A math challenge

[Robert Bonomi]

Nemo,

The 'theory' _is_ simple. figure out where the center of mass is for whatever the table shape is, and position the _ground_ points of the legs as far away as practical.

legs that slant outward from the table will provide better stability than pure vertical ones.

The -best- three-legged placement you can do with a rectangular top is two legs at opposite ends of one of the *short* sides, and the third leg in the middle of the other short side. (comment, there is a _reason_ that 'ironing boards" -- for those who remember what _those_ are -- are built that way.)

Here's the way it works: Draw the triangle that represents the leg positions. No possible pressure _inside_ that triangle can cause the table to tip, short of structural failure.

Tipping occurs when sufficient pressure is applied -outside- that triangle. The shorter the moment arm that that pressure has to work on, where the fulcrum is the nearest edge of the triangle, the more force it actually takes to reach the tipping point.

For a rectangular top, there are two possible 'maximum size' triangles. Two legs on a narrow end, and one in the middle of the other narrow end and two legs on a wide side, with the third in the middle of the other side.

Both forms provide _exactly_ the same size 'stable area' (inside the triangle), ie., half the total table surface.

*BUT* the distance from an 'unsupported' corner to the closest point _on_ the leg triangle is smaller when the 3rd leg is in the middle of the short side. Which means that it is -harder- to tip by applying pressure there.

Also, and *very* counter-intuitively, angling the two short-side legs out past the long side of the table does -more- to improve stability than does extending the single leg past the short end. Of course, doing -both- is better than doing just one (either one).

[Robert Bonomi]

FALSE TO FACT.

Regardless of the shape of the top, *or* the number of legs, anything -inside- the polygon described by the footprint of the legs is irrelevant to stability. Any vertical pressure within that polygon cannot cause the table to tip, short of induced structural failure, that is.

What matters is that which is _outside_ said polygon -- pressure 'out there' _will_ provide a 'tipping force', with the fulcrum being the nearest segment of the polygon. Best stability is attained by _minimizing_ the maximum distance from the _edges_ of the top to the polygon of the legs.

[krw]

Wrong.

Right. But it's the CENTER OF GRAVITY that must be above the base, which for a normalish table will be at the center of the table. Not that for a really weird table the CoG doesn't even have to be on the surface. The entire table can be outside the legs (think donut) and it can still be stable. The CoG

*must* be within the base for it to be stable.

Wrong. What matters is the *center* of the table (assuming a "normal" shaped table) must not be outside the base. It doesn't matter how much of the edge it outside, as long as the center of gravity is inside.

The edges don't matter at all. *ALL* of the mass acts as if it is located at the center (of gravity).

[Doug Miller]

Nonsense.

Consider a table having a top six feet square, and four legs which form a one-foot square exactly centered under the top. The table is "stable" under your definition -- that is, it won't fall over in and of itself -- but it's obviously very susceptible to being tipped by even a modest weight. Probably a single coffee cup at the edge would be sufficient.

Now consider a similar table, with the legs moved outward so that they form a

*five*-foot square exactly centered under the top. The center of gravity has not moved, yet this second table is clearly far more resistant to being tipped over. [...]

Sorry, but you're entirely mistaken, because you're considering the wrong problem.

You have completely and entirely correctly described what is necessary to ensure that a table will not fall over _on its own_.

The problem under discussion, however, is how best to make a table that will provide the greatest resistance to tipping over when pressure is applied to the portion of the tabletop that overhangs the base. These are two entirely separate questions.

How much of the edge is outside the base is the *only* thing that matters. You're right that an object can't fall over if its center of gravity is over its base. But you're ignoring the simple and obvious fact that the farther its edges extend beyond the base, the easier it becomes to tip it far enough that the center of gravity is no longer over the base.

[dpb]

[e-mail] wrote: ...

Only if the table surface is unloaded outside the supporting line(s).

It is, as at least two others have already said, also dependent on the lever arm from the edge to the line of the legs--if that is very long, it doesn't take much force irrespective of where the legs are wrt the actual COG...

Simple 1D drawing of table on end or side view (nonproportional font)...

Elbows | | | \ / COG \/ _____A__________B__ | | | | | |

---------------------- Floor/ground...

Clearly the amount of force to tip is much less at the "elbows on table" position owing to the larger distance between that applied force point and the resistance at leg A attach point as opposed to a butt resting on the edge closest to leg B.

As noted upthread, the most tipping stability assuming the legs are far enough apart to avoid the weight of the table alone being a significant factor is to minimize the maximum normal distance along the lines of support. That keeps the largest lever arm at the smallest it can be for any given size of table top.

--

[Nemo]

I think you've given an elegant definition of stability but I don't see how that fixes the positions of the legs under the table. Haven't you just reworded the original problem?

You're right, though - the solution for a rectangle is simply the projected solution for a square. For a random shape 'trickier' doesn't begin to describe it. Try 'nightmarish'!

