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I brought a closet today. It has dimension $a\times b\times c$ where $c$ is the height and $a\leq b \leq c$. To assemble it, I have to lay it out on the ground, then move it to upright position. I realized if I just move it in the way in this picture, then it would require the height of the ceiling to be $\sqrt{c^2+a^2}$.

Is this height required if we consider all possible ways to move the closet? Or is there a smart way to use less height?

Move a closet upright.

This is a image from IKEA's instruction that comes with the closet.

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If anyone is interested. I solved the real life version by assemble it standing up. –  Chao Xu Mar 24 '13 at 1:04
    
There! Your comment answers your question... –  DonAntonio Mar 24 '13 at 1:08
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It seems to be an assumption of your problem that the closet height $c$ is somewhat more than the other dimension, $a \le b$. You give the answer in your question, a minimum height to clear is $\sqrt{a^2 + c^2}$. What more, if anything, are you looking for? E.g. a proof? –  hardmath Mar 24 '13 at 1:22
    
Yes, $c$ is larger. I asked for the proof, but it was after the picture, so it's hard to see. I have moved it before the picture. –  Chao Xu Mar 24 '13 at 3:44
    
No, I just added an attribution for this. Good catch. –  Chao Xu Mar 25 '13 at 0:00
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3 Answers 3

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We are asked to prove that a minimum clearance of $\sqrt{a^2 + c^2}$ is required to move the assembled closet (box) from a "prone" position in the hallway floor to an upright position.

Let $V$ be the vertical height of the box, defined as the difference between the heights of the uppermost and lowest corners of the box in whatever position. This value is invariant with translation and with rotation about a vertical axis. In the upright position $V = c$, and depending on which face the box is lying on its prone position, the vertical height there is either $V = a$ or $V = b$.

To restate what is to be shown, in going from the prone to the upright position, the vertical height $V$ must at some intermediate position equal $\sqrt{a^2 + c^2}$, and that (as already shown by the OP's diagram), the upright position may be reached without exceeding that vertical height.

Using the above mentioned symmetries (translation and rotation about a vertical "z-axis"), we can maintain the center of the box in a fixed location and keep the long axis of the box (having length $c$) confined to a plane perpendicular to the (horizontal) width of the hall (assuming for the sake of generality that hallway width and length are arbitrarily great, so allowing movement of the box restricted only by the height of the hallway ceiling).

The leaves two degrees of freedom: the angle $\theta$ the box's long axis makes with respect to horizontal, and the angle $\phi$ around that axis which the box is rotated. For specificity let's say $\phi = 0$ designates an orientation in which the $b$ length edges are horizontal.

By standard trigonmetric arguments, it suffices to consider quarter rotations $0 \le \theta \le \pi/2$ and $0 \le \phi \le \pi/2$, for which:

$$ V(\theta,\phi) = c \sin \theta + (b \sin \phi + a \cos \phi) \cos \theta $$

For fixed angle $\theta$ the angle $\phi$ which minimizes $V$ is zero, so:

$$ V(\theta,\phi) \ge V(\theta, 0) = c \sin \theta + a \cos \theta $$

Thus when $\theta = \tan^{-1} c/a$, $V(\theta,\phi) \ge \sqrt{a^2 + c^2}$, and it follows that $\theta$ cannot reach upright angle $\pi/2$ without $V$ reaching at least $\sqrt{a^2 + c^2}$, and also that if $\phi = 0$ is maintained, then the upright position may be reached without $V$ exceeding that threshold.

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I have two solutions to this problem.

Intuitive solution

Intuitively, it seems to me that the greatest distance across the box would be the diagonal, which can be calculated according to the Pythagorean theorem:

$$h = \sqrt {l^2 + w^2}$$

If you'd like a more rigorous solution, read on.

Calculus solution

Treat this as an optimization problem.

For a box of width $w$ and length $l$ (because depth doesn't really matter in this problem), the height when the box is upright, assuming $l > w$, i

$$h=l$$

If the box is rotated at an angle $\theta$ to the horizontal, then we have two components of the height: the height $h_1$ of the short side, $w$, and the height $h_2$ of the long side, $l$. Using polar coordinates, we have

$$h_1 = w sin \theta\\ h_2 = l cos \theta$$

Thus, the total height is

$$h = h_1 + h_2 = w sin\theta + l cos \theta$$

This intuitively makes sense: for small $\theta$ (close to upright), $w sin \theta \approx 0$, and $l cos \theta \approx l$, so $h \approx l$ (and similarly for large $\theta$).

The maximum height required means we need to maximize $h$. Take the derivative:

$$\frac {dh}{d\theta} = \frac d{d\theta} (w sin \theta + l cos \theta) = w cos \theta - l sin \theta$$

When the derivative is zero, we may be at either an extremum or a point of inflection. We need to find all of these on the interval $(0, \frac \pi 2)$ (because we don't case about anything outside of a standard 90° rotation).

So, we have

$$0 = \frac {dh}{d\theta} = w cos \theta - l sin \theta\\ w cos \theta = l sin \theta$$

And, because of our interval $(0, \frac \pi 2)$, we can guarantee that $cos \theta \neq 0$, so

$$\frac w l = \frac {sin \theta} {cos \theta} = tan \theta\\ \theta = atan \left (\frac l w \right )$$

Now that we know that the maximum is at $\theta = atan \left (\frac l w \right)$. We can plug that back into our polar coordinate equation to get

$$h = w sin \theta + l cos \theta\\ h = w sin \left ( atan \left (\frac l w \right ) \right ) + l cos \left ( atan \left (\frac l w \right ) \right )$$

That's my 2¢.

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I think I didn't write the problem clearly. By moving it in the way in the diagram, then certainly it would need this much height. I was querying if we can do any kind of movement to the box, can we stand it up use less than that height. –  Chao Xu Mar 24 '13 at 17:24
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If you're treating the box as a rigid body (you can't chop it, squish it, burn it, etc.), then we can make the following statements:

  • The box starts with a rotation angle of $\theta = 0$, and you want to rotate it to get $\theta = 90^\circ = \pi/2$.
  • Because the box is rigid, it cannot be deformed; thus, its angle $\theta(t)$ must be continuous.
  • By the Intermediate Value Theorem, the box must pass through every angle between $0$ and $90^\circ = \pi/2$.
  • Thus, the box must pass through the angle where the required height is $\sqrt {a^2 + b^2}$. There is no way to "do any kind of movement to the box" such that we could "stand it up [using] less than that height."
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I think you mean to argue that the required height is $\sqrt{a^2 + c^2}$. However the orientation of the box (its rotation from an initial position laid out on the floor) involves two degrees of freedom, so it isn't quite as simple as a 1D application of the Intermediate Value Theorem to "rotation angle $\theta$". –  hardmath Mar 25 '13 at 0:39
    
Why is it 2 degrees of freedom instead of 3? –  Chao Xu Mar 25 '13 at 0:43
    
@ChaoXu: We can eliminate most degrees of freedom by symmetry. Let's fix the center of the box (closet), and measure only the vertical height (difference between highest and lowest corners). Now there are 3 degrees of freedom, but rotation around the vertical axis is a symmetry wrt vertical height, rotation around the two horizontal axes as degrees of freedom. This assumes that the width as well as the length of the hall are not limiting in regard to the size of the closet. –  hardmath Mar 25 '13 at 2:01
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