Diagonal door brace orientation

Question

I'm trying to come up with functions that describe the orientation of a brace on a door, such that there is equal interface x between the brace and all 4 door members, provided door dimensions a, b, and brace width c.

Given a, b, c, what function describes x? Or r (brace rotation around center)?

Answer 1

Edit : Here are two solutions. The first one uses analytic geometry. The second one, simpler, elaborated later, uses only trigonometry on triangles.

First solution : Taking "natural" coordinates representation wrt inner rectangle (red axes on figure below), let us consider the extreme points of the brace have coordinates :

$$A(b-x,0), B(b,x), C(0,a-x),D(x,a).$$

Straight lines $AC$ and $BD$ (brace borders) have resp. equations :

$$\begin{cases}x_1(a-x)+y_1(b-x)-(a-x)(b-x)&=&0\\x_1(a-x)+y_1(b-x)+(x^2-ab)&=&0\end{cases}$$

(where $x_1,y_1$ are the generic coordinates).

The shortest distances from the origin $O$ to lines $AC$ and $BD$ are resp. :

$$d_1=\frac{(b-x)(a-x)}{\sqrt{(a-x)^2+(b-x)^2}}, \ d_2=\frac{ab-x^2}{\sqrt{(a-x)^2+(b-x)^2}}$$

(see formula here).

Therefore the distance between the two lines $AC$ and $BD$ is the difference $d_2-d_1$, which must be equal to $c$ :

$$\frac{(a+b)x-2x^2}{\sqrt{(a-x)^2+(b-x)^2}}=c$$

Squaring it, we get the following 4th degree equation :

$$\underbrace{((a+b)x-2x^2)^2-c^2((a-x)^2+(b-x)^2)}_{f(x)}=0,\tag{1}$$

$x$ being one of its roots.

Examples :

• If we take $a=3,b=2$ and $c=0.3$ (see graphical representation of $f$ below), we get the roots :

$$x \approx -0.216, \ 0.217, \ 2.457, \ 2.542$$

The second root (and this one only) is the right one.

• If we take $a=b=3$ and $c=0.5$, we get the roots :

$$x = -\frac{\sqrt{2}}{4}, \ \frac{\sqrt{2}}{4}, \ 3 \ \text{(double root)}$$

Here as well, the second root is the right one.

Second solution : Consider the following figure featuring the north-eastern part of the internal rectangle :

Angle $\theta$ is found in 3 places, allowing us to write two equations :

$$\begin{cases}x \cos(\theta)+x \sin(\theta)&=&c\\ \tan(\theta)&=&\frac{b-x}{a-x}\end{cases}.$$

Out of these two equations, one can choose

• either to eliminate $\theta$ and keep unknown $x$. This will give (back) equation (1).

• or to eliminate $x$ and keep unknown $\theta$. This will give the following equation :

$$b \cos(\theta)-a \sin(\theta)=c \frac{\cos(\theta)-\sin(\theta)}{\cos(\theta)+\sin(\theta)}$$

that can be solved by taking half-angle formulas

$$\cos(\theta)=\frac{1-t^2}{1+t^2}, \ \sin(\theta)=\frac{2t}{1+t^2} \ \ \text{where} \ \ t:=\tan(\theta/2)$$

giving rise... to a fourth degree equation.

Answer 2

If $\theta$ is the angle that the brace makes with the vertical, then

$ \tan \theta = \dfrac{ b - x}{a - x} $

Further, if we concentrate on the upper left corner, one can see that,

$ \sqrt{2} x = \dfrac{ w }{\cos( 45^\circ - \theta ) } $

Since $\cos(45^\circ - \theta) = \dfrac{1}{\sqrt{2}} (\cos \theta + \sin \theta) $

Then,

$ w = x (\cos \theta + \sin \theta ) $

Substitute this in the first equation,

$ \tan \theta =\dfrac{\sin \theta}{\cos \theta} = \dfrac{ b - w / (\cos \theta + \sin \theta ) }{ a - w / (\cos \theta + \sin \theta) } $

And this simplifies to,

$ \dfrac{\sin \theta}{\cos \theta} = \dfrac{ b (\cos \theta + \sin \theta) - w }{ a (\cos \theta + \sin \theta ) - w } $

Cross multiplying,

$ \sin \theta ( a (\cos \theta + \sin \theta) - w ) = \cos \theta ( b (\cos \theta + \sin \theta) - w ) $

Which simplifies to

$ w \cos \theta - w \sin \theta + (a - b) \sin \theta \cos \theta + a \sin^2 \theta - b \cos^2 \theta = 0 $

Solving this equation numerically, we get $\theta$, and from which $ x $ follows.