Is it always possible to balance a 4-legged table?

Question

A perfectly symmetrical small 4-legged table is standing in a large room with a continuous but uneven floor. Is it always possible to position the table in such a way that it doesn't wobble, i.e. all four legs are touching the floor?

No tricks. No lateral-thinking. Serious question (with real-life applications too!) with a serious answer.

_{This might look like it'd fit better on Lifehacks.SE, but the answer has a nice mathematical proof/formula [depending on whether it's yes/no; I won't give the game away!] which is surprisingly simple and elegant.}

Answer 1

The answer is

yes!

Here's why:

Imagine that the table can pass through the floor. We're going to call one leg the "floating leg" - the other three are going to always be on the floor. Now, after we rotate the table a quarter turn, the floating leg is going to be above the floor if it was originally below, or vice versa. By the intermediate value theorem, it will be exactly on the floor at one point in that rotation.

More detailed proof:

Three legs of the table define a plane. Define the "offset" of a leg to be the distance above the floor if the other three legs are placed on the floor directly - negative if below, positive if above. In any arbitrary placement, the offset of any two adjacent legs will be positive and negative. (WLOG assume the one on the left is positive.) Offset is a continuous function because it's the distance from the floor. If you start with a positive offset on leg A and rotate the table 90° to the right, the offset on leg A (now in the original position of another leg) will be negative. This means that at some point, offset was 0, therefore the table was completely touching the floor.

Answer 2

Yes.

The reason is actually simpler and more intuitive than the other answers:

It's obvious that the table can always stand on three legs, because the ends of two legs define a line, and the third leg can be brought into contact by rotating around that line. Then, rotating the table around a vertical axis, the fourth leg must touch the floor before any of the other legs becomes unable to touch, because the fourth leg touching is the reason one of the other legs wouldn't be able to touch.

Answer 3

The answer is:

No.

Reason by provable test case:

We have a table that is a solid, 3 foot tall, wide, and long, cube of wood, with 4 very short stubby legs on one side (we'll say 2 inches). If standing on a continuous but uneven floor, Say, a simple hexagonal pattern of half-spherical domes (5 inch high domes) (stylized like bubble paper, of course) the domes would lift the table off the floor, preventing it from balancing. At no point could you guarantee stable footing.

...Here's some images to help illustrate the problem.

Here's a grid of bubbles covering the floor. we can surmise that a table could put its legs between these bubbles and touch the floor, but only on the condition that the floor gives space to do so. if not, then you could "balance" the table only in name, on top of the bubbles, but we all know it wouldn't last.

When we switch to a hex-based formation, we see more problems. Firstly, depending on the dimensions of the table, it might not be possible to fit the legs inbetween the bubbles anymore. (In real life, tables don't have massless, frictionful legs because that would be pretty frickin dangerous. This question has what I like to call a spherical cow error [ https://en.wikipedia.org/wiki/Spherical_cow ] - Although it might be useful in real life scenarios, in order to say that it ALWAYS will be helpful in real life scenarios you have to keep adding criteria to the problem to eliminate "edge cases" which are valid concerns for other people, you just don't realize that they count just as much as you do)

From above, the situation is a lot more clear. Square tables are going to have a hell of a time trying to balance on this kind of floor, and most often, they're just going to remain in a "good enough" position where they still slide a bit and you just have to remember not to do anything on that table that requires stability.

Final verdict:

The table can't be guaranteed to balance on the floor in any situation where the table legs have mass (and therefore volume).

Answer 4

The answer is

no.

The answers that say 'yes' seem to stop after showing how you can make all 4 legs reach the floor. However, not only is it sometimes impossible to make all 4 legs touch the floor (think floors with large troughs, mounds, or waves), even when they do, that condition alone does not mean it can be said the table is "balanced", "stable", or "doesn't wobble". Indeed, those answers accept out of hand that 3 legs of the table can touch the floor nicely, but no guarantee for such a condition is afforded anywhere in the problem statement.

Even a modestly turbulent surface that you or I wouldn't have trouble walking on won't hold a table steady unless most of the surface of each foot contacts the floor. This would be impossible if the floor was relentlessly bumpy, like a gravel road or a rocky coastline. If you want a more sterile example, consider something like z = sin x + cos y. There are other, gravel-less ways to get this too. One is where the floor is large tiles that are flat but have a variety of slants. More real-life places come to mind. Speaking of real life, the table will not experience infinite friction with the ground (or with objects placed on it), which means a limit needs to be respected on how far off level the table can be before it simply falls over (or objects slide off).

