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"Trevor G Quayle" <Tin### [at] hotmail com> wrote:
> Would a shape with a square top and circular bottom work (similar to 3 which is
> a line top and round bottom).
>
> I don't know what the exact volume is but it would seem that it would be the
> average of a cylinder and a cube: (6+8)/2=7
The problem here is that the targeted volume is (pi/3)*7, not 1*7, and the
volume of the cylinder is (pi/3)*6, not 1*6, while the cube's volume is plain
boring 1*8.
So a gradual cross-section morph from square to circle in a way that the
cross-section area increases linearly (similar to #3) would indeed give a
volume equal to the average of a cylinder's and a cube's volume, but that
wouldn't fit the bill.
Of course, a rectangular box of size 2*2*(pi*7/12), skewed by 2-(pi*7/12) to
"fill" a 2*2*2 cube, would do the job, but something tells me that it would not
be quite in line with the spirit of the puzzle ;)
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Le 15.01.2009 22:48, milco2006 nous fit lire :
>
> Here is a crazy superellipsoid shape which fits the bill of 7*pi/3.
Can you explain a bit how, I fail to understand it's construction.
Is there some ratio from green to blue/red ?
> Although I
> do have a mathematical formula which gives an exact multiple it is very nasty
> so I am not sure if this counts under your brief for the shape.
I have nothing against a superellipsoid, as long as you provide its
parameters (and hopefully they are some kind of 'nice', even a cubic
root of a fraction might match), and by the same way demonstrate that
the volume is exactly and mathematicaly 7pi/3.
(and not an approximate value close enough)
Which seems a bit complex to perform, given that the sole formula of
volume that I found on mathworld for the superellipsoid is based on the
Gamma function, which is far from simplying to 7pi/3 at some stage.
(Just prove me wrong... I have no doubt that one superellipsoid, or even
a whole family of them, can have the right volume... but the real answer
need to identify which ones!)
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Le 16.01.2009 01:21, clipka nous fit lire :
>
> Of course, a rectangular box of size 2*2*(pi*7/12), skewed by 2-(pi*7/12) to
> "fill" a 2*2*2 cube, would do the job, but something tells me that it would not
> be quite in line with the spirit of the puzzle ;)
>
>
Yes, correct mindset Clipka,
Moreover, how do you make a measurement of exactly pi*7/12... That's not
too much elegant.
Note that in the spirit of the puzzle, a prism would do... if only you
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Le 16.01.2009 18:29, Le_Forgeron nous fit lire :
> Le 16.01.2009 01:21, clipka nous fit lire :
>
>> Of course, a rectangular box of size 2*2*(pi*7/12), skewed by 2-(pi*7/12) to
>> "fill" a 2*2*2 cube, would do the job, but something tells me that it would not
>> be quite in line with the spirit of the puzzle ;)
>>
>>
> Yes, correct mindset Clipka,
> Moreover, how do you make a measurement of exactly pi*7/12... That's not
> too much elegant.
> Note that in the spirit of the puzzle, a prism would do... if only you
That does not make it easier.
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Le_Forgeron <jgr### [at] freefr> wrote:
> More than the classical cone/sphere/cylinder used to count 1,2,3.
>
> Counting from 1 to 6, here a set of increasing volumes.
> Constraint are:
> * Volume must be inside a box of size 2R
> * Each side of the box must be touched, at least by a point, even on border
>
> there such simple solid ? And what does it looks like ?
>
> Solids are:
> 1. Milk carton (berlingot) quartic
> 2. Cone
> 3. conocuneus (coin conique) quartic
> 4. Sphere
> 5. spherical top on cylinder
> 6. Cylinder
#3 could also be replaced with a paraboloid.
V=1/2pi a^2 h - a=R, h=2R
V=1/2pi R^2 2R
-tgq
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Le_Forgeron <jgr### [at] freefr> wrote:
> > Note that in the spirit of the puzzle, a prism would do... if only you
>
> That does not make it easier.
