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I've taken the liberty of posting this problem (I dare not call it a puzzle lest
I face the wrath of the Puzzle Definition Police) out of "Off-Topic" because
POV-Ray was used to produce the illustration. The problem itself is presented
in the image. If anyone wants to refesh his or he memory with respect to basic
geometry formulas, it's OK to look here:
http://www.math10.com/en/geometry/volume.html
I would not consider this "cheating" as this is not intended as a memory quiz.
The solution to the problem as posed is here:
LOOK HERE ONLY IF YOU WISH TO SEE THE SOLUTION:
http://a833.ac-images.myspacecdn.com/images01/116/l_c46789151c59aa7c2d8fa08b9022ecc0.jpg
If anyone wants to see the solution using calculus referred to in the problem,
it can be seen here: (a handy site that I have no affiliation with)
http://www.ltcconline.net/greenl/courses/106/areavolume/spherecore.htm
Happy Solving,
Mike C.
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Preview of image '6m_core_p.jpg'
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I haven't looked at the solution. This is what I'm thinking:
At first it seems that the problem is not solvable because there's too
little info. We only know that the height of the shape is 6m, but we do
not know the radius of the hole nor the radius of the sphere. These two
radii could be anything, and still the resulting object could be 6m tall.
But then the problem states that the problem is solvable. I have to
believe that this is true.
If the problem is solvable, the only conclusion is that the volume of
the shape is *not* dependent on the radii of the cylinder and the
sphere. In other words, changing the radius of the cylinder and the
sphere in such a way that the CSG shape is still 6m tall will keep the
volume the same.
If this is so then this should be so even if the radius of the
cylinder is 0, in which case the height of the sphere must be 6m
(because that's how tall the CSG shape must be).
Thus the problem is reduced to calculating the trivial volume of a
sphere of diameter of 6m, which should be something like pi*r^3*4/3 =
pi*3^3*4/3 = pi*36.
Is this correct?
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Warp <war### [at] tag povray org> wrote:
> I haven't looked at the solution. This is what I'm thinking:
.....
> Is this correct?
Yes. (I'll presume that you asked instead of checking the solution link because
you would want to keep working on it if the solution were otherwise.)
Note: the original post also contains a link to a demonstration (not by me) that
the problem IS solvable, should anyone wish to confirm the statement.
BTW, nice explanation.
Best Regards,
Mike C.
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"Mike the Elder" <zer### [at] wyan org> wrote:
> I've taken the liberty of posting this problem (I dare not call it a puzzle lest
> I face the wrath of the Puzzle Definition Police) out of "Off-Topic" because
> POV-Ray was used to produce the illustration. The problem itself is presented
> in the image. If anyone wants to refesh his or he memory with respect to basic
> geometry formulas, it's OK to look here:
>
> http://www.math10.com/en/geometry/volume.html
>
> I would not consider this "cheating" as this is not intended as a memory quiz.
>
>
> The solution to the problem as posed is here:
> LOOK HERE ONLY IF YOU WISH TO SEE THE SOLUTION:
>
http://a833.ac-images.myspacecdn.com/images01/116/l_c46789151c59aa7c2d8fa08b9022ecc0.jpg
>
>
> If anyone wants to see the solution using calculus referred to in the problem,
> it can be seen here: (a handy site that I have no affiliation with)
>
> http://www.ltcconline.net/greenl/courses/106/areavolume/spherecore.htm
>
>
> Happy Solving,
> Mike C.
the volume of a sphere = pi*(6/2)^3*4/3
but what is the purpose of the 2nd black paragraph in the image?
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"Mike the Elder" <zer### [at] wyan org> wrote
Isn't adding "this problem is solvable" more or less like writing code like
"if x = true then ..."
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I wish i hadn't scrolled down as I glimpsed the solution in the first post which
spoiled my fun of working it out the non-calculus way. However I had in my mind
gone through how to do it using calculus. I haven't looked at the solution but
I believe it will run something as follows.
Let us start by imagining we cut the cored sphere into infinitely thin slices,
then we will have little discs. Then if we add up the area of all the discs we
will get the volume.
If we represent the inner sphere as having a radius of say R and the sphere as
having a radius T then. Now we can write that the height from the axis (say x)
is given by the formula height^2 = T^2 - R^2
The discs have a radius whose square changes with proportion to the axis of
rotation (say Z), this is because the formula for a circle is x^2+y^2=r^2. Thus
the square of the outer radius of the disc is given by T^2-z^2. Notably at the
origin then we get the radius of the sphere out as z=0. This only works for
values of z between -height and height.
We thus get areas for the disc that are given by: Area = pi*((outside
radius)^2-(inner radius)^2) = pi*(T^2-z^2-R^2) = pi*(height^2-z^2)
Thus to find the volume we must intergrate from -height to height the above area
formula. With a little working we get a mess which equals 2*pi*(2/3*height^3)
or (4*pi*height^3)/3
This I presume is correct given the solution to the non-calculus version is
correct. However potentially this is the proof that the problem is solvable.
Hope noone was too bored by my long mathematical explantion and notably the
calculus involed is not too strenuous! Also thanks Mike as it was a nice little
brain teaser, even to solve the calculus way.
Regards Malcolm
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I realise my last post had one error I used height to mean something different
to on the diagram. My usage of height actually represents half of the height on
the diagram. I.e where i say height i mean 3m not 6m. Sorry for that but didn't
have both pages open at once so was kind of doing it from memory off the
diagram. thus my final solution should have read: Volume =
(4*pi*(0.5*height)^3)/3.
Apologies, Malcolm
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"somebody" <x### [at] y com> wrote:
> "Mike the Elder" <zer### [at] wyan org> wrote
>
> Isn't adding "this problem is solvable" more or less like writing code like
> "if x = true then ..."
It's not obvious (at least to most people) that knowing the height of a "cored
sphere" object is sufficient to calculate it's volume. It's an inference that
can be drawn form the fact that the information IS sufficient to find the
please use the links in the original post.)
Best regards,
Mike C.
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somebody wrote:
> Isn't adding "this problem is solvable" more or less like writing code like
> "if x = true then ..."
No. See my rationale in my post.
"This problem is solvable" actually means that you can assume
something about the problem, as I described in that post. Without
knowing that it is indeed solvable it would be quite hard to prove that
it indeed is. Would require quite complicated math.
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"Warp" <war### [at] tag povray org> wrote
> "This problem is solvable" actually means that you can assume
> something about the problem, as I described in that post. Without
> knowing that it is indeed solvable it would be quite hard to prove that
> it indeed is. Would require quite complicated math.
A problem is either solvable, not solvable, or a fields medal candidate. We
can safely ignore the last possibility here I think. If you ask me a
problem/puzzle to find a solution to, and if I trust that you are not
leading me to a wild goose chase, it means it's given that it *is* solvable.
If I don't trust you, no amount of reiterating that it's solvable will
convince me ( "if ((((x=true)=true)=true..." syndrome ).
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