Ординатура / Офтальмология / Английские материалы / Essentials of Ophthalmic Lens Finishing, 2nd edition_Brooks_2003

C H A P T E R 6 C E N T R AT I O N O F S E G M E N T E D M U LT I F O C A L L E N S E S

USING A CENTRATION INSTRUMENT WITH FLAT-TOP MULTIFOCALS

The process for centration of flat-top multifocals is done in about the same way with a centration instrument as it was done with a lens protractor, but it is easier. An example problem may help explain the process.

Example 6-5

A prescription for a flat-top lens reads as follows:

R: –2.50 –1.75 × 10

L: –2.75 –1.50 × 171

Add: +1.25

PD: 63/60

Segment height = 15 mm

Frame: A = 50

B = 38

DBL = 18

How would the left lens be centered and marked for edging?

Solution

Step 1. The practitioner verifies the lens for power and MRP location. The location of the MRP is spotted with the lensmeter. (Even though the lens has been spotted, these dots are ignored until the lens has been centered with reference to the segment. After this has been completed, the location of the dots is checked for accuracy.)

Step 2. If the instrument has blocking capabilities, a double-sided adhesive blocking pad is placed on a lens block and the block is mounted on the instrument. The paper is peeled away from the pad on the block to expose the adhesive.

Step 3. The total segment inset required is determined. In the example, total inset is as follows:

Total inset = (A + DBL) – Near PD 2

=(50 + 18) – 60

2

=68 – 60

2

=8

2

=4 mm

Step 4. The practitioner determines whether the lens must be decentered to the right or to the left. Because the lens is to be placed in the centering device face up and is a left lens, the lens must be decentered to the left (see Figure 4-3, B, and Box 4-1).

117

Step 5. The vertical reference line in the instrument is moved to the right or left by the amount of decentration calculated. In this example the vertical reference line would be moved 4 mm to the left.

Step 6. The amount of segment drop or raise required is calculated. In this case, the following calculation applies:

Segment drop = Segment height – B 2

=15 – 38

2

=15 – 19

=–4 mm

Because the number is negative, the vertical movement of the segment top is downward, indicating a segment drop.

Step 7. The lens is placed face up in the instrument and the segment is aligned between the segment border lines. Centration instruments have a variety of methods for bordering the outer edges of the segment so that the segment center will not have to be located by hand.

In this example, the left lens has been placed face up for blocking. The nasal, segment side will be to the left. The movable vertical line has been positioned 4 mm to the left of center. Marking the center of the segment is not necessary because boundary lines spaced equally to the left and right of the movable vertical line border the segment. The segment is centered horizontally when symmetrically enclosed by bordering lines. This is shown in Figure 6-5.

Step 8. The lens is moved up or down so that the segment top is at the appropriate segment drop or raise. In the example this is 4 mm below the main horizontal reference line.

Step 9. The handle of the instrument is grasped and swung into place, or the button or footswitch is pressed. This blocks the lens.

These steps are summarized in Box 6-2.

Catching Errors before the Lens is Edged

Laying out a segmented multifocal is possible using only the segment for reference. If everything in lens centration has been done correctly, the near PD and the segment height should come out exactly right when the lens is edged and placed in the frame. However, the

118

FIGURE 6-5 This left lens, placed convex side up, has a 2.5-mm distance decentration, a 1.5-mm segment inset, a 4-mm total inset, and a segment drop of 4 mm.

completed spectacles may still be unacceptable. There are several reasons why they might be rejected. The following are a few:

•The MRPs of the distance portion of the lenses may be either too high or too low.

•The MRPs of the distance portion of the lenses may not be at the same height. This most likely will cause unwanted vertical prism.

•The distance PD may be too wide or too narrow and cause unwanted horizontal prism. This happens because the segment inset has not been surfaced correctly.

•The cylinder may be at the wrong axis.

If the lens has not been properly surfaced, any or all of these errors can occur. They may occur even if the lens segment is positioned properly during centration. The primary reason errors occur is the relationship between the distance MRP and the segment have not been proofed. The following section outlines how these errors happen and how they may be prevented.

