Optimum fitting for flanges in a rectangular plate
$begingroup$
I have a $2500~text{mm}times6300~text{mm}times25~text{mm}$ (width $times$ length $times$ thickness) steel plate I want to cut flanges of diameter $235~text{mm}$ can anyone please suggest
$1)$ How many flanges would fit in this plate?
$2)$ A method of cutting circular flanges so that wastage is minimum?
$3)$ A generalized algorithm that would help me calculate this for any plate size?
(P.S: I have heard about the packing problem but i am unable to understand it)
geometry algorithms optimization circles
edited Jun 6 '13 at 13:00
Librecoin
2,385824
asked Jun 6 '13 at 11:24
kumaranando85kumaranando85
82
$endgroup$
add a comment |
$begingroup$
I have a $2500~text{mm}times6300~text{mm}times25~text{mm}$ (width $times$ length $times$ thickness) steel plate I want to cut flanges of diameter $235~text{mm}$ can anyone please suggest
$1)$ How many flanges would fit in this plate?
$2)$ A method of cutting circular flanges so that wastage is minimum?
$3)$ A generalized algorithm that would help me calculate this for any plate size?
(P.S: I have heard about the packing problem but i am unable to understand it)
geometry algorithms optimization circles
edited Jun 6 '13 at 13:00
Librecoin
2,385824
asked Jun 6 '13 at 11:24
kumaranando85kumaranando85
82
$endgroup$
add a comment |
$begingroup$
I have a $2500~text{mm}times6300~text{mm}times25~text{mm}$ (width $times$ length $times$ thickness) steel plate I want to cut flanges of diameter $235~text{mm}$ can anyone please suggest
$1)$ How many flanges would fit in this plate?
$2)$ A method of cutting circular flanges so that wastage is minimum?
$3)$ A generalized algorithm that would help me calculate this for any plate size?
(P.S: I have heard about the packing problem but i am unable to understand it)
geometry algorithms optimization circles
edited Jun 6 '13 at 13:00
Librecoin
2,385824
asked Jun 6 '13 at 11:24
kumaranando85kumaranando85
82
$endgroup$
I have a $2500~text{mm}times6300~text{mm}times25~text{mm}$ (width $times$ length $times$ thickness) steel plate I want to cut flanges of diameter $235~text{mm}$ can anyone please suggest
$1)$ How many flanges would fit in this plate?
$2)$ A method of cutting circular flanges so that wastage is minimum?
$3)$ A generalized algorithm that would help me calculate this for any plate size?
(P.S: I have heard about the packing problem but i am unable to understand it)
geometry algorithms optimization circles
geometry algorithms optimization circles
edited Jun 6 '13 at 13:00
Librecoin
2,385824
asked Jun 6 '13 at 11:24
kumaranando85kumaranando85
82
edited Jun 6 '13 at 13:00
Librecoin
2,385824
asked Jun 6 '13 at 11:24
kumaranando85kumaranando85
82
edited Jun 6 '13 at 13:00
Librecoin
2,385824
edited Jun 6 '13 at 13:00
Librecoin
2,385824
edited Jun 6 '13 at 13:00
Librecoin
2,385824
2,385824
asked Jun 6 '13 at 11:24
kumaranando85kumaranando85
82
asked Jun 6 '13 at 11:24
kumaranando85kumaranando85
82
asked Jun 6 '13 at 11:24
kumaranando85kumaranando85
82
82
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
When you can get a lot of flanges out of the rectangle, the optimal packing is hexagonal. You can see it 
. If you put one row along the long edge of your plate, you will get $lfloor frac {6300}{235}rfloor = 26$ along the bottom row. You get $lfloor frac {2(2500-235)}{235sqrt 3} +1rfloor=12$ rows. Each row has $26$, so you get $312$ of them. If you put the straight edge along the short direction, you get a first row of $10$ and each row has $29$ for a total of $290$. The first is better. When the flanges are larger compared to the sheet, it gets more complicated. With other plate dimensions, the intervening rows might be one flange short.
answered Jun 6 '13 at 12:49
Ross MillikanRoss Millikan
300k24200375
$endgroup$
$begingroup$
Can you please explain how you get the calculation for ⌊2(2500−235)2353√+1⌋=12 rows.I mean where does this ⌊2(2500−235)2353√+1⌋=12 rows come from??
