Sizing Snowmen (Snowy Shenanigans II)

When you are using bigger snowballs, more volume gets lost toward the sides.

This image shows how a lot of volume is lost to the sides, and that volume could be used to build more small snowballs higher up.

Relevant wiki: Volume of a Sphere

Let $n$ be the number of equally sized spheres in a snowman. The volume of one sphere is $V = \frac{4}{3}\pi r^3$ , so the volume distributed over $n$ spheres would be $\frac{V}{n} = \frac{4}{3}\pi r^3$ , which means each sphere would have a radius of $r = \sqrt[3]{\frac{3V}{4n\pi}}$ and a height of $h = 2r = 2\sqrt[3]{\frac{3V}{4n\pi}}$ . The total height $T$ of all the spheres stacked together would then be $T = nh$ or $T = 2n\sqrt[3]{\frac{3V}{4n\pi}}$ . So when the volume $V$ is constant, then $T = kn^{\frac{2}{3}}$ for some constant $k > 0$ , so as $n$ increases, $T$ increases.

In other words, if every child starts with the same amount of snow and makes snowmen with equally sized spheres, then the tallest snowman will be the one made with the most amount of spheres. Since Georgina is making her snowman with the most amount of spheres, her snowman will be the tallest.

Take a ball of plasticine, roll in into a long thin line, cut the line into many small sections, roll these into small balls, rest them end to end. This is longer than width of the original ball, using the same volume. Same argument applies to this problem

Let's assume that every child has a certain volume of snow $V$ and that the height of their respective snowman is denoted $h$ . Let's analyze each case :

Fraser :

The total volum of snow is shared between 2 spheres of equal radius $r_{F}$ , that we can obtain because we know that $V = 2 \cdot V_{sphere}$ :

$r_{F} = \sqrt[3]{\frac{3 V}{8 \pi}}$

Georgina :

The total volum of snow is shared between 10 spheres of equal radius $r_{G}$ , that we can obtain because we know that $V = 10 \cdot V_{sphere}$ :

$r_{G} = \sqrt[3]{\frac{3 V}{40 \pi}}$

Owen :

The total volum of snow is shared between 6 spheres of equal radius $r_{O}$ , that we can obtain because we know that $V = 6 \cdot V_{sphere}$ :

$r_{O} = \sqrt[3]{\frac{3 V}{24 \pi}}$

At this point, the height of the snowman can be defined as follows: $h = n r$ where $n$ represents the number of spheres composing the snowman, and $r$ the radius of each sphere in one specific case. The next table resume every information for every case :

Where we can easily conclude that ​Georgina's snowman is the highest.

Let's suppose for the sake contradiction, that the number of spheres do not affect the height of the stack.

In that case, what happens in the limiting case? In the limiting case, we have concentrated the entire volume of snow into a really long stack. This contradicts our assumption.

In fact, studying the limiting case above immediately yields that increasing the number of balls increase the height

This problem follows the same principle as the Dirac delta function. Maintaining the same area (in this case volume) it is possible to get higher as it gets thinner.

You want to maximize R / V = R / [(4/3) pi R^3]. This is inversely proportional to R^2, hence largest when R is smallest.

The question is the same as asking for a certain height which solution uses the less snow.

V = 4/3 pi (R/x)^3*x where x is the number of balls

V = 4/3 pi R^3*1/x^2

V = constant/x^2

=> for the same height, the smaller balls you do the less volume you use = the smaller the balls are the higher you get with the same volume of snow.

Let $V$ be the snow's volume, $r$ be each snowball's radius, $h$ be the height of the snow person, and $n$ be the number of snowballs used.