In conclusion,

there are many factors that determine how and whether or not a table wobbles on a surface. There are probably more than I realize, since I am not a physicist. As for this question, I do not think a large room with a continuous but uneven floor comes close to providing context sufficient to rule out the tendency to wobble.

Answer 5

The answer is "YES"

Place the table on the floor, and label the legs, clockwise from above (A, B, C, D) and their points of contact on the floor (1, 2, 3, and 4) respectively. Note that despite the different labels, the four legs, by symmetry, must be identical in length.

If all four legs are in contact, then the condition is satisfied.

If not, then without loss of generality we can set Leg A as not in contact with Floor Point #1.

Then rotate the table by one-quarter turn clockwise, while allowing it to be supported by the floor throughout the rotation.

Now Leg A is at Point #2. and in contact with Point #2. After all, Leg B was in contact before the rotation, and Leg A is identical to Leg B.

On the other hand, Leg D, formerly in contact with Point 4, has rotated around to Point 1, where it is out of contact, since it is replacing the identical Leg A, which was out of contact.

So Leg A made contact, while Leg D lost contact. If Leg A made contact first, then at that point all four legs are in contact. If Leg D left first, then the table is resting on two legs, and it tips over...

Modesty and honesty require that I credit Martin Gardner http://www.scientificamerican.com/article/strange-but-true-turning/

And yes, I did read and recall the original 1973 article...

Answer 6

I can't tell if anyone has posted this reference, but here is an arXiv link entitled ON THE STABILITY OF FOUR-LEGGED TABLES with the following abstract:

We prove that a perfect square table with four legs, placed on continuous irregular ground with a local slope of at most 14.4 degrees and later 35 degrees, can be put into equilibrium on the ground by a “rotation” of less than 90 degrees. We also discuss the case of non-square tables and make the conjecture that equilibrium can be found if the four feet lie on a circle.

So your answer is it has been proven for certain constraints on how irregular the ground it.

Answer 7

The correct proof

First we assume that the floor has elevation that is a continuous function of the position bounded between zero and the $p$ where $p$ is some parameter that depends on the shape of the table (see below for details), and that the table legs are lines that end in the points of a square $ABCD$. The problem is then assumed to be to position the table so that all four legs touch (and do not go through) the floor.

In the following method note that every step is physically possible in the sense that the table never intersects the floor.

First note that there is a rotational orientation of the table (based on $p$), such that for any point on the floor we can hold the table in that orientation such that one leg touches the floor at that point. This is possible because we can hold the table sufficiently tilted and above the floor, so that the end of the lowest leg is directly above the desired point, and now lowering the table results in that leg touching the floor first by the intermediate value theorem (IVT) and the tilt. We can assume that $A$ is the end of that leg touching the floor at the chosen point. Now tilt the table around the line through $A$ that is perpendicular to $AB$ and horizontal, such that $B$ goes towards the floor, until the leg ending at $B$ touches the floor by IVT. Tilt the table around the line through $AB$ until a third leg touches the floor by IVT. Note that if the third leg is not there we can tilt some more so that the last leg touches the floor by IVT.

Note that in the above subroutine it is easy to choose the initial orientation of the table such that rotating the table by any angle around the vertical will give yet another initial orientation that works. So we get a function with its inputs being the desired point and the initial rotation around the vertical and its output being a table 'position' with $ABC$ on the floor (and $D$ possibly below). It is not hard to prove that this function is continuous, given some weak conditions on the floor (see below). This is the key.