CHECKING MAJOR REFERENCE POINT HEIGHT IN THE DISTANCE PORTION

Before edging a lens, the practitioner checks the distance power. Sphere and cylinder powers should be correct. With the segment line horizontally straight in the lensmeter, the cylinder axis should be as ordered. Now the lens is spotted.

C H A P T E R 6 C E N T R AT I O N O F S E G M E N T E D M U LT I F O C A L L E N S E S

BOX 6-2

Using a Centration Device for Flat-Top Multifocals

1.Verify the lens for power and MRP location. Spot the location of the MRP with the lensmeter.

2.Place the lens block in the instrument.

3.Determine the total segment inset required:

Total inset = (A + DBL) – Near PD 2

4.Determine whether the lens must be decentered to the right or to the left. If the lens is convex side up, for a right lens decenter to the right; for a left lens, to the left.

5.Position the movable vertical line in the instrument to the right or left by the amount of decentration calculated.

6.Calculate the amount of segment drop or raise required:

Segment drop = Segment height – B 2

7.Place the lens face up in the instrument and align the segment between the segment border lines.

8.Move the lens up or down so that the segment top is at the indicated segment drop or raise.

9.Grasp the handle of the instrument and swing into place, or press the button or footswitch. This blocks the lens.

MRP, Major reference point; DBL, distance between lenses; PD, interpupillary distance.

During the centration process the vertical location of the MRP is checked. When the segment drop or raise is correct in the centration instrument, the MRP should be on the main horizontal line. If it is, the MRP height has been placed correctly. This assumes that no specific MRP height has been ordered. If the segment height is especially high or low, an MRP that is not on the main horizontal line may also be correct. The box on p. 119 MRP Placement provides an explanation of this idea.

If after centering a lens with a flat-top bifocal segment, the practitioner discovers that the MRP is 2 mm above the horizontal midline as shown in Figure 6-6, is this a problem? In some cases this may be inconsequential, even normal. But in other instances the lens will almost certainly cause the prescription to be unsuitable and should be rejected or returned before more time is wasted. These two factors are critical in the decision whether to accept or reject the lens:

FACTOR 1: Is the second lens in the pair identically surfaced?

If the second lens in the pair also has its

C H A P T E R 6 C E N T R AT I O N O F S E G M E N T E D M U LT I F O C A L L E N S E S

MRP Placement

In theory, the dispensary should specify the location of the major reference point for all single vision and multifocal lenses. In practice they do not. Therefore some choices that perhaps would better be made in the dispensary must be made in the optical laboratory.

When a specific MRP height is not given, the surfacing laboratory must make the decision on where to place it. The software program used by the surfacing laboratory provides the following two options:

1.The MRP always is placed on the 180-degree line, regardless of where the top of the segment happens to be.

2.The MRP is placed on the 180-degree line unless the segment is either high and will approach or cross the 180-degree line or especially low.

With the second option the surfacing laboratory sets maximum and minimum distances the MRP can be from the top of the segment.

Following are some typical maximum and minimum distances:

Example

A frame has a B dimension of 38 mm, and the lens has a bifocal segment height of 20 mm. Where would the distance MRP end up being for option 1 and option 2?

Solution

For option 1, the MRP would be directly on the horizontal midline of the lens, halfway between the top and bottom. Half of 38 mm is 19 mm, so the MRP is at 19 mm. This is found in the segment, 1 mm below the segment line.

For option 2, the typical measurements shown above will be used. The minimum allowable distance from the MRP down to the top of the segment is 2 mm. This means that if the bifocal segment is 20 mm high, then the MRP must be 20 mm plus 2 mm, or 22 mm, high.

MRP, Major reference point.

MRP 2 mm above the horizontal midline, then no differential vertical prism will result between the two eyes. In fact, if no vertical MRP height was specified, then with both MRPs at the same vertical height (within reasonable limits) no error has occurred.

