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

FIGURE 4-1 A lens protractor may be used in lens layout for edging but has been used more traditionally in the lens layout for surfacing process. The inner degree scale is used when the lens is placed convex side up (as when glasses are being worn). The outer scale is appropriate when the convex side is facing down.

a minimal amount of equipment. Centration may be done accurately this way, but it requires a steady hand. Historically this was done by hand with a lens protractor. A lens protractor is shown in Figure 4-1. Centration really could be done with only a sheet of millimeterruled graph paper.

THE LENS PROTRACTOR

A few lens protractors have three raised pegs that hold the surface of the lens without allowing it to come directly in contact with the face of the protractor. These pegs hold the lens steady, which keeps it from rocking, and prevent the ink spots on the lens from smearing or rubbing off.

The degree scales on a lens protractor are used mainly for marking the lens axis for surfacing and are not used in the edging process.1

1A lens protractor has two or three degree scales. The lens degree scale that increases in value in a counterclockwise direction is the scale of reference when the front surface of the lens faces up. The clockwise numbered axis scale is the reference scale when the front surface of the lens faces down. A third scale, normally 90 degrees away from the other two, would be used if the lens were turned sideways.

The intersection of the cross in the center of the protractor corresponds to the boxing center of the edged lens.

LENS CENTRATION USING A LENS PROTRACTOR

Step 1

The lens is spotted in the manner that is described in Chapter 2 for correct MRP orientation and correct cylinder orientation.

Step 2

The amount of decentration per lens required is calculated using the following formula:

Decentration per lens = (A + DBL) – PD 2

The practitioner then determines if the lens must be decentered to the right or to the left. If the lens is a right lens and is facing down, decentration “in” is to the left (Figure 4-2, A). A left lens facing down would be decentered to the right.

Note: If the right lens were facing up, the lens would be decentered to the right (Figure 4-2, B). A left lens would be decentered left (Box 4-1).

Step 3

The right lens is placed face down (front surface down) on the protractor.2 The three spots on the lens are aligned with the main horizontal line on the protractor. The central dot is placed at the intersection of the cross (Figure 4-3). Horizontally the lens is decentered the number of millimeters calculated.

In the example previously given, the required decentration was 3 mm per eye nasally—that is, 3 mm in. If this right lens is to be decentered in, it must be moved to the left 3 mm. The three dots remain on the horizontal line and the central MRP dot is 3 mm to the left of center (Figure 4-4).

2By placing the lens face down, the spot on the front surface of the lens is closer to the protractor grid surface. This helps to prevent parallax error. Parallax is the phenomenon that occurs when someone attempts to align visually an object with a scale when the object is not directly in contact with the scale. If the viewer’s eye is not on a line directly perpendicular to the correct reading on the scale above or below the object, an error results. The most common example of a parallax error occurs when the front seat passenger attempts to read the speedometer of a car. Parallax error also can be prevented by using a centration device. A centration device such as a marker/blocker allows the lens to be positioned face up and then blocked while still in the instrument.

Step 4

The lens is decentered vertically if an MRP height is ordered. Exactly how this decentering is done is explained later in this chapter. In the example given here, no vertical decentration is required.

Step 5

Using the horizontal protractor line as a guide, a long horizontal line is hand drawn through the central part of the lens. (Alternatively a flexible ruler may be used to help in drawing a line if the lens has been reference dotted ahead of time as seen in Figure 4-4. Figure 4-5 shows a lens being marked with a ruler.)

A shorter, vertical line is drawn at the center of the protractor. Figure 4-6 shows the vertical line. This line

should only be about 10 mm long. The difference in length between the two lines gives an immediate visual cue as to the orientation of the lens. The intersection of the two crossed lines shows exactly where the lens will be blocked. The center of the marked cross indicates the future boxing center of the edged lens. These steps are summarized in Box 4-2.

