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Thread: hyper-rectangular box volume

  1. #1
    Newbie mnh001's Avatar
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    hyper-rectangular box volume

    Since the hyper-volume of a hypercube of side L is L^4 and the surface volume
    is 8L^3, what about a hyper rectangular box? I can see the surface volume would
    be 8(L*W*H) but what would the hyper volume be?
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  2. #2
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    Re: hyper-rectangular box volume

    wouldn't there be a fourth dimension to your hyper rectangular box?

    Call it $H^\prime$

    I'm no expert but I'd imagine the hypervolume is then $L*W*H*H^\prime$
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  3. #3
    Newbie mnh001's Avatar
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    Re: hyper-rectangular box volume

    Yup, that's why it's a hyper box, it is 4 dimensional. But with the hypercube
    (tesseract) it's easier to see because each of the hypercube's edges is the
    same length, so there's no confusion that hyper volume is L^4. But with the
    hyper rectangular box each of the edges are a different length so which one
    do you use for the hyper volume?

    From what I understand, each 'surface' of a hypercube is a 3D cube. So if it's
    not a hypercube but a hyper rectangular box then each 'surface' of it would be a
    3D rectangular box. And if L, W and H are each different... that's where my
    confusion comes in about the hyper volume.
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  4. #4
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    Re: hyper-rectangular box volume

    Quote Originally Posted by mnh001 View Post
    Yup, that's why it's a hyper box, it is 4 dimensional. But with the hypercube
    (tesseract) it's easier to see because each of the hypercube's edges is the
    same length, so there's no confusion that hyper volume is L^4. But with the
    hyper rectangular box each of the edges are a different length so which one
    do you use for the hyper volume?

    From what I understand, each 'surface' of a hypercube is a 3D cube. So if it's
    not a hypercube but a hyper rectangular box then each 'surface' of it would be a
    3D rectangular box. And if L, W and H are each different... that's where my
    confusion comes in about the hyper volume.
    Romsek answered that confusion. For surface volume, it would be $2(L\cdot W \cdot H + L\cdot W\cdot H' + L\cdot H\cdot H' + W\cdot H\cdot H')$. When $L = W = H = H'$, you get the special case: $2(L^3+L^3+L^3+L^3) = 8L^3$. For hyper volume, you have $L\cdot W\cdot H\cdot H' = L^4$.
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  5. #5
    Newbie mnh001's Avatar
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    Re: hyper-rectangular box volume

    Oh, so even the 3D boxes that make up the surfaces have a 4th component?
    Wow, hard to wrap the head around.
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    Re: hyper-rectangular box volume

    Quote Originally Posted by mnh001 View Post
    Oh, so even the 3D boxes that make up the surfaces have a 4th component?
    Wow, hard to wrap the head around.
    Not exactly. They have 3 components chosen from a possible 4. The different "surfaces" will have different sets of 3.
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  7. #7
    Newbie mnh001's Avatar
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    Re: hyper-rectangular box volume

    Hmm, ok I think. Let's try an example.

    If I have a 3D box that is 3 by 5 by 9 and I translate it up to the 4th
    dimension, the L, W and H will still be 3, 5 and 9, yes? So then what
    will the H' be? Or do I not translate up but assign the edges of the
    4D box, say 3 by 5 by 9 by 11?
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    Re: hyper-rectangular box volume

    "Picturing" higher dimensional objects is typically easiest with simple objects, like simplices:

    https://en.wikipedia.org/wiki/Simplex
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  9. #9
    Newbie mnh001's Avatar
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    Re: hyper-rectangular box volume

    Interesting page. Simple? Not really. Kinda above my level. But I did get what I needed from the previous posts. Thanks.
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    Re: hyper-rectangular box volume

    Quote Originally Posted by mnh001 View Post
    Hmm, ok I think. Let's try an example.

    If I have a 3D box that is 3 by 5 by 9 and I translate it up to the 4th
    dimension, the L, W and H will still be 3, 5 and 9, yes? So then what
    will the H' be? Or do I not translate up but assign the edges of the
    4D box, say 3 by 5 by 9 by 11?
    $H^\prime$ can be whatever you want it to be provided it's positive.
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    Re: hyper-rectangular box volume

    Quote Originally Posted by romsek View Post
    $H^\prime$ can be whatever you want it to be provided it's positive.
    Awww! Where's the fun in that?

    -Dan
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  12. #12
    Newbie mnh001's Avatar
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    Re: hyper-rectangular box volume

    Yes, I figured on that. But aww gee, can't I have a hyperbox of -1? That guy in the poem had a cube of -1.
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    Re: hyper-rectangular box volume

    Quote Originally Posted by mnh001 View Post
    Yes, I figured on that. But aww gee, can't I have a hyperbox of -1? That guy in the poem had a cube of -1.
    Short answer: yes.
    Longer answer: Typically, negative measures require a generalization of volume that result from Lesbesgue integration over a signed measure, which itself is a generalization of the traditional nonnegative measure. I will not go into much more details on it, as Measure Theory is a rather complex subject already.
    Last edited by SlipEternal; Nov 7th 2017 at 08:12 AM.
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