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 $w \le width$.