$endgroup$
– kumaranando85
Jun 6 '13 at 12:56
$begingroup$
For $n$ rows and circle diameter $D$ you get a total hight of $D*(1 + (n-1)*sqrt(3)/2)$. This results in 12 for your example.
$endgroup$
– Axel Kemper
Jun 6 '13 at 13:16
$begingroup$
If you look at the figure, the spacing between the centerlines of the rows is $frac {235sqrt 3}2$ (draw the equilateral triangle that makes up one sixth of the hexagon.) You need one half disk above and below the centerlines, so that is where the deduction of $235$ comes from.
$endgroup$
– Ross Millikan
Jun 6 '13 at 13:18
$begingroup$
@RossMillikan I am just trying to understand the problem and trying different sizes can you please confirm how many flanges would fit if the rectangle is 1700 mm×4750 mm×25 mm (width × length × thickness).Thanks for your help.
$endgroup$
– kumaranando85
Jun 8 '13 at 15:59
$begingroup$
@kumaranando85: the thickness doesn't matter-we are assuming the plate is the proper thickness. Again, there are two ways to lay out the flanges. If a straight row is along the $4750$ direction, you would get $lfloor frac {4750}{235} rfloor=20$ along the bottom. Since there is not a half flange worth of excess, alternate rows would have $19$. You get eight rows, so $156$ total.
$endgroup$
– Ross Millikan
Jun 8 '13 at 17:21
add a comment |
$begingroup$
If you arrange the flanges in a rectangular grid, you'll get 26 rows and 9 columns. This is a total of 234 flanges. If you arrange the flanges in a staggered fashion (aka "hexagonal close packed"), you'll get 26 rows and 12 columns or a total of 312 flanges.
There is no generalized algorithm known (to me) which can do much better. Packing researchers seem to be focused on packing circles in squares rather than in rectangles.
There is an online circle packing calculator available for hexagonal packings. Note that you have to scale dimensions by 235 as the calculator assumes unit diameters.
You might have to add a safety margin to the 235mm to take into account the cutting width.
answered Jun 6 '13 at 12:46
Axel KemperAxel Kemper
3,49111418
$endgroup$
$begingroup$
I am just trying to understand the problem and trying different sizes can you please confirm how many flanges would fit if the rectangle is 1700 mm×4750 mm×25 mm (width × length × thickness).Thanks for your help.
$endgroup$
– kumaranando85
Jun 8 '13 at 16:01
$begingroup$
What about using the online calculator mentioned above and find it out on your own?
$endgroup$
– Axel Kemper
Jun 8 '13 at 17:05
add a comment |
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2 Answers
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active
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votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When you can get a lot of flanges out of the rectangle, the optimal packing is hexagonal. You can see it . If you put one row along the long edge of your plate, you will get $lfloor frac {6300}{235}rfloor = 26$ along the bottom row. You get $lfloor frac {2(2500-235)}{235sqrt 3} +1rfloor=12$ rows. Each row has $26$, so you get $312$ of them. If you put the straight edge along the short direction, you get a first row of $10$ and each row has $29$ for a total of $290$. The first is better. When the flanges are larger compared to the sheet, it gets more complicated. With other plate dimensions, the intervening rows might be one flange short.
answered Jun 6 '13 at 12:49
Ross MillikanRoss Millikan
300k24200375
$endgroup$
$begingroup$
Can you please explain how you get the calculation for ⌊2(2500−235)2353√+1⌋=12 rows.I mean where does this ⌊2(2500−235)2353√+1⌋=12 rows come from??
$endgroup$
– kumaranando85
Jun 6 '13 at 12:56
$begingroup$
For $n$ rows and circle diameter $D$ you get a total hight of $D*(1 + (n-1)*sqrt(3)/2)$. This results in 12 for your example.
$endgroup$
– Axel Kemper
Jun 6 '13 at 13:16
$begingroup$
If you look at the figure, the spacing between the centerlines of the rows is $frac {235sqrt 3}2$ (draw the equilateral triangle that makes up one sixth of the hexagon.) You need one half disk above and below the centerlines, so that is where the deduction of $235$ comes from.
