this thread is for Rpenner

Uh huh, I see you said maybe I could divide circumference by diameter. That deserves an unignore for you.

 

Log in or Sign up to hide all adverts.

Thanks.

 

Log in or Sign up to hide all adverts.

show us how. with values

 

Log in or Sign up to hide all adverts.

Step one, find a perfect natural disk of some macroscopic size. The Earth? They tried that in the original definition of the meter and botched the measurement and Newton told us that the Earth is not spherical. The Moon? Not on my budget. A sand dollar? Nope. A hydrogen atom's electron wave function? Neither macroscopic or circular -- the ground state wave function has a spherical symmetry but no hard edge.
Step two, give up and find a perfect manufactured disk. They don't exist. How about U.S. Coins? After all 31 U.S. Code § 5112 specifies what their mass and diameter must be. But then 31 U.S. Code § 5113 ruins it by saying "The Secretary of the Treasury may prescribe reasonable manufacturing tolerances for specifications in section 5112 of this title" which is evidence that even the US Legislature recognized (at some point) that manufacture of a perfect disk is not without problems. Even then these are not disks, but approximate cylinders.
Step three, measure the circumference exactly. Well here we have another problem because no measurement has ever been exact. Counting is exact, but measuring is estimating the ratio between one physical magnitude and another. So details matter. That the cylinder is curved makes the problem tougher and the procedure matters.
Mark the edge of the disk and roll it without slipping. How do you know you rolled it in a straight path without slipping?
Wrap a string/thread/chain about it, mark the overlap and measure the straightened string/thread/chain. How do you know you haven't stretched it out of shape by making it conform to the curve of the cylinder? Aren't you measuring how long a physical object is outside the cylinder, which is not exactly the same thing as the cylinder? Does the overlap point add to the length in some systematic way?
Step four, measure the diameter. How do you know your two points are a diameter?
Step five, take the ratio. This is either going to be a ratio of two rational quantities which is itself a rational number and therefore not pi, or its going to be a ratio of two estimated quantities (hopefully with error budgets) and thus it too will only be a (hopefully good) estimate of pi.

Say I have a disk which is between 779220779.20 and 779220779.25 carbon-carbon bond lengths in diameter and between 2447994275.50 and 2447994275.55 carbon-carbon bond lengths in circumference. Thus we estimate pi is between 9791977102/3116883117 and 48959885511/15584415584, which is a good estimate, but lots of numbers are between those ratios.

2447994275.50 /779220779.25 = 9791977102/3116883117 < 1770721/563638 < 208341/66317 < pi < 312689/99532 < 417037/132747 < 48959885511/15584415584 = 2447994275.55/779220779.20

Another way to write this estimate is \(\tilde{\pi} = 3.141592653573 \pm 0.000000000133 = 3.141592653573(133)\) which is in good agreement with the tabulated decimal value of pi, 3.141592653589793+.

Physicists know how important analysis of measurement errors are to the testing of physical theory, but since a circle has a mathematical definition, pure mathematics is of far greater utility in determining the 20th digit of pi than physical measurement.

\(\pi = 16 \tan^{-1} \frac{1}{5} - 4 \tan^{-1} \frac{1}{239} = \lim_{n\to\infty} \sum_{k=0}^{n} \frac{(-1)^k}{2k+1} \left( \frac{16}{5^{2k+1}} - \frac{4}{239^{2k+1}} \right)\)
This infinite sum only requires 8 terms to bound pi more tightly than the previous thought experiment on estimation.

pi = 3804/1195 - 72810068/1706489875 + 12476980230684/12184551018734375 -712697588470600764/24359780855939418203125 + 13569999650349009580148/14908446881486951862650390625 - 2325395850082757774745401724/78061994479096482923424854931640625 + 132828936352577206862383654376604/131742566878013892304309920002620849609375 - 2529107224465187544395498503569499028/72358338102298380214562374427593322601318359375 + ...
≈ 3.183263598326 - 0.042666569000 + 0.001023999999 - 0.000029257143 + 0.000000910222 - 0.000000029789 + 0.000000001008 - 0.000000000035 + ...

