Calculating Slab-Off: Ideal Way to Find Reading Depth?

[Chad Sobodash]

I've read in different places that people calculate the reading depth using all sorts of means. One place says to always use 8mm from the OC, because it's a statistic probability that most people will look down that far to read. The Laramy-K OpenOptix guide says to use ((0.5 * B) - seg) + 5) = depth. Are either of these correct?

My gut says to use ((OC - seg) + (0.5 * seg)) = depth. This way, you're going from the OC instead of the B measurement, and assuming most likely the center of the segment for the point of reading.

Please correct me so I can put it in my bench manual and never stumble again!

I'm scheduled to take the ABO on May 24th, and I couldn't be more nervous! I've been told I most likely won't have a question involving this on my exam, but I feel it's something I should have a definite knowledge of before I can call myself a "real" optician (I'm still officially an optician, only because this is a de-certified state).

Thanks in advance for the help. Out of curiosity, has anyone considered making an IRC channel for opticians? I have the server space, so I'd be willing to go ahead and do this if there's an interest. I love to idle on IRC servers for quick questions regarding this-and-that.

[gmc]

For purposes of the ABO exam, they will give you a reading depth, usually 10mm. The round number makes the calculation easier.

[MakeOptics]

((0.5 * B) - seg) + 5) = depth

1/2 the B in the equation is an assumption for the DRP (or OC in your equation), it's a fairly accurate assumption but won't hold true in every scenario.

DRP - Seg, this is the drop from the DRP to the segment line, this is a measure on route to the NRP.

+5mm, this is a FT28 specific reference again an assumption won't hold true in every scenario but accurate enough. This 5mm is computed by subtracting the height of the segment by 1/2 the segment width (@ widest point), this is before edging on the blank. A FT28 is 28mm wide and the segment height is 19mm, that means that the OC of the segment is 14mm from the edge and the top is cut off but it takes an additional 5mm from the top to get to that OC.

A more accurate non-assuming formula would not be useful to opticians in the field but would be more useful to programmers creating programs.

Depth = DRPHeight - SegFitHeight + (SegHeight - SegWidth/2)

In this case ignore your gut, the measure is going to be too high, a good rule of thumb would be 10mm but if you have to compute use

Depth = OC - Seg + 5

[MakeOptics]

If you are trying to get super accurate and fancy you could use the depth computed above:

True DRP to NRP distance = SQRT(Depth^2 + Inset^2)

That's the actual distance traveled not the depth.

[Robert Martellaro]

I've read in different places that people calculate the reading depth using all sorts of means.

The reading depth is not a calculation, it's an observation. Have the client hold a reading card or mirror in their typical relaxed near task position. Use a marking pen or tape to locate the reading level. Don't forget about VI (vertical imbalance) on the primary (distance gaze), positioning the vertical distance OC (if distance OC is at, or below the primary gaze) as needed. The reading depth is essentially the distance from the distance OC to the reading level.

For purposes of the ABO exam, they will give you a reading depth, usually 10mm. The round number makes the calculation easier.

You might also need to know where the slab-off prism goes- base up on the back, on the most minus or least plus (on the vertical meridian) lens. There are semifinished reverse slabs- same as above but everything is...reversed.

[MakeOptics]

The reading depth is not a calculation, it's an observation.

I disagree, in this case the NRP is molded into the lens design. The control comes with the OC placement and the fit height of that segment.

[Chad Sobodash]

A more accurate non-assuming formula would not be useful to opticians in the field but would be more useful to programmers creating programs.

Depth = DRPHeight - SegFitHeight + (SegHeight - SegWidth/2)

In this case ignore your gut, the measure is going to be too high, a good rule of thumb would be 10mm but if you have to compute use

Depth = OC - Seg + 5

Thanks so much for the help. MakeOptics, you totally answered my question.

You might also need to know where the slab-off prism goes- base up on the back, on the most minus or least plus (on the vertical meridian) lens. There are semifinished reverse slabs- same as above but everything is...reversed.

Yup, I got that down just fine. It's calculating the ideal reading depth from given measurments that was giving me trouble.

[MakeOptics]

Thanks so much for the help. MakeOptics, you totally answered my question.

For the test KISS applies.

