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

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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


Here is a crazy superellipsoid shape which fits the bill of 7*pi/3. 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 mainly
constructed this as a point of interest that such a shape can be found and
built easily with pov csg. I am still hoping I can find a simpler shape which
wouldn't have required quite so much work to build.

Malcolm

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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

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

-tgq

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"Trevor G Quayle" <Tin### [at] hotmailcom> 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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