$endgroup$
– Ross Millikan
Jun 6 '13 at 13:18
$begingroup$
@RossMillikan I am just trying to understand the problem and trying different sizes can you please confirm how many flanges would fit if the rectangle is 1700 mm×4750 mm×25 mm (width × length × thickness).Thanks for your help.
$endgroup$
– kumaranando85
Jun 8 '13 at 15:59
$begingroup$
@kumaranando85: the thickness doesn't matter-we are assuming the plate is the proper thickness. Again, there are two ways to lay out the flanges. If a straight row is along the $4750$ direction, you would get $lfloor frac {4750}{235} rfloor=20$ along the bottom. Since there is not a half flange worth of excess, alternate rows would have $19$. You get eight rows, so $156$ total.
$endgroup$
– Ross Millikan
Jun 8 '13 at 17:21
add a comment |
$begingroup$
When you can get a lot of flanges out of the rectangle, the optimal packing is hexagonal. You can see it . If you put one row along the long edge of your plate, you will get $lfloor frac {6300}{235}rfloor = 26$ along the bottom row. You get $lfloor frac {2(2500-235)}{235sqrt 3} +1rfloor=12$ rows. Each row has $26$, so you get $312$ of them. If you put the straight edge along the short direction, you get a first row of $10$ and each row has $29$ for a total of $290$. The first is better. When the flanges are larger compared to the sheet, it gets more complicated. With other plate dimensions, the intervening rows might be one flange short.
answered Jun 6 '13 at 12:49
Ross MillikanRoss Millikan
300k24200375
$endgroup$
$begingroup$
Can you please explain how you get the calculation for ⌊2(2500−235)2353√+1⌋=12 rows.I mean where does this ⌊2(2500−235)2353√+1⌋=12 rows come from??
$endgroup$
– kumaranando85
Jun 6 '13 at 12:56
$begingroup$
For $n$ rows and circle diameter $D$ you get a total hight of $D*(1 + (n-1)*sqrt(3)/2)$. This results in 12 for your example.
$endgroup$
– Axel Kemper
Jun 6 '13 at 13:16
$begingroup$
If you look at the figure, the spacing between the centerlines of the rows is $frac {235sqrt 3}2$ (draw the equilateral triangle that makes up one sixth of the hexagon.) You need one half disk above and below the centerlines, so that is where the deduction of $235$ comes from.
$endgroup$
– Ross Millikan
Jun 6 '13 at 13:18
$begingroup$
@RossMillikan I am just trying to understand the problem and trying different sizes can you please confirm how many flanges would fit if the rectangle is 1700 mm×4750 mm×25 mm (width × length × thickness).Thanks for your help.
$endgroup$
– kumaranando85
Jun 8 '13 at 15:59
$begingroup$
@kumaranando85: the thickness doesn't matter-we are assuming the plate is the proper thickness. Again, there are two ways to lay out the flanges. If a straight row is along the $4750$ direction, you would get $lfloor frac {4750}{235} rfloor=20$ along the bottom. Since there is not a half flange worth of excess, alternate rows would have $19$. You get eight rows, so $156$ total.
$endgroup$
– Ross Millikan
Jun 8 '13 at 17:21
add a comment |
$begingroup$
When you can get a lot of flanges out of the rectangle, the optimal packing is hexagonal. You can see it . If you put one row along the long edge of your plate, you will get $lfloor frac {6300}{235}rfloor = 26$ along the bottom row. You get $lfloor frac {2(2500-235)}{235sqrt 3} +1rfloor=12$ rows. Each row has $26$, so you get $312$ of them. If you put the straight edge along the short direction, you get a first row of $10$ and each row has $29$ for a total of $290$. The first is better. When the flanges are larger compared to the sheet, it gets more complicated. With other plate dimensions, the intervening rows might be one flange short.
answered Jun 6 '13 at 12:49
Ross MillikanRoss Millikan
300k24200375
$endgroup$
When you can get a lot of flanges out of the rectangle, the optimal packing is hexagonal. You can see it . If you put one row along the long edge of your plate, you will get $lfloor frac {6300}{235}rfloor = 26$ along the bottom row. You get $lfloor frac {2(2500-235)}{235sqrt 3} +1rfloor=12$ rows. Each row has $26$, so you get $312$ of them. If you put the straight edge along the short direction, you get a first row of $10$ and each row has $29$ for a total of $290$. The first is better. When the flanges are larger compared to the sheet, it gets more complicated. With other plate dimensions, the intervening rows might be one flange short.
