Maths geniuses - how many

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Maths geniuses - how many
Posted on: 14.11.2011 by Arcelia Siebeneck
I've got a coke can piggy bank that I'm using to save up
Arcelia Siebeneck
14.11.2011
Originally Posted by MrSteve81
Isn't a
Arcelia Siebeneck
14.11.2011
Originally Posted by MrSteve81
Isn't a
Arcelia Siebeneck
14.11.2011
Originally Posted by MrSteve81
Isn't a
Arcelia Siebeneck
14.11.2011
Originally Posted by MrSteve81
Isn't a
Arcelia Siebeneck
14.11.2011
Originally Posted by MrSteve81
Isn't a
Arcelia Siebeneck
14.11.2011
Originally Posted by MrSteve81
Isn't a
Arcelia Siebeneck
14.11.2011
Originally Posted by MrSteve81
Isn't a
Arcelia Siebeneck
14.11.2011
Originally Posted by MrSteve81
Isn't a
Shana Minsk
14.11.2011		Isn't a
Arcelia Siebeneck
14.11.2011		excellent - what I really wanted to know was if I can save
Arcelia Siebeneck
15.11.2011		yeah - I guess it's one thing if all the
Arcelia Siebeneck
14.11.2011		I've got a coke can piggy bank that I'm using to save up
Arcelia Siebeneck
14.11.2011
Originally Posted by MrSteve81
Isn't a
Shana Minsk
14.11.2011		Isn't a
Arcelia Siebeneck
14.11.2011		excellent - what I really wanted to know was if I can save
Arcelia Siebeneck
15.11.2011		yeah - I guess it's one thing if all the
Gilma Marchini
14.11.2011
Originally Posted by Arbite
pi x r^2 x height for the volume of the coin. Divide the volume of the can by this. Then times the overall number by ~.75-.8

The last part accounts for all the spaces there will be. Should give you a rough estimate.
This is pretty wrong. Years of professor layton video gaming tells me that there has to be a gap between coins. I'm no math expert, but challenge accepted.

Four coins side-by-side will form a square of sorts, which would be 56.8mm long in each leg. A can is 63.5mm in diameter, with a bit of trig we discover this square would be wider from one corner to the opposite than the can is in diameter. This means that the coins cannot be laid flat in an optimal matter. I'll do some more math to decide how many can fit in that matter later, but for now I have calculus homework to do.
Ngan Ernestine
14.11.2011
Originally Posted by Arbite
pi x r^2 x height for the volume of the coin. Divide the volume of the can by this. Then times the overall number by ~.75-.8

The last part accounts for all the spaces there will be. Should give you a rough estimate.
looks like solid advice to me
Arcelia Siebeneck
14.11.2011		I've got a coke can piggy bank that I'm using to save up
Arcelia Siebeneck
14.11.2011
Originally Posted by MrSteve81
Isn't a
Johnetta Olewine
14.11.2011		You should put all your coins in a milo tin.

I'm going to drive past the birthplace of milo this weekend actually. http://en.wikipedia.org/wiki/Milo_(drink)
Shana Minsk
14.11.2011		Isn't a
Arcelia Siebeneck
14.11.2011		excellent - what I really wanted to know was if I can save
Gilma Marchini
14.11.2011		In an ideal mathmatical world my in-boring-lecture guesstimate is 128 + 16. So 144. IF stacked perfectly and the can is 121.92mm in height, 63.5mm in diameter, and isn't bent in the process; this ignores indents on the top or bottom. Guessing for the dents I would say 132.

There are 11 layers of of two wide coins and 10 layers of 1 deep coins (5 on each side of the 11). This is repeated 4 times along the height of the can. There would be enough space in the top of the can to stack eight coins, sixteen if your persistent. With the dent in the bottom, which I can't find numbers for, I would imagine that the space for the 16 coins would be shattered, maybe reducing it to 4 additional coins, landing you at about 132.

This is assuming you could even position all these coins via the small hole in a coke can, but if I was given a few hours and a cloth glove I believe I could pull it off.
Leeanna Ayla
15.11.2011		If I guess right do I get the can full of coins?
Arcelia Siebeneck
15.11.2011		yeah - I guess it's one thing if all the
Gilma Marchini
14.11.2011
Originally Posted by Arbite
pi x r^2 x height for the volume of the coin. Divide the volume of the can by this. Then times the overall number by ~.75-.8

The last part accounts for all the spaces there will be. Should give you a rough estimate.
This is pretty wrong. Years of professor layton video gaming tells me that there has to be a gap between coins. I'm no math expert, but challenge accepted.

Four coins side-by-side will form a square of sorts, which would be 56.8mm long in each leg. A can is 63.5mm in diameter, with a bit of trig we discover this square would be wider from one corner to the opposite than the can is in diameter. This means that the coins cannot be laid flat in an optimal matter. I'll do some more math to decide how many can fit in that matter later, but for now I have calculus homework to do.
Ngan Ernestine
14.11.2011
Originally Posted by Arbite
pi x r^2 x height for the volume of the coin. Divide the volume of the can by this. Then times the overall number by ~.75-.8

The last part accounts for all the spaces there will be. Should give you a rough estimate.
looks like solid advice to me
Lilliana Perris
14.11.2011		Never seen a 2 Pound coin myself.....
Danial Sawn
14.11.2011		pi x r^2 x height for the volume of the coin. Divide the volume of the can by this. Then times the overall number by ~.75-.8

The last part accounts for all the spaces there will be. Should give you a rough estimate.

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