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01/23/2003 01:11:22 PM · #1 |
"If A equal success, then the formula is A equals X plus Y and Z, with X being work, Y play, and Z keeping your mouth shut."
- Albert Einstein
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01/23/2003 01:26:25 PM · #2 |
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01/23/2003 02:35:33 PM · #3 |
A = Y + B , B being lots of Beer
Message edited by author 2003-01-23 14:35:48.
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01/23/2003 03:06:13 PM · #4 |
| So for large values of B, the approximation is that A = B ? |
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01/23/2003 03:26:48 PM · #5 |
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01/23/2003 05:48:06 PM · #6 |
A + B/C - (4A + B) = ZZZzzzzzZzzzZZZzz
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01/23/2003 05:52:18 PM · #7 |
Originally posted by Moondoggie:
A = (B + Z) - X |
Ah, but the flaw is that (B + Z) = a large mess on the floor, over your clothes etc. |
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01/23/2003 05:53:00 PM · #8 |
Nooooooooooooooooo math! Stopppppp!!
*falls over and dies* |
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01/23/2003 06:08:24 PM · #9 |
Originally posted by Gordon:
Originally posted by Moondoggie:
A = (B + Z) - X |
Ah, but the flaw is that (B + Z) = a large mess on the floor, over your clothes etc. |
No, only if Z preceeds B. As in: (Z + B) = X
Message edited by author 2003-01-23 18:08:51. |
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01/23/2003 09:41:32 PM · #10 |
| It must be mid week again! |
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01/24/2003 01:42:14 AM · #11 |
| I would say that A ~ -dB/dX :). |
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01/24/2003 03:12:54 AM · #12 |
Originally posted by Moondoggie:
...No, only if Z preceeds B. As in: (Z + B) = X |
Now you try and ignore the commutativity of addition...(B + Z) = (Z + B)
It might also be interesting to compare with Edison's view that Success = (10% Inspiration) + (90% Perspiration)
Message edited by author 2003-01-24 03:15:39. |
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01/24/2003 07:08:55 AM · #13 |
Originally posted by GeneralE:
Originally posted by Moondoggie:
...No, only if Z preceeds B. As in: (Z + B) = X |
Now you try and ignore the commutativity of addition...(B + Z) = (Z + B)
It might also be interesting to compare with Edison's view that Success = (10% Inspiration) + (90% Perspiration) |
Why no Y? We need more Ys! I am inspired to perspire, if we can just have more Ys. Y = Play. |
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01/24/2003 10:43:01 AM · #14 |
Originally posted by GeneralE:
Originally posted by Moondoggie:
...No, only if Z preceeds B. As in: (Z + B) = X |
Now you try and ignore the commutativity of addition...(B + Z) = (Z + B)
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It isn't always commutable anyway, even in proper maths. And as I'd want at least a 6 pack of beer, I think that makes it an matrix, so all commutative bets are off. |
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01/24/2003 10:59:06 AM · #15 |
| Hmm...yes, perhaps a center-weighted matrix would hit the spot (trying to somehow relate this to photography) |
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