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  #1781  
Old 06.04.2018, 17:12
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Re: Ask a Scientist

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What could influence the weight-bearing strength of these clever, flat-pack stools (and table)?
http://www.vij5.nl/portfolio/strap-stool/
http://www.vij5.nl/portfolio/strap-table/
For parts and assembly, please scroll down.

Since I'm specifically interested in remaining with wood and not adding metal, and in retaining the quick to assemble feature, and flat storage, I suppose I need answers about the ways the force is distributed, to do with the angle of the legs, for example, or adding triangles or cross-beams, etc. Thanks.
I presume you want to know how to make them stronger, rather than an analysis of their failure mode.

Not much. They're a "house of cards" design who's integrity depends on them not wobbling. The straps are the only restraining element which holds them together (Neat idea).

They appear to be well made and robust. It's odd that they don't mention load bearing capacity. Judging by the pics I would say somewhere north of 200kg for the chairs, about 150kg (tops) for the table.

How long they last will depend on how close the tolerances are on the nubs and holes. If they are loose, or they get damaged/worn the F(up) will rapidly give way to mg.
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  #1782  
Old 06.04.2018, 18:42
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Re: Ask a Scientist

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How long they last will depend on how close the tolerances are on the nubs and holes. If they are loose, or they get damaged/worn the F(up) will rapidly give way to mg.
I think it's also a question of how deep the holes (how long the nubs) are. The deeper the hole, the less of a lever sideway forces have on the nubs/hole and the surrounding wood. IMHO the holes' shallowness makes them overly sensitive to sideways forces.

Additionally, the thin wood "bridge" on the bottom and top rim of the holes provides poor guidance for the nub, and less resistance than one might be used to.

I think they're worn(?) loose easily and quickly, especially by children.
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  #1783  
Old 07.04.2018, 19:12
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Re: Ask a Scientist

Thanks very much, JagWaugh and Urs Max. Very helpful!
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  #1784  
Old 08.04.2018, 20:18
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Re: Ask a Scientist

http://pd.chem.ucl.ac.uk/pdnn/symm1/trans1.htm

"....The question arises as to the choice of parallelogram for our two-dimensional unit cell. The black lattice lines are a result of choosing the unit cell shown in green. However, one could equally choose the unit cells shown in red, cyan or blue.

The unit cell shown in cyan is clearly identical to the green one, but has a different choice of origin.

Crystallographers usually choose an origin that is convenient to describe the contents of the unit cell. Frequently the origin corresponds to a point of high symmetry within the unit cell: for example, it is often sited on an axis or where two or more axes intersect.

Crystallographers also choose unit cells which contain the smallest number of lattice points. The unit cell shown in blue contains two lattice points (one at the origin corner and one in the middle) whereas the others all contain one. ..."

Can someone please explain the use of terms, here?
Why ist the unit cell in blue said to contain TWO lattice points, not FIVE?
When does the term "origin corner" apply?
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  #1785  
Old 08.04.2018, 20:45
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Re: Ask a Scientist

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http://pd.chem.ucl.ac.uk/pdnn/symm1/trans1.htm

"....The question arises as to the choice of parallelogram for our two-dimensional unit cell. The black lattice lines are a result of choosing the unit cell shown in green. However, one could equally choose the unit cells shown in red, cyan or blue.

The unit cell shown in cyan is clearly identical to the green one, but has a different choice of origin.

Crystallographers usually choose an origin that is convenient to describe the contents of the unit cell. Frequently the origin corresponds to a point of high symmetry within the unit cell: for example, it is often sited on an axis or where two or more axes intersect.

Crystallographers also choose unit cells which contain the smallest number of lattice points. The unit cell shown in blue contains two lattice points (one at the origin corner and one in the middle) whereas the others all contain one. ..."

Can someone please explain the use of terms, here?
Why ist the unit cell in blue said to contain TWO lattice points, not FIVE?
When does the term "origin corner" apply?
I'm no scientist, but allow me to look into my crystal ball...

Looking at the illustration, I think that the 'lattice points' (atoms, molecules, whatever they are) in a 2d representation are not given full value - hmm, in other words, the black lattice lines bisect the lattice points (in 2D, they would divide the lattice point into 4.) Each corner of the unit cell contains 1/4 of a lattice point. This is why a primitive cell contains (a total of ) 1 lattice point, while the blue cell contains 2 ( 4x 1/4 lattice points from the corners, plus 1 full lattice point in the middle, fully contained in the cell)

In a 3D lattice, each lattice point in a corner would only have 1/8 of the lattice point contained in any particular cell.

the 'origin corner' would just be an arbitrary starting point from which to measure, or maybe describe, a unit cell.??

