The diagram above is of a lens that has +1.00 diopters of power on the 180 or horizontal meridian, and +2.00 diopters of power on the 90 or vertical meridian. If you look at this lens in the focimeter (lensometer) you will get a nice clear image at +1.00D and another nice clear image at +2.00D, and in-between you will get fuzzy images of varying amounts. Below +1.00D or above +2.00D you will get fuzzy images that will rapidly become no image, just blur.

Remember the images that we got with a toric lens when we were talking about the focal length? For a refresher click here.

For our lens with one power of +1.00D and another power of +2.00D, there are two focal lengths where the image will be clear but distorted; one where it is out of focus but not distorted, and a range of distances where it is both fuzzy and distorted. There will be times, especially when we are talking about prism, that it will be helpful to determine the approximate power at a meridian other than the major meridians.  If the power at other meridians was actual power, we would have sharp focal lines. We do not have sharp focal lines anywhere except for the two major meridian powers. The powers that we are going to compute are useful theoretically, but they are NOT ACTUAL POWERS.

Lets look at the soda can that we used a few weeks ago. If you cut this can in half horizontally you get a perfect circle. If you cut it vertically you get straight lines. If you cut it diagonally you get something that is between the circle and the straight line: it will never have MORE curvature than the circle, it will be 'between' the circle and the line.
 
 

Suppose that the surface to the left has +1.00 D of power on the horizontal and +2.00D of power on the vertical. The cuts between these two powers are going to have curvature between +1.00 and +2.00, but they are not circles, so they are not perfect power. They will not have less curvature than +1.00 nor will they have more curvature than +2.00. The curves will vary between the major curves. But since they are not circular, they will be approximate curves.

Yes?

So, we can approximate the power on the lens in any meridian, not just the major meridians. There is a general rule of thumb for approximate powers. It starts with two facts that you already know:

What does this mean? 0 degrees away from the axis of the Rx is where there is only sphere power. When we have the Rx of +1.00 +1.00 x 180, on the 180 meridian, which is 0 degrees away from the axis of 180, none of the cylinder power is present. We have only sphere power on this meridian: the power on the 180 is +1.00 + (0)(+1.00) = +1.00D.

90 degrees away from the axis of the Rx +1.00 +1.00 x 180 is on the 90th meridian. On the 90th meridian we have all of the cylinder power of +1.00. So on the 90th we have +1.00 + (1)(+1.00) = +2.00D power. This is exactly the same thing that we learned when we talked about compound lenses.

Now we will add a third line to this rule: 1/2 way between the axis and 90 degrees away from the axis will have 1/2 of the cylinder power. So,

45 degrees away from the axis of 180 is 45. 45 degrees away from the axis of 180 is also 135. On both the 45 meridian and the 135 meridian, we have 1/2 of the cylinder, or (.50)(+1.00) = +0.50D of cylinder added to the sphere. So, on the 45 meridian and on the 135 meridian, we have +1.00 +0.50 = +1.50D of power.

Our next steps are at 30 and 60 degrees away from the axis. These are NOT halfway steps from 0 to 45.

On the 30th meridian we have (.25)(+1.00) = +0.25 cylinder power to add to the sphere power. So we have +1.00 + (0.25)(+1.00) = +1.25D power. Same for the 150 meridian. The 150 is 180 - 150 = 30 degrees away from the axis of 180, so there is (0.25)(+1.00) cylinder to add to the +1.00 sphere, for +1.25D of power.

On the 60th meridian we are 60 degrees away from the axis, so we have the sphere plus .75 times the cylinder: +1.00 + (0.75)(+1.00) = +1.75D of power.

Lets look at another one. Rx -4.00 -2.00 x045.

On the axis 45 we have 0 times the cylinder power plus the sphere power: -4.00 + (0)(-2.00) = -4.00D. Right? This you already knew.

On the 135 meridian, which is 90 degrees away from the axis of 45 we have 1 times the cylinder power plus the sphere power. -4.00 + (1)(-2.00) = -6.00D. This you also already knew.

45 degrees away from the axis of 45 is 90 in one direction and 0 degrees in the other direction. So, on the 0/180 meridian and also on the 90 meridian we have 0.5 times the cylinder power plus the sphere power. -4.00 + (0.50)(-2.00) = -5.00D of power.

30 degrees away from the axis of 45 is 15 degrees and also 75 degrees. So, on the 15 degree axis and also on the 75 degree axis we have 0.25 times the cylinder plus the sphere. -4.00 + (0.25)(-2.00) = -4.50D of power.

That is the general 'rule-of-thumb' that you will need when you take the ABO exam, where calculators are not allowed. Next we are going to use the formula, where you will be able to find the approximate power on ANY meridian. This you will need throughout this course. You can always refer back to the 'rule of thumb' above in order to see if the answers you are getting are reasonable.

What we did above is say "the sphere power is in effect all over the lens, a portion of the cylinder is in effect at any meridian depending on how far the meridian is from the axis."

So in our formula we will find the portion of the cylinder that is in effect, and add that to the sphere power.

