and simultaneously, BTP WPD Management would re-set the WPD of the other BTPs in the ATM to NORMAL WPD), and that would cause v3 (the ATM VELOCITY in instant 3.)

In subsequent instants, the exact same thing would happen, with the ATM traveling at CONSTANT VELOCITY, v3, according to Newton’s Second Law of Motion, NOT Newton’s First Law of Motion.

In other words, Newton’s First Law of Motion is really only a limited-circumstance case of Newton’s Second Law of Motion, describing the situation when the source of the base rate of acceleration ceases to act, BUT THE SAME ACCELERATION CYCLE CONTINUES, so that the increased WPD MAGNITUDE Number of the WINNING MSP BTPs inside of the ATM is not re-set, and continues to act, causing motion.

In our case, the source of the base rate of acceleration was (FfSet + NORMAL WPD) of the two WINNING MSP BTPs in the Highest Circle of MACOM SheLL acting-on the MACOM when the two BTPs in the Lowest Circle of MACOM SheLL were STEERED AWAY from the MACOM; then what happened after the UN-STEERING is that the (FfSet + NORMAL WPD) was BALANCED, and the only source of UNBALANCED NET FORCE acting-on the MACOM was the increased WPD MAGNITUDE Number of the two WINNING MSP BTPs in the Highest Circle of MACOM SheLL.

Whoop whoop! We have EXPLAINED NEWTON’S FIRST LAW OF MOTION, and we have found *a critical link* between the micro realm and the macro realm!

This might seem like no big whoop, but IMO it’s actually THE BIGGEST WHOOP IN PHYSICS since 1687, when Isaac Newton published his Laws of Motion: we have OFFICIALLY EXPLAINED *OBSERVATIONS* OF MASSIVE OBJECTS EXPERIENCING CONSTANT-VELOCITY MOTION IN THE ABSENCE OF EXTERNAL UNBALANCED FORCE APPLICATION, and the explanation is that INSIDE OF THE ATM at the COM of the massive object, THE WINNING MSP BTPs acting-on the MACOM do not cease acting—they CONTINUE TO HAVE INCREASED WPD MAGNITUDE Numbers as compared to the NORMAL WPD for those ATMs—until the object experiences NET FORCE in a NEW DIRECTION!

And NOT ONLY THAT, but we have also explained how and why DECELERATION occurs, to wit: a NUF (which is acting-in the opposite direction as the WINNING MS) or a SNUF has a magnitude greater than the first-instant base rate of acceleration caused by the WINNING MS.

End of Q&A 7!

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Q&A 8

Q: In the ATM Accelerating Down a 3DADL Diagram, FmackATMPower (not labeled) would be in the 3-d Down direction on the ATM SHELL/Compass (and also on the MACOM SheLL Compass, but we have to keep in mind that the two compasses are different, having different diameters and different ENERGY-MAIL BOXES at different 3-d locations of the QOL); but this raises a question about the corresponding Emack that causes FmackATMPower—“EmackATMPower,” let’s call that Emack, which is actually comprised of the WPD Tax Payments 🔫 made during the Pit-Stop Time-Slot by each of the WINNING MSP BTPs upon being GLOWED-UP with a new internal force of mass (3-d WOPR Stream Spectra 🐳) corresponding to the new WPD MAGNITUDE Number—and the question is this: does EmackATMPower remain “split” between the two MSP BTPs in the Highest Circle of MACOM SheLL?

A: No, that would not make sense, because an ATM can only move in ONE DIRECTION at any given instant, so an ATM’s “POWER WPD,” aka ATMPower (comprising FmackATMPower and EmackATMPower) must be expressed in opposite directions along the same line, on different sides of the ATM.

See MATH 3 🧮:

And remember that the ATM SHELL/Compass has ENERGY-MAIL BOXES (locations of the QOL) in every 3-d direction on the ATM SHELL/Compass, which would enable the Emack from each BTP in the ATM to be AUTOMATICALLY DEPOSITED INTO AN ASSOCIATED ENERGY-MAIL BOX of the ATM SHELL/Compass.

