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# Wired Equator

### #61

Posted 19 May 2008 - 12:15 PM

### #62

Posted 08 September 2008 - 06:59 PM

### #64

Posted 01 May 2010 - 04:02 AM

the formula u used to calculating radius was r=C*(pi)/2Well, the way I make it is

C= 40000000M therefore (Cx3.142857143/2=r) the Radius is 62857142.86 M

C= 40000000 + 10 is C = 40000010M therefore the radius is 62857158.57 M

62857142.86 - 62857158.57 is 15.71428571M.

Your answer of 1.6 M would be correct if the wire were only 1 M longer.

Or have I missed it completely?

check the maths book dude its C/2*(pi)

### #65

Posted 25 June 2010 - 03:07 AM

lol that's what my cousin saidWouldn't a flea jump over, rather than creep under the wire?

### #66

Posted 08 July 2010 - 11:17 PM

This was the question I saw.

Lord Fimbleton lost 6 meters of his cricket fence, how much shorter would you expect the boundary, which was 50m away from the centre?

Only 1 meter shorter. So the radius is now 49 meters

Now the Gods also hold a cricket match, a very vast cricket ground, billions of miles across, but they have also lost 6 meters of fencing, how much shorter would their ground be?

1 meter. So the radius will be 1 meter shorter.

You would think that, for the second one, it would be less, but it isn't

This is why:

We know that the perimeter for a circle is P= 2pi*r

Lets do this simultaneously

Old P=2pi*Old r 1

New P=2pi*New r 2

2-1

New P-Old P=2pi*(New r-Old r)

6=2pi*(New r-Old r)

6/(2pi) = New r-Old r

which is approx. 1 meter. So the numbers don't matter. Doesn't matter how big or small it was to begin with. Don't let that deceive you.

I recommend to anyone who enjoys these, or similar, puzzles to buy and read 'Why do buses come in threes?' by Rob Eastway and Jeremy Wyndham. And their other books too. They are very good, interesting and enjoyable to read.

### #67

Posted 08 July 2010 - 11:24 PM

Interestingly enough, it doesn't matter what the original diameter of the object is. If you add 10 meteres to the original circumference x, the change in radius will be 1.6 meters.

Here's the algebra: Solving for change in radius (R2-R1)

Knowns:

x1 = 2*Pi*R1

x2 = 2*Pi*R2

x1 = x2 - 10

Solution:

R1 = x1 / (2*Pi)

R2 = x2 / (2*Pi)

so by substitution, R2 - R1 = x2 / (2*pi) - x1 / (2*pi)

-or-

R2 - R1 = (x2 - x1) / (2*Pi)

then

R2 - R1 = [x2 - (x2 - 10)] / (2*Pi)

The x2's cancel out. The result, regardless of x2 or x1 is simply:

R2 - R1 = 10 / (2*pi) or 1.6

Crazy to think that whether the object is as small as marble, or as big as the galaxy if you add 10 to the circumference measuring tape, you can pass a small child underneath anywhere.

I know. It's quite mind blowing to think that. I posted a similar thing (I didn't look through all the posts). I saw this in a book (Why do buses come in threes) and it just goes to show just how wrong gut feelings can be, and how logic and maths can show it.

### #68

Posted 08 July 2010 - 11:36 PM

VERY VERY Well Said! That's what I said minus all the educated words..lol

LOL! But yep...exactly. I also stated (and someone else did too) a similar illusion. Where you can remove or add the same amount to both a small and large object and still have the same difference.

e.g. a tennis ball(A) and tennis ball(B) being 7 cm bigger in circumference and a planet(A) and planet(B) being 7 cm bigger in circumference. They will both have the same difference in radius, but you wouldn't expect that as you would expect the planet to have a smaller change, because it's a bigger object.

What people need to remember is that the radius will increase/decrease by the same amount, regardless of the size of the object, BUT the change will be less significant in the larger object, as it was already huge.

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