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?)

 

[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.

 

[haruspex]

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?

Yes. Can you see how to do that?
You know the immersed volume of each ball, so you can calculate the height the water will reach relative to the bottom of the ball. Find the volume of the hexagonal prism with that height containing the ball and subtract the immersed volume to find the volume of the surrounding water. Dividing that by the area of the hexagon to find the original height of the water surface relative to the bottom of the floating ball.

 

[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.