Calculating Vertical Centration of

Lenses Using the Boxing System

When no preference is expressed on the order form about MRP height, the height is presumed to be in the

C H A P T E R 4 C E N T R AT I O N O F S I N G L E V I S I O N L E N S E S

standard position. In most cases this standard height will be at the level of the boxing center of the edged lens. This means that the three dots on the lens remain on the 180-degree line of the lens centration device throughout centration.

BOX 4-1

Lens Decentration Direction

The direction of decentration depends on whether the lenses are facing up or down. With a centration device, most of the time the lens will be facing up. It is facing up so that it can be blocked in the centration device. So if the lens is a right lens, inward decentration is to the right: “right/right.” The left lens decentration direction will be “left/left.” By-hand decentration with the lens face down is just the opposite.

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SPECIFICATION OF VERTICAL

PLACEMENT

For edging a lens, the MRP location must be given in terms of how far it is above or below the horizontal midline of the lens. However, when the order arrives from the dispensary, MRP height is not specified this way. Instead MRP height is the distance from the lower line of the box enclosing the shape of the lens up to the MRP location. The laboratory must convert from MRP height to MRP raise (or drop).

To make this conversion from MRP height to MRP raise (or drop), the vertical dimension of the frame must be known. Then half the vertical dimension of the frame (the B dimension) is subtracted from MRP height to find MRP raise or drop. (Note: MRP height almost always results in “raise.” An MRP drop may indicate a mismeasurement.)

Example 4-2

An order specifies an MRP height of 25 mm. The frame has a vertical dimension (B) of 46 mm. What will the MRP raise be?

Solution

Vertical decentration is calculated as follows:

Vertical decentration = MRP height – B 2

FIGURE 4-5 Before centration instruments were used, lenses were centered on a lens protractor, then marked by hand, sometimes using a ruler, sometimes freehand.

In the example, because MRP height is 25 mm and the frame B is 46 mm, then the following is true:

Vertical decentration = 25 – 46 2

= 25 – 23 = 2 mm

Because the MRP height is greater than half the B dimension, the vertical decentration is positive and the lens is moved up by 2 mm. The height above (or below) the horizontal midline of the lens may be visualized from Figure 4-7.

VERTICAL COMPENSATION FOR PATTERN ERRORS

As discussed in the previous chapter, some patterns may not have the center of rotation of the pattern at the boxing center. This may happen because the pattern is old and uses a different standard or because it is a homemade pattern and was made incorrectly.

When the rotational center of patterns are above the boxing center, the difference between these two locations must be compensated for. If the rotational center is above the boxing center, the distance between these two centers must be subtracted from the calculated MRP raise. (If the rotational center is below the boxing center, the difference is added to the MRP raise.)

Example 4-3

In this example, vertical decentration is 2 mm above the boxing center. A pattern has its mechanical center (center of rotation) 3 mm above the boxing center. How will the MRP raise have to be compensated?

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FIGURE 4-6 A marked cross on a lens has a short vertical and long horizontal component. This provides an immediate visual cue on where the 180-degree midline is located and where the boxing center of the edged lens will be. However, it is at the future location of the boxing center of the lens once the lens has been edged. Because of decentration, the cross is no longer at the geometrical center of the uncut lens blank. Is the lens concave side up or convex side up?*

*Answer: The lens is convex side up.

Solution

A pattern with its center of rotation 3 mm above its boxing center will cause the MRP to be 3 mm above the boxing center of the edged lens. Therefore 3 mm are subtracted from the normally calculated vertical decentration.

+2 mm – 3 mm = –1 mm

In other words, when this particular pattern is used, the MRP must be positioned 1 mm below the horizontal

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BOX 4-2

Centration Steps Using a Protractor

1.The lens is spotted.

2.Decentration per lens is calculated, and direction of decentration is determined.

3.The lens is placed face down on the protractor. The three lens dots are aligned on the 180-degree line with the center dot at the correct decentration.

4.If vertical decentration is required, the vertical decentration is calculated and the lens is decentered up (or down) as required.

5.Mark the newly found boxing center of the lens with a cross.

180-degree reference line to make the lens center properly.

