Here is an attempt based on my experiences with furniture moving. The long dimension a=4.3
will surely be horizontal. One of the short dimensions, call it b
will be vertical, the remaining dimension c
will be horizontal. The box must be as "short" as possible during the passage at the corner. So, one end of the box will be lifted:
We calculate the projection L = x1 + x2
of the lifted box onto the horizontal plane. Now we move the shortened box around the corner. Here is an algorithm as a Python program (I hope it is readable):
# hallway dimensions: height = 2.5 width = 1.5 def box(a, b, c): # a = long dimension of the box = 4.3, horizontal # b = short dimension, 0.2 (or 0.07), vertical # c = the other short dimension, horizontal d = math.sqrt(a*a + b*b) # diagonal of a x b rectangle alpha = math.atan(b/a) # angle of the diagonal in axb rectangle omega = math.asin(height/d) - alpha # lifting angle x1 = b * math.sin(omega) # projection of b to the floor x2 = a * math.cos(omega) # projection of a to the floor L = x1 + x2 # length of the lifted box projected to the floor sin45 = math.sin(math.pi/4.0) y1 = c * sin45 # projection of c to the y axis y2 = L / 2 * sin45 # projection of L/2 to the y axis w = y1 + y2 # box needs this width w ok = (w <= width) # box passes if its width w is less than the # the available hallway width print "w =", w, ", pass =", ok return ok def test(): # 1) try 0.07 as vertical dimension: box(4.3, 0.07, 0.2) # prints w= 1.407, pass= True # 2) try 0.2 as vertical dimension: box(4.3, 0.2, 0.07) # prints w= 1.365, pass= True test()
So, the box can be transported around the corner either way (either 0.2 or 0.07 vertical).
Adding Latex formulae for the pure mathematician:
$ \begin{align*} d= & \sqrt{a^{2}+b^{2}}\\ \alpha= & \arctan(b/a)\\ \omega= & \arcsin(height/d)-\alpha\\ L= & x_{1}+x_{2}=b\sin\omega+a\cos\omega\\ w= & y_{1}+y_{2}=\frac{c}{\sqrt{2}}+\frac{L}{2\sqrt{2}} \end{align*} $
The box can be transported around the corner if $w \le width$.