Geometry of a stool leg

Question

I am building a stool with cross-braced legs and I am trying to figure out how to properly compute the size of the sides. I made the schema below:

• ABCD is a rectangle of length L and width l;
• the inner shape is the leg, so GH is parallel to EF;
• w is the width of the leg.

I am looking for a and b such as: BG = BE = b and DH = 2xDF = 2a.

My intuition is that for a given L, l and w, there should a unique a and b that satisfy the constraints. I first tried to calculate the area of ABCD in two ways:

1. A_1 = L.l
2. A_2 = sum of the area of all the inner right triangles and rectangle.

Then solve for A_1 = A_2. But A_2 becomes rapidly quite hairy so I don't think it is the right approach. Any suggestion?

PS: for those curious about the stool, you can see it on this picture.

Answer 1

There are three relationships that must be satisfied. The first condition is that $\triangle FAE \sim \triangle GCH$. This implies $AF \cdot CH = AE \cdot CG$, or $$(\ell - a)(L - 2a) = (\ell - b)(L - b). \tag{1}$$ The second is a Pythagorean relationship, namely $$\begin{align} CH^2 + CG^2 &= (HJ + JG)^2 \\ &= \left(HJ + \sqrt{EG^2 - EJ^2}\right)^2 \\ &= \left(HJ + \sqrt{EB^2 + BG^2 - EJ^2}\right)^2, \end{align}$$ and if we denote $d = HJ = IE$, this becomes $$(L - 2a)^2 + (\ell - b)^2 = \left(d + \sqrt{2b^2 - w^2}\right)^2. \tag{2}$$ Similarly, the final relationship arises from right $\triangle FAE$: $$(\ell - a)^2 + (L - b)^2 = \left(d + \sqrt{5a^2 - w^2}\right)^2. \tag{3}$$ Taken together, $(1)$-$(3)$ is a system of three equations in the unknowns $a, b, d$, given the dimensions $L, \ell, w$. However, their solution is not straightforward. It leads to a high-degree polynomial in $a$ and $b$ that does not appear to have a simple closed-form solution. As such, a numerical method may be required. For instance, $L = 17$, $\ell = 9$, and $w = 1$ yields the approximate solution $a \approx 0.552646$, $b \approx 0.741606$.

Answer 2

The stool leg is bounded between two parallel lines that are separated by $w$. If $m$ is the slope of these lines ($m$ is negative), then

$ y = m x + K_1 $

Substitute the point $(L , b)$, this will give

$K_1 = b - m L $

For the bottom line, its equation is

$ y = m x + K_2 $

Substitute $(L - b , 0 )$. This will give

$ K_2 = m (b - L ) $

The distance between the two lines is $w$, so

$ w = \dfrac{ - m L + b - m (b - L) }{ \sqrt{1 + m^2} } = \dfrac{ b (1 - m) } { \sqrt{1 + m^2} }$

Note the the slope is related to $L, H, a, b $ as follows

$ m = \dfrac{ b - H }{L - 2 a} = \dfrac{ a - H }{ L - b } $

These are actually two equations:

$ m (L - 2 a) = b - H $

and

$ (b - H) (L - b) = (L - 2 a)(a - H) $

These two equations are quadratic in $a, b, m $, however the first, when squared to remove the square root becomes

$ w^2 (1 + m^2) = b^2 (1 - m)^2 = b^2 (1 - 2 m + m^2 ) = b^2 - 2 b (bm) + (bm)^2 $

we need to get rid of the $(mb)$ term, and for that, note that

$ m = \dfrac{ a - H }{ L - b } $

Therefore,

$ m L - m b = a - H $

i.e.

$ m b = m L - a + H $

Substitute this, you get

$ w^2 (1 + m^2) = b^2 - 2 b ( m L - a + H ) + (m L - a + H)^2 $

Now it is quadratic.

You can solve these equations using Mathematica Online, or using SAGE

I solved using This SAGE worksheet, where $L, H, w$ are as in the example given by heropup, i.e. $L = 17$ , $ H = 9 $ and $ w = 1$. As is shown on the worksheet, the only valid solution is $ a = 0.552646 $ and $ b = 0.741606 $ and $ m = -0.51957 $.

As another example, suppose $L = 10, H = 20, w = 4$, then the solution is $a = 1.798091, b = 3.102028, m = -2.63873 $

Answer 3

With Analytic Geometry.

Calling

$$ \cases{ p_1 = (0,h-a)\\ p_2 = (L-b,0)\\ q_1 = (2a,h)\\ q_2 = (L,b) } $$

we have two lines

$$ \cases{ L_1\to p = p_1 + \lambda_1 \vec v_1\\ L_2\to q = q_1 + \lambda_q \vec v_2\\ \vec v_1 = \frac{p_2-p_1}{\|p_2-p_1\|}\\ \vec v_2 = \frac{q_2-q_1}{\|q_2-q_1\|}\\ } $$

and as $L_1$ and $L_2$ are parallel, we have:

$$ \cases{ \vec v_1\times\vec v_2 = \vec 0\\ (p_1-q_1)\cdot \vec v_1^{\top} = w \ \ \text{where}\ \ \ \vec v_1^{\top}\ne \vec 0,\ \ \ \vec v_1^{\top}\cdot\vec v_1 = 0\\ } $$

Now, as

$$ \cases{ \vec v_1 = \frac{1}{\sqrt{(h-a)^2+(L-b)^2}}(L-b, a-h)\\ \vec v_2 = \frac{1}{\sqrt{(h-b)^2+(L-2a)^2}}(L-2a,b-h) } $$

then

$$ \cases{ \vec v_1\times \vec v_2 = \vec 0\Rightarrow 2 a^2-a (2 h+L)+b (L-b+h)=0\\ (p_1-q_1)\cdot \vec v_1^{\top} = w\Rightarrow \frac{a (2 a+b-2 h-L)}{\sqrt{(h-a)^2+(L-b)^2}}+w=0 } $$

so we have to solve for $\{a,b\}$

$$ \cases{ 2 a^2-a (L+2 h)+b (L-b+h)=0\\ \frac{a (2 a+b-2 h-L)}{\sqrt{(h-a)^2+(L-b)^2}}+w=0 } $$