I was on the bus on the way to uni this morning and it was raining quite heavily. I was sitting right up near the front where I could see the window wipers doing their thing. It made me think "what is the best configuration of window wipers for cleaning the maximum area of window?"

Rather than dealing with rectangles of arbitrary size, it is good to start by looking at the specific case of a square.

To abstract the problem into a mathematical format, I'll restructure it like so:

What is the maximum proportion of a square that can be packed with non-overlapping circle-sectors with origins on the edge of a square?

However, $($while still sitting on the bus$)$ I realised that the answer is always "the whole square". This is because you can recursively define configurations of sectors that get closer to each other with a smaller angle, as shown below:

The way to get around this "solution" is by adding another restriction: The sectors should be uniquely defined by their origin and radius. This is possible if "every sector must have the maximum angle possible where the radius doesn't leave the square". For example, even the first case for the sequence I gave is no longer allowed. The two sectors have angles of $\frac{\pi}{4}$, but with the new rule the arcs must both have angles of $\frac{\pi}{2}$ as their radii stay in the square for this angle. However, in this case the two sectors overlap, showing that it is an illegal configuration. Here are some examples of allowed configurations:

Clearly various "infinite" things happen, as these all resemble apollonian gaskets. It seems like a calculus problem but I've spent too much time dealing with topology that I'm not really sure where to start!

Once the square case is covered, it's natural to generalise to rectangles and parallelograms, and eventually to any polygon. Once all polygons are "solved", any two dimensional shape bounded by a closed curve may be solvable as well! Any thoughts and discussion will be appreciated.

EDIT A number of people (mainly friends making fun of me at uni) have pointed out that real windshield wipers do overlap, and that's fundamental to their effectiveness. I realise this, but for the purpose of mathematical abstraction I felt that "supposing there is no overlap" creates an ideal level of complexity.

Some more precision Let $P$ be a polygon with boundary $\partial P$. (First suppose $P$ is a square, but it will definitely be interesting to generalise.) Let $o_i$ be points in $\partial P$, and $r_i \in \mathbb{R}$. Then each pair $o_i, r_i$ uniquely determines a circle $C_i$ centred at $o_i$. Label each of the points where $C_i$ intersects $\partial P$, and draw straight lines from these points to $o_i$. This splits the circle into sectors. Now remove every sector that encloses a point outside of $P$. The sector(s) you are left with depend only on the starting point, radius, and polygon. Given that you can have as many starting points as needed, what is the greatest proportion of the polygon (in terms of area) that can be packed by these sectors which do not overlap?

Suggested variations

  1. What is the optimal solution where you try to maximise the area which is cleaned but minimise the number of wipers?
  2. It seems like the 4th diagram is probably the solution to the square. But what if only finitely many wipers can be used? What is the solution for each $n$?
  3. What if you do in fact let the sectors overlap? (This will completely change the nature of the problem as it isn't meaningful unless the dynamics of the wipers is considered.)
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    Windshield wipers allow overlap. Do you allow overlap you in your problem? – John Alexiou Apr 13 '17 at 15:02
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    [This](http://www2.stetson.edu/~efriedma/mathmagic/0112.html) could be somehow related... – Watson Apr 13 '17 at 17:59
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    @ja72 Although real windshield wipers allow for overlaps, for simplification my problem has no overlaps. – Harambe Apr 13 '17 at 22:03
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    Your first example of a legal configuration is the the greedy algorithm. Conjecture: it's the best you can do in the limit; the proportion should be computable. Is it the best you can do with a fixed number of wipers? – Ethan Bolker Apr 15 '17 at 23:52
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    This lovely and all but what exactly is the question? Maximize the area by circles one edge of a square, having properties explained above, (provided the constraints admit an optimal solution.) – marshal craft Apr 20 '17 at 01:41
  • I wonder if the 4th drawing is the optimal solution. – marshal craft Apr 20 '17 at 01:47
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    It could also be good if wipers didn't cloud your vision completely. How about a min-max or bilevel problem where you seek to minimize the number of wipers and at the same time maximize the area covered? – Dominique Apr 24 '17 at 14:37
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    The wording of the "extra restrictions" is ambiguous currently. It would be great if you could provide a single, mathematically precise, condition. – pre-kidney Apr 29 '17 at 19:07
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    Some wipers displace parallelly on pantograph mechanisms, – Narasimham Apr 29 '17 at 21:37
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    An attempt at rigorously solving this could be done by solving (bounding?) a recurrence. There are only so many types of regions that can be encountered, if we assume (and I think can prove) that all regions built iteratively will have 3 sides. Then we have one straight side from the square of length we'll call 1, and the two other sides with some curvatures $b$ and $c$ and length $d$ and $e$. This can be slightly complicated if two sides are from the square. Then maximizing over sectors to add reduces it to smaller regions. – Alex Meiburg Jul 26 '17 at 20:59
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    Wow cool problem!! I would love to see a bus with wipers like that. Looks like there's 3 classifications of arc to deal with, based on where the arc's centre is and where its two arc corner points are: all three points on the same line gives us a semi-circle, centre shares side with one arc corner, or all three points are on different sides of the square. Seems like a proof of this would be very difficult, considering the trouble Fan Chung and her husband went to for a square packing problem that seems much simpler: http://www.sciencedirect.com/science/article/pii/S0097316509000272. – Colm Bhandal Aug 09 '17 at 12:58
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    This problem is driving me insane. I'm losing sleep. I wonder if the following idea is of any use whatsoever. Let's say we look for "normal" configurations of wipers- patterns. Possibly for a finite number of wipers. Then prove that these normal configurations are always the maximum- you can't do any better. For fixed $n$ number of wipers, we only have the ability to vary radii and centre points of circles. There definitely exist "local maxima" or maybe "local plateaus" where there is no legal change to any centre point or radius that gives us more area. Is the global max a local max too? – Colm Bhandal Aug 11 '17 at 10:17
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    Some more tiny trivial points I'll make that'll probably be less than useless. It should be easy to prove that there can be no section of the perimeter of the square that doesn't belong to the interior or a wiper arc, for a maximal configuration. Also, another way of classifying arcs is "normal" ones that don't dissect the square and "strange" ones that cut the square into two or more pieces. LHS image all normal. RHS image 3 strange, rest are normal. Finally, I wonder if induction is of any use to us, or at the very least the consideration of the cases for finite #wipers, for ideas. – Colm Bhandal Aug 11 '17 at 10:30
  • take any arrangement of square+arcs, make 4 copies, mirrored and fliipped, and wrap around a torus. now consider each intersection between circle arcs as a point. triangulate the points. an arrangement without 'partial arcs' will be a DeLaunay triangulation, dual of Voronoi diagram. i wonder if there is some theorem there that would help after that. – don bright Jun 14 '18 at 08:42
  • well i botched that comment but the general idea is that an arrangement with 'full circles' when wrapped on a torus will have a special relationship with the delaunay triangulation of the points.. a relationship that an arrangement of 'partial circles' wont have. im stuck there. – don bright Jun 14 '18 at 08:50
  • You are assuming that these windscreen wipers are sectors. Why can't the wiper bar be translated across the windshield? – yolo Sep 29 '18 at 14:31

