Yes, it is true. A trestle table has four "legs". The base is defined by the points touching the floor. How they get there is immaterial, at least until the table moves (the base may change as it's tipped).

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Right. A rigid structure will transmit force in a straight line from the load point to the support point, assuming no other constraints. This would be true even if the line goes through empty space.

Bea, you're wrong, I'm afraid. I

But I'm glad I'm not the only one who's having a problem with the answer!

Nemo

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On 26 Jun 2010 00:41:06 GMT, snipped-for-privacy@nusquam.rete wrote:

I guess I'm not seeing your problem. Spread the three legs as far as possible such that the COG, projected onto the floor, is furthest from the lines drawn between the feet. For a square the answer is obvious. A rectangle isn't much harder to see. For random shapes it gets trickier. ;-)

I guess I'm not seeing your problem. Spread the three legs as far as possible such that the COG, projected onto the floor, is furthest from the lines drawn between the feet. For a square the answer is obvious. A rectangle isn't much harder to see. For random shapes it gets trickier. ;-)

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

Tain't easy, is it!?

Nemo

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On 6/26/2010 6:25 PM, snipped-for-privacy@nusquam.rete wrote:

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.

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.

<snip>

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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

----------------------------------------------------------- Posted using Android Newsgroup Downloader: .... http://www.sb-software.com/android -----------------------------------------------------------

far as

the lines

rectangle isn't

don't see

reworded the

the

describe it. Try

possible, you end

is

along

with a leg

more

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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snipped-for-privacy@nusquam.rete wrote:

http://mathworld.wolfram.com/EquilateralTriangle.html

Scroll 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.

http://mathworld.wolfram.com/EquilateralTriangle.html

Scroll 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.

On 26 Jun 2010 22:25:55 GMT, snipped-for-privacy@nusquam.rete wrote:

Calculus.

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

Yes, in this case it really is.

Calculus.

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

Yes, in this case it really is.

Nemo,

The 'theory'

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

Here's the way it works: Draw the triangle that represents the leg positions. No possible pressure

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

Also, and

On Jun 23, 1:40 am, snipped-for-privacy@nusquam.rete wrote:

Not so much math as physics. For an object to stand without falling over, its center of gravity must be above its base. The further inside the base the COG is the more stable it will be (you have to tip it far enough to get the COG outside the base).

For your table, assume the mass is in the top (ignore legs). If it's large compared to its thickness, ignore the thickness. The COG is then pretty much the center of the top. The problem then is to place the legs so this point is the furthest from the lines connecting the legs (the "base"). In the case of a square, this is easy (pick two corners and the middle of the opposing side). For a rectangle, I think it becomes clear if the rectangle is exaggerated; two adjacent corners on a long side and the opposite center.

Not so much math as physics. For an object to stand without falling over, its center of gravity must be above its base. The further inside the base the COG is the more stable it will be (you have to tip it far enough to get the COG outside the base).

For your table, assume the mass is in the top (ignore legs). If it's large compared to its thickness, ignore the thickness. The COG is then pretty much the center of the top. The problem then is to place the legs so this point is the furthest from the lines connecting the legs (the "base"). In the case of a square, this is easy (pick two corners and the middle of the opposing side). For a rectangle, I think it becomes clear if the rectangle is exaggerated; two adjacent corners on a long side and the opposite center.

Good explanation. Thanks. I particularly like the lines between support points, which are obviously (well, obvious once you suggested them!) directly above the axes of rotation if the table were to tip.

--

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

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

alexy wrote:
...

The problem w/ square/rectangular top on three lets still is, of course, that the lever arm for applying a tipping force is longer from the corners at the one-legged end. Minimize the normal (perpendicular) distance form the corners to the lines between the support points; the greatest of those is the highest moment arm and the point most prone to tip. The circular top has no "longest" length in any preferred direction; the maximum is the same in each of symmetric views.