Take a look at the blue cell again. If you could pick it up and move it to any other spot, it would always contain a total of 2 lattice points, no matter where the origin corner may be.

does that make sense?
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Last edited by pilatus1; 08.04.2018 at 21:07.
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  #1786  
Old 08.04.2018, 21:45
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Re: Ask a Scientist

Thank you, Pilatus1! Yes, as a non-scientist I can follow your reasoning which seems to make sense. Thanks very much.
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  #1787  
Old 16.04.2018, 14:35
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Re: Ask a Scientist

This question is about measuring weight. More strictly correct: mass, but please just let me use the colloquial "weight".

Person A stands on the scale and weighs 60 kg.
Person B stands on the scale and weighs 80 kg.

They put a plank P across the scale, to increase the surface area.
The plank P is 60cm long and weighs 4 kg.
A and B stand really close to each other, on the plank, and weigh themselves together.
Total weight = (A + B + P) = 60 + 80 + 4 kg.

Okay till there.

Next they used a plank Q, 1m long, and rest it across not just one scale in the middle, but two scales, one at either end of Q.
A stands on one end of Q and B stands on the other end of Q.

What – if anything at all – will be the meaning of whatever the two scales measure?
Will the plank distribute the weight evenly across the two scales?
If so, will the total weight shown on the scales still add up to the total of (A + B + Q)?

What would be the difference in the significance, (if indeed there is any any) of the measurements shown on the two scales, if a plank R were used, which is 2 metres or plank S which is 4 metres long?

From see-saws we know that moving towards a fulcrum can change the leverage and the relative apparent weight on either side. That’s not what I’m asking here. I’m assuming, at all times, that the two scales are positioned evently at either end of the plank.

Thanks for any explanations.
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  #1788  
Old 16.04.2018, 14:44
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Re: Ask a Scientist

The sum of the two displayed values will always equal the sum of the 3 weights. With the two scales equidistant from the end/center of the plank each scale will display the weight of the person+ 1/2 the weight of the plank (presuming constant density of the plank).
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  #1789  
Old 16.04.2018, 14:45
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Re: Ask a Scientist

Okay, admit it, you got this neighbour's kid who wants you to do it's homework
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  #1790  
Old 16.04.2018, 14:56
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Re: Ask a Scientist

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The sum of the two displayed values will always equal the sum of the 3 weights. With the two scales equidistant from the end/center of the plank each scale will display the weight of the person+ 1/2 the weight of the plank (presuming constant density of the plank).
Thank you.

What happens if five people stand along the 4 metre plank?
If they have different weights, presumably that would go against your assumption of constant density.

Presumably it would also so do, to varying degrees, if
  • A, B, C, D and E stood evenly spaced along the plank, and
  • A, B and C stood bunched up near one end and D and E bunched up near the other?

It is still true that the sum of the two values (one shown on each scale), will equal A + B + C + D + E + S?
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  #1791  
Old 16.04.2018, 14:58
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Re: Ask a Scientist

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Okay, admit it, you got this neighbour's kid who wants you to do it's homework
Nope. On the contrary! The question arose from a discussion amongst a few intelligent, educated adults who couldn't remember which part of their school physics they were supposed to be trying to access.
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  #1792  
Old 16.04.2018, 15:05
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Re: Ask a Scientist

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It is still true that the sum of the two values (one shown on each scale), will equal A + B + C + D + E + S?
This is always true. ("S" is the weight of the plank, I presume.)

Think of it this way: The weight of the system is only bearing on two points: Scale 1 and Scale 2.

One measures the corner weights of a racecar by putting a scale under each wheel, one after the other. The 4 values indicate the total weight, the deltas of those 4 values indicate how well balanced the weight distribution is left to right. On a few cars I work on we've moved stuff (like the battery) around to get better balance.
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  #1793  
Old 16.04.2018, 15:14
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Re: Ask a Scientist

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This is always true.
Great. Then what, please, does the density have to do with it, as you mentioned in your assumption?

