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On Jun 24, 7:48 pm, snipped-for-privacy@host122.r-bonomi.com (Robert Bonomi) wrote:

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).

t*> o*

th*> e*

someone*> leans*

My*> solution*

nar*> row*

shaped*> leg*

po*> int of*

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__ _do_ __have a math problem. It's the one I
issued in the math challenge, and I'm still looking for a purely mathematical
solution. This means that solutions involving trailer jacks, anti-gravity
devices, jet-packs, trained seals, skyhooks or even something as far-out as
levelling off my patio are completely off-topic.

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

Nemo

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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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Nemo,

The 'theory'__ _is_ __simple. <grin> 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). <grin>

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.

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

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

basilisk

#### Site Timeline

- posted on June 25, 2010, 12:45 pm

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).

- posted on June 26, 2010, 12:41 am

t

th

someone

My

nar

shaped

po

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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- posted on June 26, 2010, 12:57 am

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. ;-)

- posted on June 26, 2010, 10:25 pm

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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- posted on June 26, 2010, 10:46 pm

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.

- posted on June 27, 2010, 5:13 am

<snip>

far as

the lines

rectangle isn't

don't see

reworded the

the

describe it. Try

possible, you end

is__ _not_ __

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

----------------------------------------------------------- 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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- posted on June 27, 2010, 12:51 pm

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.

- posted on June 27, 2010, 4:54 am

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.

- posted on June 26, 2010, 3:13 am

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

- posted on June 23, 2010, 1:15 pm

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.

- posted on June 23, 2010, 5:36 pm

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.

- posted on June 23, 2010, 5:51 pm

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.

--

- posted on June 23, 2010, 1:38 pm

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

- posted on June 23, 2010, 1:44 pm

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.

- posted on June 23, 2010, 2:05 pm

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

- posted on June 23, 2010, 5:48 pm

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

- posted on June 23, 2010, 6:22 pm

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.

- posted on June 23, 2010, 6:39 pm

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

- posted on June 23, 2010, 8:11 pm

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

Nonny

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On most days,

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- posted on June 23, 2010, 8:16 pm

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

basilisk

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