How to find the tangent length of a tear drop of a given perimeter

Solution 1:

In this answer, we give an equation, which expresses $BC$ in terms of $L$ and $r$. Please note that this equation is valid if and only if $L\ge\left(4+\pi\right)r$. However, the accuracy of of the value of $BC$ is not that good for $L\le 8\pi r$. Because we do not have much experience in AutoCAD, the expression for $BC$ is given in two different formats, hoping at least one of them can be used to create an AutoCAD statement with the help of supported math functions. We would like to emphasize that if you decide to use the results obtained from those formulae in a design or a construction, you should do it at your own risk.

We provide you with a graph showing the variation of the percentage relative error with the perimeter $L$ of the tear-drop. According to this graph, the accuracy of the value of $BC$ increases drastically with the increasing $L$. But, we do not know for sure whether this accuracy suffices or not for your work, even in the mentioned range of $L$ (i.e. 20 to 30 feet).

Assuming that AutoCAD allows defining and calculating a dimensionless variable before using it in a formula, we first express $BC$ in terms of $L$ and $r$ concisely as shown below. $$BC=\left(1+\frac{1}{\phi^2}\right)\left(\frac{L}{2}+r\Big\{\tan^{-1}\left(\phi\right)-\pi\Big\}\right)-\frac{r}{\phi}$$ $$\text{where}\qquad\enspace \phi = \frac{\left(L-\pi r\right)+\sqrt{\left(L-\pi r\right)^2-16r^2}}{4r}\quad$$

If that is not possible, we need to resort to the following lengthy formula to determine $BC$ in one go. $$BC = \left(1+\frac{8r^2}{\left(L-\pi r\right)^2-8r^2+\left(L-\pi r\right)\sqrt{\left(L-\pi r\right)^2-16r^2}}\right)$$ $$\qquad\qquad\enspace\mathbf{\LARGE{\times}}\left(\frac{L}{2}+r\Biggl\{\tan^{-1}\biggl(\frac{\left(L-\pi r\right)+\sqrt{\left(L-\pi r\right)^2-16r^2}}{4r} \bigg) -\pi\Biggr\} \right)$$ $$\mathbf{\LARGE{-}}\frac{4r^2}{\left(L-\pi r\right)+\sqrt{\left(L-\pi r\right)^2-16r^2}}\qquad\qquad\qquad\qquad$$

$\bf{\text{Derivation of the Formula}}$

We start by setting up the corresponding Newton-Raphson formula. Note that we use $x$ in place of $BC$ throughout our description. We define a function $f\left(x\right)$ using the $6^{\text{th}}$ equation given in OP’s problem statement. $$f\left(x\right)=x+r\left\{\pi-\tan^{-1}\left(\cfrac{x}{r}\right)\right\}-\frac{L}{2} \tag{1}$$

The first derivative of the above function is given by, $$f’\left(x\right) = \cfrac{x^2}{r^2+x^2}.\tag{2}$$

Using (1) and (2) we can set up the following Newton-Raphson formula. $$x_n = x_{n-1} -\cfrac{ f\left(x\right)}{ f’\left(x\right)}=\left(1+\cfrac{r^2}{x_{n-1}^2}\right)\Bigg(\frac{L}{2}+r\bigg\{\tan^{-1}\biggl(\cfrac{x_{n-1}}{r}\biggr)-\pi\bigg\}\Bigg)- \frac{r^2}{x_{n-1}} \tag{3}$$

We found out that this formula converges rapidly. If we could find an educated guess for $x_{\text{n-1}}$, which could deliver a very accurate value for $x_\text{n}$, when substituted in (3), we are in business.

$\bf{\text{Finding an Educated Guess}}$

Now, we need to rewrite (1) in the following format and, then, express the $\text{tan}^{-1}\text{-term}$ using one of the series expansions given below it.

$$x_\text{n-1}=\cfrac{L}{2} - r\left\{\pi-\tan^{-1}\left(\cfrac{x_\text{n-1}}{r}\right)\right\} \tag{4}$$

$$ \tan^{-1}\left(\cfrac{x_\text{n-1}}{r}\right) = \cfrac{\pi}{2} - \cfrac{r}{x_\text{n-1}} + \cfrac{r^3}{3x_\text{n-1}^3} - \cfrac{r^5}{5x_\text{n-1}^5} + \ldots\quad \text{where}\quad x_\text{n-1} \ge r \tag{5a}$$

$$ \tan^{-1}\left(\cfrac{x_\text{n-1}}{r}\right) = \cfrac{x_\text{n-1}}{r} - \cfrac{x_\text{n-1}^3}{3r^3} + \cfrac{x_\text{n-1}^5}{3r^5} + \ldots\quad \text{where}\quad x_\text{n-1} \le r \tag{5b}$$

If $L \ge \left(\dfrac{3}{2}\pi+2\right)r$, we know that $ BC\ge r$. Therefore, we use (5a) to replace $\text{tan}^{-1}\text{-term}$. Since we are looking for a guess, we retain the first two terms of (5a) and discard the rest to obtain, $$ \tan^{-1}\left(\cfrac{x_\text{n-1}}{r}\right) \approx \cfrac{\pi}{2} - \cfrac{r}{x_\text{n-1}}. $$

We plug this approximation of $\space\tan^{-1}\left(\cfrac{x_\text{n-1}}{r}\right)\space$ in (4) to get the following quadratic equation in $x_{\text{n-1}}$. $$2x_\text{n-1}^2 - \left(L-\pi r\right)x_\text{n-1}+2r^2=0$$ This equation has two positive real roots, i.e. $$x_{\text{n-1}}=\frac{\left(L-\pi r\right)\pm \sqrt{\left(L-\pi r\right)^2-16r^2}}{4}.$$

The larger of the two roots is selected as the educated guess to be inserted into (3). For the sake of convenience, we express this in the following form, where $\phi$ is a dimensionless quantity. $$\phi = \dfrac{ x_{\text{n-1}}}{r} =\frac{\left(L-\pi r\right) + \sqrt{\left(L-\pi r\right)^2-16r^2}}{4r} \tag{6}$$

Please note that (6) introduces a constrain $L\ge \left(4+\pi\right)r$ on the range of validity of $\phi$. Because of that, the range of validity of the equation given below for $BC$ also shrinks from $L \ge \left(2+\dfrac{3}{2}\pi\right)r$ to $L\ge \left(4+\pi\right)r$.

To finish the derivation, let us replace all $\dfrac{x_{\text{n-1}}}{r}\text{-terms}$ of (3) with $\phi$.

$$BC= x_n = \Bigg(1+\frac{1}{\phi^2}\Bigg)\Bigg(\frac{L}{2}+r\bigg\{ \tan^{-1}\left(\phi \right)-\pi\bigg\}\Bigg)- \frac{r}{\phi} \tag{7} $$

Although (7) is valid for the range $L\ge\left(4+\pi\right)r$, probably, the values of $BC$ calculated using (7) are accurate enough for you only in the range $ L\ge 8\pi r$. It is possible to derive equations expressing $\phi$ in terms of $L$ and $r$ for $L\lt\left(4+\pi\right)r$ as well, but they all fail to yield accurate-enough values for $BC$.

$$\small{\text{Relative Error} = \left|\frac{\text{Exact Value}\space \mathbf{\large{-}} \space\text{Computed Value}}{\text{Exact Value}}\right|}$$

The (quasi) exact values were determined numerically using Newton-Raphson method, which converges very fast for this particular scenario.