 

yes I know am asking why what does it mean visually? is Pi * r^2 not just a square increased by the ratio of c/d? and r*180 gives you the area of a circle if r = 180/pi? so is this not the area of circle = the area of square?

 

you're such a doll[still vague on gender].
but you know what that was about.

 

A manhole cover will do.

A wheel is good enough. They have a nasty habit of being round.

It doesn't have to be exact. The nearest millimetre will do.

Oh FFS, find the maximum.

Yes, and you soon find out that π isn't 3.125.

 

maybe because,
" It doesn't have to be exact. The nearest millimetre will do. "
learn something about sig figs and estimated digits. then learn how a single digit will throw off the whole measurement.

and also, you never showed us. ridiculing penners comment with lack of proper procedure is not doing so.

 

well the reason I was asking is because I found this interesting I have not look over all of Rpenners math so this may not even mean anything so just try and make sense of it for me I remember Rpenner said circles with different diameters or not similar so there might be no connection but if there is let me know Rpenner. ok here goes if Pi * r^2 = X then r*180=Y then X=Y then X= a squares area... Y= a circles area

This is true for both the natural circle of radius 180/pi and this circle of different diameter and different C/d ratio ...180/3.125


so 180/3.125 =r then 57.6 *180 = 10368 then 3.125 * 57.6^2 =10368= circle one

now 180/pi =r then r*180 = 10313 approximately then pi *180/pi= 10313 approximately =circle two

 

LOL Now you are just attempting comedy, right?

Did you understand the intent of the post you are responding too? .... Don't think so!

 

this doesn't work for any other number combinations?? unless am mistaking??

 

For the example I used, surprisingly the answer is no. To the nearest millimeter, with a carbon-carbon bond length of 154 pm, the circumference is 377 mm and the diameter is 120 mm (more or less by design in my hypothetical). These bounds are insufficient to resolve 3.125 as distinct from the ratio of measured circumference to measured diameter \(376.5/120.5 < 25/8 < \pi < 377.5/119.5\).

This is of course assuming the disc (roughly CD-sized) has exactly a circular boundary. A manhole cover or wheel may not be manufactured to millimeter precision or may not retain original precision due to wear and tear.

 

For the example I used, the answer is yes. Your bicycle wheel is typically 26" or 660.4 mm in diameter. The circumference is a shade over 2074.7 mm. Scale that up by 50% and you've got 3112 mm which is in the same ball park as 3141 mm and 3125 mm. There's 16mm between the two, more than half an inch. Scale it back down and it's 11mm. You can easily distinguish π from 3.125.

 

You are saying your math tells you a circle 1 meter in diameter has a circumference more than 16 mm longer than the approximation of 3125 mm, but that doesn't really address your claim that you can reliably measure this in actuality.

You seem to concede that you need a large circle to measure it with sufficient precision so that the ratio is distinguished from 3.125.

 

Baloney. Just google on measuring pi and you soon find stuff like this: http://www.mathsisfun.com/numbers/pi.html

Measure around the edge (the circumference):

Please Register or Log in to view the hidden image!

I got 82 cm
Measure across the circle (the diameter):

Please Register or Log in to view the hidden image!

I got 26 cm
Divide:
82 cm / 26 cm = 3.1538...

 

Maybe you're ignoring the thickness of the tape measure.

 

if i was you, i would look again, especially at your 82
again, rounding error.

 

i noticed you forgot this part.
you know, the next line.

 

Show ignored content

Uhhnnn, it's just krash the troll spouting more nonsense.

 

the nonsense is that part you conveniently left out. you know, the very next line.
it's ok tho. i posted it for you.

edit-
farsight, it's no bother to me that i'm ignored. i know how much your mentally disable physics/science is ignored in real life.
anyways,
everyone else can see my comments towards you. sometimes that is more important.