LENS CENTRATION INSTRUMENTS

A centration instrument has many advantages over a lens protractor, even though the process is basically the same. These advantages are as follows:

1.The optical system of the instrument eliminates most parallax problems.

2.Internal illumination makes viewing of marks on the lens and lens segments easier.

3.The instrument contains a movable vertical reference line. This reference line may be moved left or right and makes horizontal alignment for decentration easier.

4.If the instrument just marks the lens, the system contains a lens-marking stamp that always strikes at the location of the boxing center. It gives a consistent, straight set of marks. This type of centration device is called a lens marker.

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Instruments that only mark a lens are now rare. Most lens centration devices also include a lens-blocking mechanism. Instead of stamping the lens and having to carry out an additional step of blocking the lens at the location of the stamp, the lens is blocked directly. This type of device is called a marker/blocker.

LENS CENTRATION WITH A CENTRATION INSTRUMENT

The following are the steps for use of a centration instrument for single vision lenses.

Step 1

The lens is spotted (as described in Chapter 2).

Step 2

If the instrument has blocking capabilities, a doublesided adhesive blocking pad is stuck on a lens block and the block is mounted on the instrument. Then the paper is peeled off the pad to expose the adhesive.

Step 3

The amount of horizontal decentration per lens required is calculated using the following formula:

Decentration per lens = (A + DBL) – PD 2

Step 4

The practitioner must determine whether the lens must be decentered to the right or to the left. In most centering devices the lens is face up. If the lens is a right lens and is facing up, decentration “in” is to the right. (see Figure 4-2, B). A left lens facing up would be decentered to the left (see Box 4-1).

Step 5

The position of the movable vertical reference line3 in the instrument is adjusted to the right or left by the amount of decentration calculated.

Step 6

The right lens is placed face up (front surface up) on the screen. The three spots on the lens are aligned with the horizontal line on the instrument screen.

Step 7

The center lens dot is placed on the movable vertical reference line. (The position of this line corresponds to horizontal decentration.)

3For single vision lenses, the movable vertical line is basically used as a place marker. When laying out single vision lenses, some people prefer not to use the movable vertical line at all. Instead they move the dot on the lens directly to the desired amount of decentration.

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Step 8

When the MRP height is specified, the lens is decentered up (or in rare instances, down). The amount of decentration is according to the correct number of millimeters of MRP raise (or drop).

Step 9

The handle is grasped and swung into place, or the button or footswitch is pressed. This blocks (or marks) the lens4 (Figure 4-8).

Example 4-4

A frame has an eyesize (A) of 54 mm and a DBL of 20 mm. The wearer’s PD is 66 mm. The lenses are already spotted. How must the instrument be set, and how must the lens placed in order to properly block the lens? The reader should assume that the lens is a left lens.

Solution

Lens decentration is calculated as follows:

Decentration per lens = (A + DBL) – PD 2

= (54 + 20) – 66 2

= 8 2

= 4 mm

To preset the movable vertical line in the instrument for the left lens, the practitioner should first recall in which direction the MRP should be moved. Because the wearer’s PD is smaller than the frame’s geometrical center distance or “frame PD,”5 the lenses will decenter nasally or inward. The lens is placed convex side up. Therefore the left lens is moved to the left. As Figure 4-9 shows, the movable vertical line is positioned 4 mm to the left of the central reference line.

Now the lens is placed face up in the instrument. It is aligned so that the central dot is at the intersection of the horizontal line and the movable vertical line. Figure 4-10 indicates that the other two dots must fall directly on the horizontal reference line.

The lens is blocked. The lens block is positioned at what will become the boxing center of the edged lens (Figure 4-11).

4To check for calibration, the practitioner views a marked or blocked lens through the instrument before it leaves the grid. This shows whether the mark or block is really at the origin of the instrument grid and is straight.

5“Frame PD” is equal to A + DBL.

INCREASING ACCURACY

The dots that are applied by the lensmeter can be small, as they should be. However, sometimes these dots tend to “disappear” into the black horizontal line of the centration device. One “cure” is to make the center dot larger than it might otherwise be by remarking with a marking pen.