119

FIGURE 6-6 This figure has one potential problem: The inset is correct; the axis should also be correct because the 180-degree line marked by the lensmeter is parallel to the horizontal; but the major reference point will fall above the 180 line if fabricated. This may or may not be a problem, depending on the other lens.

FACTOR 2: Are the left and right MRPs at different vertical heights? The practitioner should note how much farther the MRP is from the segment line in one lens than it is in the other lens. If the difference in heights is more than 1 mm when the power of the weaker lens is greater than or equal to ±3.375 D, the lens pair will not pass ANSI standards.

This should emphasize the importance of care in measuring lens parameters for the duplication or replacement of a single lens in a pair of lenses.

If the power of the weaker lens is less than ±3.375 D and the two MRP locations are more than 1 mm apart vertically, a problem may still exist. To find out whether a problem still exists, the two spotted lenses are placed front-to-front so that the segments are superimposed. (The practitioner should not actually touch the front surfaces of the lenses against each other, or the surfaces may get scratched.) The strongest lens should be facing outward. The center spot of the stronger lens is viewed through the weaker lens and the weaker lens is dotted at that location.

Now the newly spotted point is placed on the weaker lens in the center of the lensmeter aperture. Read the prismatic effect. If it is greater than or equal to one-

120

third prism diopter, the lens will not pass ANSI Z80.1 standards. (Without actually using a lensmeter, this prismatic effect can be predicted using Prentice’s Rule. This is done by multiplying the difference in vertical MRP heights in centimeters by the power of the weaker lens in the 90-degree meridian.)

Example 6-6

The following prescription is fabricated:

R: –7.00 D sphere

L: –5.00 D sphere

Add: +1.50

The two lenses are spotted. The right MRP is 2 mm higher than it should be. The left MRP is exactly on the horizontal midline. Is this acceptable?

Solution

Because both lenses are greater than ±3.375 D and the two MRP locations more than 1 mm apart vertically, the lens pair will be unacceptable. Unwanted vertical prism will be beyond what is tolerable. The right lens should be remade.

Example 6-7

Following is another example that is a bit more complex:

R: +1.00

L: +1.50 –1.25 × 180

Add: +2.00

In checking for correct vertical MRP placement during the centration process, the practitioner sees that the right lens is 3 mm above the horizontal midline. However, the MRP of the left lens is right on the line.

C H A P T E R 6 C E N T R AT I O N O F S E G M E N T E D M U LT I F O C A L L E N S E S

Will the lens pair prove suitable after edging? (In thinking through the question, the reader should look carefully at the prescription when considering which lens is stronger in the vertical meridian.)

Solution

The reader should look at the powers of both lenses in their 90-degree meridians (Figure 6-7). As can be seen, two different results will be obtained depending upon whether point A on the right lens or point B on the left is considered. So which lens is used to evaluate prismatic effect?

(Hint: When evaluating a lens pair for unwanted vertical prism, the practitioner uses the lens with the weakest power. It would not be “fair” to use the lens with the stronger power. What would happen if one lens has no power and the other lens has –7.00 D of power? If the zero-powered lens is spotted first, no prismatic effect will be detected, no matter where on the lens a measurement is taken. Sliding the lens pair over to the –7.00 D lens could show a great deal of vertical prism at any distance above or below the center of the lens.)

The power of the left lens is +1.50 –1.25 × 180. To make it easier to visualize, the lens powers are placed on a power cross. This is shown in Figure 6-7 and tells us that the lens with the least power in the 90-degree meridian is the left lens. This is the lens used to determine how much vertical prismatic effect is caused by the lens pair.

What is the prismatic effect for the left lens? Prentice’s Rule is used to evaluate the prismatic effect on the decentered point of the weaker lens. The formula for Prentice’s Rule is = cF. The values for c and F must be determined.

