Tennis Balls in the Ocean

[some bloke]

TL;DR

Just wanted a sanity check on my calculations of tennis balls in the ocean.

Hello all,

I indulged myself a curiosity today on the question of how much the sea levels would rise if we covered them with tennis balls.

(Don't ask why, because I don't know!)

Here's the rough breakdown of my calculations and assumptions:

1: Tennis balls are, on average, 6.7cm (0.067m) in diameter and weigh 59.2g. They are buoyant, so will displace 59.2g of water each.
2: They will tessellate in a single (magically stable) layer on the entire surface of the ocean, which for no apparent reason will stop having any waves etc. and become perfectly flat. This will essentially be a grid of hexagons, each hexagon with a ball inside.
3: The balls will be touching.
4: The density of the ocean water is 1.024.5kg/m^3

I realised that the effect would be repeated so I only need calculate for one hexagon to work out the total effect.

The area of a hexagon in which the diameter of a tennis ball can be inscribed is 0.00389m^2. I got this by breaking the hex into equilateral triangles, then splitting one to get a triangle with a known edge equal to the radius of the circle and a known angle of 60°, then used soh-cah-toa to get the edge length of the equilateral triangle, and thus the edge of the hexagon, which I used to calculate the area.

The volume of ocean water displaced by a 59.2g tennis ball is mass/density which is 5.778x10^-5 m^3

The height raised is therefore volume divided by the hex area, which is 0.0149m, or just under 1.5cm.

It'd take about 92,859,600,000,000,000 tennis balls.

The height raised makes sense, I just want to make sure I got my method correct (it's been a while since I opened the "Physics calculations" door in my brain!)

Thanks in advance!

 

[haruspex]

TL;DR: Just wanted a sanity check on my calculations of tennis balls in the ocean.

The height raised is therefore volume divided by the hex area

Not so fast. Consider e.g. if instead they were almost the same density as the water, also hexagonal in section and 1m tall.
(Btw, not all cells would be hexagonal. What other shape is likely to arise, and how many of those would there be?)

Edit: I misread post #1. Your method works fine for my 1m columns case.

 

[256bits]

TL;DR: Just wanted a sanity check on my calculations of tennis balls in the ocean.

It'd take about 92,859,600,000,000,000 tennis balls.

A Point 5 could mention the area size of the oceans so one can check that number out.

Something tells me that there isn't enough rubber trees on earth to make that many tennis balls, let alone baby's heads available to cover them in the fuzz. Agriculturally and baby-making speaking, a few of the oceans may have to realistically remain uncovered til the sands of ends of time run out. Even so, with present technology and bio-tech replacements for the components of tennis balls, a workaround solution should be available to sample test the premise on perhaps just a small lake.

 

[Roberto Pavani]

An equivalent way to see it: the sea level rise is the same as the water rise you'd observe when placing one tennis ball in a hexagonal-section glass of salt water (with the hexagon circumscribing the ball).
The ocean problem reduces to a single cell by symmetry.

Geometry: each tennis ball touch 6 other balls

 

[Roberto Pavani]

TL;DR: Just wanted a sanity check on my calculations of tennis balls in the ocean.

It'd take about 92,859,600,000,000,000 tennis balls

That would be a nice GDP boost

 

[some bloke]

Not so fast. Consider e.g. if instead they were almost the same density as the water, also hexagonal in section and 1m tall.
(Btw, not all cells would be hexagonal. What other shape is likely to arise, and how many of those would there be?)

The hexagons (Assuming they were kept vertical by neigbouring hexagons) would displace almost a meter-tall column of water, but it would be the space that it occupies. Are you saying that I need to make some relation between the water displaced upwards and the volume occupied by the ball itself in this?

I'm now picturing that hexagonal section of water in a hexagonal vial. If I put a smaller hexagonal rod into it, the water would shoot upwards because the area the displaced water has to occupy is less than the hexagonal section, because the protruding rod is in the way - like a leaking piston.

A Point 5 could mention the area size of the oceans so one can check that number out.

Something tells me that there isn't enough rubber trees on earth to make that many tennis balls, let alone baby's heads available to cover them in the fuzz. Agriculturally and baby-making speaking, a few of the oceans may have to realistically remain uncovered til the sands of ends of time run out. Even so, with present technology and bio-tech replacements for the components of tennis balls, a workaround solution should be available to sample test the premise on perhaps just a small lake.

I've used the figure of 361 million square kilometers, which is an approximation I found!

 

[some bloke]

Regarding the available resources, I found an estimate that 12.33 billion tennis balls have been produced since they were invented. By my calculations (which I now doubt due to not accounting for the balls being in the way of the displaced water) this would raise the ocean by 0.00197mm.

 

[A.T.]

...how much the sea levels would rise if we covered them with tennis balls.

Coincidentally this is already a running project of humanity, just using less specific junk.

 

[256bits]

I'm now picturing that hexagonal section of water in a hexagonal vial.

