Kevin Lee's "Weirder Than Normal/Norma? Now That's Weird" · 1:13pm Feb 25th, 2017

It's been fun working with KL on his story, Weirder Than Normal—Normal? Now That's Weird, and I've finally broke down and registered here so that I could post a few notes on this story. There's been a bit of a major change to what we're doing with it, but don't worry too much about it. It's just being done to fix an ugly bit of math we've been forced to use for a while now. Time for an upgrade.

For those who don't like math, run! — RUN NOW!

Anyone who likes a challenge, here's some fun & some history:

Okay, so Kevin already posted the abbreviated version, so there's not much point in rehashing that—too much.

The thing to keep in mind here is in this setting, Q, from ST:TNG is Discord. In case anyone might have picked up on a throwaway line Q gave in the episode "Q Who?", there was a moon that was falling out of orbit and threatening an inhabited planet. The Enterprise was dispatched to try and help. But before they could act on any plan, Q showed up, having been stripped of his powers. Now Q was assigned to assist Geordi to solve the problem, and Q told him that to fix the moon's orbit, Q would simply "change the Gravitational Constant of the Universe".

Fast forward to last year where Kevin approached me for him wanting to use some of my characters in his fimfiction. It was during our discussions where he was looking for a way to figure out how the Equestrian star system could be made to work that I recalled Q's quote from above. Thus I did my best… in short, it looks a little like this:

Essentially, the stepped pyramid shows the domains of the changed Gravitational Constants as a function of the Log of the distance between interacting objects.

It wasn't pretty, and I didn't like it, but it got the job done—for when those chapters were getting posted. The biggest problem was it's not elegant! And Q, if anything, would insist on doing things elegantly.

Essentially, I wanted a smooth, Normal Distribution curve in place of the stepped pyramid—except the ND curve would actually be the exponent of a function dependent upon the distance between the objects and I could not imagine what the resulting varying G-constants would generate.

(and YES, I did say a varying universal physical constant—remember, this is Q we're talking about! He can get away with that sort of silliness.)
The hardest part was envisioning the changes to the circular orbit velocity and acceleration formulae that would ultimately incorporate a smoothly changing G'. It was fairly straight forward work to do for the stepped pyramid domains—but again, it lacked any sort of elegance to them.

I kept fiddling with it, looking at numerous variations of the Guassian Normal Distribution until a couple months ago, I stumbled upon the Log-Logistic Distribution function and I suspected this was what I was needing. By setting the X-axis to scalular distance, it has a rather strong resemblance to the Log-Logistic Distribution with its very long tail stretching out the edges of the star system. As a result, it gave me new insight to the problem and now, I'm confident in the unveiling:

G'=G°R(e)^Δ
where Δ=(20/{[σ√(2π)]Ɛ^([Log53.5{X}-μ]²/[2σ²]}} - 0.03471457999999999425510513411766)
Note: G'= G° when Log53.5(X) <= -0.25853534485505325 & Log53.5(X) => 9.25853534485505325,
In other words, when
X<=0.35740404130617077114230241267959m & X>=10,046,726,489,536,530.659224355430293m
μ=4.5; σ²=2.25; σ=1.5; the 20 & 0.03471457999999999425510513411766 are part of the kurtosis of the ND curve, 20 adjusting the maximum peak value, and the other number sets the value of the intercept of the ND curve with the flat-line function of G° when X is at or beyond the limits given in the Note above.

Dimensions of G' are (m^(3+Δ)/(s^2*kg)), where m is meters, s is seconds, & kg is kilograms ;

Derived formulae: V'o ~ √(G'[M1+m2]/X^(1+Δ)) ; a'=G'M1/X^(2+Δ) ; F'=m2a'= G'M1m2/X^(2+Δ), where X = distance between objects' centers (which takes the place of r in the classical formulae, but I'm using X to avoid confusion with R(e).

And you're no doubt thinking, "Well, gee. That blue line's just like any other Normal Distribution curve."
And you'd be right….except…the curve being shown is this part of the equation:

=(20/{[σ√(2π)]Ɛ^([Log53.5{X}-μ]²/[2σ²]}} - 0.03471457999999999425510513411766)
…which is the exponent that R(e) is being raised to…

It's difficult to describe why this is exciting. Please bear in mind, that both the horizontal & vertical axis shown are actually two different logarithmic scales. Here's what the curve looks like when displayed on a sufficiently wide linear scale of the distances involved:

Now, it should be evident how the G' Factor is rising as Linear Distance changes--each of the tick marks in the Horizontal Scale are 1,000 kilometers.