+1.00

+1.00

3 mm

A

+1.00 D Sphere

+0.25

B

+1.50

+1.50 –1.25 x 180

C H A P T E R 6 C E N T R AT I O N O F S E G M E N T E D M U LT I F O C A L L E N S E S

The reference point on that lens is 3 mm (or 0.3 cm) away from its center. Therefore the following is true:

c = 0.3 cm

The power manifested in the vertical (90-degree) meridian is +0.25 D. Therefore the following is true:

=cF

=(.3)(0.25)

=0.075

In this case the unwanted vertical prism is 0.075 . What is the allowable amount of vertical prism?

According to ANSI Z80.1 standards, the allowable amount of prism is a maximum of one-third prism diopter. The 0.075 prism amount is definitely less than the 0.33 allowable. This is sufficiently below the maximum allowable tolerance to make the lens pair acceptable and not cause discomfort for the wearer.

CHECKING FOR INCORRECT SEGMENT INSET

Naturally the practitioner wants the distance PD and the near PD to be right once the lenses are edged. To predict whether this will happen or not, the practitioner needs to know whether the MRP is at the right position horizontally relative to the center of the segment. The correct distance between the two is equal to the segment inset. The segment inset was defined as follows:

Segment inset = Distance PD – Near PD 2

Checking to see whether segment inset will be correct is really done at the same time the lens is being positioned in the centration instrument for blocking. After the lens is positioned for blocking, the practitioner looks for the previously spotted MRP. It should be temporal to the movable vertical line by an amount equal to the segment inset.3 (The movable vertical line marks the location of the center of the segment top.)

For example, if the distance PD is 65 and the near PD 62, then the MRP should be 1.5 mm outward from the movable vertical line that goes through the center of the segment.

3Another way of checking correct MRP location is to see whether it is at the correct distance decentration. The MRP should be nasal from the stationary vertical line in the centration instrument by an amount equal to the distance decentration. This is the case because of the following equation:

Total segment inset = Distance decentration + Segment inset

121

INCORRECT SEGMENT INSET

If the MRP is found to be horizontally off compared to the segment, the practitioner needs to know whether the problem is bad enough to reject the lens. Sometimes the way the lens is centered for edging determines whether it is usable.

If the MRP is horizontally off, the lenses can be edged so that either the distance PD or the near PD will be correct, but not both. However, producing a perfectly acceptable pair of lenses still may be possible. So if it is determined that the segment inset is wrong before the lenses are edged, the practitioner may choose one of the following four options:

1.To reject the lens and have it remade

2.To elect to have a correct near PD, but an incorrect distance PD

3.To elect to have a correct distance PD, but an incorrect near PD

4.To elect to position the lens midway between near and distance, altering both PDs slightly. This may be done only if the prismatic effects resulting from this choice are insignificant.

This discussion is not about how to get by with an unacceptable lens, which should not be tolerated. Any of these choices are possible only if the resulting pair of spectacles is comfortably within ANSI prescription standards. Sometimes more than one option will allow this. How does the practitioner decide which option is best?

The Factor that Favors an Accurate Distance Interpupillary Distance

When the relationship between distance PD and near PD is affected by a segment inset that is slightly off, the centration process allows the practitioner to favor either distance or near PD accuracy. The factor that especially favors a high exactness in the distance PD is the power is the distance lens. If the distance lens has a high refractive power, any slight variation in optical center placement will cause unwanted horizontal prism.

Factors that Favor an Accurate Near Interpupillary Distance

Just as a high-powered distance prescription favors an accurate distance PD, so does a high near addition favor an accurate near PD. Even though it is easier for the eyes to tolerate horizontal prism at near power because the eyes already are converging and diverging with changes in viewing distances, high addition powers do cause unintended horizontal prism when not properly positioned.

122

BOX 6-3

Compensating for Small MRP Placement Errors

Factors Favoring Maintenance of Distance PD,

Alteration of Near PD

1.A high-powered distance prescription (in 180-degree) meridian

2.A low near addition power

3.A very wide segment

Factors Favoring Alteration of Distance PD,

Maintenance of Near PD

1.A low-powered distance prescription (in 180-degree meridian)

2.An especially high near addition power

3.An especially small segment

MRP, Major reference point; PD, interpupillary distance.

