does a bucket of half inch steel balls weigh more than a bucket of quarter inch steel balls?

Question

I've been told that a volume of steel balls regardless of the daimeter will weigh the same. How can this be proven?

Answer 1

You are correct. Assuming all the balls all have the same density then the two buckets will weight the same.

You can demonstrate this by calculating the volume of a cube with a single sphere inscribed inside with a diameter equal to the width of the cube. Now calculate the volume of the sphere and subtract it from the volume of the cube. The difference or "void" space should be about 30%.

Now do the same thing but use 8 smaller spheres inside the cube with a diameter of 1/2 the width of the cube. Again calculate the void space. You will see it is also about 30%.

You can do this with any size cube and set of spheres and you will always get about 30% void space. That is as long as you completely "fill" the cube with spheres.

There are other ways to demonstrate this fact but I remember this one because it is common knowledge in the oil business that most rock that contains oil has a pore space of 30% and I did the calculation to prove it for myself.

Answer 2

1 Inch Steel Ball

Answer 3

You have a few correct answers so far, and a lot of wrong ones.

The "void fraction" for packed UNIFORM spheres is a constant, and does not depend on the value of their diameter. The proof is simply that if you arrange the balls in layers and work it out, the diameter cancels out of the result.

The void fraction for mixed sizes is smaller, and the more mixed, the smaller.

For larger balls there are edge effects, which will get generally worse as the ball size increases, but not smoothly. Think of trying to fill a twelve-inch bucket with seven-inch balls, then think how neatly four-inch balls would fit in.

At the sizes you are asking about, the edge effects will be quite tiny, but they may spoil the ball packing enough that the bucket of larger balls could weigh just a little less than the bucket of smaller ones.

Answer 4

why would you put steel balls in a bucket? you should get a scale and steel balls and a bucket, but they could weigh the same because a 1/4 steel ball is half of 1/2 a steel ball. It would weigh about the same. Personally, I think that the 1/4 steel balls in a bucket would weigh more because if you divide the weight of 1/4 steel balls in the bucket by 2, don't take any out, the bucket is bound to more hold more of the 1/4 steel balls, because the 1/4 balls are smaller, and take up less space, allowing it to hold more 1/4 steel balls, divided by 2, it should be larger than the amount of the 1/2 steel ball's bucket.

Answer 5

A bucket full of identically sized spheres will have about 36% empty space if randomly packed. Precise stacking can improve this to 26%. This is true for any size sphere, provided the spheres are small relative to the size of the bucket. So the bucket of half-inch balls will weigh the same as the bucket of quarter-inch balls. Those who have reasoned that the smaller balls are more densely packed because the voids are smaller are neglecting to consider that the voids are also proportionately more numerous.

Answer 6

That's an interesting question.

And it may not be as simple as checking the air gaps between the balls (obviously the air gap is smaller with 1/4" balls, but there are a heck of a lot more of them - it might be a wash.)

Let's check it out:
Think of a 2 dimensional square with 1" sides (obviously, 1 sq.") We can neatly fit four 1/2" circles inside or sixteen 1/4" circles.

The area of a 1/2" D ball is 0.1964 sq". Times 4=0.7856 sq". Which leaves the space between the 1/2" balls at 0.2144 sq".

Each 1/4" ball has an area of .049sq" times 16 balls = 0.784sq". The 1/4" balls leave a space open of 0.216sq".

1/2" balls leave 0.214 the 1/4" balls leave 0.216.

And given that we are rounding off pi and the results, it's just too close to call.
My guess is that they will weigh the same.

Answer 7

No that bucket would weigh less. More of the total area is filled with steel in the case of the 1/4"

Answer 8

You were told wrong. The radius of the quarter inch steel balls is smaller, so they will pack together tighter, leaving less air in the bucket. To prove it, simply draw a bunch of circles the same size, side by side, and compare that drawing with circles that are larger. You will see a lot more space between them.

Regards,

Chris

Answer 9

The smaller the balls the more mass you will have. Imagine a ball that is the size of one steel molecule. You would have a solid piece of steel and this would have no air gaps so it would weigh much more.

Answer 10

depends on the mass of the steel balls

assuming they are all the same density (mass), the 1/4 would weigh more. But its possible the bigger balls are denser steel, and thus have more mass