Calculating Edge Thickness

[Whitwoo]

Hi everyone
I am going through a few example questions, and got to this one and couldn't work my calculator properly. Also, how often does a dispensing optician actually use this equation?

The equation used in the example is:
Fmeridian = Fcyl Sin2 angle + Fsphere

The question is:
Calculate the edge thickness along the 135 meridian of a RE -3.75/-1.75x90 plastic lens (n=1.600) if the centre thickness is 2mm. The frame details are:
Frame 58-14 Effective Diametre 60mm PD RE 31mm
The diameter along the 135 meridian is 52mm and the front curve is +4.00D

My calculator is a casio.

Any one that can help?

cheers

[lensgrinder]

Here is a link to Darryl's article "Methods for Estimating Lens Thickness"

[Jacqui]

I found a cute software program at www.pocketsoft.co.uk that I use for this.

[Memoir]

The equation used in the example is:
Fmeridian = Fcyl Sin2 angle + Fsphere

The question is:
Calculate the edge thickness along the 135 meridian of a RE -3.75/-1.75x90 plastic lens (n=1.600) if the centre thickness is 2mm. The frame details are:
Frame 58-14 Effective Diametre 60mm PD RE 31mm
The diameter along the 135 meridian is 52mm and the front curve is +4.00D

F135 = Fcyl sin2 θ + Fsph
= -1.75 (sin2 45) + -3.75
F135 = -4.625D

F = F1 + F2
F2 = F – F1
= -4.625 - +4.00
F2 = -8.625D

r1 = (n – 1)1000
F1
= 600
4
r1 = 150mm

r2 = (1 – n)1000
F2
= -600
-8.625
r2 = 69.56mm


s1 = r1 - √(r12 – y2)
= 150 - √(1502 – 262)
= 150 - √21824
= 150 – 147.73
s1 = 2.27mm

s2 = r2 - √(r22 – y2)
= 69.56 - √(69.562 – 262)
= 69.56 – √ 4162.59
= 69.56 – 64.52
s2 = 5.04mm

e = (s2 – s1) + t
= (5.04 – 2.27) + 2
e = 4.77mm along the 135 meridian

therefore, the edge thickness along the 135 meridian is 4.77mm.

[Memoir]

s1 = r1 - √(r12 – y2)
= 150 - √(1502 – 262)

the 2 after r1 and y etc in the sag formulas should be superscript (ie 'squared')

[Memoir]

My calculator is a casio.

For the power at meridian formula, try inputting the calculation with brackets.. ie..

enter:

minus, 1, decimal, 7, 5 ( -1.75 )
open bracket, open bracket ( (( )
sin, 4, 5, close bracket, x2, close bracket ( sin45)2) )
plus, minus, 3, decimal, 7, 5, (+-3.75)

equals...

maybe? :p LOL

[tmorse]

e = (s2 – s1) + t
= (5.04 – 2.27) + 2
e = 4.77mm along the 135 meridian, therefore, the edge thickness along the 135 meridian is 4.77mm.

All great stuff. In some textbooks this last formula can be best remembered as:
Tc + Tp = s1 + s2
Thickness of center + Thickness of periphery(edge) = sag of 1st surface
+ sag of second surface of lens

('c' comes before 'p' in alphabet, and '1' comes before '2')

[lensgrinder]

F135 = Fcyl sin2 θ + Fsph
= -1.75 (sin2 45) + -3.75
F135 = -4.625D

F = F1 + F2
F2 = F – F1
= -4.625 - +4.00
F2 = -8.625D

r1 = (n – 1)1000
F1
= 600
4
r1 = 150mm

r2 = (1 – n)1000
F2
= -600
-8.625
r2 = 69.56mm


s1 = r1 - √(r12 – y2)
= 150 - √(1502 – 262)
= 150 - √21824
= 150 – 147.73
s1 = 2.27mm

s2 = r2 - √(r22 – y2)
= 69.56 - √(69.562 – 262)
= 69.56 – √ 4162.59
= 69.56 – 64.52
s2 = 5.04mm

e = (s2 – s1) + t
= (5.04 – 2.27) + 2
e = 4.77mm along the 135 meridian

therefore, the edge thickness along the 135 meridian is 4.77mm.

Keep in mind that you are decentering the lens inward by 5 mm this will increase y.
You can use the law of cosines to find this measurement.


a = decentration = 5
b = half diameter @ 135 = 26
C = angle of half diameter = 135



This will make sag 2 6.8 and sag 1 3.03 which will make the edge thickness 5.8 @ 135