h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

Entry: How long is a piece of string? - A19120583
Author: gavinorr - U3235914

An example of the truth which many people don't believe, even after it hs been proven.

Hi Gavinorr,

I was already aware of this interesting problem in perception. I'll leave it for others to say whether they can follow your explanation.

Two things I'm not sure about:

1. The title, which isn't particularly descriptive of the problem (I'll try to think of some alternatives)

2. The final section "HOW TO PUT THIS FACT TO SOME USE", doesn't really describe a practical use - only a contrived rehash of the problem as a competition. Again, I'll see if I can think of a practical application.

Icy

A: twice as long as half a piece

Apart from at rather facetious statement, I liked the entry but agree with icy's comments

The pub paragraph was the only way I could think to close the piece, but I agree it's a bit naff, so I've removed it.

I could close with a comment that although the ratio between the "extension" and the increase in the circumference is fixed, the increase in the area of the circle is different, since this is based on " pi r squared" rather than "2 pi r". Similarly for the volume of the sphere, since it is "1.333 pi r cubed".

I might also add a table showing circumference, area and volume to demonstrate (a graph would be better, but I can't do them).

Hi gavinorr,

I'm not much of a mathmetician, so it took me a while to figure out where exactly the string was - but I got there in the end. I really don't know how you would word it to make it any clearer.

I did spot one typo though >>(or the hight above the surface)>> height.

I can't be more help.

Fixed the typo and re-worded the article to be more of a "puzzle" than a "presentation".

Still need to change some of the paragraphs to headers, and might add some history on the puzzle (which has been around since Euclid's time, apparently).

effectively it's a differentiation problem. If c= 2*pi*r, then dr/dc = 1/2pi, which is independent of r. Simple, eh?

It is, but it is amazing how many people refuse to believe it, even when they know the formula, and can follow the progression of the equations. They keep looking for something that you have done wrong (like dividing by zero, or somesuch).

A piece of string is as long as you think it is

>>it is amazing how many people refuse to believe it<<
Yes, as simple as the result is, I did have to stare stupidly at it for a while before I was convinced.

You are posing a 2 dimensional problem, but using 3 dimensional language that had me confused at first. Here is what I mean:
'position it so that it is at the same distance from the surface of the basketball all the way around'
I first pictured the string as above or below the ball (why else magically suspend it?). An easy fix would be to say '...from the circumference of the basketball...' instead of the surface.
I would suggest replacing further instances of 'surface' with 'circumference' as well.

In footnote 4, you could link 'great circle' to A9948432

Regarding the title, I agree with ladye_seagull in that the length is a given in the problem. It seems to me that the question you are posing is 'how high is a piece of string'

>>and might add some history on the puzzle<< (post #7)
I look forward to that - I love this kind of thing

Good puzzle

I agree that I could improve the visualisation by calling the piece "How high is a piece of string?", but it just woudn't be the same!

Maybe I can leave the title but change the presentation of the question to use circumference instead of surface, or something similar.

got to

I'm not very good at Maths but I found the diagnostics helpful!

How are we getting on with this? Changes made?

Hmmm...

I think this is still a bit dry and could be improved in terms of both style and content. One thing especially bothers me - the Entry is full of very simple maths but still needs to have a table at the end with just a bit more simple maths in it. There's potential here but it's not being reached.

Alex

Some changes made (including the title).
I think I'm pretty much done, but I'll have another look first chance I get.

Table removed, headers added. I still think it's a bit long winded but am not sure if I can cut it down and still prove the point.

Any road up, enough from me for tonight - I'm off to bed

Ok, just some nitpicking from me now...

The headers should be in mixed case (eg Some Information)

There is a pi symbol available in GuideML: pi --> &pi;

Try to space out your equations to make them more readable:

i.e. C=2Pir --> ie C = 2&pi;r
i.e. r=c/2pi --> ie r = c/2&pi;

Try not to use so many brackets - for instance:

THEN r2 = (C1+25)/(2pi)
-->
THEN r2 = (C1+25)/2&pi;

r2 = (C1/(2pi)) + (25/(2pi))
-->
r2 = C1/2&pi; + 25/2&pi;

(Please ignore the fact that the C1 in this post has inevitably turned blue)


You can specify pi to as many decimal places as you like, it makes no difference to the end result.
-->
You can specify pi to as many decimal places as you like - provided you use the same vlaue each time, it makes no difference to the end result.


Otherwise this is looking quite good

Alex

What's missing from this, in my opinion, is the significance of the result.

It's a problem in perception, but in the end it's a purely academic exercise, unless there's some sort of fundamental truth resulting from it. Originally you were going to add practical applications - I tried to think of one myself, without success - but you dropped this, and there's nothing to replace it, to summarise the significance of what we've learned from this.

As far as I can tell, the point is this:

R1 = C1/2pi
R2 = C2/2pi
R2 - R1 = (C2 - C1)/2pi

--> Adding a certain value to the circumference adds a certain value to the radius, regardless of how big the sphere is.

The interesting thing is that it's counter-intuitive, but I can't think of a good way to make use of that fact.

Keep thinking.