Then during an acceleration cycle, the Universal O/S could instruct the ENERGY-MAIL BOX of each WINNING MSP BTP to *FORWARD* its Emack to a special-purpose ENERGY-MAIL BOX on the ATM SHELL/Compass (the “ATMThrust ENERGY-MAIL BOX,” let’s call it) that lies in the opposite direction along the same line—on the other side of the ATM—as DIRECTION SETFx.

In that manner, the *FORWARDED* Emacks of the WINNING MSP BTPs would AUTOMATICALLY *CONSTRUCTIVELY INTERFERE* with each other in the ATMThrust ENERGY-MAIL BOX, and thereby GENERATE EmackATMPower!

And as a result, the ATM AS-A-WHOLE would get POWER WPD, with FmackATMPower in one direction and EmackATMPower in the opposite direction along the same line on different sides of the ATM.

End of Q&A 8!

☕️☕️☕️☕️☕️☕️☕️☕️☕️☕️

Q&A 9

Q: Referring to the conclusion of Q&A 8—which is that during an acceleration cycle, an ATM has POWER WPD (with FmackATMPower in one direction and EmackATMPower in the opposite direction along the same line on different sides of the ATM)—does this mean that if the ATM is at-rest, with SETFx = 0 at 0 (meaning that MAGNITUDE SETFx = 0 and DIRECTION SETFx = 0 on the MACOM SheLL Compass), then the ATM has NO WPD?

A: No, an ATM must have WPD when it’s at rest (“REST WPD,” let’s say, comprising “FmackATMRest” and “EmackATMRest”), because the BTPs inside of the ATM SHELL/Compass are always expressing NORMAL WPD!

To create FmackATMRest, the Universal O/S could instruct each ENERGY-MAIL BOX of the ATM SHELL/Compass to *FORWARD* the Emack of the associated BTP to an ATMThrust ENERGY-MAIL BOX that lies in the opposite direction along the same line as DIRECTION SETFx = 0, which is going to be the 180-degree direction on the ATM SHELL/Compass (so let’s call it the “ATMBackDoor ENERGY-MAIL BOX”), which would then emit EmackATMRest, as illustrated in the “ATM At-Rest Diagram”:

An at-rest ATM would always express FmackATMRest at 0-degrees on the ATM SHELL/Compass and EmackATMRest at 180-degrees on the ATM SHELL/Compass.

And the MAGNITUDE of the ATM’s REST WPD would equal the total internal force of mass of the ATM (aka ATM Relative Mass) at “one” velocity.

FmackATMRest = ATM Relative Mass

End of Q&A 9!

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Q&A 10

Q: Would an ATM lose REST WPD (see Q&A9) when it enters an acceleration cycle and gets POWER WPD (see Q&A 8)?

A: Not entirely, because NOT ALL of the BTPs in the ATM can be WINNING MSP BTPs in the same acceleration cycle, so therefore when an ATM has POWER WPD, it must also have “REST-MINUS WPD” (comprising “FmackATMRestMinus” and “EmackATMRestMinus”).

That might seem strange—for one ATM to have two WPDs that can have different directions—since we know that an ATM can only MOVE in ONE DIRECTION!

But there is no contradiction, because when the ATM is “under power”—i.e., in an acceleration cycle—the POWER WPD has control over the ATM’s direction and speed.

And we can’t forget that at the beginning of every Pit-Stop Time-Slot—when all of the forces experienced by the MACOM must be accounted-for during MOTION ANALYSIS to determine SETFx (MAGNITUDE SETFx and DIRECTION SETFx)—the ATM’s “Mass BTPs” (all of the BTPs in the ATM *except for* the WINNING MSP BTPs) were included in the calculation of MAGNITUDE SETFx, so their influence is always being asserted.

