Vertical imbalance and PD problems with zero power???

[TracyElise]

Academic question where you can show off your knowledge. Please do!!!
Academic Senario: :shiner: I know it's a long one, but I am desperate!!!
the RX OD +3.00 -4.00 x120 / OS +1.50 -3.00 x045. PD 64
Frame measurements 56/14 Aviator Shape.

I make the glasses and they check out perfectly. The patient trys to get use to the new rx but felt uncomfortable. The patient returns to the dr instead of me. The dr. assistant checks the glasses and says the PD is at 74, off by 10mm and there is 1^BU prism. The patient returns to me with a note from the Dr. that says "Dear Optician...fix it."

The questions that follow the assignment are: Can you fix it? SHould you fix it? Stuff like that. (trying to get us to look deeper than the surface.)

When I did the calculations (power at 180 and 90) I found;
OD 0.00@180 and +2.00@90
OS 0.00 @ 180 & 90.
{D = DSph + DCyl (sin^)2} formula

(THE INSTRUCTOR STRICKLY WANTS US TO DEAL WITH THE FACTS GIVEN IN THE SENARIO, :finger:NO ASSUMPTIONS ARE TO BE MADE.) This is where my problem comes in.

1. Can you tell me the only way you can verify the PD of this kind of RX? I am completely stumped here, no ideas!!!

2. And why is there vertical imbalance?
[I have already submitted answers of the following... 1. Dr. assistant checked them incorrectly in the lensometer according to ANZI Standars. 2. The OC could be sitting 5mm above the patients line of sight. 3. The patient could be experiencing Anisometropic problems (difference in power between the eyes that may cause differing amounts of prism when the wearer looks off center, as in reading.), 4. The prism dial could have been bumped between lenses or frame table moved accidently. ]

The instructor says there are more reasons????:angry: WHAT ARE THEY??????
He says "There is also something else that can cause vertical imbalance especially in this RX." <direct quote. Well, I can't think of any more!!!
Can anybody else???

3. And finally, what do you percieve to be the problem? WHy? (this question is especially difficult one for me to remember the stipulation that you must to ONLY go off the facts given in the senario.:finger: )

Please let me assure you that I am not looking for an easy way out of this assignment. I take my education very seriously. But I am truly stumped. I have communicated back and forth with the instructor and I am getting nowhere with him. I've talked to all the Optician in my office, even some of the doctors and all I got was blank stares.
My brain is frying as I type this.
PLLLLEEEEAAASSEE give it a shot. Let me know if you can give ANy input into ANY of these questions. I would greatly appreciate it!!!
(can you tell I'm begging???)

Desperate for some Optician wisdom,

TracyElise:o

[Joann Raytar]

The instructor says there are more reasons????:angry: WHAT ARE THEY??????
He says "There is also something else that can cause vertical imbalance especially in this RX." <direct quote. Well, I can't think of any more!!!
Can anybody else???

The layout could have been wrong - bad box measurements. The person laying the lenses out could have measured straight down from the pupil instead of measuring to the bottom of the frame; in an aviator that can be especially noticable. Also, the person inspecting the lenses could have forgotten to start with the highest power at 90 when looking for vertical prism. A bad box measurement could also create the horizontal prism problems. I've seen folks measure aviators using the PD from the datum instead of the widest measurement across the A and go straight across instead of using the narrowest across the bridge.

[Chris Ryser]

The layout could have been wrong - bad box measurements..

I agree with Jo. Just would like to add that often mistakes are made with spheres and cyl powers that have a near zero power at 90 degrees while marking the centers off center with the lensometer.

[Spexvet]

It's possible to have prism in a plano lens, vertically and/or horizontally.

High, oblique cyl like this can make the target in the lensometer do funky things. Notice that when you move the lens horizontally, the target moves vertically, even though it may not move horizontally at all, due to plano @ 180. If the vertical imbalance is not checked at the OC horizontal point, it may look like the center is high or low (vertical prism).

IMHO, the REAL issue will have to do with cyl power or axis. More than Likely, one or the other is not exactly like his old Rx, and that messes up people like this (I'm married to one).

A less likely reason could be that they lenses are on different base curves.

The missing information, that would be helpful:
- old Rx (always critical)
- seeing how the Dr assistant checked glasses
- does the patient see well through glasses? (feeling "uncomfortable and seeing poorly are very different)
- base curves

Good luck. Hope this helps. Don't bother asking for help when you're in the simple camera or mirror units - it's been tooooooo long!