Incidentally, when you spread the three legs as far as possible, you end up with two in adjacent corners and one half-way along the opposite side. That is _not_ the most stable configuration. Another possibility is with two legs somewhere along adjacent sides and one in the opposite corner. The greatest spread here would be with a leg in each of three corners - not too stable either.

Tain't easy, is it!?

Nemo

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[alexy]

I disagree. A little thought experiment here. A table 10' square. Along one edge are traditional fixed legs, 4' apart. on the opposite edge is one trestle leg, free to pivot where it meets the table, but spreading to feet 20' apart. The table will be VERY easy to tip with pressure 5' away from the trestle leg, as it will just rotate around the attachment point. Attach that leg rigidly, and it will be VERY stable.

[J. Clarke]

each of three

Mathematically you'd want to find the position that minimizes the tipping moment for all locations. Actually doing that would be more work that I want to go into.

[Anonymous]

Let me throw in another factor here. Since this is a real table, the top and legs will have weight. You really have to factor in the mass of the table on either side of the tipping line. This means that just finding the shortest fulcrum doesn't necessarily give the greatest stability. A 1' fulcrum where the tipping line is parallel to an edge will require a lesser tipping force than if the line cut across a diagonal.

To make the answer as general as possible, I've been considering a square slab supported on three points. That way, the weight or length of the legs don't complicate matters. (But I _still_ haven't found an answer!)

Nemo

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[krw]

Nonsense yourself. Physics doesn't lie.

It is stable. However, the measure of stability, which I've stated *many* times here, is the distance from the *CENTER* (of gravity) to the edges of the base (defined as lines between the feet, whether these lines or the center are in fact within the table, or not.

Do learn to read. The edges of the table have nothing to do with stability. The stability is defined only by the center of gravity and the base. Moving your legs changed the base. *Obviously* it's more stable. The *fact* is that the stability is defined by the COG and the base.

I don't know what universe you're in. That's the way physics works in this.

Nonsense.

...and that is measured by the distance between the point on the "floor" under the COG and the base line(s), *ONLY*.

Wrong. What matters is the distance form the COG to the base. Period!

It's not a matter of falling over by itself at all. The force it requires tip the table over is determined by the distance from the COG to the base edge as the moment arm and the mass. The edge of the table top has *nothing* to do with it.

[krw]

Then it becomes "part of" the table top. If you want to get pissy, you don't mention the coefficient of friction between your load and the top (it may fall of before the table tips). Obviously I'm simplifying this a lot, for instance ignoring that the table top is three dimensional and ignoring the mass of the legs. These assumptions were made above.

You don't think a mass at the end of a lever is going to change the COG?

^ ^ A' C'

The distance between my A' and C' defines how much resistance to "Elbows" there is. Yes, A' to E' is a lever arm and that must be countered by the table's mass * A'-C'.

Huh?

[krw]

Calculus.

each of three

Largest distance between the CoG (center of the table) and the lines between the legs.

Yes, in this case it really is.

[Nemo]

rectangle isn't

reworded the

describe it. Try

possible, you end

J. Clarke, I think you're right. Mathematically, we want to find the postions of the three legs that minimize the tipping moment.

So - any suggestions?

Nemo

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[Doug Miller]

Consider these two tables.

Table A is a square, 32 inches on a side, with four legs forming a square base

30 inches on a side. Obviously the tabletop overhangs the base by 1 inch all the way around.

Table B is also a square, 32 *feet* on a side, with four legs forming a square base 30 *inches* on a side, and hence the tabletop overhangs the base on each side by fourteen feet and change.

Same base. COG is in the same place. The distance from the COG to the edge of the base is the same.

According to you, the only thing that matters is the distance from the COG to the base -- which is the same in both cases -- and hence the tables are equally stable. Table B is no easier to tip over than Table A on your planet.

[Doug Miller]

down to the two figures described by equations 18 through 21. The left figure is the one you want. That's demonstrably optimal: any rotation or translation of that triangle within the square necessarily moves the midpoint of at least one side of the triangle farther away from the nearest corner of the square and hence increases the tipping moment about that side.

[Swingman]

Obviously one where practical overrules theoretical ... in theory trolls can't live without bridges.

[krw]

The OP's question was how to optimize the legs for a (given) table.

No, according to me, the stability of a *given table* is quantified by measuring the lateral distance from the COG to edge of the base.

[Josepi]

We gotz some big bridges here...LOL

[Doug Miller]

That's correct. And I'm attempting to help you see why your approach to solving his problem is not correct, because your understanding of the problem is also not correct.

Right, that's what I said -- according to you, my two hypothetical tables A and B must be equally stable, because they each have the same lateral distance from the COG to the edge of the base.

[Doug Miller]

And let's add Table C, just like Table A -- but the legs are forty feet long.

And Table D, just like Table A, but the legs are only an inch long.

All four tables have the same lateral distance from the COG to the edge of the base. That's your phrase, Keith. And according to you, that distance is all that matters. Thus, according to you, all four are equally stable.