Length of Arc of Deltoid

Theorem
The total length of the arcs of a deltoid constructed within a stator of radius $a$ is given by:
 * $\mathcal L = \dfrac {16 a} 3$

Proof
Let $H$ be embedded in a cartesian coordinate plane with its center at the origin and one of its cusps positioned at $\left({a, 0}\right)$.


 * Deltoid.png

We have that $\mathcal L$ is $3$ times the length of one arc of the deltoid.

From Arc Length for Parametric Equations:


 * $\displaystyle \mathcal L = 3 \int_{\theta \mathop = 0}^{\theta \mathop = 2 \pi/3} \sqrt {\left({\frac{\mathrm d x} {\mathrm d \theta}}\right)^2 + \left({\frac{\mathrm d y} {\mathrm d \theta}}\right)^2} \mathrm d \theta$

where, from Equation of Deltoid:
 * $\begin{cases}

x & = 2 b \cos \theta + b \cos 2 \theta \\ y & = 2 b \sin \theta - b \sin 2 \theta \end{cases}$

We have:

Thus:

Thus:
 * $\sqrt {\left({\dfrac {\mathrm d x} {\mathrm d \theta} }\right)^2 + \left({\dfrac {\mathrm d y} {\mathrm d \theta} }\right)^2} = 4 b \sin \dfrac \theta 2 \left|{1 + 2 \cos \theta}\right|$

In the range $0$ to $2 \pi / 3$, $1 + 2 \cos \theta$ is not less than $0$, and so:
 * $\displaystyle \mathcal L = 3 \int_0^{2 \pi / 3} 4 b \sin \dfrac \theta 2 \left({1 + 2 \cos \theta}\right) \, \mathrm d \theta$

Put:
 * $u = \cos \dfrac \theta 2$

so:
 * $2 \dfrac {\mathrm d u} {\mathrm d \theta} = -\sin \dfrac \theta 2$

As $\theta$ increases from $0$ to $\dfrac {2 \pi} 3$, $u$ decreases from $1$ to $\dfrac 1 2$.

Then:

Substituting:
 * $2 \dfrac {\mathrm d u} {\mathrm d \theta} = -\sin \dfrac \theta 2$

and the limits of integration:
 * $u = 1$ for $\theta = 0$
 * $u = \dfrac 1 2$ for $\theta = \dfrac {2 \pi} 3$

we obtain, after simplifying the sign: