We are still concerned with how much correction we will need for the glasses wearer to be able to read without diplopia.  We may correct the problem with slab-off, or with one of the other options.  We may discover that there is no remaining vertical imbalance problem.

When we computed vertical imbalance at the reading level before the two lenses had the same ADD, so we could ignore the prism induced by the segment.   (We could ignore it because each segment would be providing the same amount and base direction of prism, so they would cancel each other.)

Now we have the segments inducing different amounts of prism, so we need to include this amount in our calculations.

The glasses pictured here might have the following Rx:

OD +5.00 sph,  ADD +3.00
OS +2.00 sph,  ADD +1.00

If the reading level is 8 mm below the distance OC, the prism imbalance for reading that is attributable to the distance portion of the lenses would be 2.4, with the most base up in the right lens.  But now, as you can see, when the wearer is looking down through the segment, there is also prism imbalance present from the segments.  How much?  That depends on where the person is looking through the segments.

Let us suppose that the segment is a round 22, and that the top of the segment is 3 mm below the OC of the distance part of the lens.  The segment OC is 11 mm below the top of the segment, right?  And the person is looking 8 - 3 = 5 mm into the segment.  So, the point were the wearer is looking is 11 - 5 = 6 mm above the segment OC.  So, the right segment induces 1.8 BD, and the left segment induces 0.6 BD, for a total imbalance from the segment of 1.2 BD in the right lens.

But we started with 2.4 BU in the right lens;  this 1.2 BD will actually cancel some of the BU (since we are now talking about the same lens) and the result is the wearer is going to experience a total imbalance at the reading level of 1.2 BD OD, and it is very possible that the wearer will be able to tolerate this imbalance!

Using steps that you will see in the box below, let's do this one lens at a time:

1. Determine the position of the reading level in the segment.
In this case, using the diagram above, we decided that the reading position is 6 mm above the segment OC.
2. OD:
1. Distance Rx is +5.00 on the 90th.
2. Reading level is 8 mm below distance OC, so distance prism is 4.0 BU OD.
3. Segment power is +3.00.  Reading level is 6 mm above segment OC, so reading prism is 1.8 BD OD.
4. Combining 4.0 BU and 1.8 BD in the same lens will give us 2.2 BU OD.
3. OS:
1. Distance Rx is +2.00 on the 90th.
2. Reading level is 8 mm below distance OC, so distance prism is 1.6 BU OS.
3. Segment power is +1.00.  Reading level is 6 mm above segment OC, so reading prism is 0.6 BD OS.
4. Combining 1.6 BU and 0.6 BD in the same lens will give us 1.0 BU OS.
4. Combining 2.2 BU OD and 1.0 BU OS gives 1.2 BU OD.

On the other hand, this wearer may have been wearing FT40s, where the segment OC is on the line.  Now, the person is looking 5 mm into the segment, so the reading level is 5 mm below the segment OC, and the segment imbalance is 1.0 BU in the right lens.  The total imbalance from distance and segment becomes 3.4 BU OD, making the imbalance from the distance Rx worse.

Again, using steps that you will see in the box below, let's do this one lens at a time:

1. Determine the position of the reading level in the segment.
In this case we decided that the reading position is 5 mm below the segment OC.
2. OD:
1. Distance Rx is +5.00 on the 90th.
2. Reading level is 8 mm below distance OC, so distance prism is 4.0 BU OD.
3. Segment power is +3.00.  Reading level is 5 mm below segment OC, so reading prism is 1.5 BU OD.
4. Combining 4.0 BU and 1.5 BU in the same lens will give us 5.5 BU OD.
3. OS:
1. Distance Rx is +2.00 on the 90th.
2. Reading level is 8 mm below distance OC, so distance prism is 1.6 BU OS.
3. Segment power is +1.00.  Reading level is 5 mm below segment OC, so reading prism is 0.5 BU OS.
4. Combining 1.6 BU and 0.5 BU in the same lens will give us 2.1 BU OS.
Combining 5.5 BU OD and 2.1 BU OS gives 3.4 BU OD.


Do you see the difference between these two diagrams?  In both cases the stronger plus distance lens also has the stronger plus reading ADD, so just changing the type of bifocal segment will either make the imbalance worse or better.  Which would a Franklin do -- make this situation better or worse?  What would an Ultex do?  What would a FT28 do?

