#15
  
Let's say the earth is smooth all the way around at the equator.  Let's also say that if I string a line around the equator it would be snug to the surface of the earth, correct?

Now, If I cut that string and add just 20 feet to it you would grant me that it would now be lose around the equator, correct?  (sorta like a belt that's too big for your gut).



Here's the question: is it lose enough for a person to be able to crawl under it?  Remember, I'm just adding 20 ft to a circle that's 7K (approx) mile in diameter.

Yes/no?  Also prove your answer one way or the other.
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#16
  'nother math problem Dumb_Polack Let's say the earth ...
C'mon, no one wants to take a stab at it???
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#17
  'nother math problem Dumb_Polack Let's say the earth ...
The circumference of the earth is 24,901 miles = 131,477,280 feet.

The diameter of the earth would be then 418,505,015 feet.

Add 20 feet to the circumference = 131,477,300 feet.

The new diameter is 418,505,021 feet.

If you are under 6 feet tall you could walk under it.
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#18
  RE: 'nother math problem Cooler The circumference of...
(07-12-2019, 10:55 AM)Cooler Wrote: The circumference of the earth is 24,901 miles = 131,477,280 feet.

The diameter of the earth would be then 418,505,015 feet.

Add 20 feet to the circumference = 131,477,300 feet.

The new diameter is 418,505,021 feet.

If you are under 6 feet tall you could walk under it.

Close but not quite. Simpler method is that if you add 20' to the circumference you gain 20/pi  to the diameter. where pi  is about 355/113 or 3.1416. That's 6.4', so the radius goes up half that or 3.2'. If your child is 3' tall, he/she could walk underneath the string all the way around the globe.
Rip to width. Plane to thickness. Cut to length. Join.
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#19
Big Grin    RE: 'nother math problem Edwin Hackleman [quote='Cooler' pid=...
(07-12-2019, 11:14 AM)Edwin Hackleman Wrote: Close but not quite. Simpler method is that if you add 20' to the circumference you gain 20/pi  to the diameter. where pi  is about 355/113 or 3.1416. That's 6.4', so the radius goes up half that or 3.2'. If your child is 3' tall, he/she could walk underneath the string all the way around the globe.

I figured to pull up the string in one place for my needs only. Big Grin

The question was could a person crawl under it (which implied one person at one location).  I was not worried about children in Australia walking under the string, only if I could walk under the string.
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#20
  RE: 'nother math problem Cooler The circumference of...
(07-12-2019, 10:55 AM)Cooler Wrote: The circumference of the earth is 24,901 miles = 131,477,280 feet.

The diameter of the earth would be then 418,505,015 feet.

Add 20 feet to the circumference = 131,477,300 feet.

The new diameter is 418,505,021 feet.

If you are under 6 feet tall you could walk under it.

The math is correct.

If the rope is equally suspended above the earth, it would be half the 6 feet or 3 feet.  So you'd need to do the limbo to walk under it - or be very short in stature.

It would be easier to step over the rope than to walk under it.  Just let the extra 20 feet fall to to the surface.
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Allan Hill
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#21
  'nother math problem Dumb_Polack Let's say the earth ...
I don't think you could fit under it.  Just because the circumference is 6 feet more, spread out over that distance its probably less than an inch at any particular point.
"Oh. Um, l-- look, i-- i-- if we built this large wooden badger" ~ Sir Bedevere
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#22
  RE: 'nother math problem Large Wooden Badger I don't think you co...
(07-12-2019, 10:57 AM)Large Wooden Badger Wrote: I don't think you could fit under it.  Just because the circumference is 6 feet more, spread out over that distance its probably less than an inch at any particular point.
I didn't do the math.  I relied on a circumference/diameter calculator online. 

The diameter grew by 6 feet, not the circumference.
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#23
  'nother math problem Dumb_Polack Let's say the earth ...
Yes you can!  Here's the proof.  from 4th grade math you'll remember that:

C= 2 (pi)r 

Now, the C will increase by 20 ft, correct?  That'll cause the r to increase by (x), right?  So, let's write that down:

C +20 = 2* pi(r+x)

Factor it out and you get:
C +20 = 2*pi+r +2*pi*x

but from above C = 2*pi*r, correct?  So I'll just substitute C on the LH side of the equation for this and you get:


2*pi*r +20 = 2*pi*r + 2*pi*x

now let's drop the 2+pi*r from both sides of the equation and you're left with:

20 = 2*pi*x, right?   move the 2 to the other side by dividing and you get:

10 = pi*x     Let's call pi = 3.141  so dividing 10 by that and you get approx 3.18 ft, more than enough room for someone to crawl under


As a side note, since the r factors out, it doesn't matte the size of your circle (i.e. you could tie a string around the universe (assuming it is round)) and you'd get the same results.

Pretty slick, eh?
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#24
  'nother math problem Dumb_Polack Let's say the earth ...
(07-12-2019, 10:55 AM)Cooler Wrote: The circumference of the earth is 24,901 miles = 131,477,280 feet.

The diameter of the earth would be then 418,505,015 feet.

Add 20 feet to the circumference = 131,477,300 feet.

The new diameter is 418,505,021 feet.

If you are under 6 feet tall you could walk under it.

Hmmm...don't think I've ever seen a circle with the diameter larger than the circumference. Smile
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