answered Jun 6 '13 at 12:49
Ross MillikanRoss Millikan
300k24200375
answered Jun 6 '13 at 12:49
Ross MillikanRoss Millikan
300k24200375
answered Jun 6 '13 at 12:49
Ross MillikanRoss Millikan
300k24200375
answered Jun 6 '13 at 12:49
Ross MillikanRoss Millikan
300k24200375
300k24200375
$begingroup$
Can you please explain how you get the calculation for ⌊2(2500−235)2353√+1⌋=12 rows.I mean where does this ⌊2(2500−235)2353√+1⌋=12 rows come from??
$endgroup$
– kumaranando85
Jun 6 '13 at 12:56
$begingroup$
For $n$ rows and circle diameter $D$ you get a total hight of $D*(1 + (n-1)*sqrt(3)/2)$. This results in 12 for your example.
$endgroup$
– Axel Kemper
Jun 6 '13 at 13:16
$begingroup$
If you look at the figure, the spacing between the centerlines of the rows is $frac {235sqrt 3}2$ (draw the equilateral triangle that makes up one sixth of the hexagon.) You need one half disk above and below the centerlines, so that is where the deduction of $235$ comes from.
$endgroup$
– Ross Millikan
Jun 6 '13 at 13:18
$begingroup$
@RossMillikan I am just trying to understand the problem and trying different sizes can you please confirm how many flanges would fit if the rectangle is 1700 mm×4750 mm×25 mm (width × length × thickness).Thanks for your help.
$endgroup$
– kumaranando85
Jun 8 '13 at 15:59
$begingroup$
@kumaranando85: the thickness doesn't matter-we are assuming the plate is the proper thickness. Again, there are two ways to lay out the flanges. If a straight row is along the $4750$ direction, you would get $lfloor frac {4750}{235} rfloor=20$ along the bottom. Since there is not a half flange worth of excess, alternate rows would have $19$. You get eight rows, so $156$ total.
$endgroup$
– Ross Millikan
Jun 8 '13 at 17:21
add a comment |
$begingroup$
Can you please explain how you get the calculation for ⌊2(2500−235)2353√+1⌋=12 rows.I mean where does this ⌊2(2500−235)2353√+1⌋=12 rows come from??
$endgroup$
– kumaranando85
Jun 6 '13 at 12:56
$begingroup$
For $n$ rows and circle diameter $D$ you get a total hight of $D*(1 + (n-1)*sqrt(3)/2)$. This results in 12 for your example.
$endgroup$
– Axel Kemper
Jun 6 '13 at 13:16
$begingroup$
If you look at the figure, the spacing between the centerlines of the rows is $frac {235sqrt 3}2$ (draw the equilateral triangle that makes up one sixth of the hexagon.) You need one half disk above and below the centerlines, so that is where the deduction of $235$ comes from.
$endgroup$
– Ross Millikan
Jun 6 '13 at 13:18
$begingroup$
@RossMillikan I am just trying to understand the problem and trying different sizes can you please confirm how many flanges would fit if the rectangle is 1700 mm×4750 mm×25 mm (width × length × thickness).Thanks for your help.
$endgroup$
– kumaranando85
Jun 8 '13 at 15:59
$begingroup$
@kumaranando85: the thickness doesn't matter-we are assuming the plate is the proper thickness. Again, there are two ways to lay out the flanges. If a straight row is along the $4750$ direction, you would get $lfloor frac {4750}{235} rfloor=20$ along the bottom. Since there is not a half flange worth of excess, alternate rows would have $19$. You get eight rows, so $156$ total.
$endgroup$
– Ross Millikan
Jun 8 '13 at 17:21
$begingroup$
Can you please explain how you get the calculation for ⌊2(2500−235)2353√+1⌋=12 rows.I mean where does this ⌊2(2500−235)2353√+1⌋=12 rows come from??
$endgroup$
– kumaranando85
Jun 6 '13 at 12:56
$begingroup$
Can you please explain how you get the calculation for ⌊2(2500−235)2353√+1⌋=12 rows.I mean where does this ⌊2(2500−235)2353√+1⌋=12 rows come from??