Just ottomh w/o actually doing the geometry, seems to me the disparity is greatest for the square and gradually decreases as the L/W ratio increases for the rectangle. I'd look at the size required/wanted for the purpose and even if don't want round, potentially look at cutting a 45 or similar off a long corner if it were overall, square. Of course, a mockup from just ply and simple legs would make it much simpler to get an actual feel for just how unsteady any given size would feel.

--

The problem w/ square/rectangular top on three lets still is, of course, that the lever arm for applying a tipping force is longer from the corners at the one-legged end. Minimize the normal (perpendicular) distance form the corners to the lines between the support points; the greatest of those is the highest moment arm and the point most prone to tip. The circular top has no "longest" length in any preferred direction; the maximum is the same in each of symmetric views.

Just ottomh w/o actually doing the geometry, seems to me the disparity is greatest for the square and gradually decreases as the L/W ratio increases for the rectangle. I'd look at the size required/wanted for the purpose and even if don't want round, potentially look at cutting a 45 or similar off a long corner if it were overall, square. Of course, a mockup from just ply and simple legs would make it much simpler to get an actual feel for just how unsteady any given size would feel.

--

On Jun 23, 2:40 am, snipped-for-privacy@nusquam.rete wrote:

Substituting an inherently less stable form for a slightly wobbly table is a poor trade-off. All you need is to have one leg of the four adjustable.

R

Substituting an inherently less stable form for a slightly wobbly table is a poor trade-off. All you need is to have one leg of the four adjustable.

R

On Wed, 23 Jun 2010 06:38:26 -0700 (PDT), RicodJour

An easier solution should the three leg choice be preferable is to forgo the square/rectanble table and make it round.

An easier solution should the three leg choice be preferable is to forgo the square/rectanble table and make it round.

wrote:

Maybe a matchbook or a rolled up napkin under one leg.

It works at some of the better restaurants I have been in.

Seriously, this group is great, there is no question that can't be asked and get realistic and practical answers.

Larry C

Maybe a matchbook or a rolled up napkin under one leg.

It works at some of the better restaurants I have been in.

Seriously, this group is great, there is no question that can't be asked and get realistic and practical answers.

Larry C

On Jun 22, 11:40 pm, snipped-for-privacy@nusquam.rete wrote:

The simpliest solution is to make a small wedge of the wood of your choice and slip it under the leg that appears to be the shortest when the table is standing on the flagstore. It will make it solid. If the table is moved a different leg will appear to be the shortest, so use the wedge on that leg.

Al

The simpliest solution is to make a small wedge of the wood of your choice and slip it under the leg that appears to be the shortest when the table is standing on the flagstore. It will make it solid. If the table is moved a different leg will appear to be the shortest, so use the wedge on that leg.

Al

On Jun 22, 11:40 pm, snipped-for-privacy@nusquam.rete wrote:

If the flags don't have sharp discontinuities where they meet (i.e. if the surface is uneven but not abrupt) then you can rotate the four-leg table into an orientation where all four legs are on the ground.

Alternately, the table can be made to flex slightly to take up the irregularity; this means your table top cannot be glass or stone... The 'adjustable leg' solution might be simplified with a good McPherson strut.

If the flags don't have sharp discontinuities where they meet (i.e. if the surface is uneven but not abrupt) then you can rotate the four-leg table into an orientation where all four legs are on the ground.

Alternately, the table can be made to flex slightly to take up the irregularity; this means your table top cannot be glass or stone... The 'adjustable leg' solution might be simplified with a good McPherson strut.

Make two triangular tables that can be arranged in the size rectangle you want. preserves balance and stability and gives you the shape you want.

basilisk

The idea is good, but I suspect that the 6 resulting legs would hinder the use of the table with chairs.

Nonny

--

On most days,

it’s just not worth

On most days,

it’s just not worth

Click to see the full signature.

Why? The legs will all still be at the corners.

basilisk

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