With segmented multifocals a limited viewing area exists for near vision. If the segment is small, the viewing area is small. Moving the segment away from its intended location takes off viewing space from one side or the other. If the near PD is off it will cut down that viewing area either temporally or nasally. So, if the segment is small, it should be accurately placed with a correct near PD. Box 6-3 summarizes factors that favor either the distance or the near PD.

Favoring Distance or Near Interpupillary Distance?

The following examples help the reader understand the logic used to decide whether to favor the distance PD or the near PD, or whether to split the difference.

Example 6-8

The following is prescribed:

R: +5.50 sphere L: +5.50 sphere PD: 69/65 Add: +1.00

Flat-top 28 segment: segment height 19 mm Frame: A = 53

B = 42 DBL = 20

The right lens is positioned for blocking as shown in Figure 6-8. Although the segment height and total segment inset is correctly positioned so that the lens will cut out with the near PD correct, the MRP is incorrectly positioned. Is the lens usable? If so, should distance or near PD be favored?

C H A P T E R 6 C E N T R AT I O N O F S E G M E N T E D M U LT I F O C A L L E N S E S

35

28

25

22

FIGURE 6-8 After the near segment is centered as indicated by the prescription, the position of the major reference point (MRP) should be checked to assume accuracy of the finished interpupillary distance (PD). In this instance, the required total inset is 4 mm, which is correct. Yet the MRP is incorrect because a 2 mm decentration per lens is indicated and only 1 mm appears. Correcting the position of the MRP to achieve the correct distance PD will throw off the near PD.

Solution

To answer the question, the practitioner must know how far off the MRP is from where it should be in relation to the segment. In the example, the distance decentration should be as follows:

Distance decentration = (A + DBL) – PD 2

= (53 + 20) – 69 2

= 73 – 69 2

= 4 2

= 2 mm per lens

For the distance PD to equal 69 mm, the center dot of the right lens should have a 2-mm distance decentration. Unfortunately this lens shows only 1 mm. So even if the left lens is accurate, the distance PD will be off 1 mm. It will end up being 70 mm instead of 69 mm. Left this way, the near PD will be a correct 65 mm. Which is more important in this prescription, distance or near PD accuracy?

The distance power in this prescription is high. If the distance PD is wrong, unwanted horizontal prism occurs at distance. The amount of unwanted prism will be as follows:

=cF

=(0.1)(5.5)

=0.55 base out

Now what would happen if the practitioner decided to make the distance PD correct at the expense of throwing off the near PD? To do this the near PD would have to be decreased by 1 mm. If the near PD is intentionally decreased to 1 mm less than it should be, this will decrease the distance PD back to the required 69 mm. Doing this will eliminate unwanted horizontal prism in the distance portion. But what will happen if the near PD is 64 mm instead of 65 mm?

If the near PD is off by 1 mm, two optical effects will be produced at near, as follows:

1.The prismatic effect will change. But because of the low add power, that amount of change is small. The

exact amount is found with Prentice’s Rule ( = cF), where c is the distance the segment has been moved from its proper location and F is the power of the add. Therefore the prism produced by intentionally making the near PD too small is only as follows:

=(0.1)(1)

=0.10 base in

2.The other effect from allowing the near PD to be wrong is that the temporal field of view through the near segment will be slightly reduced at near. However, this point becomes less significant with sufficiently large segment sizes because still plenty of room exists for near viewing through the segment.

Therefore which option should the practitioner

choose in this case: correct distance PD or correct near PD? In making the final choice, the practitioner should look at ANSI Standards for prescription lenses. The ANSI tolerance for near PD for multifocals is ± 2.5 mm. For an ordered near PD of 65 mm, anything between 62.5 mm and 67.5 mm is considered acceptable.

ANSI tolerances for the distance PD are also

± 2.5 mm. But in this prescription, the unwanted prismatic effect produced by a wrong distance PD is more than that produced by a wrong near PD. Therefore the best choice would be to modify the near PD so that the distance PD will come out right.

