# Thread: SAG of astigmatic lenses.

1. ## SAG of astigmatic lenses.

Hello everyone.

I'm taking a day release course in Ophthalmic lenses and I'd appreciate some clarification with a few areas of SAG if anyone can.

I can calculate edge thickness of round and shaped lenses easily enough but I got a bit lost during the lecture when it came to astigmatic lenses.

I have a lens (n=1.5) -4.00/-2.00 x90 edged 50x40mm oval with an axial thickness of 2mm. I have to find the vertical and horizontal edge thickness.

I know my principal powers are -4.00 and -6.00 but that's where I blank out.

2. You would use the sag approximtion formula in this case since you were not given a base curve.

$\ sag \approx {({d \over 2})^2 \times D \over 2000(n-1) }$
where
d is the diameter in mm
D is the power
n is the refractive index

Now use each power in the major meridians (ie -4.00 X 90, -6.00 X 180) and use 50mm for your power at 180 and 40mm for your power at 90.

$\ sag \approx {({50 \over 2})^2 \times 6.00 \over 2000(1.5-1) }$

$\ sag \approx {{3750 \over 1000}$

$\ sag \approx 3.75$

Now add this to your axial thickness and you have 5.75 @ 180

$\ sag \approx {({40 \over 2})^2 \times 4.00 \over 2000(1.5-1) }$

$\ sag \approx {{1600 \over 1000}$

$\ sag \approx 1.6$

Add this to your axial thickness and you have 3.6 @ 90

3. That's a great help, many thanks.

If you can clarify one thing though. How do you decide which power to use for each diameter?

4. Your lens power was -4.00 -2.00 X 90. You will produce all of your cylinder power 90 degrees away from the axis, so at 90 you only have your sphere power and at 180 you have your sphere power plus your cylinder power which is -6.00.
The measurements were 50X40 50mm horizontally and 40mm vertically. The power horizontally is -6.00, the power vertically is -4.00.

5. So if I'd said the lens was 40X50mm oval you'd quite simply use -4.00 with 40mm vertical and -6.00 with 50mm horizontal?.

6. Originally Posted by Mymyopia
So if I'd said the lens was 40X50mm oval you'd quite simply use -4.00 with 40mm vertical and -6.00 with 50mm horizontal?.
40mm would be the horizontal measurement with 6.00 D in the 180th meridian.
50mm would be the vertical measurement with 4.00 D in the 90th meridian

7. I feel like such a idiot for asking but I need help once again. I've got a grip on the previous SAG type questions but I've got stuck once again.

We were given the question....A lens +4.00/+2.00 x90 is edged 48X42 oval. If the thin edge thickness is 1.4mm and n=1.498 calculate the thick edge substance if made on a -5.00 sphere curve.

I've transposed my RX into +9.00dc x180 +11.00dc x90
-5.00 ds

I know which powers and diameters give me my thick and thin edge but I can't get the correct answer (3.46mm)

I've noticed my problem is never how to calculate SAG it's how to apply it. I promice this will be the last time I ask.

8. removed

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•