Present-day science recognizes this fact, too, with reference to an object’s “inertia,” which is often described as an object’s “resistance” to attempts to change the object’s speed or direction with external force.

“Inertia,” however, is both a vague concept (it’s non-specific) AND an ambiguous concept (it’s confusing, because Newton’s First Law of Motion is also often referred-to as the law of inertia.)

So we are not going to continue talking about an object’s “inertia” or an object’s “resistance to changes in motion”; we are going to be referring-to an object’s REST WPD (if the MAGNITUDE OF THE NET FORCE experienced by object is zero), and an object’s REST-MINUS WPD (if the MAGNITUDE OF THE NET FORCE experienced by the object is > 0).

And to repeat how REST-MINUS WPD works: not all BTPs in an ATM are WINNING MSP BTPs contributing to MAGNITUDE SETFx during an active acceleration cycle, when an ATM has POWER WPD; the Mass BTPs in the ATM—the BTPs that are NOT WINNING MSP BTPs contributing to MAGNITUDE SETFx in the active acceleration cycle and therefore aren’t “carrying their own weight”—continue to express NORMAL WPD, but they are all “dead weight” (merely passengers of the MACOM) in the ATM that has POWER WPD. And we’ve already deduced that the Emack of the WINNING MSP BTPs is being *FORWARDED* to the ATMThrust ENERGY-MAIL BOX, where it is emitted to create POWER WPD (FmackATMPower and EmackATMPower) for the ATM as-a-whole, ergo, it stands to reason that when an ATM has POWER WPD, the Emack of the Mass BTPs can simply continue to be *FORWARDED* to the ATMBackDoor ENERGY-MAIL BOX, where it would be emitted to create REST-MINUS WPD (FmackATMRestMinus and EmackATMRestMinus).

{FmackATMRestMinus = (FmackATMRest, which is ATM Relative Mass) MINUS (Atomic Number MULTIPLIED BY number of WINNING MSP BTPs)}

Now we can UPDATE the original ATM Accelerating Down a 3DADL Diagram—let’s call the updated diagram “ATM Accelerating Down a 3DADL Diagram, Showing POWER WPD and REST-MINUS WPD”—to depict the two different types of ATM WPD that co-exist.

First, NOTICE the POWER WPD, which comprises FmackATMPower and EmackATMPower.

Because DIRECTION SETFx is 180-degrees on the MACOM SheLL Compass, the ATM Thrust ENERGY-MAIL BOX and the emitted EmackATMPower are at 0-degrees on the ATM SHELL/Compass, which causes FmackATMPower at 180-degrees.

RECALL that only the two WINNING MSP BTPs in the Highest Circle of MACOM SheLL are contributing to MAGNITUDE SETFx in the active acceleration cycle, so therefore only their Emack is being *FORWARDED* to the ATMThrust ENERGY-MAIL BOX to create EmackATMPower.

Second, NOTICE the REST-MINUS WPD, which comprises FmackATMRestMinus and EmackATMRestMinus.

Remember that all of the Mass BTPs in the ATM continue to express NORMAL WPD, so their Emack continues to be *FORWARDED* to the ATMBackDoor ENERGY-MAIL BOX, where it is emitted to create EmackATMRestMinus, which causes FmackATMRestMinus at 0-degrees on the ATM SHELL/Compass.

End of Q&A 10!

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Q&A 11

Q: What would happen if the ATM (let’s call it “ATM A”) in the ATM At-Rest Diagram and the ATM (let’s call it “ATM B”) in the ATM Accelerating Down a 3DADL Diagram, Showing POWER WPD and REST-MINUS WPD, collided along the same 3DADL, as illustrated in the “ATM Collision On 3DADL Diagram”?

A: The only way to find out is to do the MOTION ANALYSIS (between the 🐢 icons):

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First, we simply look at the scene and do some “red flag” 🚩spotting, looking for atypical or special issues.

This case wins all the medals in the Atypical and Special Issues Red Flags Olympics.

