Number of dimples on a golf ball

[johnred]

Homework Statement

A golf ball has a diameter of 3cm and a dimple has a diameter of 0.3cm. Give a (reasonable) estimate for the largest number of dimples that can fit on the golf ball. Hint: The number of dimples is lower than 100.

Relevant Equations

Area of the ball = 4*pi*R^2
R=1.5
r=0.15
The area of a dimple is the area of a spherical cap.

The approximate number of dimples is 1/(1-√(1-r^2/R^2))

 

[jedishrfu]

I got 400 by simply dividing the sphere's area by the area of the dimple:

$4\pi R^2 / (\pi r^2) = 4 r^2 / r^2 = 4 * 1.5^2 / 0.15^2 = 400$ dimples

 

[WWGD]

I got 400 by simply dividing the sphere's area by the area of the dimple:

$4\pi R^2 / (\pi r^2) = 4 r^2 / r^2 = 4 * 1.5^2 / 0.15^2 = 400$ dimples

But aren't there parts of the sphere that are not dimpled?

 

[jedishrfu]

Quite true, my upper limit is a rough estimate, a finer estimate would consider that And then there’s the bin packing issue and whether the sphere can be completely covered with dimples that touch one another.

 

[WWGD]

Quite true, my upper limit is a rough estimate, a finer estimate would consider that And then there’s the bin packing issue and whether the sphere can be completely covered with dimples that touch one another.

Interesting Combinatorial questions.

 

[haruspex]

To get a lower bound, you could divide the surface into $\frac{\pi R}{2r}$ latitudinal bands of width 2r and calculate how many dimples would fit in each.

 

[johnred]

Quite true, my upper limit is a rough estimate, a finer estimate would consider that And then there’s the bin packing issue and whether the sphere can be completely covered with dimples that touch one another.

Indeed, we are looking for the upper limit.

The ball be completely covered with dimples that touch one another (but never overlap). The problem states that there are overlaps with 100 hemispherical dimples (of diameter 0.15cm). As we want to avoid the overlaps, I assume the number of dimples should be lower than 100. Therefore, there cannot be 400 dimples.

 

[haruspex]

Indeed, we are looking for the upper limit.

The ball be completely covered with dimples that touch one another (but never overlap). The problem states that there are overlaps with 100 hemispherical dimples (of diameter 0.15cm). As we want to avoid the overlaps, I assume the number of dimples should be lower than 100. Therefore, there cannot be 400 dimples.

Using my approach in post #6, I seem to have fitted 258 in.

 

[hutchphd]

I have in my hand a golf ball. The dimples are sized approximately as described and there are many more than 100 on the surface. The problem is misstated.
No I will not count them...

 

[Vanadium 50]

Since the estimates are coming out at a few hundred, and the problem states fewer than 100, why not write 99 and put a box around it?

 

[haruspex]

Strange that it says <100, yet a formula is given,
"The approximate number of dimples is 1/(1-√(1-r^2/R^2))"
that yields 200.

 

[epenguin]

I got 400 by simply dividing the sphere's area by the area of the dimple:

$4\pi R^2 / (\pi r^2) = 4 r^2 / r^2 = 4 * 1.5^2 / 0.15^2 = 400$ dimples

Whatever you get by that calculation you can then reduce by about 10%, because the area between touching circles is about that (fraction of area occupied by close packed touching circles π√12)
https://en.wikipedia.org/wiki/Circle_packing

We were only asked for a rough calculation.

An exact number for not very large balls is a more difficult problem. The answer won't be much different. I think I might be able to get something better given enough time.

There is a big difference between your answer and the given one. Maybe we have not been given the right specifications. The dimples on the golfball do not touch each other do they? Maybe there is a specification dimple diameter and distance between dimples? But that is only a matter of the parameters, doesnot change the geometric question.

How do they make golf balls anyway? At first sight it seems impossible.

 

[haruspex]

Whatever you get by that calculation you can then reduce by about 10%, because the area between touching circles is about that (fraction of area occupied by close packed touching circles π√12)
https://en.wikipedia.org/wiki/Circle_packing

We were only asked for a rough calculation.

An exact number for not very large balls is a more difficult problem. The answer won't be much different. I think I might be able to get something better given enough time.

There is a big difference between your answer and the given one. Maybe we have not been given the right specifications. The dimples on the golfball do not touch each other do they? Maybe there is a specification dimple diameter and distance between dimples? But that is only a matter of the parameters, doesnot change the geometric question.