2 Answers2


You have made the assumption that all windscreen wipers require a pivot therefore over complicating things. An optimal solution would be that of two windscreen wipers with the height of the window translating instead of rotating from the middle to the two sides. But say you still want there to be pivots. A smart idea could be three wind screen wipers like below (you must have alternating wiping for overlaps). Blue for circumference covered and red for the wipers. enter image description here

Your 4th diagram does seem optimal but How would the driver see?

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Very interesting question indeed! So for now the goal is to prove your fourth diagram to be optimal. I am most likely not going to be able to prove this by myself, but I have some observations that I strongly believe to be useful in a proof. I also have some conceptual ideas to continue the proof and maybe even finish it, so my main goal is to inspire other people to continue or even finish the proof.

Proposition 1: Consider the area under a monotone increasing curve $f:[0,1]\to[0,\infty)$, so the set $$S:=\{(x,y):0\leq x\leq 1,0\leq y\leq f(x)\}.$$ Consider placing at most $n$ upper-semicircles with their centers on the $x$-axis. Then there is a configuration of largest area and one can be found by repeatedly placing the largest possible semicircle on the right.

Proof of Proposition 1: First note that the space of possible configurations can be modeled as a compact subset of $\mathbb{R}^{2n}$ by considering the centers and radii as parameters. Then the corresponding area is a continuous function, so it attains a maximum.

Consider a configuration of largest area. Number the semicircles $1$ to $n$ from right to left. Note that we can assume without loss of generality that the semicircles are then ordered from long to short. If not, we can simply keep swapping them until they are, because $f$ is monotone increasing.

Assume for the contrary there exists a semicircle that is not as large as possible. Let $i$ be minimal such that semicircle $i$ is not as large as possible. If $i=n$ we immediately get a contradiction. Let $r$ and $s$ be the radii of semicircles $i$ and $i+1$. We can increase $r$ and decrease $s$ by some $\varepsilon>0$. Because $r\geq s$ this gives a configuration of larger area, so again we have a contradiction. $\square$

Corollary 2: If infinitely many semicircles are allowed, there is still a configuration of largest area and one can be found by the same procedure.

Corollary 3: In the case $f(x)=hx$ the largest possible area with infinitely many semicircles is $$\frac{\pi h}{8\sqrt{h^2+1}}.$$ Actually finding this formula is just a lot of algebra, but feel free to verify.

The next logical step would be to consider the case where we have a right angled triangle and we can only place window wipers with their pivot on one of the legs and radius at most the distance to the hypotenuse. This becomes more difficult, not only because the window wipers on different legs influence each other, but also because not all wiping areas have to be semicircles anymore. From here on, let me just lay out some ideas to continue or even finish the proof.

First, let me note that I suspect this triangle case to be optimally solved by placing a quarter circle in the corner, but only if the width and height are within some margin of each other. For the general case, I suspect we want to split into two cases. With or without non-semicircular wiping areas.

First consider the case with non-semicircular wiping areas. Consider the largest non-semicircular wiping area. The rest of the triangle is then split into three regions. Two of them can be at least approximated by the case of Corollary 3 and the final region is again the case we are now trying to solve. Since we are only trying to show a specific configuration is optimal, my idea was to somehow bound the total area of the arbitrary configuration in question using these ideas.

Then for the case without non-semicircular wiping areas. I highly suspect this to never be optimal. For example, if we can place a quarter circle in the corner that does not cut any of the semicircles, then placing this instead of all the semicircles now inside it is obviously a large improvement. However I am having trouble generalizing this argument.

I want to finish off by noting that the conjectured optimal solution has an area of $$\frac{\pi}{2}\left(2-\sqrt{2}+8\sum_{n=1}^\infty\frac{1}{(\sqrt2+2n)^2(\sqrt2+2+2n)^2}\right)\approx0.9697926466...$$ in a unit square. More generally, the largest possible area in Proposition 1 for the function $f:[0,w]\to[0,\infty)$ given by $f(x)=1-\sqrt{x^2+1}$ is $$\frac{\pi}{8}\sum_{n=1}^\infty\frac{1}{\left(n+\frac1w-1\right)^2\left(n+\frac1w\right)^2}$$ if $0<w\leq1$ defines up to where you can place the semicircles.

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