$\frac {4}{3} \pi r^{3} n = V$

$2rn = h$

Therefore,

$r = \frac {h}{2n}$

which means

$\frac {4}{3} \pi (\frac{h}{2n})^{3} n = V$

Thus,

$\frac {4}{3} \pi \frac {h^{3}}{8n^{3}}n = V$

and

$\frac {\pi}{6} \frac {h^{3}}{n^{2}} = V$

This means

$h^{3} = \frac {6Vn^{2}}{\pi}$

and

$h = \sqrt[3]{\frac{6Vn^{3}}{\pi}}$

which can be written as

$h = \sqrt[3]{\frac{6V}{\pi}} \sqrt[3]{n^{2}}$

Because $\sqrt[3]{\frac {6V}{\pi}}$ is constant, we know that $h$ is directly proportional to $\sqrt[3]{n^{2}}$ . Since this number is always positive, Georgina is correct.

This volume problem can be modeled with cubes. The greater the side of each cube, the more snow will be used for width and depth. To maximize the height, the width and depth need to be minimized. Of course, in the real world, the narrow one will have stability issues and if the environment is not cold enough it will also melt faster.

Let's denote the heights of the three snowmen $h_1,\:h_2$ and $h_3$ . One can deduce that $h_n=2nr_n$ , where $r_n$ is the radius of the spheres of the $n^{th}$ snowman. Additionally, the volume of the $n^{th}$ sphere is given by $V_n=\frac{4}{3}\pi r_n^3$ . This implies that $r_n=\sqrt[3]{\frac{V_n}{\frac{4\pi}{3}}}=\sqrt[3]{\frac{3V_n}{4\pi}}$ . Additionally, $V_n=\frac{V}{n}$ , whereas $V$ is the volume each child receives, so $r_n=\sqrt[3]{\frac{3V}{4n\pi}}$ . The total height is therefore:

$h_n=2n\sqrt[3]{\frac{3V}{4n\pi}}$

Applying this formula for all our cases (though this is not strictly necessary as the cube root is a slow-growing function compared to the double):

$h_1=4\sqrt[3]{\frac{3V}{8\pi}}=2\sqrt[3]{\frac{3V}{\pi}}, \:\:\:h_2=20\sqrt[3]{\frac{3V}{40\pi}}, \:\:\:h_3=6\sqrt[3]{\frac{V}{\pi}}$

Cubing each height, we get:

$\underbrace{h_2^3}_{600\frac{V}{\pi}}>\underbrace{h_3^3}_{216\frac{V}{\pi}}>\underbrace{h_1^3}_{24\frac{V}{\pi}}$

And therefore the tallest tower is the second one. In conclusion, having more spheres is equivalent to building a taller snowman.

If the size of the ball is irrelevant, and all balls are the same height, then the column of 10 snowballs will be encapsulated entirely by the snowman made of two snowballs. But that means the snowman of two snowballs has more volume, which is not allowed. As that is not allowed, then the difference in volume must be converted to more balls, and as such, the more balls, the higher.

I thought about isoperimetric theorem, that you have, in this case, the same volume (the same amount of snow), so with less radio the perimeter will be bigger

If you fell into a sausage machine you might make a long sausage, but if you fell in a spaghetti machine you’d make a really long noodle.

You could answer this problem in two ways, but this is the solution that I found the simplest:

The square-cube law states that if the height of a cube is increased by 10, the surface area is increased by 100, and the volume is increased by 1000.

Using this concept we can ask this other question:

Which would have less volume? A cube with height 50 cm? Or 10 cubes with height 5 cm?

FOR CUBE WITH HEIGHT 50 CM: It is equal to a 5cm high cube with it's height multiplied by 10 meaning: Height=50 cm Surface area=15000 cm² Volume=125000 cm³

FOR 10 CUBES WITH HEIGHT 5 CM: A cube with height 5 cm has a surface area of 150 cm², and a volume of 125 cm³ If there are 10 of them then the volume would be: 10×125 cm³ = 1250 cm³

Therefore 10 cubes with 5 cm height have less volume than a a cube with 50 cm height.

Tying this with the snowman question, we can make this analysis:

You can make more of a smaller 3 dimensional object like a cube or sphere( or any 3d object affected by this law) than of a large 3d object, given a set volume.