[Darryl Meister]

I may not entirely follow the problem, since your first assertion is that they "check out perfectly." After all, if they check out perfectly, you should not have any vertical imbalance. And, since they are single vision lenses, vertical imbalance at near shouldn't be an issue, either. However, it is entirely possible that the doctor's assistant measured the lenses incorrectly, especially given the amount of oblique cylinder involved (which, as the other posters have already pointed out, will induce a significant amount of vertical prism if the PD is off).

[TracyElise]

I may not entirely follow the problem, since your first assertion is that they "check out perfectly." After all, if they check out perfectly, you should not have any vertical imbalance. And, since they are single vision lenses, vertical imbalance at near shouldn't be an issue, either. However, it is entirely possible that the doctor's assistant measured the lenses incorrectly, especially given the amount of oblique cylinder involved (which, as the other posters have already pointed out, will induce a significant amount of vertical prism if the PD is off).

But the PD is not off and there is no vertical imbalance, the Dr. Assistant checked the glasses incorrectly(0 power at 180 and 0 power in one eye at 90). And when I submitted the idea that the glasses could be sitting 5mm above the line of sight, the instructor blows me down with the fact that I am making assumptions. I can only go off the information given, unfortunately. I have submitted answers concerning adjustments, base curves, vertex distances, pantoscopic tilt, face form, everything that I thought could POSSIBLY be wrong. To this he replies that I am making assumptions.
He is going to send me to the crazy house if I am not carefull.:angry:

[QDO1]

It's possible to have prism in a plano lens, vertically and/or horizontally.

Yes, you can order a plano prism - they would, for example, be plano prisims in a testing trial set.

Perhaps a look at the design philosophy of XYZ optics in Oakley sunspecs might point you in the right direction for your original question.

In this senario you will get both horizontal and vertical prism if the centres are set too high or low. The centres might be aligned to each other, when checked on the focimeter, but if they are not alligned for position of wear, the patient will be presented with prism. You will also need to adjust the height to compensate for the frontal angle of the frame etc.

An inexperienced glazer might induce some prisim in the job, when blocking up the pair of lenses, as stock lenses might have a little prism in them. an experienced glazer would make sure that any prism in the right lens was offset by any prisim in the left lens

In the example it sounds like the job was ordered on datum, and the DR checked for position of wear

[lensgrinder]

Greetings,
We know that with a +3.00 -4.00 @ 120 our full cylinder power is @ 30. So we have a -1.00 @ 30. We have 1^ BU. This is like a resultant prism question. We can resolve this in to vertical and horizontal components, except we do not need the horizontal component. If we move this lens horizontally the lens will move vertically in the 30th meridian. Think of a right triangle. What we are going to figure for is the hypotenuse. Which will be the total amout we will move this lens to induce 1^ of BU prism.
We know the angle is 30 and since we know the lens has a diopter of prism in the 90th meridian with +2.00 D in the same meridian, how much do we need to move this lens to create that?
1^=2 X f (Prentice's Rule)
f=.5cm or 5mm

Now we know our vertical component is 5mm which is opposite our angle. So to find the hypotenuse we will use sine=Opposite/Hypotenuse

sin30=5/Hpotenuse
Hypotenuse=5/0.5
Hypotenuse=10
Or 10mm. So it is possible for a lens with no power @ the horizontal meridian to have prism vertically.

This was figured using the right eye since no information was given as to whether the lens had 0.5^BU in one eye and 0.5^ BD in the other. So I used 1^ BU just in the right.
Hope this helps.

[TracyElise]

I agree with Jo. Just would like to add that often mistakes are made with spheres and cyl powers that have a near zero power at 90 degrees while marking the centers off center with the lensometer.

Is this marking before or after edging the lens? If there is not power in the 90 or 180, how can it be marked wrong? The oc will be centered in the lensometer regardless. Right?

[TracyElise]

The layout could have been wrong - bad box measurements. The person laying the lenses out could have measured straight down from the pupil instead of measuring to the bottom of the frame; in an aviator that can be especially noticable. Also, the person inspecting the lenses could have forgotten to start with the highest power at 90 when looking for vertical prism. A bad box measurement could also create the horizontal prism problems. I've seen folks measure aviators using the PD from the datum instead of the widest measurement across the A and go straight across instead of using the narrowest across the bridge.

I tried this answer but the instructor was looking for something else. Can you think of anything else?