Example 6-9

In another example, a pair of glasses is ordered using the same frame as before. This time the prescription is somewhat different:

35

28

25

22

FIGURE 6-9 This high near addition lens was supposed to be made for a 70/64 interpupillary distance. But after positioning the segment, the spotted major reference point location is checked for accuracy. Distance decentration should have been 1.5 mm per lens. Instead it is 4 mm per lens.

R: +0.75 –0.50 × 90

L: +0.75 –0.50 × 90

PD: 70/64

Add: +4.00

Flat-top 28: segment height 19 mm

Frame: A = 53 mm

B = 42 mm

DBL = 20 mm

The practitioner lays out the right lens for edging. The total segment inset is 4.5 mm. However, checking the position of the MRP for the distance unearths an error. Instead of getting the needed distance decentration of 1.5 mm, the decentration is 4.0 mm (Figure 6-9). This is 2.5 mm different from what was ordered. Is the lens usable? If it is, either the distance or the near PD (or both) will have to be slightly off. Which should it be?

Solution

In evaluating what to do, the practitioner first should look at unwanted prism. Because the error occurs in the horizontal meridian alone, the power of the lens in that meridian causes unwanted prism. What would the lens look like in front of the eye if it were edged this way? See Figure 6-10. Because the distance PD will be wrong, the eye looks through a point 2.5 mm away from the MRP. But according to Prentice’s Rule

124

the amount of unwanted prism that occurs is only as follows:

=cF

=(0.25 cm)(0.25 D)

=0.0625 base in

This is a small prismatic effect. But displacing the near portion changes the prismatic effect at near by the following:

=(0.25 cm)(4.00 D)

=1.00 base out

It also narrows the near field of view. Therefore the best choice is to alter the distance PD so that the near PD will be correct. Even though the distance PD appears to be off considerably because distance power is small, unwanted prism is very low. It is well within ANSI standards.

In the previous examples, one example favored altering the near PD. The other favored the distance PD. Not all examples are clear cut. Many times the answer lies in between the two, with both being altered somewhat.

In evaluating lens pairs like this, the practitioner is trying to make lenses work that will not in any way compromise vision for the wearer. The point is not to try to sneak a reject lens through the system and use it.

Preventing Expensive Mistakes

The accurate spotting of lenses is of extreme importance. When any doubt exists as to the correctness of the original spotting, it should be double-checked.

+0.75

2.5 mm

+0.25

Nose

+0.75 –0.50 × 90

FIGURE 6-10 Imagining or sketching out a power cross on an eye can help in understanding how prism is figured using Prentice’s Rule. The horizontal meridian of the lens containing the +0.25 D lens power is used for calculations.

C H A P T E R 6 C E N T R AT I O N O F S E G M E N T E D M U LT I F O C A L L E N S E S

During the centration process, the practitioner takes notice of the location of the spotted MRP. The two previous examples were instances in which the prescription looked like it might fail, but because of giving attention to MRP location, failure was prevented. However, when the MRP is not where it should be, the glasses are likely to fail ANSI standards. Edging a pair of lenses that were surfaced incorrectly wastes time and may disqualify the finishing laboratory from credit for the cost of the surfaced lenses.

TILTED 180-DEGREE LINE

After centration the segment may be positioned properly, but the previously spotted 180-degree line is tilted. Following are some factors that indicate how serious a problem this might be.

If the lens is a sphere and the central MRP dot is located at the proper vertical height and inset, no problem exists. An example of this is shown in Figure 6-11. A spherical lens has the same power in all meridians. Because no cylinder exists, any rotation of the lens has no effect on refractive power.

If the lens is a cylinder or spherocylinder, rotation is important. The cylinder axis will be correct only if the 180-degree line is at 180 degrees. For example, in

FIGURE 6-11 In the lens spotting process, the correct axis is set in the lensmeter. Afterward, the lens is rotated until it conforms to the specified axis, then spotted along the 180-degree line. While centering this lens, the practitioner discovers that when centration is “complete,” the spotted 180-degree line is no longer on the 180. This indicates an axis error during surfacing.

