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Opthalmic
Frames The formula for determining the smallest possible lens blank which will work for any given frame and PD combination is as follows: Minimum Blank Size (MBS) = (GCD – PD) + ED In the examples on the previous page, the ED is the same as the A measurement since the frame illustrated is round. Therefore, the theoretically smallest lens size that can be used in Example A is 44 mm. The theoretically smallest lens which can be used in Example B is equal to 48 mm. Example A: Example B: These minimum blank sizes are considered to be theoretical for two reasons:
It is necessary to
calculate vertical decentration when determining where the line of a bifocal
segment is positioned relative to the datum line of the frame. This may
be expressed in the following formula. In the example to the left, since the B measurement = 50 mm the datum line is 25 mm from the bottom most portion of the lens. The seg height is 22 mm. Subtracting half the B measurement from the seg height results in : 22 – 25 = –3. Therefore the segment line is decentered 3 mm below the datum line, or would be commonly referred to as seg = �3 below.� If the seg height were say, at 28 mm, the result would be + 3 mm positioning the seg line 3 mm above the datum line or �3 above.�
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