We see three red flags 🚩:

Red Flag (i), ATM B is trying to be a Moving Service Provider and a Moving Service Recipient at the back door 🚪 (180-degrees) on the ATM B SHELL/Compass;

and

Red Flag (ii), ATM B is giving ATM A an EmackATMBRestMinus TFG smack 😘 in the FmackATMARest;

and

Red Flag (iii), ATMs on the loose in the universe cannot tip UD, like BTPs on the loose in the universe can, but nevertheless, we need to assign positive/negative values to the MAGNITUDES OF THE PHYSICAL FORCES they are experiencing, so that we can find THE MAGNITUDE OF THE NET FORCE (MAGNITUDE SETFx).

Begin RED FLAG 🚩ANALYSIS

Red Flag (i), ATM B is trying to be a Moving Service Provider and a Moving Service Recipient at the back door (180-degrees) on the ATM B SHELL/Compass.

We RECALL (see Q&A 4) that THE BACK DOOR 🚪of an ATM will always be the only MSR in any collision that occurs at the back door, and neither ATM in the collision can become a Moving Service Provider or a Moving Service Recipient vis-a-vis the other, and therefore the ATMs are going to end-up STUCK TOGETHER 🫂, linearly MOTIONLESS, until they experience a change in DIRECTION SETFx.

Significance of Red Flag (i) to the MOTION ANALYSIS:

We don’t have to *account-for* either of the Moving Services (FmackATMBPower and FmackATMARest) that ATM B and ATM A, respectively, are trying to provide to each other, because THE BACK DOOR 🚪of ATM B will be the only MSR in the collision, but we do still need to determine the ATTEMPTED MOTION, if any, of ATM A and ATM B.

Red Flag (ii), ATM B is giving ATM A an EmackATMBRestMinus TFG smack 😘 in the FmackATMARest.

EmackATMBRestMinus will be COURTESY-FORWARDED by the ATM A ENERGY-MAIL BOX at 0-degrees on the ATM SHELL/Compass to the ATM A BackDoor ENERGY-MAIL BOX, where it will CONSTRUCTIVELY INTERFERE with EmackATMARest, and therefore give ATM A a WPD Goose 🪿, which we’re calling “Goose Minus” (equal in magnitude to FmackATMBRestMinus).

FmackATMBRestMinus = (FmackATMBRest, which is ATM B Relative Mass) MINUS (Atomic Number of ATM B MULTIPLIED BY number of WINNING MSP BTPs in ATM B)

Significance of Red Flag (ii) to the MOTION ANALYSIS:

We have to count every Goose 🪿 of a TFG-receiver ATM’s Fmack as an EXTERNAL physical force experienced by the ATM.

Significance of Red Flag (iii) to the MOTION ANALYSIS:

We know that forces balance one another when they have the same magnitude but act-in opposite directions, ergo, we can simply DECIDE to continue the convention of assigning a positive value to the magnitude of a physical force that is trying to get an ATM to move in “forward gear” (i.e., toward any degree in the top half of the ATM SHELL/Compass), and assigning a negative value to the magnitude of a physical force that is trying to get an ATM to move in “reverse gear” (i.e., toward any degree in the bottom half of the ATM SHELL/Compass), then we can simply ADD THE MAGNITUDES OF THE PHYSICAL FORCES TOGETHER to figure-out their BALANCE.

RECALL, however (see Topic 4), that this is a simplified approach to MOTION ANALYSIS that we’re using while we build ATMs from lone BTPs on the loose in the universe and get the ATMs moving without collisions; WPD differences won’t be “competing” until after all FfSet has been BALANCED, and FfSet balances *INSIDE OF ATMs*, so therefore we’re not worrying right now about how to deal with *different-magnitude forces* acting-in “West gear” (trying to cause “Western tilting” of the MACOM’s 423DCTCC) and “East gear” (trying to cause “Eastern tilting” of the MACOM’s 423DCTCC).

