# Thread: Realtionship between lens curve and diameter?

1. ## Realtionship between lens curve and diameter?

Hello,

I know that there is a fixed relationship between the curve of a lens as measured with a lens clock (+8.00, -4.50, etc) and the radius of the sphere which would result if you were to continue drawing the curve, but I don't know what that formula is. In the picture below, the lens is the yellow section, with a curve of +8. I need to find r. Any help would be appreciated.

2. A geneva lens clock is depressed .01mm for each graduation on the dial. It can be used as a thickness guage on a contact lenses placed convex down on a table. This should let you know how much the center pin is depressed for each mark on the scale relative to the immobile outter pins. The diameter of any sphere is twice that of it's radius.

Chip

Chip

3. Assumption is the geneva lens clock is calibrated for 1.53 tooling.

n=index, in this case, 1.53
D= diopter of the curve

Radius = (n-1)/D= .06625 meters or 6.625 centimeters.

4. I know that there is a fixed relationship between the curve of a lens as measured with a lens clock (+8.00, -4.50, etc) and the radius of the sphere which would result if you were to continue drawing the curve, but I don't know what that formula is. In the picture below, the lens is the yellow section, with a curve of +8. I need to find r.
The exact equation for the surface radius r is given by:

$r=\frac{x^2 + s^2}{2s}$

where s is the sagitta or height of the curve measured by the lens clock and x is the separation of the pins, typically around 10 mm. If we assume a standard tooling index of 1.530, the exact equation for the surface power F153 is given by:

$F_{1530}=\frac{1060s}{x^2 + s^2}$

Note that this equation would result in a non-linear scale on the face of the lens clock. However, since the pin separation is relatively narrow, the approximate sag formula can be used to produce a linear scale, which is fairly accurate up to a surface curves of 15 diopters:

$F_{1530}=\frac{1060s}{x^2}$

So every 0.1 mm of surface height s is roughly equal to 1.00 D of surface power at a refractive index of 1.530.

Best regards,
Darryl

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