The Creation Equation

[D J]

You seem to be making a big deal about the "enter planet A and get planet A+3".

It's meaningless. All the author had to do was raise some planets to one exponent and other planets to another exponent, then just find the slope of the line that most closely matched the resulting points.

Again, it's a trick of mathematics, and I'm sure you could do the same thing, with the proper exponents, and have "enter planet A and get A+2 or even A+1".

Can you do this with the proper exponents?

"The orbit of Mars produces the orbit of Uranus."

"Try repeating the calculation with the mean orbit of Mercury as an input, you get Mars out."

You should have a result within the 0.2% range of accuracy given by the equation.

[wedgebert]

Orion, you're starting to take on a very Yul-like trait of not responding to peoples questions/statements and just posting the same thing over and over again.

But here's a quick equation that will relate the orbit of any planet to the orbit of the next planet in line.

X = Orbital distance of the first planet

pi + (pi * (2x + e)^log(1)) _ _ _ pi^log(1)
----------------------------------- = --------------
_ _ _ e^ln(1) + 1 _ _ _ _ _ _ _ _ log(10)

(ignore the white _'s, they're just for spacing)

Raise the first planet to the exponent log(1) + ln(1) and you'll get the next planet in line raised to the exponent log(ln(1) + 1)

But does this equation prove anything other than I know how to manipulate mathematics?

[D J]

I know than there are many ways to obtain the results on a straigh line IE linear progression.

The creation equation do more than this.

http://hometown.aol.com/astroequation/Observations.htm

The way he arrive to the equation based on his evaluation of a monument construct 3000 BC ie Stonehenge(Babel) is astonishing.

B)Edges of features of monument ground plan conform to proportionality of Solar System orbits, perihelion and aphelion, raised to appropriate exponent.

C) Stonehenge is circular format graph, in all essentials it is identical and equivalent to Solar System Cartesian format graph.

How does peoples from that epoch were able to obtain this knowledge?
http://www.geocities.com/godisreal_uk/babel.htm

[wedgebert]

The Solar System graph and the main Equation are founded on the universal mathematical constant PI , which means that the equation is a 'pure math' equation. The equation works with any units, even inches, provided the input value is less than p .

Huh? Using PI makes it PURE math? What does he mean by pure mathematics? And how would using a constant alter the form of mathematics? It doesn't, unless that constant is specific to a particular branch of math.

The first PI symbol, on the left of the equation, represents the input number, it also represents the upper limit for that number. Any number equal to or less than PI may be put into the equation in this place.

(Note that he has the symbol for PI where I have written PI.

This guy has no clue about math. You can't just use the symbol PI as a variable, especially not when you use PI elsewhere as the actual value of PI.

When he says the first PI must be less than or equal to PI, does he mean PI as in 3.141 or PI as in the input number?

He is also amazed that Earth is unity (unity raised to any power is unity) when measured in AUs, yet the AU is determined solely by Earth's distance from the sun.

Even after looking at his diagrams for Stonehenge, I don't see how he can came to those conculsions. Here's a http://www.astro.queensu.ca/~hanes/p...Topic_015.htmlquick pagehttp://that dispels a few myths abou...quation.<br />

[D J]

The Solar System graph and the main Equation are founded on the universal mathematical constant [i]PI[/i , which means that the equation is a 'pure math' equation. The equation works with any units, even inches, provided the input value is less than p .

Huh? Using PI makes it PURE math? What does he mean by pure mathematics?

Note than he use " " meaning than this statement is exagerated.

http://hometown.aol.com/astroequation/Observations.htm

He is also amazed that Earth is unity (unity raised to any power is unity) when measured in AUs, yet the AU is determined solely by Earth's distance from the sun.

I have pointed the part you don`t have reading:* *
When it is applied to the solar system it works only with astronomical units as a unit of measure, and this raises a problem, *since this unit is supposedly an arbitrary unit based on the orbit of the Earth. It is not supposed to be of cosmic significance.*
*There is no reason why any of the planets should be 'unity' on the scale of zero to p. The fact that Earth is unity is a mystery, except and unless the system uses astronomical units as a fundamental cosmic constant !*

When he says the first PI must be less than or equal to PI, does he mean PI as in 3.141 or PI as in the input number?