When we are ready to contemplate ATM 🏧 DEMOLITION DERBIES—when we’re ready to analyze INTER-ATM 🏧 INTERACTIONS—then we will have to figure-out how to account-for “West gear” and “East gear” EXTERNAL physical forces experienced by ATMs.

End of RED FLAG 🚩 ANALYSIS

Now we’re ready to determine SETFx of both ATMs.

Let’s begin with finding SETFx of the MACOM in ATM A (“SETFxA”).

We see that ATM A has REST WPD, so we know that each BTP in ATM A has the same NORMAL WPD, and SETFx = 0 at 0-degrees (we can either say “0-degrees on the MACOM SheLL Compass” or “0-degrees on the ATM SHELL/Compass”; I think it makes more sense to refer-to the ATM SHELL/Compass when we’re talking about collisions between different ATMs.)

Per Red Flag (i), we don’t have to count FmackATMARest in the MAGNITUDE SETFxA accounting, because the calculation of MAGNITUDE SETFxA will already account-for the internal force of mass (all of the Mass BTPs) of ATM A.

To determine MAGNITUDE SETFxA, we only have to count EXTERNAL forces experienced by ATM A.

And per Red Flag (ii), we have to count Goose Minus = FmackATMBRestMinus as an EXTERNAL force that ATM A is experiencing at 0-degrees on the ATM A SHELL/Compass.

FmackATMBRestMinus = ATM B Relative Mass MINUS (ATM B Atomic Number MULTIPLIED BY the number of WINNING MSP BTPs contributing to ATM B’s POWER WPD)

We are assuming that both ATMs are Hydrogen, which has Atomic Number 1 and a Relative Mass of 128,882 BTPs; and from Q&A 6, we know that there are TWO WINNING MSP BTPs in ATM B’s active acceleration cycle.

Goose Minus = FmackATMBRestMinus = 128,882 MINUS (1 MULTIPLIED BY 2) = 128,880 wave-length force

Per Red Flag (iii), Goose Minus is trying to get ATM A to move in “forward gear,” so we would assign a positive value to the magnitude of Goose Minus.

MAGNITUDE SETFxA = + (128,880)

DIRECTION SETFxA = 0-degrees on the ATM SHELL/Compass

Now we have to step into the shoes of the Universal O/S and go through the Decision-Tree🎄(see Q&A 7):

DIRECTION SETFxA did not change, so the decision for ATM A should be Option 2, STAY at rest.

But let’s PRETEND that we were WORRIED about IGNORING the change in MAGNITUDE SETFx—the UNBALANCED EXTERNAL FORCE (MAGNITUDE SETFxA = 128,880 at 0-degrees)—so we went-ahead and selected Option 3, GO chuck out the prior rest and begin a new acceleration cycle, doing new Acceleration Cycle Math between the 🧮 icons below, TO FIND-OUT WHAT WOULD HAPPEN.

🧮🧮🧮🧮🧮🧮🧮🧮🧮🧮

We don’t have any WINNING MSP BTPs acting-on the MACOM to speak of, but we know:

SETFxA = 128,880 wave-length force,

and Hydrogen Relative Mass = m = 128,882 BTPs

Substituting, we find:

v1 = 128,880/128,882 = .99998 = 1 wave-length force per BTP per instant

In other words, ATM A remains at-rest:

SETFxA = 0 at 0-degrees on the ATM A SHELL/Compass

AND THAT CHECKS-OUT, because INSIDE OF ATM A, THE MACOM IS NOT EXPERIENCING ANY UNBALANCED PHYSICAL FORCE, and as an at-rest TFG-receiver, ATM A would be EXPECTED to experience…wait for it…MORE REST.

So in fact, ATM A is NOT in an acceleration cycle, and it’s reassuring to see that we can’t “use math” to create acceleration that does not exist!

🧮🧮🧮🧮🧮🧮🧮🧮🧮🧮

Lesson learned: When you’re at-rest, you can’t get any MORE at-rest!