The problem is the circle packing problem.
How many circles can fit is a specific area?

For a circle of radius r, 1 circle will fit in a square of side dimension r, or within an equilateral triangle of side length r. packing density - 0.785. Within a hexagon of side length r, the density increases to 0.9069. The bounded shape and size will have an effect of the packing and the packing density.

For a finite area with side dimensions greater than r and of arbitrary shape, the hexagonal packing will most likely not have the greatest density, at least along the edges for large finite shapes. For example, a square with sides 2r, 3 balls fit within using hexagonal packing. That can be increased to 4 with I think it is called box packing. or square packing. ( the top layer arranged one on top of the bottom, rather than in the dip between the balls as in hexagonal packing ).

One might find that around coastal areas, the boundary, the most dense packing is something other than expected. How far that extends into the main body of ocean and the error produced would have to be determined.

Strictly speaking hexagonal packing for circles for some is that where there is no circle ( ball ) placed within the centre of the 6 balls forming the perimeter of the hex. Triangular packing has a ball in the centre. But I do not think we should not quibble about that, since we know what is meant.

 

[Roberto Pavani]

border effects goes like L, area goes as L^2 so the inperfect packing error is negligible on ocean scale

 

[some bloke]

I hypothesise that with the majority of the ocean being non-coastal, it would benefit most from the balls being arranged in a hexagonal grid, consisting of rows which are offset by r for each row, allowing them to pack into a hexagonal arrangement, in the same way as honeycomb.

I can see what you're saying - if, for an extreme, there were a long thin bay which fit precisely 1 tennis ball wide, which was not aligned with the grid we've created, then dropping the grid of tennis balls onto it and removing any which aren't in the ocean would leave the bay entirely devoid of tennis balls. However, I do not believe that the surface area estimate I have used for the ocean is accounting for such fine details anyway (and would likely vary significantly depending on where the high tide is at the time of measuring!), so I don't think that such details would have a significant impact on the final estimate.

On your statement on which shapes fit what, I don't believe you are accurate; a circle of radius r will fit in a square of edge size 2r, or an equilateral triangle of side length r/(sqrt(3)/2).

As far as I am aware, the most efficient way to tessellate circles is in the same way as a hex grid (each circle touches 6 others). This would be the most efficient until the edges, where it only concerns the area of the ocean which is within 1 tennis-ball of the shore, where irregular packing would likely be more effective due to obstacles.

 

[Baluncore]

Déjà vu.

 

[Roberto Pavani]

My previous answer may have been too short. The coastline effect gives a correction that scales with coastline length (L), whereas the total ball count scales with ocean area (L²). Even a wildly generous coastline estimate changes the result by only ~10⁹ balls out of ~10¹⁷ (~10⁻⁸ relative error).

 

[Roberto Pavani]

TL;DR: Just wanted a sanity check on my calculations of tennis balls in the ocean.

The height raised is therefore volume divided by the hex area, which is 0.0149m, or just under 1.5cm.

A slightly simpler way to see it: since Archimedes guarantees each ball displaces exactly $m/\rho$ of water regardless of geometry, and the hex cell is the area "owned" by each ball, the rise is simply:

$\Delta h = \frac{m_{\text{ball}}}{\rho_{\text{water}} \cdot A_{\text{hex}}}$ (for floating objects)

No need to worry about immersion depth or spherical cap geometry, Archimedes does all the work. The shape matters only for determining the packing arrangement (hence the hex area), not for the displaced volume.

 

[Roberto Pavani]

Probably I'm wrong but my naive idea is to fill the "hole" generated by the submerged part of the ball with water, the amount of water needed is given by mass of the ball divided by water density, now there are no more balls on the ocean but just "extra water" (or extra water in the exagonal glass for symmetry reason) in this case if I remove that extra water how much the level will diminuish? exactly the amount will increase without extra water but putting back the ball on the glass.
And from that the computation:
Δh = m_ball / (ρ_water · A_hex)

 

[haruspex]

Probably I'm wrong but my naive idea is to fill the "hole" generated by the submerged part of the ball with water, the amount of water needed is given by mass of the ball divided by water density, now there are no more balls on the ocean but just "extra water" (or extra water in the exagonal glass for symmetry reason) in this case if I remove that extra water how much the level will diminuish? exactly the amount will increase without extra water but putting back the ball on the glass.
And from that the computation:
Δh = m_ball / (ρ_water · A_hex)

Apologies - I misread your method. It’s fine, and it is the same as in post #1.

 

[256bits]

My previous answer may have been too short. The coastline effect gives a correction that scales with coastline length (L), whereas the total ball count scales with ocean area (L²). Even a wildly generous coastline estimate changes the result by only ~10⁹ balls out of ~10¹⁷ (~10⁻⁸ relative error).

Previous answer was more to the point for a realistic earth, with the error given here of correct magnitude.