You think about the Franklin and the Ultex.  You know enough about both to be able to guess, based on what we just did with the FT40 and the round 22.

Let's look at the FT28.  The segment OC for a FT28 will be 4-5 mm below the top of the seg.  Where is the wearer looking?  5 mm below the top of the seg.  How much prism is present due to the ADD at 5 mm below the top, if the segment OC is 5 mm below the top?  None.  So the wearer will be left with the imbalance that was induced by the distance Rx, or 2.4 BU OD.

How do you decide which lens to do the slaboff or reverse slaboff on?  Well, slaboff used to be most minus or least plus.  But the segment may, under specific circumstances, change that.  So, slaboff creates BU so you do it in the lens with the least BU or the most BD.  [When looking at just the distance, that is the same as most minus -- most BD -- or least plus -- least BU]  Reverse slaboff creates BD, so you order it in the lens with the least BD or the most BU. [Ditto.]

By the way, did you decide that a Franklin would act just like the FT40 that is pictured here, and make this particular situation worse?  And did you decide that the Ultex would make the situation even better than the round 22?  In fact, in this case, it will actually change the imbalance to the other direction, but also neutralize enough to eliminate the vertical imbalance problem.

What about a PAL?  The optics of the PAL are said to be just like the Franklin and FT40, in that the segment OC is considered to be 'on the line'.  The only difference is that you have a seg above instead of a seg drop.  Well, it gets more complicated than that.  First, the distance OC is considered to be at the dot on the line with the manufacturer's etchings -- unless there is yoked prism . . .  We consider the seg OC to be 'on the line' because it does not create an image jump and because it does create the kind of extra thickness in the distance portion that Franklins create.  But when we get to determining where the seg OC actually is in comparison to the reading circle . . . well, it is getting more complex.  Try the lenses.  If the wearer is having a problem, have the refractionist try some prisms over one lens while the wearer looks down, and come up with a slaboff amount by trial-and-error.  That is just my recommendation.  I have not found this particular situation in any of my textbooks.  If any of you have a better suggestion, let me know.

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What if the left lens had the stronger bifocal add?  For the moment, let's stay with that same distance Rx, but switch the ADDs.

OD +5.00 sph,  ADD +1.00
OS +2.00 sph,  ADD +3.00
Let's keep the reading level at 8 mm below distance OC, and the seg drop at 3 mm below OC.  So, the wearer is still looking 5 mm into the segment to read.

The imbalance due to the distance is still 2.4 BU OD.  Now, with the round 22, the imbalance due to the reading ADD is still 1.0 BD, but in the OS this time.  So we have compounded the imbalance even more, to a total of 3.4.

In this case, it is the Franklin-type optics that will improve the imbalance at reading level -- so a FT40 would also make it better, not worse.  Your turn to draw.  Just make rough sketches to demonstrate to yourself that the segment OC on the bifocal line will reduce the total imbalance instead of increasing it.  And do please make a diagram.  That is the best way to see what 'should' be happening in these examples.

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Well, now, I'll bet that you want me to give you steps to follow.  So, here they are:
 

If you learned to do bicentric grind one lens at a time, here is the way to do unlike ADDs:  Determine the reading position in the segment: Subtract the amount the segment is below the distance OC from the reading level to determine how far into the segment the reading position is.  If the segment top is above the distance OC (VERY rare) add the seg above to the reading level. Determine the position of the segment OC in the segment, with respect to the top of the segment. Make a drawing to determine where the reading position is with respect to the segment OC, and come up with a number of mm that the reading position is from the segment OC.  This is where a diagram is vital. OD: Determine the distance power in the 90th meridian. Use Prentice's' rule and the reading position below distance OC to find the distance prism.  Make sure you determine base direction. Use Prentice's' rule on the segment reading position and the OD ADD to determine the reading prism.  Make sure you determine base direction. Combine b. and c.: If the base directions in b. and c. are the same, add the prism amounts and keep the prism direction. If the base directions in b. and c. are different, subtract the prism amounts and keep the direction of the larger amount. OS:  repeat #2 for the OS distance Rx and ADD. Combine OD and OS: If the base directions are the same, subtract the prism amounts and keep the prism direction for the lens with the larger amount. If the base directions are different, add the prism amounts.  The base will be either BU for the lens that was originally up, or BD for the lens that was originally down. Regular slab-off creates BU prism, so you do it in the lens with the most BD or the least BU. (We cannot use the most minus least plus, because the segment may change that) Reverse slab-off creates BD prism, so you do it in the lens with the most BU or the least BD.