Baddly quoted or misinterpreted
The equation works with any units, even inches, *provided the input value* is less than PI .

[wedgebert]

Why is he exagerating the statement 'pure math'? What is impure math? How does using π (found it, or something close to it!), make it any more pure?

I didn't misquote him, and it's his fault if I misinterpeted it because he doesn't understand how to write mathematical equations.

Here's a copy of his text with the actual PI symbol (π)

[( π*^(9/4π).Ln30) - F]*^(2π/3) = 20π

The first π symbol, on the left of the equation, represents the input number, it also represents the upper limit for that number. Any number equal to or less than π may be put into the equation in this place

So the contraint for the equation is π &lt; π. Huh?

[beskeptical]

The equation may or may not tell you something about solar system formation. If it is a coincidence that the planets orbit where they do, we won't know until we are able to detect other solar systems more precisely. If there is a mechanism behind the order, which makes sense anyway, then it may become more and more apparent as we learn more about solar system formation. To say it couldn't be a natural process is silly.

Some astronomer has been recording the rotation of the asteroids in the asteroid belt for several years. He found they rotate in unison! He gathered more and more data expecting to find his original data was wrong, but he continued to find the same thing.

Well, now someone has come along with an explanation. The solar heating of the sides of the asteroids facing the Sun cause the continual rolling at the same rate from asteroid to asteroid.

So who's to say we know all there is to know about solar system formation? If there is some regularity that shows up as a mathematical relationship, that's one more piece of data to plug into our model. Given that it was described years ago as Bode's law, it hardly seems like a 'discovery'. And, it doesn't seem to have quite the precision attributed to it by the person here.

[AstroSmurf]

Whoah, this thread is still going? I thought the matter was trivial.

Anyway, what we're dealing with is a hypothesis that the planetary orbits can be related with the statement:

f(x) = [x^a * b + c]^d

The parameters of the equation are a,b,c and d, which can be determined from the following four relationships:

f(Mercury) = Mars
f(Venus) = Jupiter
f(Earth) = Saturn
f(Mars) = Uranus

(I am of course referring to the semimajor axis of these planets)
Given that we have four parameters and four data points, this should be solvable to a very high precision. Could someone with a good mathematics package please do this and post the results? Use AU for measurements, so we can compare directly with the stated values.

[informant]

The Solar System graph and the main Equation are founded on the universal mathematical constant [i]PI[/i , which means that the equation is a 'pure math' equation.

Huh? Using PI makes it PURE math? What does he mean by pure mathematics?
[...]
Why is he exagerating the statement 'pure math'? What is impure math? How does using &lt;pi> (found it, or something close to it!), make it any more pure?

I think the explanation is in his next sentence:

The equation works with any units, even inches, provided the input value is less than p.

His wording is not the best, but that seems to be what he means.

[SeanF]

When it is applied to the solar system it works only with astronomical units as a unit of measure . . .

The equation works with any units, even inches, *provided the input value* is less than PI .

Aren't these contradictory? Does it only work with AUs, or does it work with any units?

[AstroSmurf]

When it is applied to the solar system it works only with astronomical units as a unit of measure . . .

The equation works with any units, even inches, *provided the input value* is less than PI .

Aren't these contradictory? Does it only work with AUs, or does it work with any units?

Define "works" :P
You could of course stuff anything into the equation, but you won't get very meaningful results. I suppose he means that it doesn't work for Neptune or Pluto, but he doesn't want to put it quite that way :P

He is of course wrong in calling this a 'dimensionless' equation; it isn't. To make it work with other units, you need to convert the constants in the equation - and they have very strange units of measurement:

x^(9/4pi) * C - F == y^(3/2pi)
where
x = the semimajor axis of a planet in the inner system, in AU
y = the semimajor axis of a planet in the outer system, in AU
C = proportionality constant, stated to be 3.401197 AU^(2/3)
F = proportionality constant, fitted to be 0.500772 AU^(3/2pi)

If you want to go easy on yourself, use x/L and y/L instead of x and y, where L = 1 AU in whatever length unit you're using (light-seconds? :wink. In this case, it actually becomes a dimensionless equation - not that this will impress me or anything.