Perform this subroutine once to get $ABC$ on the floor. To be precise we simply get any three legs on the floor, and then rotate the table such that those legs are the ones that actually have endpoints being $ABC$. Then choose any path $P$ from $A$ to $B$ on the floor. Drag the table on the floor such that $A$ follows $P$. We claim that we can do so with $B,C$ remaining on the floor. This follows from the subroutine's continuity since we can keep the rotation around vertical fixed and just use the subroutine on the points along the path. In fact this is what can actually happen physically if you constrain the table's movements. Now drag the table on the floor such that $A$ remains where it is while $B$ goes to where $C$ originally was. Again we claim that we can keep $B,C$ on the floor in the process of dragging. This is because the subroutine applied to all possible rotations around the vertical gives all possible table 'positions' where $A$ is in that place and $B,C$ are on the floor. In particular, a quarter turn around the vertical gives a table 'position' where $A,B$ are at the original locations of $B,C$, and in that case $D$ is below the floor. By IVT, there is a smallest rotation less than that quarter turn such that the subroutine will give a table 'position' with $D$ exactly on the floor. This means that we can drag the table to move $B$ 'around' $A$ keeping $B,C$ on the floor because of the subroutine's continuity, and before a quarter 'turn' we would have found the solution.

Table parameter

Clearly if the table top is much wider than the distance between legs, the method can fail to work simply because the side of the table top hits the floor. In the worst case the legs are so short and the table top has some downward protrusions that reach the plane through the square! In that case certain floors would make the problem impossible. However, as long as there is a plane that completely separates the leg ends from the table top, there will be some suitable parameter $p > 0$ for the above method to work.

Similarly if the legs point outwards, the middle part of the legs may hit the floor if we use the method. Again, for reasonable tables there is some suitable parameter $p$ that works. (No I'm not going to rigorously prove this.)

Floor conditions

Also, to ensure that the subroutine works, any sphere with centre on the floor and radius $AB$ must intersect the floor at points that are at unique angles from the centre. I'm not sure whether this condition can be removed without making the solution non-constructive, in the sense that we can easily prove existence of a solution but there may not be a physical method (like the one above) to obtain it systematically from any random starting position.

In any case, this condition is fulfilled by various simpler conditions such as having Lipschitz constant at most $1$.

Answer 8

Answer:

The answer is no, there exist at least one counter example of floor that prevent the table to be placed, don't get me wrong, it is perfectly possible that the chance to that floor to exists is so low that don't really matters, but I'll show that exists a whole family of floors preventing table placement.

Strategy:

So given I gave a counter-example, no, it is not always possible to place the table

Proof:(false)

Assume the floor seen from aboxe is just a square/rectangle, one of the corners of the square is almost (I give a small delta so I can claim the floor to be continuos) centered on a "hole" and the vertical floor section is 1/x.

More Info:

You can get a table touching on 4 legs if it is a non-squared shape (in example something like a piece of cake), since the table leg's are placed around corners of a rectangle/square, the 4th leg can't however be placed correctly. I builded this answer around Kevin answer, he sais it is possible, however I used his construction to find out that actually is it not possible (you were so near to the solution XD).

The family of floors

All non-linear polar constructed floors, monotonic 2d functions windows that are contained within a quadrant, don't allow the table to be placed. (the family can be actually bigger)

Reason

You can always find 2 points on the surface to get a line, however you place the line, if you turn the table to get the third leg touchin the 4th leg will not be touching.

What's new

Other answers give either a wrong answer (Yes), or mention cases that are either not clear where the solution lies (need constraints, and readers have to figure them out) or have no intuitional proof. I think the one I give here is very easy to reason about.

EDIT:

The construction still holds, but needs some adaptation, as noted in the comments, I can still place legs of the table on 2 elevation rings, however If I skew the floor along 1 axis, and I cut out simmetric planes, I'm no longer able to place the table:

Conjecture (true, but need proof)

Given 2 ellypses of different diameter (same center and not circles), except for pairs centered on a simmetry axis, there are no pair of parallel lines whose intersections with the 2 ellypses are the corners of a rectangle.

In the end:

We need also to "stretch" the floor in one direction and cut out simmetry axis in additiction to the family of floors I constructed above.

Answer 9

I'm not entirely sure if this counts as lateral thinking, but couldn't you just chop off a leg? Looking at the answers the problem seems to be one leg connecting with the floor preventing the opposite leg doing so.

Answer 10

You may be asking the wrong question. As stated, no it isn't always possible to position an 'ideal table' on an uneven floor and guarantee that all four legs have firm contact with the ground. The real question is it is possible to always distort a floor such that an ideal table will wobble.

Start with an 'ideal floor' (perfectly smooth and level) made from soft wood, place your ideal table anywhere. No wobble. Now take a hammer and knock out a small uneven plug from the floor under one of the legs. Now the table wobbles. Move table and if doesn't wobble, repeat hammer knock out. Repeat until you have a floor upon which there is no position where the table doesn't wobble.

