Effective strength of prescription glasses for arbitrary angle I am trying to understand the optics of prescription glasses.
Prescriptions have a spherical and a cylindrical (with associated axis) component specified in dioptres which are roughly additive. For illustrative purposes lets assume these are -2.0 diop spherical and -1.0 diop cylindrical at 90°.
From my understanding, for one axis (0°), only the spherical component is effective (-2.0 diop) since the cylindrical component has constant thickness in this direction. For the other axis (90°) the glasses have the effective strength of -3.0 diop since the cylindrical component is effective.
My question is: What is the effective strength for an arbitrary angle? Is this simply a sinusoidal relationship or is is more complicated?
 A: Blessedly, the diopter measurement used for corrective lenses is the same as the definition of curvature used in mathematics:
$$D = \frac{1}{R} $$
Where $D$ (denoted $\kappa$ in mathematical parlance) is the curvature measured in diopters and R is the radius of curvature. For reference, a flat surface has $R =\infty$ and $D=0$. Obviously a spherical lens has the same curvature at any angle. And a cylindrical lens will have a curvature that goes from $D\rightarrow [0,1/R]$ as the angle from the cylinder's axis $\theta \rightarrow [0,90^\circ ]$.
Curvature can be calculated, and need not be constant over a curve. Let's do this for a green curve across a cylinder's surface like shown:

We need to write down the parameterization (denoted $\gamma(\lambda)$ ) of the green line. It is:
$$\gamma(\lambda) = \begin{pmatrix} R \cos{\pi \lambda} \\ R \sin{\pi\lambda} \\ R \tan{\theta}(1-2\lambda) \end{pmatrix} $$
Where $\lambda \rightarrow [0,1]$. The curvature of a parameterized curve in $\mathbb{R}^3$ is: $$D = \frac{||\gamma' \times \gamma'' ||}{||\gamma' ||^3} $$
where $\gamma' = \frac{d\gamma}{d\lambda}$ and $\gamma''= \frac{d^2\gamma}{d\lambda^2}$, the first and second derivatives respectively, '$\times$' is the cross product, and $||...||$ is the norm (magnitude).
I handed this off to $\mathrm{Mathematica}$ at this point, and got the following for $D$:
$$D = \frac{1}{R+ \Large{4R \tan{\theta}^2 \over \pi^2 } } $$
So yes, the Diopter value does change for the cylindrical component for various angles (note it no longer depends on $\lambda$, meaning the green curves are constant in curvature given any $\theta$). A plot of this for $R=1$ shows:

You mentioned a $D=-1.0$ at $90^\circ$. I plotted this for positive $R$, so simply flip it over for the negative variety. 