0 2 1

0 3 1

100

9

0

8

0

7

0

60

5 0

35

28

25

22

Cylinder

(should

4 0

10

5

0

5

10

Figure 6-12, a tilt of 5 degrees in the 180-degree line results in an error of the cylinder axis of 5 degrees. The amount of error considered to be within tolerance depends upon the power of the cylinder (see Appendix B for these tolerances). Because a flat-top bifocal segment cannot be rotated to allow for correction of the cylinder axis, the 180-degree line must not be at an angle that exceeds accepted standards.

POSITIONING FLAT-TOP TRIFOCALS

No difference exists between the basic procedure for centration of trifocals and that of bifocals. Remembering the following points helps prevent confusion.

Only the top line of the trifocal is used in segment positioning. (The lower line can be ignored.) Therefore if a segment height is 19 mm, the uppermost line will be at 19 mm.

The top of the trifocal is much more likely to be above the horizontal midline than the top of a bifocal. For a segment raise, the major reference point may fall within the segment, depending on how the surfacing laboratory ground the lens. If this is the case, spotting

of the MRP will not be accurate, because of the prismatic effect exerted by the segment add power.

If the lens has a segment raise and there is no MRP height given by the practitioner, the surfacing laboratory automatically may raise the MRP above the trifocal segment top (see MRP Placement box on p. 119). If this is the case, the dot indicating MRP positioning should still have the correct horizontal inset. However, it will be vertically higher by an amount equal to the segment raise plus this laboratory-determined amount. Both lenses will have MRPs of equal vertical height.

Curve-Top Segment Lenses

Curve-top segments are positioned nearly identical to that of flat-top segments. Following are the differences:

1.Segment height is judged from the highest point of the curved upper portion.

2.Horizontal orientation is gauged by aligning both corners of the segment top with the same horizontal reference line.

126

The following problem illustrates this procedure.

CURVED-TOP SEGMENT CENTRATION

Example 6-10

The reader should correctly center a left lens for the following order:

R: +1.00 +1.75 × 78

L: +0.75 +2.00 × 109 PD: 62/59

Add: +2.25

Curve-top 25 segment; segment height 17 mm Frame: A = 48

B = 39 DBL = 18

Solution

Before placing the lens in the marking device, the practitioner should figure distance decentration, total inset, and segment drop. (Distance decentration is used to check for accuracy of MRP placement in surfacing.)

Total inset will be used to preset the centration device and can be figured two ways:

+ Segment inset

Distance decentration = (A + DBL) – Distance PD 2

= (48 + 18) – 62 2

= 66 – 62 2

= 4 2

= 2 mm per lens

Segment inset = Distance PD – Near PD 2

=62 – 59

2

=3

2

=1.5 mm per lens

C H A P T E R 6 C E N T R AT I O N O F S E G M E N T E D M U LT I F O C A L L E N S E S

Therefore

Total inset = 2 mm + 1.5 mm = 3.5 mm per lens

Both methods yield the same results.

Now the movable vertical line in the marking device is moved 3.5 mm to the left. The line is moved to the left because the lens is placed face up in the unit and blocked on the front side. This is shown in Figure 6-13.

The grounds for acceptance or rejection of a curvetop multifocal blank are the same as those for flat-tops.

Round-Segment Lenses

Some bifocal lenses have segments that are perfectly round. The segment diameters may vary from 22 to 38 mm. These lenses were popular many years ago but have steadily decreased in popularity. They are a versatile segment style because they are not restricted in orientation. They may be rotated without making the segment appear tilted like a flat-top segment would appear if it were rotated.

Round-segment lenses also are used less now because at first, they are harder to process. The person doing centration for edging or surfacing must have more of an understanding of the optical properties of the lens.

FIGURE 6-13 For a curved-top segment, the lateral corners of the curved upper segment border are equidistant from the horizontal reference line. They both must be at the same level.