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One measures the corner weights of a racecar by putting a scale under each wheel, one after the other. The 4 values indicate the total weight....
Do you mean wheel1 + wheel2 + wheel3 + wheel4 = total weight... even when, while measuring wheel1, the car is resting on the ground on the other three wheels at the same time?

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, the deltas of those 4 values indicate how well balanced the weight distribution is left to right. On a few cars I work on we've moved stuff (like the battery) around to get better balance.
The "deltas of those 4 values"... mmmh, oh, um....
Please, Jag, have mercy on those lesser mortals whose expertise lies in languages, social work, psychology, farming and medicine (the aforementioned group recently discussing this).
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  #1794  
Old 16.04.2018, 15:18
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Re: Ask a Scientist

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Great. Then what, please, does the density have to do with it, as you mentioned in your assumption?


If the plank of wood (the beam) is of uniform density , then the load that it bears is a uniform distributed load - of 4 kg. 2kg to each scale. If one end were made of ironwood (dense) and the other of balsa (not very dense), the load would not be distributed evenly across the scales. But it would still weigh 4kg and the total on the scales would still read 4kg (with just the beam - not the people standing on it).

I'm trying to approach your question from a construction/engineering perspective but the idea of using scales at each end as supports makes it confusing because a scale is a spring loaded device, which enters other factors into the equation. It would be easier to imagine if they mentioned the supports as fixed platforms..
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Last edited by pilatus1; 16.04.2018 at 17:00. Reason: stupid autocorrect!
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  #1795  
Old 16.04.2018, 15:27
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Re: Ask a Scientist

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Great. Then what, please, does the density have to do with it, as you mentioned in your assumption?

If the plank has a total weight of 10, but it is really thin on one end, and thick on the other, then it's own center of gravity will not be at it's midpoint. The 10 will be distributed between the scales as the COG (center of gravity) is shifted.

Do you mean wheel1 + wheel2 + wheel3 + wheel4 = total weight... even when, while measuring wheel1, the car is resting on the ground on the other three wheels at the same time?

Indeed I do.


The "deltas of those 4 values"... mmmh, oh, um....
Please, Jag, have mercy on those lesser mortals whose expertise lies in languages, social work, psychology, farming and medicine (the aforementioned group recently discussing this).

"Deltas"=Difference. If you weigh the front axle and read 750 on the left, and 752 on the right you have a delta of 2.
Answers in red.

Here is a brain teaser for your group:

The task is to measure the density of a lump of metal, say the size of a golf ball. All you have to work with is:

A glass or cup.
An accurate scale.
Some water (As much as you want).
Some string (As much as you want).

You have nothing to directly measure the volume of the lump of metal, nor it's dimensions.

Last edited by JagWaugh; 16.04.2018 at 15:40. Reason: forgot some red
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  #1796  
Old 16.04.2018, 15:43
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Re: Ask a Scientist

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Answers in red.

Here is a brain teaser for your group:

The task is to measure the density of a lump of metal, say the size of a golf ball. All you have to work with is:

A glass or cup.
An accurate scale.
Some water (As much as you want).
Some string (As much as you want).

You have nothing to directly measure the volume of the lump of metal, nor it's dimensions.
distilled water?
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  #1797  
Old 16.04.2018, 15:46
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Re: Ask a Scientist

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distilled water?
Let's say yes. Or at least water of a known density.
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  #1798  
Old 16.04.2018, 15:48
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Re: Ask a Scientist

I wonder if Pythagoras had a scale next to his bathtub..
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  #1799  
Old 16.04.2018, 15:52
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Re: Ask a Scientist

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Answers in red.

Here is a brain teaser for your group:

The task is to measure the density of a lump of metal, say the size of a golf ball. All you have to work with is:

A glass or cup.
An accurate scale.
Some water (As much as you want).
Some string (As much as you want).

You have nothing to directly measure the volume of the lump of metal, nor it's dimensions.
fill the cup to the brim and weigh it =m1
put the ball in the cup and so let some water out, ensure the ball is completely under water.
get the ball out without losing any more water
weigh the cup again = m2
and calculate the weight of water lost m3 =m1 - m2
weigh the ball alone = m4

Density = m4/m3

now weigh the cup again and calcuate the weight of water lost

Tie the string around your finger to remind you not to answer such questions in the future.
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Old 16.04.2018, 15:52
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Re: Ask a Scientist

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I wonder if Pythagoras had a scale next to his bathtub..
No but Archimedes did
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