# Thread: What is the thin-edge substance of the lens?

1. ## What is the thin-edge substance of the lens?

The lens, 1.75DS / +1.75DC Χ 180 is made in glass, n = 1.56, in toric form
with a 4.00D base curve and edged 40mm round. The thick-edge thickness
is 3.14mm. What is the thin-edge substance of the lens?

I bealive i have to find the sag of +1.75; -1.75 and -4.00 (of course without the sign, but then i don't know the formula to find the thin edge subsatnce.

2. The fact that the base curve is "-4.00" suggests that this is a plus-cylinder configuration, with a toric surface on the front, and a concave sphere on the back (NOT your normal configuration these days) - not that it matters, other than, you may need to keep that in mind to keep your signs straight. In any case, because this is a compound lens (meaning, it has a non-zero cylinder power), there will be points of maximum and minimum thickness around the perimeter.

The two powers in the major meridia are -1.75 and plano (not -1.75 and +1.75; we have to produce the total power in the cylinder meridian, not the cylinder power).

The simple power formula is Power = Front + Back (note that the back is almost invariably negative). Refactor that to Front = Power - Back.

Front(sph) = -1.75 - (-4.00) = 2.25
Front(cyl) = 0.0 - (-4.00) = 4.00

The edge thickness in any meridian at 40mm is figured thus:

ET = (Sag@40(back) - Sag@40(front)) + CT

Since we know the ET in the thickest meridian (the one where the 2.25 curve obtains), solve for CT using that:

CT = 3.14 - (Sag@40(4.00) - Sag@40(2.25))
CT = 3.14 - (1.44 - 0.81)
CT = 2.51

For the ET in the thin meridian:

ET = (Sag@40(back) - Sag@40(front)) + CT
ET = (Sag@40(4.00) - Sag@40(4.00)) + 2.51
ET = 2.51

Of course, in this case, we were really done once we figured the CT, because the ET in the thin meridian is the same as the CT (because the front and back curves are the same).

All this assumes that they're looking for thin lens calcs, which I'd expect to be the case.

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