Instead of increasing the size of the dots, an easy way to eliminate the problem of disappearing dots while simultaneously increasing accuracy is to position the dots 0.5 mm above the 180 line until all decentration positioning is completed. As a last step, when everything is completed, lower the lens back down 0.5 mm so that the dots disappear into the line.6

centering devices use a system that visually compares the blank to the required size. Two main systems exist.

CENTRATION UNITS THAT USE EFFECTIVE DIAMETER CIRCLES

The first system is a series of circles of known diameter that appear on the screen of the instrument (Figure 4-12). These show possible effective diameters (ED). The effective diameter of a frame is on the frame package or package insert, is listed in a frame reference catalog, or can be measured. (See Chapter 3 for how to measure the ED.)

When properly positioned, the decentered lens blank must enclose completely the circle of the same diameter as the effective diameter of the frame (Figure 4-13).

Ensuring Proper Size of the Lens

Blank

Nobody wants to intentionally waste a lens. To help avoid edging a lens that is not quite large enough, some

6Thanks to Glenn Herringshaw of Bloomington, Ind., for contributing this hint.

CENTRATION UNITS THAT USE

SUPERIMPOSED PATTERNS

A second system used in centration devices to ensure that a lens is large enough superimposes the shadow of the pattern on the screen of the centering device. The scale used in decentration, the lens blank, and the shadow of the pattern are seen at the same time.

This system works best when the pattern is exactly the same size as the frame. If the pattern size is the same size as the frame, the shadow of the pattern must be enclosed completely by the decentered lens blank. If it is not, any part of the pattern found outside of the lens blank will end up as “air space” (Figure 4-14). (This is the system used on many patternless edgers. The edged lens shape is drawn over the unedged lens on the screen.)

This pattern shadow system still can be used with patterns somewhat larger or smaller than the finished lens, but compensation must be made. To compensate, the size of the pattern must be known. This can be measured directly.7

7Pattern size can be calculated by adding the absolute value of the “set number” printed on the pattern to 36.5 mm.

Example 4-5

The eyesize of a particular frame is 50 mm and the pattern size is 46 mm. How much larger in every direction will the edged lens be?

Solution

The eyesize minus the pattern size is 50 minus 47, or 3 mm. This means that the lens will be larger by half that amount in every direction. The lens will be three halves or 1.5 mm larger than the pattern in every direction.

Knowing this, the practitioner can make a visual estimate with the use of the pattern shadow system, even if the pattern is smaller than the frame’s eyesize. For the lens to be large enough, the edge of the lens blank must clear the pattern shadow by half the difference between pattern size and eyesize—in this

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C H A P T E R 4 C E N T R AT I O N O F S I N G L E V I S I O N L E N S E S

case, 1.5 mm. If this amount of clearance is not present, the lens will not cut out.

Example 4-6

A lens has an eyesize of 58 mm. The pattern is measured and found to be 52 mm. The wearer’s PD is 68 and the frame DBL is 18, giving a 4 mm decentration per lens. When the lens is prepared for edging and viewed through the instrument, it appears as shown in Figure 4-15. Will the lens be large enough to cut out?

Solution

Answering the question requires finding the minimum distance from the edge of the pattern to the edge of the lens. To do this, the pattern size is subtracted from the eyesize and divided by 2:

The practitioner must add 3 to the superimposed pattern shadow in every direction to simulate edged lens size. Applying this to Figure 4-15, it is evident that less than 3 mm is between the pattern edge and lens edge. This means that the lens will not cut out.

MINIMUM BLANK SIZE

The smallest lens blank that can be used for a given prescription lens and frame combination is called the minimum blank size (MBS). The MBS depends on the following:

•The effective diameter of the frame

•The amount of decentration of the optical center away from the boxing center

Eyesize – Pattern size

2

or

58 – 52 = 3 mm

2

If no decentration occurred and the lens optical center was positioned exactly at the boxing center, the minimum blank size would be equal to the effective diameter (Figure 4-16).

Minimum blank size also must allow for extra size required because of lens decentration. When 1 mm of