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tikay
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« Reply #13455 on: November 08, 2011, 06:45:26 PM » |
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Yeah, but that's not it.
Told you that you were being argumentative. |
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All details of the 2016 Vegas Staking Adventure can be found via this link - http://bit.ly/1pdQZDY (copyright Anthony James Kendall, 2016). |
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Woodsey
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« Reply #13456 on: November 08, 2011, 06:47:00 PM » |
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Its was probably used as a resting point for shelter in bad weather, and people/animals just caused wear and tear over the years leaning/rubbing on it.
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RED-DOG
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« Reply #13457 on: November 08, 2011, 06:48:36 PM » |
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duck/swan guano causing detrioration?
Colder... |
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The older I get, the better I was.
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RED-DOG
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« Reply #13458 on: November 08, 2011, 06:49:19 PM » |
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Its was probably used as a resting point for shelter in bad weather, and people/animals just caused wear and tear over the years leaning/rubbing on it.
Colder.... Another clue | Click to see full-size image. |
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The older I get, the better I was.
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technolog
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« Reply #13459 on: November 08, 2011, 06:50:55 PM » |
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The rope towing the barges?
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It's better to be looking at it than looking for it.
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Woodsey
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« Reply #13460 on: November 08, 2011, 06:51:51 PM » |
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The rope towing the barges?
I was thinking that but was wondering how? |
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RED-DOG
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« Reply #13461 on: November 08, 2011, 07:00:17 PM » |
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The rope towing the barges?
Yes! The bridge is on the apex of a bend. and the rope, impregnated with grit, rubs the sandstone away. They wern't really worried about the damage to the bridge, but the rope used to get trapped and make life difficult for the horses.  |
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The older I get, the better I was.
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technolog
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« Reply #13462 on: November 08, 2011, 07:02:22 PM » |
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Speaking of maths...
...as we were, I learnt something new today. 0.999999999999999 (recurring) = 1 |
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It's better to be looking at it than looking for it.
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RED-DOG
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« Reply #13463 on: November 08, 2011, 07:05:04 PM » |
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Speaking of maths...
...as we were, I learnt something new today. 0.999999999999999 (recurring) = 1 See now I believe you, but I don't understand it. |
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The older I get, the better I was.
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outragous76
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« Reply #13464 on: November 08, 2011, 07:21:02 PM » |
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OooooOh! So close!
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".....and then I spent 2 hours talking with Stu which blew my mind.........."
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Lucky
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« Reply #13466 on: November 08, 2011, 07:35:14 PM » |
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Speaking of maths...
...as we were, I learnt something new today. 0.999999999999999 (recurring) = 1 See now I believe you, but I don't understand it. I've seen some of the so called proof of this but I'll always believe that 0.99999999 (recurring) will always be less than 1 by an amount that is 0.00000001 (recurring). For every nine you add to the infinite number of recurring nines, we can add another 0 to the recurring zeros but there will always be an infinitesimal difference between 0.99999999 (recurring) and 1. |
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RED-DOG
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« Reply #13467 on: November 08, 2011, 07:39:24 PM » |
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Speaking of maths...
...as we were, I learnt something new today. 0.999999999999999 (recurring) = 1 . See now I believe you, but I don't understand it. I've seen some of the so called proof of this but I'll always believe that 0.99999999 (recurring) will always be less than 1 by an amount that is 0.00000001 (recurring). For every nine you add to the infinite number of recurring nines, we can add another 0 to the recurring zeros but there will always be an infinitesimal difference between 0.99999999 (recurring) and 1. Yes. it's like moving halfway to something, stopping and then moving halfway again and so on, you will never get there |
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The older I get, the better I was.
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77dave
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« Reply #13468 on: November 08, 2011, 07:42:28 PM » |
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Speaking of maths...
...as we were, I learnt something new today. 0.999999999999999 (recurring) = 1 See now I believe you, but I don't understand it. I've seen some of the so called proof of this but I'll always believe that 0.99999999 (recurring) will always be less than 1 by an amount that is 0.00000001 (recurring). For every nine you add to the infinite number of recurring nines, we can add another 0 to the recurring zeros but there will always be an infinitesimal difference between 0.99999999 (recurring) and 1. yes but 0.99999999 (recurring) + 0.00000001 (recurring) is > than 1 |
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Mantis - I would like to thank 77dave for his more realistic take on things.
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Jon MW
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« Reply #13469 on: November 08, 2011, 07:45:59 PM » |
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Speaking of maths...
...as we were, I learnt something new today. 0.999999999999999 (recurring) = 1 See now I believe you, but I don't understand it. I've seen some of the so called proof of this but I'll always believe that 0.99999999 (recurring) will always be less than 1 by an amount that is 0.00000001 (recurring). For every nine you add to the infinite number of recurring nines, we can add another 0 to the recurring zeros but there will always be an infinitesimal difference between 0.99999999 (recurring) and 1. yes but 0.99999999 (recurring) + 0.00000001 (recurring) is > than 1 b=0.9999999999.... 10*b = 9.99999999999999.... subtract b from both sides 10*b - b = 9.99999999...... - b but b=0.999999999........ so that means 9*b = 9 divide both sides by 9 b=1 QED |
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Jon "the British cowboy" Woodfield
2011 blonde MTT League August Champion
2011 UK Team Championships: Black Belt Poker Team Captain - - runners up - -
5 Star HORSE Classic - 2007 Razz Champion
2007 WSOP Razz - 13/341
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