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Posted

HI :

 

Theoretical Physics performed this one value

for Planck's Length :

1.616 252 (81) x 10^-35 m .

 

We depict below a geometrical approximation and graphic .

( not matter the ...10^-n issue )

planklent.bmp

Posted

Now we introduce an approximation

to the Planck time .

It's = 5.391 x10^-44 s .

 

(see formula below)

planck time.JPG

Posted

you do realise that the planck length is derived from physical constants right?

 

and that your 'approximation' is no where near the actual value and is hence useless?

 

also the lower formula you posted is actually 0.0053915156 and not 5.391*10^-44

 

seriously, have you ever done maths before?

Posted

Why not just pick a completely random number? Would that not be less computationally intensive for the same degree of accuracy?

Posted

It's great that after real science discovered a value of a constant (using real math and real physics) you come and play with it visually.

 

But if your theory holds any merit, you should be able to actually derive new constants (and show the *DERIVATION* if the current one) with it without using guesswork. "Pretty numbers" work post-hoc, they don't work towards actually discovering something new, as you clearly showed with your error and "correction".

Posted

951.4439 isn't plancks constant either

 

you also make no mention of units.

 

do you even check your own work with the real value at all?

 

also how are you 'deriving' these?

Posted

Are you going to answer the questions and critiques people put forth, or are you just here to post more and more numerical images?

 

We're a discussion forum, and we asked you a bunch of questions.. I think you should relate to them before you give us more numbers that might make no sense if the basics are irrelevant.

Posted

4135648.89014674 that definitely isn't planck's constant

 

you say you've taken note, why not show that and stop firing out what are essentially random number combinations which are completely useless.

Posted

There's a [math]\pi[/math] missing which reflects the symmetry of the universe. It was mentioned in some forgotten notes of Tesla.

Posted

Both 'Pi' and 'e' (Euler's number)are basics in science.

below , the start-point numbers I use to apply(not randomness)and an approximation to 'e' .

the series (1+1/2+1/16+1/32) means movement , run...

start point numbers.JPG

numero e.jpg

Posted

wow, you actually got one thats reasonable accurate. finally.

 

a much more awesome approximation can be found here although its a bugger to calculate http://www.futilitycloset.com/2010/05/12/pandigital-approximations/

 

takes forever to evaluate numerically but it's good for 18,457,734,525,360,901,453,873,570 decimal places.

 

of course, if you are going to that much effort, then you are better calculating the true value from the start.

Posted

Hmm... was it by sheer coincidence that you actually got fairly close that time?

 

I mean, it's not as good an approximation as the much more obvious [imath]\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{8!}[/imath] but at least this time you have actually given an approximation rather than just a very poorly written completely random number.

Posted

Just to see how to think about this 'problem', acoe, are you doing this as a joke-exercise (in which case, NICE! you came close) or is this supposed to show some post-hoc correlation with numerology (in which case, fail... answer the previous points we made, and give an appropriate and fully accurate number).

Posted
wow, you actually got one thats reasonable accurate. finally.

 

a much more awesome approximation can be found here although its a bugger to calculate http://www.futilitycloset.com/2010/05/12/pandigital-approximations/

 

takes forever to evaluate numerically but it's good for 18,457,734,525,360,901,453,873,570 decimal places.

 

of course, if you are going to that much effort, then you are better calculating the true value from the start.

 

Heh. The second one has a reason:

 

(1 + 9^(-4^(7*6)))^(3^(2^85)) = (1+(3^2)^(-(2^2)^(7*6)))^(3^(2^85))

 

= (1+3^(-2*2^(2*7*6)))^(3^(2^85)) = (1 + 3^(-2^(1+2*7*6)))^(3^(2^85))

 

= (1 + 3^(-2^85))^(3^(2^85)) = (1 + 1/(3^(2^85))^(3^(2^85))

 

In general, the limit of (1+1/x)^x as x goes to infinity is e. As 3^(2^85) is huge, that means you're going to be really, really close to e.

=Uncool-

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