[D J]

Why is he exagerating the statement 'pure math'? What is impure math? How does using ? (found it, or something close to it!), make it any more pure?

I didn't misquote him, and it's his fault if I misinterpeted it because he doesn't understand how to write mathematical equations.

Here's a copy of his text with the actual PI symbol (?)

[( ?^(9/4?).Ln30) - F]^(2?/3) = 20?

The first ? symbol, on the left of the equation, represents the input number, it also represents the upper limit for that number. Any number equal to or less than ? may be put into the equation in this place

So the contraint for the equation is ? &lt; ?. Huh?

I think you are confusing things for nothing.
There is two statements in his claims about the equation one apply to any array of objects that were suitably distributed,
the other is Astronomy and the unit used in Astronomy is AU until you can say than Astronomers use feet and inches in their calculation I don`t know why you insist to confuse things.
i don`t bother if the equation can be used for other application than the one we are discussing here ie Astronomy
As for his use of symbol or difficulty of language I think the author native language is not english.Whatever he write P or Pi we know than he means Pi.
Edited
If you have again difficulty understanding the equation here some explanations who can help
http://www.geocities.com/godisreal_uk/primer.htm

[informant]

Whatever he write P or Pi we know than he means Pi.

Are you sure? It doesn't seem so to me.

[D J]

Verified yourself
http://www.geocities.com/godisreal_uk/primer.htm

[informant]

A circle of circumference three (arbitrary units) has a radius of (3/2p) and an area of (9/4p).

These two related nonrandom values serve as exponents in the production of an equation and a graph of the orbits of the Solar System.

So p stands for the constant pi, right? I mean the ratio of the circumference of a circle to its diameter.

The first 'p' symbol, on the left of the equation, represents the input number, it also represents the upper limit for that number. Any number equal to or less than 'p' may be put into the equation in this place.

Now wait just a moment there. Now p is a variable!? :???:

By the way...

'F' is the offset constant, calculated to ten decimals.

F = 0.50077209786.....

What, exactly, is an 'offset constant'? What does it mean physically?
And, if F is a constant and p is a constant, then the ‘equation’ is no equation at all. It's just a proposition!
If F is a constant and p is a variable, then the 'equation' merely defines p; it does not relate two quantities!

Edited.

P.S. (Corrected.)

Mercury gives Mars, Venus gives Jupiter, Earth gives Saturn and Mars gives Uranus.

If I understand the author correctly, the equation would best be written in the iterative form:

d_(n+3) = ( ln 30 * (d_n)^(9/(pi*4)) - F )^(3/(pi*2)) (equation 1)

d_n is the “mean orbit” of a planet in Astronomical Units.
d_(n+3) is the “mean orbit” of the third next planet in Astronomical Units, counting from the Sun.
F is a constant given by
F = ln 30 * pi^(9/(pi*4)) - (20 * pi)^(3/(pi*2)) (equation 2)
The use of the same symbol for a constant and a variable is extremely confusing and terribly sloppy math - terribly sloppy thinking, as a matter of fact!

Note added later: There are really two equations on the page: one that defines F (equation (2)), and another that relates the orbits of the planets (equation (1)).

[wedgebert]

It's no use trying to make sense of what the author is saying. He uses the terms constant and variable loosely and often has a symbol as both in the same equation.

Something like π^π where the first π is a variable and it has to be less than π (π is supposed to be the Pi symbol).

Of course it has to fit the constraint 0 ≤ π &lt; π

Good look meeting that since you don't know if it's supposed to really be

1: 0 ≤ X &lt; 3.141
2: 0 ≤ 3.141 &lt; 3.141
3: 0 ≤ X &lt; X
4: 0 ≤ 3.141 &lt; X

I would assume the first one is correct, but the author doesn't seem to understand mathematics so any of the them might be what he means. Not that 2&amp;3 are impossible contraints, but cannot be ruled out because of the way the equation is written.