Given a randomly sized (table width, depth) and a randomly uneven floor, you might be able to place the table somewhere it doesn't wobble but you might not. Therefore it isn't always possible.

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Update:

I decided to update my answer since people seem to have misunderstood.

I'm not suggesting that the question ask was wrong. I simple thought that reframing the question might help people see the answer.

I'm not suggest that the floor is continuously moving. The floor is "uneven" there are lots and lots and lots of ways that a floor can be uneven. I simply started with an even floor and made it uneven to make a point.

How do I know that my "repeat until you have ..." loop is guaranteed to terminate? Given the problem statement we know there are a finite number of positions available for the table on the floor (there must be some distance, no matter how small that defines a "new" position for the table). Lets call the number of possible configurations n. Take the set of all uneven floors, in that there has to be at least one floor where there is only one position where the table wobbles (if there were no positions then the floor would be even). There must also be a floor where there are only two positions where the table wobbles. One where there are only three positions, etc. Therefore, there must be at least one floor that has only one position where the table table wobbles. Start with that floor. Take hammer and 'unlevel' the floor under one of the legs, call that point p. In order for the loop to never end the change of p must lead to a new position which is a rotation around point p where the table does not wobble, but there is still only one position on the floor where the table doesn't wobble. Hit p again, repeat. Since there are only n possible positions of the table in the room you cannot repeat the procedure more than n times. Since you are not changing any other points you will by eventually have a floor where there is no position where the table doesn't wobble. Therefore, there is at least one floor where you can not position the table in a wobble free position. Thus it is not ALWAY possible to position the table so that it doesn't wobble.

Answer 11

I don't think you can solve this question infallibly unless you approach it using a theoretical approach specifically in multivariable calculus. If you reword your question into a mathematically approachable model, then the question goes:

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Let $m > 0$ be the greatest slope that is acceptable for a "realistic floor." Let $l$ and $w$ be the length and width of the chair.

Define $S$ to be the set of all continuous surfaces such that for each surface $S_i$, $|{dz \over {dx}}|,|{dz \over {dy}}| < m$ at every point in $S_i$.

Define the set $P_{A,B}$ be all the possible contact points (x,y,z) between a continuous surface $A$ and a plane $B$, both of which are in $\mathbb R^3$ (the 3rd dimension).

Does $S$ have the property that there always exists a plane $B_i$ such that $P_{S_i, B_i}$ contains 4 contact points that form a rectangle of dimensions $l, w$ when connected?

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I simply don't have the knowledge to derive and justify the result, and I don't think that kind of infallible solution would be easy to get here; math.stackexchange would be more appropriate.

I have only posed more questions so far, so to make up for this (kind of), I will hypothesize that the answer to the question is no; if I assume that $S$ does have said property for dimensions $l,w$, then the same property must be applicable for really large dimensions and really small dimensions (e.g. where $l,w \in (0, \infty)$ ), and I simply find that unfeasible.

Answer 12

The answer is certainly no, the best way to think about this is in terms of basic linear algebra, or if math is not your strong point just in terms of a 3d graph.

Your questions translates to "for any four points in 3d space, does there always exist a single 2d plane containing them all". For our problem: the four points are the bottoms of the four legs and the 2d plane is the floor.

We can certainly put any three points in the same plane by just drawing lines between them and defining that as the plane, but we will not always be able to add the fourth (in fact almost every other point will not be in this plane).

To visualise this consider the four points of the form (x,y,z) in Cartesian (normal) coordinates:

• a = (0,0,0) i.e. this signifies x=0, y=0, z=0
• b = (1,0,0)
• c = (0,1,0)
• d = (1,1,1)

Clearly, a,b and c all lie in the (x,y)-plane, we will call this the floor. However, d is by definition NOT in the (x,y)-plane as the x value is set to zero. It lies exclusively in the (y,z)-plane and will always be shifted away by one unit no matter how much you rotate the points.

Try this out by plotting each of these points in the following link:

point/graph plotter

Rotate the graph around and line 3 points up into the same plane, the third will always appear a unit of 1 to the side.

I hope this helps.

Edit:

I changed the points to the "table-like" layout