Now the Decision-Tree is done, and we can confidently select Option 2, STAY at rest, and we see that the Universal O/S will instruct BTP WPD Management to ENABLE BUPKIS, so then during the Pit-Stop Time-Slot, BTP WPD Management would re-set the WPD of all WPDs in ATM A to NORMAL WPD to ENABLE ATM A to DO BUPKIS in the On-Track Time-Slot.

ATM A WILL NOT MOVE OR ATTEMPT TO MOVE in the On-Track Time-Slot.

Now let’s find SETFx of the MACOM in ATM B (“SETFxB”).

We see that ATM B has a POWER WPD (FmackATMBPower and EmackATMBPower) *and* a REST-MINUS WPD (FmackATMBRestMinus and EmackATMBRestMinus.)

This means that ATM B has an active acceleration cycle, so we *DO* have to count FmackATMBPower in the MAGNITUDE SETFxB accounting, because that *is* UNBALANCED NET FORCE that ATM B is experiencing.

And FmackATMBPower is trying to get ATM B to move in “reverse gear,” so we would assign a negative value to the magnitude of FmackATMBPower.

But per Red Flag (i), we *DO NOT* have to count FmackATMBRestMinus in the MAGNITUDE SETFxB accounting, because the internal force of mass (all of the Mass BTPs) of ATM B are accounted-for in ATM B’s present-instant velocity.

MAGNITUDE SETFxB = -FmackATMBPower

FmackATMBPower = (the total number of WINNING MSP BTPs) multiplied by (present-instant ATM B velocity)

From Q&A 6, we know:

there were TWO WINNING MSP BTPs,

andTerminal Velocity (v2, reached in instant 2) = 5,410,276 m/s

Substituting, we find:

MAGNITUDE SETFxB = -{(2) x (5,410,276)} = -10,820,434 wave-length force

DIRECTION SETFxB = 180-degrees on the ATM B SHELL/Compass

SETFxB = 10,820,434 wave-length force at 180-degrees

Now we have to step into the shoes of the Universal O/S and go through the Decision-Tree🎄(see Q&A 7):

DIRECTION SETFxB did not change, so the decision for ATM B should be Option 1, STAY in the active acceleration cycle.

But per Red Flag (i), the fact that the ATM B’s BACK DOOR 🚪 is the only MSR in the collision PREVENTS MOVEMENT of ATM B along the 3DADL.

ATM B, however, does not know that!

ATM B will CONTINUE TO ATTEMPT TO MOVE at Terminal Velocity, with FmackATMBPower = 10,820,434 wave-length force.

🐢 🐢 🐢 🐢 🐢 🐢 🐢 🐢 🐢 🐢

End of Q&A 11!

☕️☕️☕️☕️☕️☕️☕️☕️☕️☕️

Q&A 12

Q: Are ATMs sufficient to create matter, or are both tooters *and* hooters necessary to create the universe as we know it?

The thrust of the question is this: we’ve made the case for a “bottomless” ATM, but we could, in theory, also make a “topless” ATM, which we could call a “TATM” 🦥, as illustrated in the “First TATM Diagram” and the “TATM At-Rest Diagram”.

The First TATM Diagram and the TATM At-Rest Diagram might seem odd with upside-down labels, but it’s important to emphasize that the sole structural difference between a TATM and an ATM is that a TATM is “topless” and an ATM is “bottomless.”

And we might not even need TATMs 🦥!

But it’s better to make them now just in case we need them later.

Either way, we definitely don’t need to RE-WRITE ATM HISTORY!

In theory, everything we’ve said about ATMs is also applicable to TATMs.

All we need to do is make FOUR NOTES about TATMs, then we can simply START USING THE TATMs *IF* WE FIND A NEED FOR THEM.

NOTE 1 about TATMs.

Every ATM and TATM Backbone comprises two BTPs: one Mothership Atomic COM (“MACOM”), and one Atomic Mass BTP, aka “BTP1.”