Keep It Simply Stupid, the majority of test takers cannot perform overly complex math and since no calculator is supplied you can assume basic addition, subtraction, multiplication, and division is the most difficult math that can be expected on the test. For computing the power in any given meridian don't try and compute the values with trig functions use the difference in axis as the percentage of the cyl to add to your sph to come up with total power:

Example:

-2.00 -1.00 x 045, what is the power along the 90th meridian:

90 - 45 = 45

0.45 * -1.00 = -0.45

-2.00 + -0.45 = -2.45

This is not accurate but accurate enough for the test and real life dispensing scenarios.

Decentration will come up that requires math, you might have a few segment drop or below center questions that require basic math.

Best tip I can give you is wear a pair of glasses so that you have something to physically feel when taking the test that way you don't need to visualize anything. Most optician are taught all rule of thumb measures and don't really know any better their entire careers. For you learn the rule of thumbs to pass the test and then learn the reasons behind the rules of thumb.

[Chad Sobodash]

For the test KISS applies.

Keep It Simply Stupid, the majority of test takers cannot perform overly complex math and since no calculator is supplied you can assume basic addition, subtraction, multiplication, and division is the most difficult math that can be expected on the test. For computing the power in any given meridian don't try and compute the values with trig functions use the difference in axis as the percentage of the cyl to add to your sph to come up with total power:

Example:

-2.00 -1.00 x 045, what is the power along the 90th meridian:

90 - 45 = 45

0.45 * -1.00 = -0.45

-2.00 + -0.45 = -2.45

This is not accurate but accurate enough for the test and real life dispensing scenarios.

Decentration will come up that requires math, you might have a few segment drop or below center questions that require basic math.

Best tip I can give you is wear a pair of glasses so that you have something to physically feel when taking the test that way you don't need to visualize anything. Most optician are taught all rule of thumb measures and don't really know any better their entire careers. For you learn the rule of thumbs to pass the test and then learn the reasons behind the rules of thumb.

I thought that the general rule for cylinder power at axis was 30/150=25%, 45/135=50%, 60/120=75%. The reason being that if we take the function f(c) = c * ((sin(i))^2), then using trigonometric values:

sin(90)=(√4)/2
sin(60)=(√3)/2
sin(45)=(√2)/2 or 1/(√2)
sin(30)=(√1)/2

So, the example would be:

-4.50 -1.00 x 090, what is the power at 045?

f(c) = -1.00 * (((√2)/2)^2) = -1.00 * (2/4) = -1.00 * 0.50 = -0.50 = c

Power at 045 is -5.00.

[mshimp]

I thought that the general rule for cylinder power at axis was 30/150=25%, 45/135=50%, 60/120=75%. The reason being that if we take the function f(c) = c * ((sin(i))^2), then using trigonometric values:

sin(90)=(√4)/2
sin(60)=(√3)/2
sin(45)=(√2)/2 or 1/(√2)
sin(30)=(√1)/2

So, the example would be:

-4.50 -1.00 x 090, what is the power at 045?

f(c) = -1.00 * (((√2)/2)^2) = -1.00 * (2/4) = -1.00 * 0.50 = -0.50 = c

Power at 045 is -5.00.

Way over thinking this for the test..... take MakeOptics advise and keep it simple....I could of "looked" at your example and told you the power in the 45 degree rather than using the trig functions. Also using 10mm drop for calculating slab works fine.

[Chad Sobodash]

Way over thinking this for the test..... take MakeOptics advise and keep it simple....I could of "looked" at your example and told you the power in the 45 degree rather than using the trig functions. Also using 10mm drop for calculating slab works fine.

It's not for the test, it's just the proof of what it actually is. So MakeOptics' example is incorrect, because it's 50% of the cylinder power at 045, not 45%. I don't have to use the function because I know 30/150=25%, 45/135=50%, and 60/120=75%. But I know this because of the function.

[Robert Martellaro]

The reading depth is not a calculation, it's an observation...The reading depth is essentially the distance from the distance OC to the reading level

I disagree, in this case the NRP is molded into the lens design. The control comes with the OC placement and the fit height of that segment.

Maybe you've misunderstood me, or I'm misunderstanding you. Knowing the position of the NRP or near OC is not relevant in determining the reading depth, except for maybe PALs, nor is the NRP or near OC used in calculating the prism slab, unless the segments differ in design or power.

