This is one of the more important concepts you should know for your ABO and one that is difficult because we don’t use it all the time. We transpose prescriptions from plus to minus cylinder. We calculate minimum blank size. We use Prentice’s Rule during final inspection. We don’t, however, tend to calculate the value of a 30° cylinder at axis 90° every day. It just rarely comes up. Yet we need to know it, and not just for our optician Certification.

# Calculating cylinder power in any meridian

Believe it or not it’s not as hard as it sounds. It’s best to start with the basics. Let’s take a simple Rx: **-1.00 -2.00 x 180**.

The sphere power (-1.00) corresponds to the axis (180) which means that on the 180° meridian the power of the Rx is -1.00. No part of the cylinder is felt at this meridian at all. It is the *axis* of the cylinder, not the meridian of that cylinder. The meridian at which that cylinder if felt is 90° away from that axis.

When we transpose that prescription to plus cylinder: **-3.00 +2.00 x 90 ** we see that the new sphere power includes ALL of the cylinder -1.00 +(-2.00). The cylinder and the sphere combined make up the *total power *of the Rx at 90°.

But what happens in between 90° and 180°? Obviously, the further away from the 90° meridian you go, the less the cylinder is felt. The closer you go to the 90° meridian, the more the cylinder is felt. If you go *half way *between 90° and 180° you should feel *half* of the cylinder’s power. Right?

Right!

Half of 90° would be 45°. At axis 45° the **-2.00 **cylinder power of this Rx should be **-1.00**. The total power at axis 45° should be -1.00 +(-1.00) or **-2.00.**

So, now we can calculate half. But what about ¼ or ¾?

Well, instead of dividing 45° in half again (which doesn’t divide neatly) we think about geometry and all those triangles we had to work with. You remember those triangles don’t you? I’m sure you do… You have the concept of the 45-45-90 triangle, which we already used to figure out half of 90°, and then the 30-60-90 triangle, which will give us those other fractions.

One quarter (¼) of the cylinder’s power is felt 30° away from its axis. If our hypothetical Rx has a -2.00 cylinder at axis 180° only ¼ of that cylinder would register 30° away from that axis. At axis 30° the power of the cylinder should be -2.00 ÷ 4, or **-0.50**. The total power of the Rx at 30° would be -1.00 (the sphere) + (-0.50), or **-1.50**.

Three quarters (¾) of the cylinder’s power is felt 60° away from its axis. That -2.00 cylinder at 180° we’ve been working with would show ¾ of its power 60° away from 180°. That would be **3 times** the ¼ that we’ve already found (-0.50) or **-1.50**. The total power of the Rx at 60° would be -1.00 (the sphere) + (-1.50), or **-2.50**

Cool, huh? But wait! There’s more… What happens when you **over **90°? I mean, there may be situations where you might want to do that. (really!) What if you want the power of the cylinder at axis **120°**, or **135°**, or even **150°**?

Well, the concept is the same. In this Rx the full cylinder is felt at axis 90°. So if you go over 90° you are still going towards 180°! The farther towards 180° you go, the less cylinder power there is going to be.

If you go 30° higher than 90° ( 120°)it’s the same as being 60° away from 180°. It’s ¾ of the cylinder power.

If you go 45° higher than 90° (135°) it’s the same as being 45° away from 180°. It’s ½ of the cylinder power.

If you go 60° higher than 90° (150°) it’s the same as being 30° away from 180°. It’s ¼ of the cylinder power.

## A formula for calculating power in any meridian

There is a mathematical formula for doing this too, and it comes in handy. It may be hard to reason the answer out every time if the Rx doesn’t come with a convenient 180° or 90° axis. If you are given an Rx of **+5.00 -2.00 x 70** and you are asked to find the total power at axis **130°** a formula might be just what you want. Here it is in a nutshell:

** Dc´= Ds + Dc (sin²θ)**

**Dc´= the power in the meridian in question **(in this case it is 130°)**
Ds = the sphere power of the Rx **(+5.00)

**Dc = the cylindrical power**(-2.00)

**θ = the difference between the axis of the Rx and the meridian in question**(130° – 70° = 60°)

A neat little table would be really useful about now:

θ | sin² θ | θ |
---|---|---|

30° | ¼ | 150° |

45° | ½ | 135° |

60° | ¾ | 120° |

**D _{130 }**

sin²θ = ¾

_{=}+5.00 -2.00 (sin²θ) θ = 130 – 70 = 60sin²θ = ¾

** D _{130 }= +5.00 + (-2.00)(¾)**

** D _{130 }= +5.00 + (-1.50)**

** D _{130 }= +3.50**

For the Rx **+5.00 -2.00 x 70**, the power in the 130° meridian is **+3.50**.** **

“That would be 3 times the ¼ that we’ve already found (-0.50) or -1.75. The total power of the Rx at 60° would be -1.00 (the sphere) + (-1.75), or -2.75.”

**the total power would be -2.50, as 3 times -0.50 is -1.50, not -1.75.

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