Part a -- Prism notation.

We have been talking about prism with the base up, down, in or out. When we talked about horizontal imbalance from the glasses optical centers not being correct we talked about prism base in and out even though we know that in many glasses the eyes are above where the optical centers should be. When we talked about vertical imbalance in anisometropia we talked about prism base up or down, even though we know that the eyes converge from the distance pupillary distance when we read. When we are talking about UNWANTED prism we generally use just up, down, in or out because we need only determine if a problem exists as a result of poor optical center placement.

For PRESCRIBED prism the prism base will very likely be in other directions. The base of a prism can be anywhere on the lens, not just on the 90th or 180th meridians.

There are two ways that we can describe where the prism base is to be. You saw an example of the first way of showing prism base direction when we talked about splitting prism. If we were given permission to split the prism for the Rx
        pl 3BU
        pl 4BI
we would end out with
        pl 1.5BU & 2BI
        pl 1.5BD & 2BI
How do we make this? How do we do final inspection on this?

The second method of showing prism base direction is going to look suspiciously like cylinder power, but with a twist. It will show the total amount of prism and give a meridian for the base direction. For the example above, it may end out looking like
        pl 2.5 @ 037
        pl 2.5 @ 217
or, just to make life more interesting, it could look like
        pl 2.5BU&I @ 037
        pl 2.5BD&I @ 037

We will start by discussing notation itself. I would like for you to read pages 120-124 in the Optical Formulas Tutorial. Just read. Do not worry about the formula yet, read for understanding of the notation.  If you have Brooks Understanding Lens Surfacing, read pages 5-10, left side.

OK, first I want to make sure that you understand the 180 vs 360 notation. Remember way back in the beginning of the first theory class we said that we do not use cylinder axis greater than 180 because what happens in the bottom of the lens is just the mirror image of what happens in the top of the lens? Well, with prism this is just not so. Prism base up at the 90th meridian will be very different from prism base down at the 270th meridian!

So, here are some exercises.  You will need the diagram on the top of page 121 in the textbook. (Well you don't HAVE to use it; draw one out yourself is an even better idea.)

Write in 360 notation, and tell me if it is OD or OS:

1. 6.7 BU&I @ 077
1. Since this is an UP and something we do not adjust the axis of the prism base, so it is?
77 degrees
2. Look at the diagram on page 121. What quadrant is 77 degrees in?
It is in quadrant I.
3.  For which eye is quadrant I U&I?
Quadrant I is U&I for the OD.
 So, this is a right lens. So, the answer is:
1. OD 6.7 @ 077.

 
2. 3.5 BU&I @ 153
1. If it is base down and something add 180 to the axis.  If it is base up and something do not.
2. Look at the diagram on page 121, or better yet, draw it for yourself.  What quadrant is the axis in?
3. Which eye is that quadrant for the combination of bases in the original problem?

 
3. 3.75 BD&I @ 127

Use the same steps.
1. Since this is a DOWN and something we need to have an axis that is between 180-360, so we have to add 180 to the base direction. This gives us what?
127 + 180 = 307 degrees.
2. Look at the diagram on page 121. What quadrant is 307 degrees in?
It is in quadrant VI.
3.  For which eye is quadrant VI D&I?
Quadrant IV is D&I for the OD.
So, this is a right lens. So, the answer is?
1. OD 3.75 @ 307.

 
4. 4.6 BD&I @ 029
 The steps are:
1. If it a DOWN and something, add 180 to the meridian.. This is the new meridian. If it is an UP and something write the meridian as is.
2. Which quadrant is the meridian in?
3. For which lens does that quadrant have the given direction?

 
5. 7.6 BD&O @ 032

 
6. 7.1 BD&O @ 113

 
7. 2.8 BU&O @ 100

 
8. 6.5 BU&O @ 004

 
9. 5.0 BI&U @ 045

 
10. 5.0 BI&D @ 053

 
11. 4.1 BI&D @ 014

 
12. 1.8 BI&U @ 056

 
13. 1.5 BU @ 090

 
14. 3.0 BI @ 180

 

 
 

1. OD 6.7 @ 077
1. OK, in 360 notation you have the meridian precisely where it actually is. So, what quadrant is 77 degrees in?
quadrant I.
2.  For the right lens, what direction is quadrant I?
 it is U&I.
 So, the result is:
1.  [OD] 6.7 BU&I @ 077.

 
2. OS 3.5 @ 153
1. Since there is no base direction showing this is in 360 notation and the meridian precisely where it actually is. So, what quadrant is 153 degrees in?
quadrant II.
2.  For the left lens, what direction is quadrant II?
 It is U&I.
 So, the result is:
1.  [OS] 3.5 BU&I @ 153.

 
3. OD 3.75 @ 307

 Steps:
1. If no base direction is showing you are forced to assume that this is 360 notation, and the meridian is exact. What quadrant is the given meridian in?
2. For the lens indicated (right or left) what base direction do you have in that quadrant?
3. If the meridian given is greater than 180, such as in this problem, subtract 180 from it. The result is the new meridian.
The answer to this problem is [OD] 3.75 BD&I @ 127.
 
4. OS 4.6 @ 209

 
5. OD 7.6 @ 212

 
6. OS 7.1 @ 293

 
7. OD 2.8 @ 100

 
8. OS 6.5 @ 004

 
9. OD 5.0 @ 045

 
10. OS 5.0 @ 233

 
11. OS 4.1 @ 194

 
12. OD 1.8 @ 056

 
13. OD 1.5 @ 090

 
14. OS 3.0 @ 180