FIRST OFF
the balls will be arranged in between each other as in the diagram below to allow more space
the balls will have a certain number going across as you can only fit so many going along the x plane, since this is a cube the x and z will have the same amount of balls, the y axis however will be will have more as the balls going upwards will be zig zaggin along since the balls will be stacked in between each other as in figure one
the example amount i will use is a 5 by 5 box, so 5 balls across x and z on the first level
on the second level there will be a 4 by 4 grid as to put the balls in between on the second level on will have to be sacrificed by x and z
on the third level there will be a 5 by 5 grid again and on the 4th a 4 by 4
so the easiest way to work this out is to do the first 2 levels and multiply them by the overall length
to do this we need to work out the height of the segment which is hidden in between the balls on the second level, or how much height we save by putting the ball in between
here is a diagram
we do this by finding the length x of the triangle, do this by suing the equation
1.25 sin 30 = 1.23503953
minus the answer by the radius to get the height of the segment which is 0.01496 mm
so now the height of 2 levels is 5mm - 0.01496mm = 4.98504mm
divide that amount by 1m height 1000 / 4.98504 = 200.6
so there are 200.6 sets of 2 levels stacked upwards so 401.2 balls up
now to work out z and x
1000 / 2.5 = 400
first level amount = 160 000
second level amount = 399 x 399 = 159 201
so in the first 2 levels we have 319 201
we know there are 200.6 sets of 2 levels
so 200.6 x 319 201 = 64 031 720.6
or 64,031,720 balls
cant be bother to work out volume, jsut work out the volume for one and times it by the amount
the balls will be arranged in between each other as in the diagram below to allow more space
the balls will have a certain number going across as you can only fit so many going along the x plane, since this is a cube the x and z will have the same amount of balls, the y axis however will be will have more as the balls going upwards will be zig zaggin along since the balls will be stacked in between each other as in figure one
the example amount i will use is a 5 by 5 box, so 5 balls across x and z on the first level
on the second level there will be a 4 by 4 grid as to put the balls in between on the second level on will have to be sacrificed by x and z
on the third level there will be a 5 by 5 grid again and on the 4th a 4 by 4
so the easiest way to work this out is to do the first 2 levels and multiply them by the overall length
to do this we need to work out the height of the segment which is hidden in between the balls on the second level, or how much height we save by putting the ball in between
here is a diagram
we do this by finding the length x of the triangle, do this by suing the equation
1.25 sin 30 = 1.23503953
minus the answer by the radius to get the height of the segment which is 0.01496 mm
so now the height of 2 levels is 5mm - 0.01496mm = 4.98504mm
divide that amount by 1m height 1000 / 4.98504 = 200.6
so there are 200.6 sets of 2 levels stacked upwards so 401.2 balls up
now to work out z and x
1000 / 2.5 = 400
first level amount = 160 000
second level amount = 399 x 399 = 159 201
so in the first 2 levels we have 319 201
we know there are 200.6 sets of 2 levels
so 200.6 x 319 201 = 64 031 720.6
or 64,031,720 balls
cant be bother to work out volume, jsut work out the volume for one and times it by the amount