$endgroup$
– kumaranando85
Jun 6 '13 at 12:56
$begingroup$
For $n$ rows and circle diameter $D$ you get a total hight of $D*(1 + (n-1)*sqrt(3)/2)$. This results in 12 for your example.
$endgroup$
– Axel Kemper
Jun 6 '13 at 13:16
$begingroup$
For $n$ rows and circle diameter $D$ you get a total hight of $D*(1 + (n-1)*sqrt(3)/2)$. This results in 12 for your example.
$endgroup$
– Axel Kemper
Jun 6 '13 at 13:16
$begingroup$
If you look at the figure, the spacing between the centerlines of the rows is $frac {235sqrt 3}2$ (draw the equilateral triangle that makes up one sixth of the hexagon.) You need one half disk above and below the centerlines, so that is where the deduction of $235$ comes from.
$endgroup$
– Ross Millikan
Jun 6 '13 at 13:18
$begingroup$
If you look at the figure, the spacing between the centerlines of the rows is $frac {235sqrt 3}2$ (draw the equilateral triangle that makes up one sixth of the hexagon.) You need one half disk above and below the centerlines, so that is where the deduction of $235$ comes from.
$endgroup$
– Ross Millikan
Jun 6 '13 at 13:18
$begingroup$
@RossMillikan I am just trying to understand the problem and trying different sizes can you please confirm how many flanges would fit if the rectangle is 1700 mm×4750 mm×25 mm (width × length × thickness).Thanks for your help.
$endgroup$
– kumaranando85
Jun 8 '13 at 15:59
$begingroup$
@RossMillikan I am just trying to understand the problem and trying different sizes can you please confirm how many flanges would fit if the rectangle is 1700 mm×4750 mm×25 mm (width × length × thickness).Thanks for your help.
$endgroup$
– kumaranando85
Jun 8 '13 at 15:59
$begingroup$
@kumaranando85: the thickness doesn't matter-we are assuming the plate is the proper thickness. Again, there are two ways to lay out the flanges. If a straight row is along the $4750$ direction, you would get $lfloor frac {4750}{235} rfloor=20$ along the bottom. Since there is not a half flange worth of excess, alternate rows would have $19$. You get eight rows, so $156$ total.
$endgroup$
– Ross Millikan
Jun 8 '13 at 17:21
$begingroup$
@kumaranando85: the thickness doesn't matter-we are assuming the plate is the proper thickness. Again, there are two ways to lay out the flanges. If a straight row is along the $4750$ direction, you would get $lfloor frac {4750}{235} rfloor=20$ along the bottom. Since there is not a half flange worth of excess, alternate rows would have $19$. You get eight rows, so $156$ total.
$endgroup$
– Ross Millikan
Jun 8 '13 at 17:21
add a comment |
$begingroup$
If you arrange the flanges in a rectangular grid, you'll get 26 rows and 9 columns. This is a total of 234 flanges. If you arrange the flanges in a staggered fashion (aka "hexagonal close packed"), you'll get 26 rows and 12 columns or a total of 312 flanges.
There is no generalized algorithm known (to me) which can do much better. Packing researchers seem to be focused on packing circles in squares rather than in rectangles.
There is an online circle packing calculator available for hexagonal packings. Note that you have to scale dimensions by 235 as the calculator assumes unit diameters.
You might have to add a safety margin to the 235mm to take into account the cutting width.
answered Jun 6 '13 at 12:46
Axel KemperAxel Kemper
3,49111418
$endgroup$
$begingroup$
I am just trying to understand the problem and trying different sizes can you please confirm how many flanges would fit if the rectangle is 1700 mm×4750 mm×25 mm (width × length × thickness).Thanks for your help.
$endgroup$
– kumaranando85
Jun 8 '13 at 16:01
$begingroup$
What about using the online calculator mentioned above and find it out on your own?
$endgroup$
– Axel Kemper
Jun 8 '13 at 17:05
add a comment |
$begingroup$
If you arrange the flanges in a rectangular grid, you'll get 26 rows and 9 columns. This is a total of 234 flanges. If you arrange the flanges in a staggered fashion (aka "hexagonal close packed"), you'll get 26 rows and 12 columns or a total of 312 flanges.