The total volume of snow, allotted to each one of the three, is a given (constant): $V_{tot}$ . If somebody decides to make 'N' number of snowballs of diameter 'D', N times the volume of each snowball should be the whole volume $V_{tot}$ : $V_{tot} = \frac{N}{8} \pi D^3$ . The game, however, is to maximise your aggregated height 'H', which is the number of snowballs 'N' times the diameter 'D' of each ball : $H = N \times D$ . From the earlier equation (on the constraint of the total volume), H turns out to be, $H=ND = {\bigg( \frac{8 V_{tot}}{\pi} \bigg)}^{\frac{1}{3}} N^{\frac{2}{3}}$ . So, $H \propto N^{\frac{2}{3}}$ . More number of snowballs; greater the height. And hence, Georgina is correct!

We could rearrange the volume of a sphere equation and work backwards, but it works to prove that if you end up with the same total height, the combined radii would be the same. For instance, if you envision that each person wanted to build a tower 60 inches tall, Fraser would make 2 spheres with a radius of 15 inches each, Georgina would make 10 spheres with a radius of 3 inches each, and Owen would make 6 with a radius of 5 inches each. When plugging these values in and finding the total volumes needed, we get cubic inches of snow of 9000pi, 360pi and 1000pi respectively. Thus it takes way more snow to make the same size tower with 2 or 6 than it does 10 spheres.

You can think of this as concentrating the volume towards its length only, means more will be the radius more volume wastage.

Also if we just think of length then ideally we can create infinite length

As the mass and density of the snow remains constant and is isotropic, hence the volume is directly proportional to its size of the snow sphere . Also mass is directly proportional to the volume. So Fraser's snow man will have the greatest chance of collapsing, then of Owen. However as the size of snowballs of Georgina is smallest, the effective weight needed to collapse is very less. Hence this snowman can withstand the maximum compression. Also it will be better if the lower snowballs have a little bigger volume than the top.

If you take out from a sphere the shape and size of two spheres with half the radius, then there will be a lot of leftover snow. Therefore, you need less volume for the same height using smaller spheres. That means you have leftover volume, which can be made into new snowballs to increase height.

The number of spheres with a radius $R$ in a snowman with fixed volume $V$ is just $\frac{V}{\frac{4}{3}\pi R^3}$ and so the height is the height of each sphere $2R$ times this number of spheres $\Rightarrow h = \frac{V}{\frac{2}{3}\pi R^2}$ . So, we can see that the larger each sphere the shorter the snowman, which means Georgina is correct.

Suppose we have total amount of snow of volume $k$ and is distributed among $n$ balls, each of radius $r$ . Then, we must have $\frac{4}{3}n\pi r^3=k$ . Thus, since the total height of the snowman is $h=2nr$ , we have $h=\frac{3k}{2\pi r^2}$ . The snowman is higher if the radii of the balls are smaller, which holds if there are more balls.

In this case we are only interested in increasing the height of the snow man as much as we can with the help of spheres having only a limited amount of snow so the less the ready of the snow the less the snow is wasted which doesn't contribute to increase the height of the snow so the less the radius of the spheres the more would be contributed to only increasing the height of the ☃️ but you have to done it very fast otherwise the ice will melt 😁

Imagine one sphere as 2 units in diameter, the other as 1. The volume of a sphere with twice the diameter contains the cube, or 8× the volume as a sphere with 1× the diameter. Since each child has the same amount of snow to work with, in this example, the smaller spheres would stack up 4× higher. Another mental visualization of this cube law is with 1" cubes. 8 of them will form a 2" cube.

Fun extra question: What would the ratio of the heights of these three snowmen be?

Volume of a cube is proportional to the cube of the radius. If you double the radius of the cube, the volume increases by a factor of 8. Therefore, to maximize radius given a constant volume, fewer cubes are better than more cubes. For example, if you have 32*pi/3 cubic meters of snow, this would allow one sphere with a radius of 2, or 8 spheres with a radius of 1.