# A math challenge

On Sun, 27 Jun 2010 12:41:23 GMT, snipped-for-privacy@milmac.com (Doug Miller) wrote:

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.
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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.
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Miller) wrote:

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.
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On Sun, 27 Jun 2010 19:05:37 GMT, snipped-for-privacy@milmac.com (Doug Miller) wrote:

Wrong. I never said any such thing. My point was only that the stability of a table is relative only to its base and COG.
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Of course you did. You said it before, and you said it again just now: "the stability of a table is relative only to its base and COG". Hence two tables having identical bases, with their centers of gravity in the same place, must be (according to you) equally stable -- regardless of any other factors such as the size of the tabletop.
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snipped-for-privacy@att.bizzzzzzzzzzzz wrote:

You're correct as far as your definition of the problem. But there's the ancillary problem of a single fly alighting on the nether edge of a table whose legs are WAY inside the perimeter.
Come to think on it, I've never seen a dining room table held up by a single 1" pole attached at its center of gravity.
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"HeyBub" <> wrote in message

And the COG is changing as items are placed on or removed from the table in normal service. That's the concern, eih? phil
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Bingo.
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On 6/26/2010 11:42 PM, snipped-for-privacy@att.bizzzzzzzzzzzz wrote:

Obviously one where practical overrules theoretical ... in theory trolls can't live without bridges.
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Last update: 4/15/2010
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We gotz some big bridges here...LOL
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We gotz some sucker bait floating here...LOL
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Faked headers from the cyber bully gang.
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Could I explain the post I would gladly do so. I am a moron just as are you all so I am lost.
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snipped-for-privacy@att.bizzzzzzzzzzzz 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.
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located

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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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?
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Setting an object on the edge of a tabletop doesn't change the center of gravity of the *table*, no.
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dpb wrote:

Elbows on the table = soup in the lap.
At least that's what my granny taught me.
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snipped-for-privacy@host122.r-bonomi.com (Robert Bonomi) wrote:

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.

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Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

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On Thu, 24 Jun 2010 19:48:24 -0500, snipped-for-privacy@host122.r-bonomi.com

And the trestle table like the one I built, often has some type of cross member attachment joining those two legs into a solid structure.
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