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Windows 7: Number Crunching

17 Oct 2009   #21

Windows 7 Ultimate x64

As far as it goes, I made that problem up on the spot based on something from my calculus book. I haven't the faintest clue what the real answer would be.
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17 Oct 2009   #22

Linux CENTOS 7 / various Windows OS'es and servers

Quote   Quote: Originally Posted by Dwarf View Post
Quote   Quote: Originally Posted by Zen00 View Post
Since you all are enjoying math then how about this one?

A bucket has a hole in it. The water flows out of that hole at a rate of 2 gallons per minute. How big is the hole?
It doesn't matter how big the hole is. The hole is actually the missing bottom of the bucket. So, in effect, the bucket is a tube. In order to get water flowing out at 2 gallons a minute, then water needs to be flowing in at the same rate. This condition will be met so long as the diameter of the bucket is equal to or greater than that of the pipe supplying the water.
Actually that's not quite true although you've got the "tube" bit right -- the volume of this is essentially the volume of cylinder - so it's relatively simple maths to work out the cylinder dimensions to solve the problem. (Again assuming an incompressible liquid - not a gas - and assume normal atmospheric pressure). There's no EXACT answer as some calculus will be required but my answer is a good practical correct one.

Imagine an infinitely large bucket that already contains a huge volume of water. A 100 gallon one is a good example here.

At 2 gals a minute it should take around around 50 mins to empty the bucket with no water coming in -- we can ignore the effects of the water pressure here as it will be relatively constant over the lifetime of this problem.

You could have larger bucket which is supplied at > 2 gals a min ( or < 2 gals a min).
The outflow doesn't depend on the rate of inflow (until the bucket is empty of course).

Actually to SUPPLY the bucket at 2 gallons a minute you've got a similar problem - the answer is the same as the pipe needs to be 1/2 inch approx.

(Ok you could have a 6 inch diameter pipe or anything > 1/2 inch SUPPLYING the water but I assume the problem meant what's the SMALLEST size that would do the job and we are considering the OUTFLOW).

a 1/2 inch hole will do the trick -- try it and see with say a 5 litre gas can - drill a 1/2 inch hole in it - fill it up with water and see how long it takes to empty. Convert US gals to litres and you should come up with the magic answer 2.2 gals / min requires a 1/2 inch diameter pipe. (Of course if you increase the pressure significantly you might get a bit more but I assume here that the water is flowing at approx 1 Bar (1 Atm) pressure.

(And this is AFTER I've just come back from the pub - now having a look again at Man Utd's win today -- back at the top of the league).

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17 Oct 2009   #23

Windows 7 Ultimate x64, Mint 9

2 + 2 = 5....

My System SpecsSystem Spec

17 Oct 2009   #24

XP_Pro, W7_7201, W7RC.vhd, SciLinux5.3, Fedora12, Fedora9_2x, OpenSolaris_09-06

Quote   Quote: Originally Posted by Barman58 View Post
For the original question

Count down the digits on left hand 10-9-8-7-6 then add the right hand

6 + 5 = 11
Ya can't do that - you're starting with an arbitrary number, 10 in this case.

You could start with any arbitrary number to count down, say 25. Then the countdown would be:
25, 24, 23, 22, 21.

Of course we can't add 5 more fingers of the other hand to the 21, to yield 26...

The solution to get the correct amount of digits would be:

(n2 - n1) + 1 = amount of digits.

(25 - 21) + 1 = 5, just as the other (10) countdown:
(10 - 6) + 1 = 5...

Then and only then, we can add the 5 more to get the ? how-ever many...

I've seen snags like this with 'buffer-sizes', where the +1 wasn't done:

(Last_index - First_index) is always short by one,
even when using the 'c' base index of [0]:

(4 - 0) = 4, but to get all the items, we need the + 1...
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