Mathematical Marginalia
Classical and Quaint Topics in Mathematics

The Mover's Problem

The following problem is a very simple variant of the Moving Sofa Problem and a slightly more complicated variant of the Moving Ladder Problem. A cabinet, board, cupboard, wardrobe or generally a piece of furniture of cuboid shape shall be moved around a corner of L-shape. The problem is much easier as the sofa problem, because the footprint of the cabinet is a rectangle, while the footprint of the sofa may be of any shape and is the searched quantitity. It's a bit more complicated as the ladder problem, because the ladder is idealized as a line segment of width zero. In Germany the problem is also known as the Schrankproblem or the Möbelpacker-Problem. You find a description together with a nice animation on this German site. Unfortunately, I could not find an equally instructive English site (perhaps because there are too many different words like cabinet, board, cupboard, wardrobe etc.). If you know such a site, please drop me a note, and I will gladly add a link.

However, the problem presented here has been a real-life problem of my family. A sideboard should be moved from one room to another through an L-shaped corridor with the constraint that the piece can only be pushed, because it is far too heavy to be lifted. So it's essentially a 2-dimensional problem of pushing a rectangle through an L-shaped polygon.

Let's specify the problem more formally:

A sideboard of length $s$ shall be pushed through an L-shaped corridor. The legs of the corridor are of width $u$ and $v$ respectively (see fig.1). What is the maximal width of sideboard, that allows the movement without getting stucked.

The L-shaped corridor has six corners. To avoid confusion, we will speak only of the corner with an interior angle of 270° as the corner. The other corners are of no special interest. To maximize the distance between one side of the sideboard and the corner, we push the sideboard as close as possible along the wall opposite to the corner. As an idealization, let's assume that two of the edges of the sideboard touch these walls at every moment.

Abb.1: The sideboard during its movement
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To simplify the calculations, we assume that the corner of the corridor is the origin of a Cartesian coordinate system. Then the opposite walls are given by the paraxial lines $x=u$ and $y=v.$ Hence the problem is equivalent to the problem to calculate the minimal distance of a line segment of length $s$ to the origin, if the endpoints of the line segment $(x_1,y_1 )$ and $(x_2,y_2 )$ lie upon the lines $x=u$ and $y=v$ respectively (see fig.2).

We parameterize the line segments (or rather the one line segment in different positions) by the distance $t$ of one of its endpoints $(x_1,y_1 )$ to the point with coordinates $(u,v).$ At first we will deduce the general formula for the distance of a line $g$ (on which the line segment with length $s$ is located) from the origin.

The line $g$ is given by the general linear function $y=mx+c.$ A line perpendicular to $g$ has the slope $–1/m.$ If this line goes through the origin, its equation is $y=-1/m x.$ We calculate the point of intersection $(x_s,y_s )$ of $g$ with this perpendicular through the origin: $$ \begin{align*} -\frac{1}{m} x_s &= mx_s+c\\ x_s &= -\frac{m}{1+m^2} c\\ \end{align*} $$ Inserting into $y_s=-\frac{1}{m} x_s$ we have $y_s=\frac{1}{1+m^2} c.$

And by Pythagoras the distance is: $$ d^2= x_s^2+y_s^2= \frac{m^2 c^2}{(1+m^2 )^2} +\frac{c^2}{(1+m^2 )^2} = \frac{c^2 (1+m^2 )}{(1+m^2 )^2} = \frac{c^2}{1+m^2} $$ $$ d=\frac{c}{\sqrt{1+m^2 }}\quad \quad(1) $$

Fig.2: Idealized Picture
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Now we calculate the slope and the y-intercept of the line $g.$ The slope $m$ is given by $$ m= \frac{y_2-y_1}{x_2-x_1 }= -\frac{\sqrt{s^2-t^2 }}{t} $$

The numerator is again a consequence of Pythagoras' theorem, the negative sign results from the fact that $y_2 \lt y_1.$ The denominator follows immediately from the picture. The line goes through the point $(u-t,v),$ hence we can calculate the y-intercept $c$: $$ \begin{align*} v &= m(u-t)+c\\ c &= v-m(u-t) \\ &= v+ \frac{\sqrt{s^2-t^2 }}{t} (u-t)\\ \end{align*} $$