If a certain minimum of beauty is *not* a prerequisite, then I have something:
Make a prism of this...
The "ears" are constructed with the same radius as the "body" of the shape; the
triangular cutout at the bottom left must have a height of R*(7*sqrt(3/4)-6)
(which *can* be done with compass & straightline).
The fact that the total area of this shape is 7*R^2*pi/6 should be evident... ;)
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Le 16.01.2009 21:00, clipka nous fit lire :
> Le_Forgeron <jgr### [at] freefr> wrote:
>>> Note that in the spirit of the puzzle, a prism would do... if only you
>> That does not make it easier.
>
> If a certain minimum of beauty is *not* a prerequisite, then I have something:
Ok... you are better than me... a miminum of beauty should be a
requisite. Definitively.
Any 'nice' base for such prism ?
Or a thought on another shape ?
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Le_Forgeron <jgr### [at] freefr> wrote:
> Ok... you are better than me... a miminum of beauty should be a
> requisite. Definitively.
>
> Any 'nice' base for such prism ?
> Or a thought on another shape ?
Hm... I thought about coming up with something via polar co-ordinates - but
everything remotely simple seems to want a term including the inverse of pi,
the square root of it, or other such nasty stuff... nothing that seems to get
away without a pi term.
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> Can you explain a bit how, I fail to understand it's construction.
> Is there some ratio from green to blue/red ?
Firstly let me apologise as the term superellipsoid is slightly incorrect as it
is really more of a rounded box in the strictest sense. The reason I left it
split into blue and red and green is that this splits the shape down into 8
parts of spheres (green), 12 parts of cylinders (red), 6 small boxes (blue) and
one big box (which is inside). This identifies the different parts of which the
shape is composed.
> I have nothing against a superellipsoid, as long as you provide its
> parameters (and hopefully they are some kind of 'nice', even a cubic
> root of a fraction might match), and by the same way demonstrate that
> the volume is exactly and mathematicaly 7pi/3.
As I mentioned the different parts are varying shapes. To calculate the volume
one merely requires to add up the volumes of the whole or part shapes. To get
an exact formulation I used 'l' to be the length between the centres of the
corner spheres and 'r' to be their radius. We also know that l+2*r=2*R which
allows us to construct a cubic equation solely in l. This can be solved to find
required values and thus the shape can be constructed. The solution is not just
a little nasty, but rather very nasty but does exist and is expressible using
cube roots and fractions of polynomials in pi.
The nastiness of the solution somewhat detracts from the problem at hand
somewhat. So although I have presented a shape that has the correct volume it
is not a very satisfying solution.
> (and not an approximate value close enough)
The value can be calculated exactly due to the ability to split the shape into
standard geometrical sections. And as the solution to the cubic can be given in
a series of constants and mathematical operators it can technically be
calculated without approximation.
I am still keen to find a more natural shape that fulfils your conditions
however I hope this explanation is helpful.
Malcolm
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Le 16.01.2009 20:04, Trevor G Quayle nous fit lire :
> Le_Forgeron <jgr### [at] freefr> wrote:
>> Solids are:
>> 1. Milk carton (berlingot) quartic
>> 2. Cone
>> 3. conocuneus (coin conique) quartic
>> 4. Sphere
>> 5. spherical top on cylinder
>> 6. Cylinder
>
> #3 could also be replaced with a paraboloid.
>
> V=1/2pi a^2 h - a=R, h=2R
> V=1/2pi R^2 2R
>
> -tgq
>
Yes. Well seen.
Nevertheless, I have a slight preference for ruled surface when
possible. (1,2,3 & 6 are ruled surfaces, 4 is obvious, and the one I'm
most dissatisfied with is 5). That's personal, so no problem if your
favorite suite uses a paraboloid, it's just that recognizing x2 vs x4
only by the eye is not obvious for me, whereas a ruled surface is less
ambiguous.
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