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[EDIT2]

Following the help I got in the comments + answers I found

$$\begin{align} a &\leq \sqrt{H^2-b^2} \\[4pt] a &\leq \sqrt{h^2-b^2} \end{align}$$

as the answer with this process https://i.stack.imgur.com/Z6mwd.jpg

I need to develop a method to know if the box can pass or not enter image description here]

I need to find a method to know if I can turn my box to keep moving in the hallways, when I give $a$, $b$, $h$ and $H$ (in meter) can show if it will pass or not.

For example $a=3, b=1.5, h=2.1, H=2.1$

I'm really stuck with that and I though that someone here could help me (if you want to do it with voice call or anything I can give you my Discord in private)

I tried to draw a isosceles triangle with the side a of my box to calculate his height of the triangle but I'm stuck with the $h$ and $H$

Blue
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Jérôme Vial
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  • Welcome to stackexchange. Please edit the question to show us what you tried and where you are stuck. What course is this for? Have you solved similar problems? – Ethan Bolker Sep 19 '18 at 15:15
  • No we didn't, I tried to use a isoceles triangle but I'm stuck because of the two different height (h and H) – Jérôme Vial Sep 19 '18 at 15:29
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    Although in this special case the problem can be answered, as is done by @Andrei, this is related to a famously difficult problem known as the "furniture movers problem" or the moving sofa problem. – Lee Mosher Sep 19 '18 at 16:02
  • Your answer demands that the diagonal of the box can pass down each hallway. That is sufficient but not at all necessary. For a square box down a balanced hallway so $a=b, h=H$ you are asking $\sqrt 2 a \lt H$. I would ask $3a \lt 2\sqrt 2 H$, which is much more generous. – Ross Millikan Sep 19 '18 at 19:14
  • See https://math.stackexchange.com/questions/51671 and https://core.ac.uk/download/pdf/55829517.pdf and https://demonstrations.wolfram.com/MovingACouchAroundACorner/ – cgiovanardi Sep 19 '18 at 20:08

2 Answers2

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The problem can be solved if you think what is the condition that the box is just stuck. That means that the box cannot be translated or rotated. Say you have the pivot point $P$ next to the $90^\circ$ label. Your upper corner of the box is just touching the top side of hallway with the width $h$. Then you cannot translate any longer in the upper direction. Similarly, the left most corner will touch the left wall of hallway $H$. The condition of just barely being able to rotate is that the top corner is right above $P$, and the left corner is exactly left of $P$. You now get two right angle triangles. Using Pythagoras' theorem you can calculate the length from $p$ to the right most corner and to the bottom corner. You can then rotate/move the box if the sum is less than $a$. $$\sqrt{H^2-b^2}+\sqrt{h^2-b^2}\le a$$

Andrei
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  • I can understand most of the development except I can't really see the two right angle triangles, could you help me a bit more ? – Jérôme Vial Sep 19 '18 at 15:52
  • The corners of the box. Let's call the corners of the rectangle $U,R,D,L$ for up, down, left, and right. Than one of your triangles id $PUR$, with the right angle at $R$, and the other is $PLD$ with right angle at $D$ – Andrei Sep 19 '18 at 15:56
  • Ok I did the graph again and now I see my two right angle triangles and I'll try to understand more the final function you found. Thank you for the really good quality answer + edit – Jérôme Vial Sep 19 '18 at 16:10
  • After trying and trying to understand how to found your function, I can't. Is it too much to ask for a graph ? or a draw or anything I could look on ? – Jérôme Vial Sep 19 '18 at 16:21
  • Put the labels I've suggested on your figure. Then $PR^2=PU^2-UR^2=h^2-b^2$. $PB^2=PL^2-LB^2=H^2-b^2$. $PR+PB=RB=a$. – Andrei Sep 19 '18 at 17:20
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It is much simpler when $H$ and $h$ are the same, as in your example. The limiting case is then when the box is at $45^\circ$. If you extend the inner edge of the box to hit the outer wall it is $a+2b$ long so the inner edge will be $\frac {a+2b}{ 2}$ from the corner. The distance between the corners is $H\sqrt 2$ so you need $$\frac {a+2b}{ 2} \le H\sqrt 2\\a+2b \le 2\sqrt 2H$$ which is not the case with your data but you are close. This is asking $6 \le 5.93$

For the general case with $h \neq H$ refer to the drawing below. It shows the box going around the corner rotated by an angle $\theta$. We need to have the point on the box in line with the bottom wall of the horizontal hallway clear the corner. The point where the extension of the inside edge of the box hits the top wall is $a \sin \theta +\frac b{\cos \theta}$ out from the corner. To have the box clear the corner we need $$a \sin \theta + \frac b{\cos \theta}-h \tan \theta \lt H$$ as $\theta$ goes from $0$ to $90^\circ$. This calls for taking the derivative, setting to zero, and so on. I don't see a nice form that we can just

enter image description here

Ross Millikan
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  • h and H won't be the same, they can be different and I think your method only work when they are the same right ? – Jérôme Vial Sep 19 '18 at 17:15
  • That is correct. It makes the computation simpler and in many cases the two hallways are the same. If they differ it is harder to find the limiting point were the box is as far out of the corner as possible. It can also be a check for a full solution. It has to reduce to this when $h=H$. I don't think the other solution is right because of this. – Ross Millikan Sep 19 '18 at 17:19
  • Yes in the end I found that $$a <= {\sqrt{H^2-b^2}}.$$ $$a <= {\sqrt{h^2-b^2}}.$$ – Jérôme Vial Sep 19 '18 at 18:09