There is no generalized algorithm known (to me) which can do much better. Packing researchers seem to be focused on packing circles in squares rather than in rectangles.
There is an online circle packing calculator available for hexagonal packings. Note that you have to scale dimensions by 235 as the calculator assumes unit diameters.
You might have to add a safety margin to the 235mm to take into account the cutting width.
answered Jun 6 '13 at 12:46
Axel KemperAxel Kemper
3,49111418
$endgroup$
$begingroup$
I am just trying to understand the problem and trying different sizes can you please confirm how many flanges would fit if the rectangle is 1700 mm×4750 mm×25 mm (width × length × thickness).Thanks for your help.
$endgroup$
– kumaranando85
Jun 8 '13 at 16:01
$begingroup$
What about using the online calculator mentioned above and find it out on your own?
$endgroup$
– Axel Kemper
Jun 8 '13 at 17:05
add a comment |
$begingroup$
If you arrange the flanges in a rectangular grid, you'll get 26 rows and 9 columns. This is a total of 234 flanges. If you arrange the flanges in a staggered fashion (aka "hexagonal close packed"), you'll get 26 rows and 12 columns or a total of 312 flanges.
There is no generalized algorithm known (to me) which can do much better. Packing researchers seem to be focused on packing circles in squares rather than in rectangles.
There is an online circle packing calculator available for hexagonal packings. Note that you have to scale dimensions by 235 as the calculator assumes unit diameters.
You might have to add a safety margin to the 235mm to take into account the cutting width.
answered Jun 6 '13 at 12:46
Axel KemperAxel Kemper
3,49111418
$endgroup$
If you arrange the flanges in a rectangular grid, you'll get 26 rows and 9 columns. This is a total of 234 flanges. If you arrange the flanges in a staggered fashion (aka "hexagonal close packed"), you'll get 26 rows and 12 columns or a total of 312 flanges.
There is no generalized algorithm known (to me) which can do much better. Packing researchers seem to be focused on packing circles in squares rather than in rectangles.
There is an online circle packing calculator available for hexagonal packings. Note that you have to scale dimensions by 235 as the calculator assumes unit diameters.
You might have to add a safety margin to the 235mm to take into account the cutting width.
answered Jun 6 '13 at 12:46
Axel KemperAxel Kemper
3,49111418
answered Jun 6 '13 at 12:46
Axel KemperAxel Kemper
3,49111418
answered Jun 6 '13 at 12:46
Axel KemperAxel Kemper
3,49111418
answered Jun 6 '13 at 12:46
Axel KemperAxel Kemper
3,49111418
3,49111418
$begingroup$
I am just trying to understand the problem and trying different sizes can you please confirm how many flanges would fit if the rectangle is 1700 mm×4750 mm×25 mm (width × length × thickness).Thanks for your help.
$endgroup$
– kumaranando85
Jun 8 '13 at 16:01
$begingroup$
What about using the online calculator mentioned above and find it out on your own?
$endgroup$
– Axel Kemper
Jun 8 '13 at 17:05
add a comment |
$begingroup$
I am just trying to understand the problem and trying different sizes can you please confirm how many flanges would fit if the rectangle is 1700 mm×4750 mm×25 mm (width × length × thickness).Thanks for your help.
$endgroup$
– kumaranando85
Jun 8 '13 at 16:01
$begingroup$
What about using the online calculator mentioned above and find it out on your own?
$endgroup$
– Axel Kemper
Jun 8 '13 at 17:05
$begingroup$
I am just trying to understand the problem and trying different sizes can you please confirm how many flanges would fit if the rectangle is 1700 mm×4750 mm×25 mm (width × length × thickness).Thanks for your help.
$endgroup$
– kumaranando85
Jun 8 '13 at 16:01
$begingroup$
I am just trying to understand the problem and trying different sizes can you please confirm how many flanges would fit if the rectangle is 1700 mm×4750 mm×25 mm (width × length × thickness).Thanks for your help.
$endgroup$
– kumaranando85
Jun 8 '13 at 16:01
$begingroup$
What about using the online calculator mentioned above and find it out on your own?
$endgroup$
– Axel Kemper
Jun 8 '13 at 17:05
$begingroup$
What about using the online calculator mentioned above and find it out on your own?
$endgroup$
– Axel Kemper
Jun 8 '13 at 17:05
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