arc lengthcirclesgeometrytrigonometry
I am struggling to find help on how I can calculate x given I only have the arc height (h) and arc length L (please see below diagram).
I understand that if I have the arc width/chord length w, then I can work out r by:
r= (h/2) * (w²/8h)
which then may aid my quest to calculate x, but I am also struggling to calculate that. Any help would be most grateful!
Using the notation of the figure you have linked to, we have
\begin{equation} R \sin \frac{\theta}{2} = \frac{a}{2} \end{equation}
we can also write
\begin{equation} \theta = \frac{s}{R} = \frac{2 s \sin \theta/2}{a} \end{equation}
From this equation, you can solve for $\theta$.
Once you have solved for $\theta$, you have
\begin{equation} h = R - R \cos(\theta/2) \end{equation}
Since $R = a/(2 \sin \theta/2)$, we have
\begin{equation} h = \frac{a}{2 \sin \theta/2} \left( 1 - \cos\left(\frac{s \sin\theta/2}{a}\right)\right) \end{equation}
Assuming that your hole is indeed circular, it can be done like this.
Of course we'll call the radius $r$. Also, I'm using $w$ as half the measured width. In this diagram, $$ \sin \theta = \frac{w}{r} \quad\quad\quad \cos\theta = \frac{r - h}{r} $$
One way to combine these expressions is by using the identity $$ \sin^2\theta +\cos^2\theta = 1 $$
After some simplification, this produces $$ r = \frac{w^2 + h^2}{2h} $$
We can also find the angle $\theta$ now $$ \sin \theta = \frac{w}{r} \\ \theta = \arcsin\frac{w}{r} \\ \theta = \arcsin\frac{2hw}{w^2+r^2} $$
Combining these together we can find the arclength $l$ $$ l = r(2 \theta) \\ l = 2 \frac{w^2 + h^2}{2h} \arcsin\left(\frac{2hw}{w^2 + h^2}\right) \\ l = \frac{w^2 + h^2}{h} \arcsin\left(\frac{2hw}{w^2 + h^2}\right) $$
For $h = 5$ and $w = 11$, $l = 24.915...$
Best Answer
With respect to chord AB of the circle O, POQ is a diameter perpendicular to it (cutting it at R). Let $\angle RBQ = \alpha$.
It should be clear that $\angle P = \alpha$ and the red marked angles are all equal to $2 \alpha$.
First, we have to find the relation between arc AB’s length (L) and chord AB’s length (2s) when they both subtend the same central angle $\angle AOB = 4\alpha$:-
If r is the radius of that circle, then $L = 4r \alpha$, and $2s = 2(r \sin (2 \alpha))$
Eliminating r from the above, we have $s = \dfrac {L \sin (2\alpha)}{4 \alpha}$
Since $\tan \alpha = \dfrac {h}{s}$, we have $L \sin (2 \alpha) \tan \alpha – 4 h \alpha = 0$
It can be simplified slightly to $L \sin^2 \alpha – 2 h \alpha = 0$, but is still transcendental. Seek help from WolframAlpha to find $\alpha$. Once it is known, the rest is easy.