Inserting the values of $m$ and $c$ into (1) we get an equation of $d$ as function of the parameter $t$: $$ \begin{align*} d &= \frac{c}{\sqrt{1+m^2}} = \frac{c}{\sqrt{1+\frac{s^2-t^2}{t^2}}} = \frac{ct}{s}\\ d &= (v+ \frac{\sqrt{s^2-t^2 }}{t} (u-t) )\frac{t}{s}\\ d &= \frac{vt}{s}+ \frac{\sqrt{s^2-t^2 } (u-t)}{s}\\ \end{align*} $$

This function may be plotted, inserting our real-life data $v=1.1, u=1.35, s=2.1$ (all in meters), e.g. by Geogebra. You will locate the minmum roughly at $1.6$ with the corresponding value of $d \approx 0.68.$ Hence, our sideboard with width $61$cm could be moved around the corner.

Fig.3: Plot by Geogebra
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You see, that at $t=0$ the value of $d$ is $1.35$ – that's what's expected, because the line segment $s$ lies on the line $x=u$ in this case. Not by the picture, but by the equation you can verify that at $t=s$ the distance is $v.$ In this case, the line segment $s$ lies on the line $y=v.$ For values $t>s$ the square root becomes imaginary, hence the domain of definition of the function is $0 \le t \le s,$ negative values of $t$ make no sense.

Of course, we should at least try, to calculate the maxima and minima algebraically. Hence we calculate the derivative of $d$ with respect to $t$, applying both the product rule $(fg)' = f'g+fg'$ and the chain rule $(f(g(t))'= f' (g(t)) g' (t)$ for the second summand: $$ d'= \frac{1}{s}(v-\frac{t}{\sqrt{s^2-t^2 }} (u-t)-\sqrt{s^2-t^2}) $$ Setting $d'(t_m )=0$ we get: $$ v-\sqrt{s^2-t_m^2 }= \frac{t_m}{\sqrt{s^2-t_m^2}} (u-t_m ) $$ Here we substitute $x=\sqrt{s^2-t_m^2}$ and find: $$ \begin{align*} v-x &=\frac{\sqrt{s^2-x^2 }}{x}(u-\sqrt{s^2-x^2})\\ vx-x^2 &= u\sqrt{s^2-x^2}-s^2+x^2\\ vx-2x^2+s^2 &= u\sqrt{s^2-x^2}\\ (vx-2x^2+s^2 )^2 &= u^2 (s^2-x^2 )\\ 4x^4-4vx^3+v^2 x^2-4s^2 x^2 &+ 2s^2 vx+s^4-u^2 s^2+u^2 x^2 = 0\\ 4x^4-4vx^3+(v^2-4s^2+u^2 ) x^2 &+ 2s^2 vx+s^4-u^2 s^2=0\\ \end{align*} $$

This is an equation of fourth degree, which could be solved algebraically in principal. However, the calculations would be cumbersome, to say the least. But we may employ a CAS system like wxMaxima to find an approximate solution, which would be more accurate than the solution by plotting in any case. In our special case (length of the sideboard $s=2.1,$ width of the corridor $u=1.35$ and $v=1.10$) wxMaxima computes a minimum at $t_m=1.5877$ with corresponding $d$ of $0.6761,$ in pretty good agreement with the values read from the Geogebra plot (see fig.4).

Abb.4: Snapshot wsMaxima
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Here you have to guess, that the 3rd zero is the only relevant one. Because of $0 \lt t_m \lt s,$ we have $0\lt x \lt s,$ hence the negative zeroes may be ignored. The zero at $x= 1.950$ belongs to the maximum at -0.779 with $d= 1.57,$ taking the negative value of the square root in the expression marked by %i25.

After we knew that our sideboard would have found its way through the corridor, doubts came up, whether it would look as nice at the new location, as we had imagined. As a matter of fact, the sideboard is still at its old place.