This answer here though is incomplete for scaling in that for a finite surface body such as the earth, an increase in land mass area will decrease the ocean area, and vice-versa. Accordingly, the coastline length increases as ocean area decreases which is what one would be interested in for the discussion of balls covering the wet area, if the wet area changes in time.
The coastline length, being fractal in nature, scales somewhere between 1L to 2L, with L being the unit of measure in this case the diameter D of a ball, or a hex dimension if so chosen. In fact, if L << D, the coastline length can approach infinity as an asymptote. Since a unit of measure L << D doesn't affect the number of balls that can be placed upon the wet surface, a choice such as this does end up being immaterial for the problem at hand.

It reminds me of a mathematical function where the area under the curve is finite while the length of the curve is infinite. The paradox is usually presented as a painting problem, where one can paint the area between a wall formed from the curve and the xy axis, but never have enough paint to cover the wall itself.
https://en.wikipedia.org/wiki/Gabriel's_horn

 

[GiorgioPastore]

....

The height raised is therefore volume divided by the hex area, which is 0.0149m, or just under 1.5cm.

You should take into account that this is just an approximation for the height change. A more accurate calculation should account for the geometry of the region occupied by the displaced water. Using the formula for the volume of a spherical cap ( https://en.wikipedia.org/wiki/Spherical_cap ) the exact calculation is not difficult.

 

[haruspex]

You should take into account that this is just an approximation for the height change. A more accurate calculation should account for the geometry of the region occupied by the displaced water. Using the formula for the volume of a spherical cap ( https://en.wikipedia.org/wiki/Spherical_cap ) the exact calculation is not difficult.

It is not an approximation. Archimedes' Principle is exact and depends only on the volume, not details of the shape.

 

[GiorgioPastore]

It is not an approximation. Archimedes' Principle is exact and depends only on the volume, not details of the shape.

Archimedes' Principle gives the exact volume of water displaced by a submerged or partly submerged body. Where the displaced water goes depends on the geometry. Do you think that the level of the free surface of a fluid in a cylindrical container varies in the same way as a function of the degree of submersion of a rigid body, if the geometry of the body is a cylinder or a sphere (keeping the same volume for both)?

 

[haruspex]

Archimedes' Principle gives the exact volume of water displaced by a submerged or partly submerged body. Where the displaced water goes depends on the geometry. Do you think that the level of the free surface of a fluid in a cylindrical container varies in the same way as a function of the degree of submersion of a rigid body, if the geometry of the body is a cylinder or a sphere (keeping the same volume for both)?

Read post #15 carefully. The argument goes like this:
With a ball floating inside its hexagonal prism, there are three regions of interest above the level of the bottom of the ball, B:

1. The volume of water, $V_w$
2. The volume of ball in the water, $V_i$
3. The volume of ball above the water, $V_b$

$V_i$ is determined by the mass of the ball.
If the height of the water above B is H and the base area of the prism is A then $V_i+V_w=AH$.
If we now remove the ball, the water level returns to its original level. Let that be h. above B. The water above B has volume $Ah$, but it is the same water that was above B previously, so $V_w=Ah$.
Hence $A(H-h) = V_i+V_w-V_w = V_i$. The answer to the question posed is $H-h$.

 

[Roberto Pavani]

My naive method was based on reversing the problem: instead of 'how much does the ocean rise when we add balls?', I asked 'how much does it fall when we remove them?'
Which is trivially m/(ρ·A).

 

[GiorgioPastore]

...
If the height of the water above B is H ...

Then you are assuming that the balls are completely submerged in water. That would be a simpler case. But with tennis balls, they are only partially submerged, which requires a more thorough analysis.

 

[haruspex]

Then you are assuming that the balls are completely submerged in water. That would be a simpler case. But with tennis balls, they are only partially submerged, which requires a more thorough analysis.

No, it does not make that assumption. I am defining H to be the height of the water above the base B of the ball. The top of the ball is higher yet.

 

[GiorgioPastore]

Read post #15 carefully. The argument goes like this:
With a ball floating inside its hexagonal prism, there are three regions of interest above the level of the bottom of the ball, B:
...

Now I see my mistake. I was considering the change of level from the bottom of the submerged object. In such a case, objects with the same mass but different shapes may have different distances from the bottom to the water's free level. However, such a distance is not directly related to the change in water level.

Here is how I would recast the explanation.

Let's consider a prismatic or cylindrical container of sectional area $A$, and let $V_{displ}$ be the volume of the liquid displaced when an object floats in it. Now, in the absence of the body, we can divide the total volume of the water into a volume $V_0=A*H_0$ and a remaining volume $A*h = V_{displ}$. Notice that, for a given value of the sectional area, $h$ doesn't depend on the shape of the object.

Now, the effect of the floating object is equivalent to adding another volume of water $V_{displ}$ on top of the previous free level, with a net increase of the water level of $V_{displ}/A$, independent of the object shape. The key point is to use the height of the volume of fluid not affected by the "displacement" as the reference level.