Maths Quiz.

[Cook$]

That's true.
But shush, get back to counting your balls.

 

[Ainsley]

I've got it (after a trip to the local ball bearing shop) - but just give me a few days, I'm counting them out now

Edit.....make it Friday, think I miss counted.

 

[FrontlineArtist]

I've got it (after a trip to the local ball bearing shop) - but just give me a few days, I'm counting them out now

Edit.....make it Friday, think I miss counted.

Do you have a ego 2? you could use that to count them.

 

[Skeet]

So did we work this one out?

Are you all shamed by my practical application of basic Maths getting pretty much, the correct answer?

Hmmm?

I did PM Robbo to come have a look, but he doesn't seem to have had time yet.

 

[s1im]

Well, here's goes GCSE maths

I also noted that each sphere would likely interlock and not space itself out evenly..........

tbh this is the closest to being correct. the only way you will accurately get the solution to this is by imputing this into a piece of fluidworks software that will work this out for you, and even then it would have to work out multiple computations for it to give yo the max balls, there is not a std math equation that would work this out,
I would bet money that bon, bid not have the correct number. he may be close but not correct, unless he has some very expensive tools at his disposal!

 

[admin]

tbh this is the closest to being correct. the only way you will accurately get the solution to this is by imputing this into a piece of fluidworks software that will work this out for you, and even then it would have to work out multiple computations for it to give yo the max balls, there is not a std math equation that would work this out,
I would bet money that bon, bid not have the correct number. he may be close but not correct, unless he has some very expensive tools at his disposal!

id bet that also

 

[Dark Warrior]

After doing my homework, working it out and then reading through the thread, I concur with MsHell.
Saved me writing all calculations down as it is already there.
Next time I'll read the rest of the thread first lol

 

[Skeet]

tbh this is the closest to being correct. the only way you will accurately get the solution to this is by imputing this into a piece of fluidworks software that will work this out for you, and even then it would have to work out multiple computations for it to give yo the max balls, there is not a std math equation that would work this out,
I would bet money that bon, bid not have the correct number. he may be close but not correct, unless he has some very expensive tools at his disposal!

I must say, that as it is entirely self serving...I would have to agree with you

 

[PaintballWizard]

Calculating spheres (EDIT: apologies for the stars, my phone put them in for some reason )

Right then..... Having had a good look at this thread, I had a think, outside the box. (see what I did there ) rather than calculating how Many can you fit in, how about how many you CAN'T fit in?

Bear with me. I'm on iPhone so no pictures sorry. You'll just have to imagine. Oh and (2) means squared and (3) cubed.

To do it via this method, we need two things. The volume of the box, the volume of the empty space, and then we can subtract the two, and divide it by the volume of a ball bearing to find out how many will fit

So let's get the easy bit out of the way and work out the volume of the box. I'm going to work all my units in mm to make it easier. Now there's 100 cm in a meter and 10mm in a cm, so 1m = 1000mm.*
The box is 1m x 1m x 1m so that's 1000mm(3) = 1,000,000,000 cubic mm.

Okay, now it gets more complicated. We need the volume of the empty space.

Take a 2.5mm ball bearing and place it in a 2.5mm(3) box. The sides, top and bottom of the sphere will meet the confining Walls of the box at an inifinte point. Now, what are you left with? 8 spaces around the corners. Now what are we going to do with them? Calculate the space.*

2.5mm diameter ball.
2.5mm box

Volume of ball = (4/3)pi r(3)
R = 1/2 D = 1.25mm
so.... Plug that in to a calculator:
= 8.18123086872

Volume of box = 2.5mm(3)
= 15.625

Okay, so wasted space is*15.625 -**8.18123086872 =*7.4437691328.*

Now that is the total area of blank space inside the cube. There are 8, equal spaces so*7.4437691328/8 to find the volume of each individual space. =*0.9304711416.

Okay, now we can apply that to the box and ball bearing problem.

But where is this going to help us?
In each corner, there is a ball. As such, this creates a space, in the corner, which we know the volume of now.*

Now I'm going to finish my paper round (writing it at the same time) and get some grub and my laptop to do some pictures for your viewing pleasure

 

[BDN]

There appears to be two answers to the contents, a theoretical one and a reality one.

Theoretical:- 1000mm divided by 2.5mm = 400 ball bearings per layer

box being 3 dimensional, 400 x 400 x 400 = 64 000 000 ball bearings.

However, try to pack the little blighters to achieve 400 per layer won't be easy. The result
would be as illustrated previously ie bottom layer 400 ball bearings, next layer a few less as
the ball bearings will roll into the gap between the balls of the lower layer.

The volume of each ball is 6.5449 cubic mm

times 64 000 000 (for the theoretical amount) = 418879013.3 cmm