How do they make golf balls anyway? At first sight it seems impossible.

Allowing a hexagonal footprint for each dimple is not the half of it. Since six will not fit snugly around one, there will be a lot more gap. As I posted, I got something around 250. Ok, it was a lower bound, but I think the exact answer will be closer to 250 than 350.
Wrt the significance of the "<100" hint, note that the approximate formula in post #1 yields 200. The 100 just seems to be a blunder.

 

[256bits]

Seems like a blunder.

I get 198 as a lower bound.
Taking 1/4 of a diameter, we can have 8 on-centre dimples fitting on the quarter diameter as a whole number.
Taking an equilateral triangle of sides 7 ( ie at each apex is 1/2 diameter of a dimple ) we can fit a maximum of 15 whole dimples within the interior.
Fitting 8 of these equilateral together to make a solid with equilateral faces of length 7,
we can see each of the 3 apexes is shared by 4 other triangles = 3/4 of a dimple per apex.
Each side shares 1/2 of a dimple
or per triangular area - 3 sides per triangle x 6 dimples per side x 1/2 dimple area = 18/2 dimples = 36/4

Apex 3/4 dimple
sides - 36/4 dimples
Perimeter total - 39/4 dimples
Interior - 15 dimples
Total per triangular face = ( 15(4)+39 ) / 4 = 99/4 dimples

Eight such triangular faces per solid = 8 x 99/4 == 198 dimples per octohedron solid.

Blowing up the solid to make a sphere, increases the surface area.
Can we add a few more whole dimples?
If so, how many whole dimples?

 

[haruspex]

I might have it up to 306.
I consider a band of dimples with centres at arc distance 2kr from a point P, where r is the radius of a dimple measured along the surface of the sphere. (This is a tiny fraction more than the r given in the question.)
Let C be the centre of one such dimple, and Q be where the next one around will touch it if they do so. PQC forms a 'spherical' triangle. Angle PQC is a right angle.
With O as the centre of the sphere, angle QOC is r/R, angle POC is 2kr/R.
Let angle QPC be α.
By the spherical sine rule, $\sin(\alpha)=\frac{\sin(r/R)}{\sin(2kr/R)}$.
Each dimple in the band subtends angle 2α at P, so the number that will fit in the band is π/α, rounded down.

I start with one dimple centred at P. then a band centred at 2r, then 4r etc. This gives me band counts 5, 12, 17,..., 31, ..., 3, totalling 306.
It seems likely that there will be places where it would be better to stagger the dimples in the band slightly, making it a little wider, in order to squeeze one more in.

 

[haruspex]

In case anyone is still interested I this, I had a go at an upper bound.
Not exactly rigorous, but...

Consider a dimple and its 5 or 6 nearest neighbours. 5 can touch it, a sixth can't.

I assume without proof that I can construct a network of 'straight' lines joining dimple centres such that it consists almost entirely of hexagonal and pentagonal faces, the vertices being dimple centres, and each face surrounding one dimple.

Taking the sphere to be unit radius and each dimple radius 0.1 (measured along the surface), the minimum area of a pentagon is 0.08754..., and that of a hexagon is 0.1055.
A pentagon contains an area of dimples equivalent to 2.5 whole dimples, while a hexagon equates to 3 whole dimples.

Putting this together, a maximum number of dimples over the whole 4π is 358.

 

[jedishrfu]

This would be an interesting topic for Grant Sanderson's 3blue1brown youtube channel.

 

[epenguin]

In case anyone is still interested I this, I had a go at an upper bound.
Not exactly rigorous, but...

Consider a dimple and its 5 or 6 nearest neighbours. 5 can touch it, a sixth can't.

I assume without proof that I can construct a network of 'straight' lines joining dimple centres such that it consists almost entirely of hexagonal and pentagonal faces, the vertices being dimple centres, and each face surrounding one dimple.

Taking the sphere to be unit radius and each dimple radius 0.1 (measured along the surface), the minimum area of a pentagon is 0.08754..., and that of a hexagon is 0.1055.
A pentagon contains an area of dimples equivalent to 2.5 whole dimples, while a hexagon equates to 3 whole dimples.

Putting this together, a maximum number of dimples over the whole 4π is 358.

Congratulations! Damn you! The thought by which I said in #12 I might be able to do it was similar to yours.

It seems to me I can fit 252 dimples onto this ball, not yet sure if not more. Quite a lot less than your maximum - on the other hand both are much more than the 100 that the question claims is more than the maximum. However others disbelieve this.