What is the acceleration magnitude of a simple pendulum? I'm trying to find the acceleration vector of a simple pendulum. The vector is labeled a in this image from Wikipedia:

Trying to solve it, I've split the acceleration into perpendicular components: radial (along the string) and tangential (along the velocity vector). I've found that:
$a_\textrm{tangential} = d \ddot\theta$
$a_\textrm{radial} = d \dot\theta^2 + g \cos\theta$
where d is the length of the rod and g the acceleration from gravity. However, combining these two causes an awful expression, which makes me think that I've got something wrong:
$|a| = \sqrt{ a_\textrm{tangential}^2 + a_\textrm{radial}^2 } = \sqrt{ d^2 \ddot\theta^2 + ( d \dot\theta^2 + g \cos\theta )^2 }$
What is the actual magnitude of the acceleration vector of a simple pendulum?
My goal is to find an expression for the acceleration magnitude ($\sqrt{a_x^2+a_y^2+a_z^2}$) as measured by a three-axis accelerometer inside the bob.
Edit: I just realised that since $\ddot\theta = -\frac{g}{d}\sin\theta$, we can simplify:
$|a| = \sqrt{ d^2 \ddot\theta^2 + ( d \dot\theta^2 + g \cos\theta )^2 } = \sqrt{ g^2 + d^2\dot\theta^4 + 2dg\dot\theta^2 \cos\theta }$
 A: If $l$ is the pendulum length the radial acceleration is simply the centripetal acceleration
$$a_\text{rad}=l\dot\theta^2.$$
The tangential acceleration is, as you say,
$$a_\text{tan}=l\ddot\theta.$$
The expression for the resultant acceleration is not too bad if you are using the usual small angle approximations to treat the pendulum.
A nice little bit of trivia: for small angles:
Maximum tangential acceleration = $g\left(\frac Al\right)$
But maximum radial acceleration = $g\left(\frac Al\right)^2$
In which $A$ is the amplitude, measured as an arc length.
You ought not to have included $g\cos\theta$ in your expression for $a_\text{rad}$. In this context, $g$ is not an acceleration, but the gravitational field strength. So the radial force component on the pendulum bob is (Tension in thread – $mg \cos \theta$).
A: the position vector to the mass point is
$$\mathbf R=\begin{bmatrix}
  x(t) \\
  y(t) \\
\end{bmatrix}=\left[ \begin {array}{c} d\sin \left( \varphi  \left( t \right) 
 \right) \\ d\cos \left( \varphi  \left( t \right) 
 \right) \end {array} \right] 
$$
from here you obtain the acceleration $\mathbf{\ddot{R}}$
$$\mathbf{\ddot{R}}=\begin{bmatrix}
  \ddot x(t) \\
  \ddot y(t) \\
\end{bmatrix}=
\left[ \begin {array}{c} d\cos \left( \varphi  \right) 
\ddot\varphi -d\sin \left( \varphi  \right) {\dot\varphi }^{2}
\\ -d\sin \left( \varphi  \right) \ddot\varphi -d
\cos \left( \varphi  \right) {\dot\varphi }^{2}\end {array} \right]
$$
substitute $\ddot\varphi=-\frac{g}{d}\,\sin(\varphi)$ from the equation of motion and obtain the magnitude $~a=\sqrt{\ddot x(t)+\ddot y(t)}$
$$a=\sqrt { \left( \sin \left( \varphi  \right)  \right) ^{2}{g}^{2}+{d}^{
2}{\dot\varphi }^{4}}
$$