[Cougar]

The use of the same symbol for a constant and a variable is extremely confusing and terribly sloppy math - terribly sloppy thinking, as a matter of fact!

Oh, it's worse than that. It is completely unacceptable. It makes this so-called equation total gibberish. To use the symbol pi as a variable? Excuuuse me? After that, I see no sense in further discussion. Talking to "Orion" is like talking to a robot anyway, one that needs a serious upgrade to its language module.

[Grey]

...until you can say than Astronomers use feet and inches in their calculation I don`t know why you insist to confuse things.

Actually, as Phil would say, "Real astronomers use kilometers. Hardcore astronomers use centimeters."

[wedgebert]

Pfft, hardcore astronomers take the easy route.

I use nothing but Planck lengths.

[russ_watters]

One word: garbage.

It makes this so-called equation total gibberish.

Hmm..... So what have we accomplished in the 4+ pages since the first reply? Is it garbage or gibberish? Make up your damn mind already.

You guys are tenacious, I'll give you that. But sometimes I wonder why you bother. This is what I call "self-debunking."

[D J]

[quote="wedgebert"]

(Note that he has the symbol for PI where I have written PI.

The author explain why p goes trough on some browser so that is not an error made by the author.
http://members.aol.com/astroequation/
The math requires 'symbol' font, if you do not have it, the Greek letter 'pi'[p] will appear as a 'p'. In this event please take 'p' to be the mathematical constant 3.141592...!

[D J]

The use of the same symbol for a constant and a variable is extremely confusing and terribly sloppy math - terribly sloppy thinking, as a matter of fact!

Does the equation works or not?

Oh, it's worse than that. It is completely unacceptable. It makes this so-called equation total gibberish. To use the symbol pi as a variable? Excuuuse me? After that, I see no sense in further discussion.

Talking to "Orion" is like talking to a robot anyway, one that needs a serious upgrade to its language module.

First note than English is not my native language.
Maybe you need to use glasses because you are the only one seen a u in the equation.

[tjm220]

First note than English is not my native language.
Maybe you need to use glasses because you are the only one seen a u in the equation.

Where did Cougar mention seeing a 'u' in the equation? :-k

[informant]

The author the why of the p goes trough on some browser so that is not an error made by the author.
http://members.aol.com/astroequation/
The math requires 'symbol' font, if you do not have it, the Greek letter 'pi'[p] will appear as a 'p'. In this event please take 'p' to be the mathematical constant 3.141592...!

The problem is with meaning, not notation. Whatever he chooses to call them, he uses two symbols for four concepts (five, really). Sloppy, sloppy work.

The use of the same symbol for a constant and a variable is extremely confusing and terribly sloppy math - terribly sloppy thinking, as a matter of fact!

Does the equation works or not?

I can’t answer that without first knowing what it means. Sadly, the author makes it hard for the reader to even know what the equation means.
This is usually a sign that the author is more interested in obfuscation than in being understood.

[russ_watters]

Does the equation works or not?

As said before, its not even an equation, its a relation. It only has one variable: F. So all it does is define a value for F. Near as I can tell, (if I plugged it into my calculator correctly) it says F=-1 (could someone else verify that?), which of course is meaningless.

[D J]

He explain even how to use the equation to calculate an orbit and the use of the (in) -the famous mysterious variable- who is not Pi but the mean orbit of a Planet .
Go to
Math lesson three.
To Calculate an Orbit

http://www.geocities.com/godisreal_uk/primer.htm

[( p(9/4p).Loge30) - F] = 20p (3/2p)

We need to change the equation around a bit, but it is still the same equation.

[( in(9/4p).Ln30) - F](2p/3) = out
Don't Panic !

We now have an input variable (in) and a result, or output variable (out).
The right hand side exponent has been moved over to the left, and when you do this you need to invert it. (Math rules O.K.!)

Enter the mean orbit of Mars in Astronomical Units. (1.5236915)
Press the "Raise to power" button
Enter the value of (9/4p) , this is 0.716197243...
Press the [=] button
The answer should be 1.35204255....
Multiply this by Loge30 (3.401197382)
Press [=] button
Subtract F, (0.50077209786) you should have 4.097791483
Press "Raise to power" button
Enter value of (2p/3) This is 2.094395102
Press [=] and the answer should be 19.18321602

The value obtained, 19.18321602, is the mean orbit of Uranus in Astronomical Units.
The orbit of Mars produces the orbit of Uranus.
It may seem a long winded way of relating the orbits of planets, but ALL the orbits are inter-related by this one single Equation.
Try repeating the calculation with the mean orbit of Mercury as an input, you get Mars out.

But this concludes the primer, which was intended to give a little practice in manipulating the functions and values. The full power of the Creation Equation is revealed in the mathematical sections, which involves calculating all the orbits.
Practice manipulating the numbers, and when you are ready, go to Mathematics Part One
And good luck !

[D J]

Maybe the graphic form may help you...
http://hometown.aol.com/astroequation/mathone.htm

...understand what is explained here
http://www.geocities.com/godisreal_uk/primer.htm

[informant]

Near as I can tell, (if I plugged it into my calculator correctly) it says F=-1 (could someone else verify that?), which of course is meaningless.

I’m glad you asked that, russ. When I went to check, I realized that I had misunderstood the equation – again, because of the author’s sloppy notation. It turns out that ^(9*pi/4) is really ^(9/(pi*4)). :roll:
I will correct my equation above.

From what I can tell the constant F is given by

F = ln 30 * pi^(9/(pi*4)) - (20 * pi)^(2/(pi*3)).

[D J]

Near as I can tell, (if I plugged it into my calculator correctly) it says F=-1 (could someone else verify that?), which of course is meaningless.

I?m glad you asked that, russ. When I went to check, I realized that I had misunderstood the equation ? again, because of the author?s sloppy notation. It turns out that ^(9*pi/4) is really ^(9/(pi*4)). :roll:
I will correct my equation above.

From what I can tell the constant F is given by

F = ln 30 * pi^(9/(pi*4)) - (20 * pi)^(2/(pi*3)).

Nah! You should try lesson two
http://www.geocities.com/godisreal_uk/primer.htm

Lesson Two
To determine the value of 'F'

[( p(9/4p).Loge30) - F] = 20p (3/2p)

Don't panic !
We've done the white bits already !


In lesson one, in the yellow rectangle, we worked out the value of p(9/4p)
We found it to be 2.270164712....
We also discovered the value of Loge30, it is 3.401197382.....
The little dot between them, on the equation, means 'multiply'
So we must multiply these two partial results together.
This gives us 7.721278275... as the value for the part of the equation shown in white.
Remember this number, it replaces the part shown in white.

Now let us solve the values for the right hand side of the equation.

Enter the number p
Multiply by 20, press [=]
You should get 62.83185307...
Press the "Raise to power" button
Enter the value of (3/2p) , this is 0.477464829...
Press the [=] button
The answer should be 7.220506177....

We now rewrite the equation, replacing the symbols with the resulting numbers.
We get:-
7.721278275 - F = 7.220506177


It is a simple matter of subtraction to find the value of F.
7.721278275 - 7.220506177 = F , which means that

F = 0.50077209786....

We have now solved the whole equation, and ended up with a situation where we know ALL the values.
You may be a little puzzled, the equation appears to tell us nothing apart from the value of F.
Don't panic ! We are nearly there.
What we do now is work the same equation in a different way.
If we rework the equation, using the mean orbit of Mercury (in AU) instead of the first p we find that we can calculate a result that is different from 20p.
In fact we get the orbit of Mars.


Lesson three shows how this is done.

[informant]

Nah!

You're absolutely right! That should have been
F = ln 30 * pi^(9/(pi*4)) - (20 * pi)^(3/(pi*2)).
My mistake.

You should try lesson two
http://www.geocities.com/godisreal_uk/primer.htm

I have. That's how I came to slowly understand what the author was trying to say (I think). But if he had been clear - as he should - I wouldn't need to.