You want to know why we did the exact opposite of what you have learned to do in combining prism in part 2 above?  Because here we are talking about the two types and amounts of prism on the same lens.  The rules that you memorized for canceling and compounding prism are for two lenses, and is what we do in step 4.

I add this next part under duress -- the shortcut method for finding imbalance is fine for if you have like ADDS, but starts making things more complicated for this procedure.
 

If you prefer the shortcut method for finding imbalance at reading, which is to take the difference in powers and apply Prentice's' rule, here is the method to use for unlike ADDs.  I think this is the harder way to do it in this case. Determine the imbalance present based on the distance Rx and the reading position. Find the distance power in the 90th meridian for each lens. Subtract.  (Find the difference in power in the distance.) Use Prentice's' rule and the reading depth to find the amount of imbalance. Determine which lens has the most BD or the most BU.  If the Rx is antimetropic, pick one. Determine the reading position in the segment. Subtract the amount the segment is below the distance OC from the reading level to determine how far into the segment the reading position is.  If the segment top is above the distance OC (VERY rare) add the seg above to the reading level.. Determine the position of the segment OC in the segment, with respect to the top of the segment. Make a drawing to determine where the reading position is with respect to the segment OC, and come up with a number of mm that the reading position is from the segment OC.  This is where a diagram is vital. Subtract the difference in ADD powers. Use Prentice's' rule to determine the amount of imbalance that is due to the segment. Combine. If the base directions are the same and the lens with the most prism in step 1 and 2 is the same, add the imbalances. If the base directions are the same and the lens with the most prism in step 1 and 2 is different, subtract the imbalances. If the base directions are opposite and the lens with the most prism in step 1 and 2 is the same, add the imbalances. If the base directions are opposite and the lens with the most prism in step 1 and 2 is different, subtract the imbalances.

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Review in System for Ophthalmic Dispensing how to determine where the segment OC is.  Pages 453-454.  Ignore image jump -- that is another consequence of the position of the segment OC on the glasses wearer.

Now read pages 117-120 in Optical Formulas Tutorial, and follow the examples.

Now, here are a few for you.  The link for the answers is at the end.  Do one, check if you got what I got.  Do the second one, check again.  If you are not getting the right answers there is no point in doing the same thing wrong time and time again.  Once you are getting them right, then do the rest and check all of your answers.
 
 

1.	Rx:      OD +1.50  ADD +4.50     OS +0.50  ADD +2.50 Reading level is 8 mm below distance OC. segment is a round 22, set 3 mm below distance OC. segment is a Franklin, set 4 mm below distance OC. segment is a FT35, set 4 mm below distance OC.  Segment measures 21 mm from top to bottom.
2.	Rx:     OD -5.00 ADD +4.50     OS -4.00 ADD +2.50 Reading level is 8 mm below distance OC. segment is a round 22, set 3 mm below distance OC. segment is a Franklin, set 4 mm below distance OC. segment is a FT28, set 4 mm below distance OC.  Segment measures 19 mm from top to bottom.
3.	Rx:      OD pl -1.50 x180  ADD +2.50     OS +0.50 -2.00 x180 ADD +1.00 Reading level is 11 mm below distance OC. segment is a round 22, set 4 mm below distance OC. segment is a FT40, set 5 mm below distance OC. segment is a FT35, set 5 mm below distance OC.  Segment measures 20.5 mm from top to bottom.
4.	Rx      -2.00 -1.00 x090  ADD +3.50     -3.00 -2.50 x045 ADD +2.50 Reading level is 10 mm below distance OC. segment is a round 22, set 5 mm below distance OC. segment is a FT40, set 6 mm below distance OC. segment is a FT25, set 6 mm below distance OC.  Segment measures 17.5 mm from top to bottom.

The answers are here.
 

Read page 117-120 in Optical Formulas Tutorial.
Read pages 453-454 in Systems for Ophthalmic Dispensing.