Wherever BTP1 is referenced in a discussion or a diagram about ATMs, that should be understood to be a reference to the *one Atomic Mass BTP* on the ATM and the TATM Backbone.

In the TATM, BTP1 is under the MACOM, so it *seems as if* there is not an Absolute-Relative relationship between the MACOM and BTP1, because the MACOM is “riding on top” of BTP1!

But this situation is not a problem—and there actually IS an Absolute-Relative relationship between the MACOM and BTP1—because as we can see, BTP1 is “hemmed in” and cannot “drive” the MACOM in any event (BTP1 is always MASS, never a MSP in this configuration), so that the MACOM and the surrounding BTPs all become one unbreakable object BY LAW, and the COM of the object is the MACOM; ergo, BTP1 is *effectively* RELATIVE-TO the MACOM in the TATM.

In other words, the COM of the TATM is what’s relevant in this case, and that’s because the MACOM and BTP1 are no longer SEPARATE LEGAL ENTITIES, but rather, they have become ONE OBJECT with a COM to which all other BTPs are POSITIONALLY-RELATIVE.

NOTE 2 about TATMs.

A TATM is “topless,” with the overall TATM structure having a “FLOWER OF LIFE”-like geometry, *except-for* the one missing “petal” in the 3-d Up (0-degrees) direction on the ATM SHELL/Compass, because the MACOM of a TATM is NOT BEHIND THE FRONT DOOR 🚪on the TATM SHELL/Compass, and therefore (see Q&A 4) THE FRONT DOOR 🚪 of a TATM will always be the only Moving Service Recipient in any collision that occurs at the front door; a collision along a 3DADL cannot cause the TATM to be either a Moving Service Provider or a Moving Service Recipient, nor can the other TATM or ATM in the collision become a Moving Service Provider or a Moving Service Recipient.

NOTE 3 about TATMs.

In a TATM, Father Time can use FORCELESS STEERING to cause the BTPs in the Highest Circle of MACOM SheLL to tilt toward the MACOM top, which will cause the TATM to move in “forward gear,” i.e., toward a degree in the top half of the ATM SHELL/Compass.

NOTE 4 about TATMs.

WE DO NOT HAVE TO WORRY ABOUT BTP1 being a “point of vulnerability” in the TATM structure, because the only way that an ATM and a TATM could experience a HEAD-ON COLLISION with each other along a 3DADL (and cause BTP1 in the ATM and the TATM to become a Moving Service Recipient) would be if the ATM was on the bottom and the TATM was on top, and one or both of them was already in an acceleration cycle,

but in that situation, before the collision, BTP1 would have been a Mass BTP contributing to FmackATMRestMinus or FmackTATMRestMinus, respectively.

Ergo, even though BTP1 would become a Moving Service Recipient during the collision, the only possible CAUSE of a DIRECTION CHANGE as a result of the collision would be a POWER WPD, and NEVER THE REST-MINUS WPD to which BTP1 was contributing!

This means that BTP1 WOULD ALWAYS REMAIN A MASS BTP after a head-on collision between an ATM and a TATM (we never have to worry about BTP1 “driving” collisions!), and therefore THE ATM/TATM STRUCTURE IS UNBREAKABLE.

Now, having completed that “tour of duty,” we are free of historical constraints, and we can use the knowledge of ATMs and TATMs we’ve built without worry about everything falling apart like a house of cards; in fact, ATMs and TATMs are like Willy Wonka’s Everlasting Gobstoppers, which can never get any smaller or larger in all of forever after the manufacturing process is over.

End of Q&A 12!

☕️☕️☕️☕️☕️☕️☕️☕️☕️☕️

In joy,

Frank

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Article 9, No Big Whoop Part IV B. THE THEORY OF EVERYTHING IS COMING UP RAINBOWS: INTRODUCING ATMs 🏧 and TATMs🦥, Expect Mothership Day Brunch On Jesus