To do this, we need to find how far the axis we need is from the axis of the Rx, just as we did in the rule of thumb above. Lets take an example:
 

1. Rx -5.00 -4.00 x 142. Find the approximate power on the 90th meridian.
• The first step is to find out how far the meridian you want is from the axis. Subtract:  142 - 90 = 52 degrees. It does not matter whether you subtract the meridian from the axis or the axis from the meridian.
• The next step is to find the sine of the angle of the difference, in this case, sin 52. Remember that key on your calculator that I said you had to have? So far you have punched into your calculator 142 - 90 =. Now you punch the key labelled "sin". It should read '0.7880 . . .'. If it still shows 52, punch "=". If it shows something other than '0.7880 . . . ' then your calculator is not in "degree mode" and you need to get out the directions and find out how to change it. If you have a  problem with this call your instructor.  If you have no idea what I am doing here refer back to the trigonometry part of the first module of this course.
• Punch the "x" key. You should now have '0.6209 . . .' showing on your calculator. This is the decimal amount of the cylinder that is in effect 52 degrees away from the axis of the Rx.
• Punch x 4.00  =, which is the amount of the cylinder in the Rx. You now have  '-2.4838 . . .', which rounds to -2.48D of cylinder.
• Punch + 5.00  =. The display should now show '-7.48', which is the approximate lens power on the 90th meridian.

Now, first, if you had no idea what I was doing with that "sin" stuff, it is short for 'sine', it is pronounced like 'sign', it is done with angles, and you need to go to the trigonometry link above and read.  Learn all three of functions: sine, cosine and tangent, because in the next several weeks we will use all of them. Doing the exercises at the end of that lesson would be a really good idea.

Notice that I did not round the results to 1/8 diopter steps. This is not a power that you will order from the lab or record in the glasses wearer's records. It is an approximation that is found as a step to finding out something else. We do not round the oblique meridian power to eighths.

Here are the steps to follow to do this formula:

Take the difference between the axis and the meridian asked for.  [It does not matter which you subtract.]	axis - meridian  or  meridian - axis
Find the sine of this angle.	sin
Square the result.	x
Multiply times the cylinder amount.  Be sure to press the  key if the cylinder is minus.	x cylinder
Add the amount of the sphere.  Be sure to press the  key if the sphere is minus.	+ sphere

Lets do a few more of those.

2. Rx +3.25 -2.50 x 036. Find the approximate power on the 180th meridian.
• The first step is to find out how far the meridian you want is from the axis.  Subtract:  180 - 36 = 144 degrees. It does not matter whether you subtract the meridian from the axis or the axis from the meridian.
• The next step is to find the sine of the angle of the difference, in this case, sin 144. The 144 is already showing in your calculator. Punch "sin". If it does not show '0.5877 . . .   'then punch "=". If it does not show '0.5877 . . . " now the calculator needs to be switched to degree mode. Read the directions for the calculator or call your instructor.
•  Punch the x key. You should now have '0.3454 . . .' showing on your calculator. This is the decimal amount of the cylinder that is in effect 144 degrees away from the axis of the  Rx.
• Punch x 2.50  = which will give the amount of the cylinder on the 180 meridian.  You have now '-0.8637 . . .', which rounds to -0.86D of cylinder.
• Punch + 3.25 =. The display should now show '2.386 . . .' which rounds to +2.39D, which is the approximate power on the 180th meridian. You did not punch the  key because the sphere was positive, not negative.

3. Rx -4.00 -2.00 x045.  Find the approximate power on the 180th meridian.
•  The first step is to find out how far the meridian you want is from the axis.  Subtract: 180 - 45 = '135' degrees. It does not matter whether you subtract the meridian from the axis or the axis from the meridian.
• The next step is to punch "sin", and "=" if you have to. You should now have '0.7071 . . .' showing on your calculator.
• Punch the x key. You should now have '0.5' showing on your calculator. This is the decimal amount of the cylinder that is in effect 135 degrees away from the axis of the Rx.
• Punch x 2.00  = which will give the amount of the cylinder on the 180 meridian.  You have now '-1', which is -1.00D of cylinder.
• Punch + 4.00  =. The display should now show '-5.' which rounds to -5.00D, which is the approximate power on the 180th meridian.


One final example.
 

4. Rx +2.50 +1.75 x115. Find the approximate power on the 38th meridian.
• Subtract:  115 - 38 = '77' degrees.
• Punch sin, and = if you have to. You should now have '0.9743 . . .' showing on your calculator. This is the decimal amount of the cylinder that is in effect 77 degrees away from the axis of the Rx.
• Punch the x key. You should now have '0.9493 . . .' showing on your calculator. This is the decimal amount of the cylinder that is in effect 77 degrees away from the axis of the Rx.
• Punch x 1.75 = which will give the amount of the cylinder on the 38 meridian. You have now '1.6614 . . .' which is +1.66D of cylinder.
• Punch + 2.50 =. The display should now show '4.16 . . ..' which rounds to  +4.16D, which is the approximate power on the 38th meridian.


You did not punch the key at all for this problem because the sphere and cylinder were both  positive, not negative.



OK, now read the pages (62-63 / 75-77) in Optical Formulas Tutorial for the description of the formula, make sure that you understand the examples there, do the exercises on page (63 / 77), and check your answers in the back of the book. Then do these for more practice, and click on the link to see what our answers should have been.

EXERCISES:

5. What is the approximate power on the 180 meridian for the Rx +10.75 -2.00 x130?
6. The Rx is -5.50 +3.50 x 100. What is the approximate power on the 90th meridian?
7. What is the power on the vertical meridian for the Rx -15.00 -3.00 x 055?
8. If the Rx is -8.25 -1.25 x060, what is the approximate power on the 15th meridian?
9. What is the approximate power on the horizontal meridian for the Rx +1.00 -2.50 x 155?