For distances 0 - 0.35740404130617077114230241267959meters, the Grav Constant is the same as normal G°. This is only a little more than 14 inches in distance.
G' quickly rises, however.

For -1σ, the separation distance is found to be at 153,130.375meters,
while G' has reached a value of 55,553,853,883.7357
with the following dimensions: m^(5.4933377459999773933301486666862…)/(s^2*kg).

Then G' reaches its maximum at a separation distance of 59,922,789.06meters
with a value of 2,909,247,251,767,070,492,463,231.6553529
and the dimensions: m^(9.78451582535244…)/(s2*kg).

For the +1σ, the value of G' drops back down to 55,553,853,883.7357
with dimensions m^(5.4933377459999773933301486666862…)/(s2*kg),
but at a separation distance of 23,448,911,747.640625meters.

That's the nature of Log-Logistic Distribution. The σ are not symmetrically the same distance from the maximum μ. Even though the exponential value that R(e) is being raised to is a Normal Distribution curve with symmetric σ, the resulting values of R(e), and thus G', are highly skewed.

Finally, the G' drops back to merge into the flat G° function line at a separation distance of 10,046,726,489,536,400.7meters, or a bit over 10 Trillion kilometers, or 1.062 lightyears!

It's also important to remember the value of G' has to be recalculated between each body involved anytime the distance changes between them! This includes any derived calculations involving a', F', and V'o – in addition, the value of Δ used for G' must be the same Δ used for of the exponent of X in the denominators for all the derived formulae.

These changes result in numerous changes to the earlier calculations Kevin and I did for the dynamics of the system, but at least there's some sense and order in the values and it's a hell of a lot easier to plug&chug to get the final values since G' is solely a function of distance X, and a smooth function at that!

Now, for those who might be a little intimidated by that ungodly exponent, here's some good news:
While I personally like the formula,
G'=G°R(e)^((20Ɛ^-((Log53.5(X)-μ)²/(2σ²))/(σ√(2π)) - 0.03471457999999999425510513411766)
when you're looking for things like the circular velocity, acceleration, or force, you probably don't care how you get the answer, so long as it's the correct answer! Also, you might be squicking at the very idea of a "variable universal constant". That's where a little trick to save you some work will come in handy and just might make you feel a little better about both those issues…

V'o ~ √(G'[M1+m2]/X^(1+Δ)) = √(G°R(e)^Δ[M1+m2]/X(1+Δ)) = √([M1+m2]G°R(e)^Δ/X^(1+Δ))
a' = G'M1/X^(2+Δ) = G°R(e)^ΔM1/X^(2+Δ) = M1G°R(e)^Δ/X^(2+Δ)
F' = m2a'= G'M1m2/X^(2+Δ) = M1m2G°R(e)^Δ/X^(2+Δ)

All I've done here is group like terms a little closer to make the next step much easier to see…

Please note that Δ symbol, once more.
Recall, that it is all of this:

(20/{[σ√(2π)]Ɛ^([Log53.5{X}-μ]²/[2σ²]}} - 0.03471457999999999425510513411766).

Not only that, but in case you missed it before, there is an X in this term, too.

That means, whatever value you fill in for X in the denominators of the above V', a', and F' equations, you have to put that in the exponent for R(e) as well as the exponents for those Xs in all those formulae. That's THREE places where the same values have to be entered for each V'o, a', and F' you're trying to find!! Which are three places in which it's very easy to make a mistake…

Personally, I rather like the formulae above that keep terms together with their specific exponents. But if you're more interested in getting the final answer than in how you get them, then this is the trick to save you a lot of work:

V'o ~ √([M1+m2]G°[R(e)/X]^Δ/X)
a' = M1G°(R(e)/X)^Δ/X^2
F' = M1m2G°(R(e)/X)^Δ/X^2

All I've done here is group like exponents together, making a portion of the formulae have the term (R(e)/X), a much simpler fraction to calculate, which is then finally being raised to Δ, once. You don't have to deal with TWO SEPARATE numbers EACH being raised to Δ.

As a result, this lets us keep G° as an unchanging universal constant while simply adding a self-correcting variable solely dependent upon the value of a variable X along with another unchanging constant R(e). (As mentioned in the story, 53.5 and 3588242 are magic numbers.)

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