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2014-11-24 02:40 6898760 Anonymous (pix510871249.jpg 640x480 57kB)
How would I calculate the number of coins I need in order to fill up a container such as a chest with silver coins?

21 min later 6898814 Anonymous
>>6898760 by using the packing coefficient for coins

21 min later 6898815 Anonymous
>>6898760 E=mc^2

22 min later 6898817 Anonymous
>>6898760 Volume of coin. Spatial dimensions of available chest space. Divide into

22 min later 6898819 Anonymous
>>6898814 i kek'd

22 min later 6898821 Anonymous
>Put silver coins in a chest until the chest is full >Take the coins out >count the coins ???

23 min later 6898826 Anonymous
>>6898817 You have to take into account the packing coefficient you dolt. Coins cannot fill an arbitrary volume like a liquid

26 min later 6898830 Anonymous
>>6898826 Gives you an upper bound at least

38 min later 6898845 Anonymous
>>6898826 Fortunately, since coins approximate cylinders, you can just use the circle packing factor for the area of the base vs the area of the coins and multiply by the thickness.

43 min later 6898853 Anonymous
>>6898817 Doesn't quite work because the coins don't fit perfectly into a rectangular prism or square perfectly cause circles yo. It does give an absolute upper bound for the number of coins that could fit into the chest, but in reality, real results will show the maximum number of coins that fit in the chest is lower than this upper bound, even when the coins are packed in the most efficient manner.

43 min later 6898854 Anonymous
>>6898760 >find volume of container >find volume of coin >divided the former by the latter >??? basically it's: [L x W x D] (container volume) / [L x W x D] (coin volume)

43 min later 6898855 Anonymous
>>6898760 >Dimensions of chest >Dimensions of coins How many coins fit in the bottoms of chest? e.g. length x width. Multiply that by height of chest divided by thickness of coins. The autist in me was compelled to reply.

44 min later 6898859 Anonymous
>>6898845 Coins only approximate cylinders when they are stacked. When they are dumped or "flipped" they aproximate a pile of coins

48 min later 6898870 Anonymous
>>6898859 Assume they are stacked, then.

1 hours later 6898928 Anonymous
n= \frac{\int \int \int dl*dw*dh}{\int\int \int dlc*dwc*dhc} l,w,h= length, width, height of container lc,wc, hc = length, width, height of coins

1 hours later 6898938 Anonymous (1407201832840.jpg 565x700 98kB)
>>6898928 >>6898855 >>6898854 >>6898817 >all these morons assuming you can pack coins with zero free space

2 hours later 6899153 Anonymous (c2[1].jpg 640x480 131kB)
>>6898938 >not having 1940s Egyptian coins >2012

3 hours later 6899165 Anonymous
>>6899153 you still couldn't fill the borders

3 hours later 6899232 Anonymous
>>6898760 Assume your coins are round cylinders with radius r and height a, and your chest is a rectangle with width w, height h, and depth d. Then we can roughly bound the amount with: \frac{whd}{4r^2a}<N_c<\frac{whd}{\p i r^2 a}

3 hours later 6899233 Anonymous
>>6899232 That should be (whd)/(4ar^2) < N < (whd)(pi a r^2)

4 hours later 6899354 Anonymous
>>6899165 Please don't be retarded. If you measure the apothem of the coin to the center of the edge, the area formula would give an effective area which, while not equal to the actual area of the coin, would work perfectly for the calculations involved in OP's problem, as the hexagon with that effective area would indeed tessellate with zero free space.

18 hours later 6900473 Anonymous
>>6898760 Melt the coins, pour into chest, solve

19 hours later 6900540 Anonymous
>>6899354 Hexagons don't tesselate with zero free space in a finite box.

19 hours later 6900559 Anonymous
>>6898760 let r and h be the radius and height of the coin let W, L, and H be the dimensions of the chest floor(W/(2r))*floor(1+(L-2r)/(r*sqr t3))*floor(H/h)

19 hours later 6900576 Anonymous
>>6900540 OP didn't specify the shape/dimensions of the container

19 hours later 6900580 Anonymous
>>6900576 Then why did you assume it would tesselate with zero free space?

20 hours later 6900601 Anonymous
>>6900580 Because it does, except for at the edges, which we can ignore

20 hours later 6900670 Anonymous
>>6900540 depending on the size of the hexagons and of the container, boundary effects may become negligeable

20 hours later 6900688 Anonymous
>>6900559 >assuming the silver fills all available space

20 hours later 6900698 Anonymous
>>6900688 Where did I assume that you fucking illiterate twat?

20 hours later 6900710 Anonymous
>>6900601 That doesn't make sense, you can't ignore it if you are going to actually calculate a number of coins to fit into the box. >>6900670 So? That would only be true in a negligible amount of cases.

20 hours later 6900715 Anonymous
>>6900698 You're telling OP to divide the Chest's volume by the coins' volume. That would assume there is no empty space, which there will be. Faggot.

20 hours later 6900731 Anonymous
this is retarded

20 hours later 6900732 Anonymous
>>6900715 No I'm not. The volume of the chest divided by the coins volume would be WLH/(h*pi*r^2) Do you see that anywhere in what I wrote? Kill yourself for being so stupid.

21 hours later 6900740 Anonymous
>>6900732 Then where do you account for the empty space, moron?

21 hours later 6900747 Anonymous
>>6900740 Read the fucking calculation

21 hours later 6900764 Anonymous
>>6900747 no u

21 hours later 6900765 Anonymous
>>6900710 that would actually be true in an infinite amount of cases

22 hours later 6900909 Anonymous
>>6898760 no. it would be [ LWH ] / [Radius of coin]^2 [Height of coin]

22 hours later 6900914 Anonymous
>>6900909 *diameter

23 hours later 6901125 Anonymous
>>6900670 technically yes, but not for any reasonable formulation of the problem

35 hours later 6902126 Anonymous
Coins dimensions can effectively be determined as square due to the inability to fit coins in the voids of the excess area of a circular coin. Chest dimension : d x b x h Coin dimension : diameter x diameter x height Chest dimension / coin dimension = # coins

50 hours later 6904055 Anonymous (1416646689292.jpg 640x479 89kB)
Guys!,.. guys,.. no wait,.. guys! Listen,.. What if we melted the coins, then poured the liquid coins into the box, thus filling all available box space with 'coins' absolute volume. This would dispense with the need for these vastly complex and unwieldy formulas. OPs original problem does not preclude this as a possibility at all.

50 hours later 6904074 Anonymous
>>6904055 Answer is here >>6900559

50 hours later 6904096 Anonymous
>>6904055 >What if we melted the coins, then poured the liquid coins into the box Then the box wouldn't be filled with coins, it would be filled with molten silver.

50 hours later 6904120 Anonymous (wjqsPWP.jpg 720x720 103kB)
>>6902126 This. I scrolled looking for this, and now I am only here to back it as what I would do.

50 hours later 6904124 Anonymous
>>6904055 are you a mathematician ?

50 hours later 6904135 Anonymous
>>6904120 That would be a stupid way to do it. The coins can be packed tighter than that by staggering them. The correct answer is here >>6900559

51 hours later 6904162 Anonymous
>>6904120 Then you're both idiots.

54 hours later 6904387 Anonymous
>>6899165 We're obviously filling an infinitely large tub

54 hours later 6904400 Anonymous
>>6898760 take a sample size of part of the chest to see how many coins fit into that space

54 hours later 6904403 Anonymous
>>6900765 But still negligible.

1.682 0.073