Fix a Wobbly Table (with Math)

Fix your wobbly table with just a small tweak - but why does this work?
More links & stuff in full description below ↓↓↓

Featuring Matthias Kreck from the University of Bonn.

Professor Kreck points out that he was describing the intermediate value theorem, not mean value theorem.


NUMBERPHILE

Videos by Brady Haran

[numberphile]

Professor Kreck says commenters are quite right - while rightly describing the "intermediate function theorem", he mistakenly said "mean value theorem".

[numberphile]

(Hello Internet followers may recognise Knut's work)

[osmium]

This method can ensure that all 4 legs are touching the ground, but it won't ensure that the top of the table is level. With very uneven ground, you could end up with the table top so tilted that your beer slides right off! In that situation, you'd need to resort to the old "paper under the leg" technique.

[yanava]

That's a real mathematician, solving beer problems, instead of meaningless universal questions. Well done!

[Frosty-ojhw]

There's another solution as well which is to only use the minimum  number of legs required for stability in the dimension you're working on, which is (N+1) where N is the number of dimensions of your surface.

If you want a table to be stable on a 2d plane (the ground) then you need (2+1) = 3 legs, a 3 legged table can never wobble on a 2d plane, even if that 2d plane is uneven.

[NikolajLepka]

or... you know.. use a 3-legged table and never get that problem in the first place

[ziemowitowczarek]

Probably one of my favorite numberphile vids ever.

[Sniper]

I think that the mathematician in this vid is my favorite, such a nice voice to listen to. The animation was also amazing to add to it.

[Doc_Fartens]

Now my only problem is remembering this the next time I have a wobbly table.

[JohannesEckhoff]

This video is both adorable and educational, I love it.

[HolmgrenJensen]

I may be wrong, but isn't the theorem being used to show that a continuous function with a positive and a negative point must traverse zero at least once the Intermediate Value Theorem, and not the Mean Value Theorem?
The Mean Value Theorem on the other hand talks to the existence of one or more points along the curve where the differential equals the slope of a straight line between the two extreme points.

[Weisz]

Currently sitting in a coffee shop, just wanna let you know that in the time since I saw this video five years ago it has CHANGED MY LIFE FOR THE BETTER -- everyone else out here uses coasters and stuff meanwhile I ALONE know the secret.


Am I the only one? I'm shocked this video only has 1million views!

[numberphile]

If you really want to get your teeth into this!

[scottmuck]

No other Numberphile video has so directly impacted my life! Been fixing wobbly tables with this method for years!

[Jergy]

Man, what I wouldn't give to have this guy as a professor.  His accent is amazing.

[poodleofdeath]

This Theorem only works, because its assumed that the height-time funktion has to be a continuous function. This does not apply to reality, where jump discontinuities can occure

[JoshBrown]

There are a few assumptions that are necessary for this to work:
The floor is continuous (which is mentioned)
The table is square (although it seems like a half turn would be sufficient for rectangular tables)
The table leg bottom is a geometric point
The points at which the leg height is 0 are flat enough for the leg to rest on without sliding

That last assumption is the kicker. One could imagine a trench dug between legs 1 and 2 such that the points where the leg height would be 0  are nearly vertical. For similar reasons, the first assumption may very well be false. If the floor has vertical concavities (e.g. overhangs/cliffs) the leg-height would instantaneously change from one point to the next as it passes over the cliff edge.

[pacogoatboy]

I enjoyed this video with Prof. Kreck.  I hope you have the opportunity to make more videos with him in the future.  He's a wonderful storyteller/math-explainer.

[fergusmgraham]

I don't get it. I understand that in theory (assuming that you can keep all of the other three legs co-planar) the height of leg 1 will go from positive to negative. But the issue was that the ground is uneven. So in reality as you rotate the table the other legs will change elevation. This means (in my humble estimation) there does not HAVE to be a point in the rotation where the height of leg 1 goes to zero. And even if it does, one of the other legs can end up with a positive elevation. I would stick to using a coaster!

[rdoetjes]

There's only one slight problem, a bier garten is in Germany and everything stands (and is supossed to) completely straight and then your table is of centre and everybody is annoyed by this unsightly demonstration of defiance that they will throw their bratwurst and beers at you.