Astigmatism is quite a difficult concept to explain to patients and students. Searching over the internet is unlikely to lead to a satisfactory simple definition. In relatively simple terms, what it means is that the optical system focuses lines oriented in one direction at a different point to those in all other directions. In optics, these directions are called meridians. If the optical system is circular the meridians are analogous to the lines of longitude of the earth where the optical centre is the North Pole and the lines extend around the globe to the other pole.

In astigmatism, some of these lines are more tightly curved than others. The lines with the highest and lowest curvature are called the principal meridians and in regular astigmatism, these are always at 90 degrees to one another. An example of a shape which would have this effect when made as a lens is a barrel. Imagine that the optical centre is at one point on the widest point of the barrel. Lines drawn out at each clock hour would each have a different curve – the most curved would be the meridian which goes around the barrel and the least curved would be the meridian heading towards the top and bottom of the barrel. In its extreme form the barrel becomes a cylinder and the up and down meridian has no curve at all – this shape, as a lens, is termed a cylindrical lens.

You can immediately see from this analogy that the meridians can only take values from 0 to 180 degrees : 190 degrees is the same as 10 degrees.

In ophthalmology, when we consider astigmatism, we do so in one of two ways. An optical element can be described either as the sum of 2 cylinders such as 10D x 90 degrees + 20D x 0 degrees or as a sphere (normal lens) plus cylinder – in this case 10D + 10D x 0 degrees (called the sphero-cylindrical form). These forms are easy to interconvert and both completely describe the element provided that it displays regular astigmatism. Transposing the spherocylindrical form is a trivial task- in this case it is 20D – 10D x 90 degrees. Just remember that the amount of cylinder always stays the same.

The spherical equivalent of a sphero-cylindrical lens is equal to the sphere plus half the cylinder. This is the best approximation to the lens by a spherical lens only As a basic check of your arithmetic remember that when you transpose the cylinder, the spherical equivalent always stays the same. Also, if you add thin lenses in contact with one another, the spherical equivalents always add algebraically.

Addition and subtraction of thin lenses in contact with one another is easy if the principal meridians of the astigmatism are aligned. In that case, the spheres and cylinders just add together. For example, take two lenses

• 0 + 1 x 90 degrees
• 3 +2 x 0 degrees

transposing the second one we get

• 5 -2 x 90 degrees

adding together now is

• 5 -1x 90 degrees

checking the spherical equivalents we have 0.5 + 4 from the combination and the result is also 4.5D

It gets much more tricky when you have to add lenses together that are not aligned You might guess that you can use vector addition from high school physics but try this :

• 0 + 1 x 90 degrees and
• 0 + 1 x 0 degrees

Vector addition would yield 1.4 by 45 degrees but the correct answer is 1D sphere.

To understand what is going on here, have a look at the top half of this diagram I have illustrated the addition of a 2D x 0 degrees (blue) and 2D x 45 degrees (red). Along the x axis is the meridians and the y axis shows the power for each meridian. The green line is the net power. Obviously if those sine waves were 0 degrees and 90 degrees they would be completely out of phase and add to a straight line (the sphere) of 2D. ; The matrix mathematics to solve directly is hinted at below.

Any lens can be considered to be a matrix like this and if multiple lenses are in contact, the matrix elements just add up. You then need to pull the three parameters out the matrix and solve for S, C and theta. This is basic first year university mathematics.