[MakeOptics]

I thought that the general rule for cylinder power at axis was 30/150=25%, 45/135=50%, 60/120=75%. The reason being that if we take the function f(c) = c * ((sin(i))^2), then using trigonometric values:

sin(90)=(√4)/2
sin(60)=(√3)/2
sin(45)=(√2)/2 or 1/(√2)
sin(30)=(√1)/2

So, the example would be:

-4.50 -1.00 x 090, what is the power at 045?

f(c) = -1.00 * (((√2)/2)^2) = -1.00 * (2/4) = -1.00 * 0.50 = -0.50 = c

Power at 045 is -5.00.

You are correct but what happens if the example calls for 35?

It's not for the test, it's just the proof of what it actually is. So MakeOptics' example is incorrect, because it's 50% of the cylinder power at 045, not 45%. I don't have to use the function because I know 30/150=25%, 45/135=50%, and 60/120=75%. But I know this because of the function.

Again correct but why jumble your head with those rations use the difference in angle as a percentage and you get close enough to move on to the next question quickly and confidently. Remember multiple choice and 125 questions max. Keep it simple and keep trucking.

Maybe you've misunderstood me, or I'm misunderstanding you. Knowing the position of the NRP or near OC is not relevant in determining the reading depth, except for maybe PALs, nor is the NRP or near OC used in calculating the prism slab, unless the segments differ in design or power.

Robert not a misunderstanding you are correct but that doesn't make me wrong. Since measuring the reading depth is going to be highly difficult and subjective given parralax using some assumptions helps simplify and gets the job done. In a PAL if the client is not viewing through the NRP the power is going to be off making our arguments moot. The NRP in a lined segment is also going to be optimal as the reading depth in most scenarios although I can think of a few scenarios where it would be off for instance a round set or blended set, but again simplicity for me is key.

[Chad Sobodash]

You are correct but what happens if the example calls for 35?

I'd use the formula. It's unlikely they'd ask me that, since it doesn't follow neatly into the trigonometric absolute set of 25%, 50%, 75%, and will yield a value that is potentially not in quarter or eighth diopters. However, if they did, let's look at the consequences of using the method you showed on a high prescription:

-12.75 -4.00 x 025
-12.00 -4.25 x 090

Find the power at 90.

0.65 * -4.00 = -2.6
-2.6 + (-12.75) = -15.35

Whereas the actual value derived from (sin(65))^2 is 82%:

0.82 * -4.00 = -3.28
-3.28 + (-12.75) = -16.03

That's over half a diopter difference.

Again correct but why jumble your head with those rations use the difference in angle as a percentage and you get close enough to move on to the next question quickly and confidently. Remember multiple choice and 125 questions max. Keep it simple and keep trucking.

I believe that bad habits die hard, and it's better to just learn the formula and be able to apply it quickly. Maybe I'm mathematically gifted, but I find it easier to remember it this way, and more sensible. I don't notice a difference in speed doing it this way versus your way.

[MakeOptics]

I'd use the formula. It's unlikely they'd ask me that, since it doesn't follow neatly into the trigonometric absolute set of 25%, 50%, 75%, and will yield a value that is potentially not in quarter or eighth diopters. However, if they did, let's look at the consequences of using the method you showed on a high prescription:

-12.75 -4.00 x 025
-12.00 -4.25 x 090

Find the power at 90.

0.65 * -4.00 = -2.6
-2.6 + (-12.75) = -15.35

Whereas the actual value derived from (sin(65))^2 is 82%:

0.82 * -4.00 = -3.28
-3.28 + (-12.75) = -16.03

That's over half a diopter difference.



I believe that bad habits die hard, and it's better to just learn the formula and be able to apply it quickly. Maybe I'm mathematically gifted, but I find it easier to remember it this way, and more sensible. I don't notice a difference in speed doing it this way versus your way.

Then the test should be a breeze. If you can memorize the trig functions then you are better then most, to the rest axis as a percentage is a close enough trick. You could also normalize the percentage and divide the axis by 90 to get the percentage closer but again it is a linear approximation of the more accurate sin squared function. Here is a nice graph of the sin2, axis as percentage, and normalized function.

Yellow = sin(axis)^2
Blue = axis/100
Red = axis/90