Difference between revisions of "Talk:2322: ISO Paper Size Golden Spiral"

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It should be noted that the logarithmic spiral this comic implies it is would actually go outside the bounds of the paper. The leftmost point of the spiral would be about 6.4mm to the left of the left edge of the A1 sheet. [[User:Zmatt|Zmatt]] ([[User talk:Zmatt|talk]]) 18:39, 19 June 2020 (UTC)
 
It should be noted that the logarithmic spiral this comic implies it is would actually go outside the bounds of the paper. The leftmost point of the spiral would be about 6.4mm to the left of the left edge of the A1 sheet. [[User:Zmatt|Zmatt]] ([[User talk:Zmatt|talk]]) 18:39, 19 June 2020 (UTC)
 +
:This drawing (as opposed to the singular mathematical formula behind the idealised spiral) basically takes a simple quarter-oval across each distinct sheet size (with, as essentially mentioned elsewhere, the root(2) ratio between sides) alternating x/y and y/x as major and minor axes respectively. Even if it is not obviously discontinuous (x and y inflection transitions occur subtly) any derivative of the curve (as polar, say) would show jumps in gradient at each stage - probably a saw-toothy pattern whereas the true logarithmic line would deminstrate itself as a continuous function at any such level of derivation. The true spiral line followed from origin outwards would ''almost'' (not quite, because of the polar gradient) hit the 'outer edge' first in line with the ultinately recursive centre-point then withdraw again to hit the next transition slightly 'inward' of the next level out. Although it's hard to describe, as you can see from my poor attempt that's probably inadvertently fallen foul of more specialised Pure Mathematics terminology due to the Pedant's Curse... ;) [[Special:Contributions/162.158.155.240|162.158.155.240]] 22:23, 19 June 2020 (UTC)

Revision as of 22:23, 19 June 2020

It annoys me that the hover text says 11/8.5 = pi/4, when 8.5/11≈0.77272727272 and pi/4≈0.78539816339. Claiming 8.5/11 equals pi/4 would be a much more beleiveable lie. 162.158.79.37 15:29, 19 June 2020 (UTC)

The explanation says that the A series "side lengths shrink by a factor of the square root of two" but that's not true. The width of A(n+1) is half the length of A(n) as depicted. The sqrt(2) ratio referenced is between the length and width of any one piece of paper.172.69.62.124 15:35, 19 June 2020 (UTC)

The side lengths do shrink by a factor of sqrt(2): the width of A(n) is sqrt(2) times the width of A(n+1), the length of A(n) is sqrt(2) times the length of A(n+1). Your statement that "the width of A(n+1) is half the length of A(n)" is also true, but it does not contradict that each step in the A-series shrinks the sides by a factor of sqrt(2). Zmatt (talk) 16:09, 19 June 2020 (UTC)

Fixed it 162.158.74.61 15:43, 19 June 2020 (UTC)

Hi ! How come 11/8.5 = Pi/4 ? First one is more thant 1, second one is less than one... Although Pi/4 and 8.5/11 (or the reverse) are pretty similar, as usual in "let's annoy mathematicians" Randall's style...

https://xkcd.com/spiral/ --188.114.103.233 17:22, 19 June 2020 (UTC)

I understand why it annoys mathematicians (it's not the golden ratio), but why does it annoy graphics designers? Please add explanation!

It should be noted that the logarithmic spiral this comic implies it is would actually go outside the bounds of the paper. The leftmost point of the spiral would be about 6.4mm to the left of the left edge of the A1 sheet. Zmatt (talk) 18:39, 19 June 2020 (UTC)

This drawing (as opposed to the singular mathematical formula behind the idealised spiral) basically takes a simple quarter-oval across each distinct sheet size (with, as essentially mentioned elsewhere, the root(2) ratio between sides) alternating x/y and y/x as major and minor axes respectively. Even if it is not obviously discontinuous (x and y inflection transitions occur subtly) any derivative of the curve (as polar, say) would show jumps in gradient at each stage - probably a saw-toothy pattern whereas the true logarithmic line would deminstrate itself as a continuous function at any such level of derivation. The true spiral line followed from origin outwards would almost (not quite, because of the polar gradient) hit the 'outer edge' first in line with the ultinately recursive centre-point then withdraw again to hit the next transition slightly 'inward' of the next level out. Although it's hard to describe, as you can see from my poor attempt that's probably inadvertently fallen foul of more specialised Pure Mathematics terminology due to the Pedant's Curse... ;) 162.158.155.240 22:23, 19 June 2020 (UTC)