Cylinder Power in the 180-degree meridian

**Bob Price** · 01-09-2006, 10:49 PM

Apparently, the formula "D=Dcyl sin^2 Ax" isn't the most accurate for finding the power of the cylinder in the 180-degree meridian. Is there a more accurate formula that can be used, without getting too technical?

**Darryl Meister** · 01-09-2006, 11:59 PM

Is there a more accurate formula that can be used, without getting too technical?

It depends on what your application is. If you are just calculating thickness, your formula will suffice. If you are wanting to calculate prism at off-center points, you will have to get more technical.

**HarryChiling** · 01-10-2006, 07:08 AM

Will the equation gain more accuracy if measured in radians?

**Bob Price** · 01-10-2006, 11:08 AM

Originally Posted by Darryl Meister

It depends on what your application is. If you are just calculating thickness, your formula will suffice. If you are wanting to calculate prism at off-center points, you will have to get more technical.

My goal is to be able to figure out everything on a workticket without the use of our computer software. Right now I'm using a graphing calculator and hope to upgrade to an IPac. Hopefully this will allow me to have access to an Excel spreadsheet (that I can carry around) where I can combine all of the formula's into one organized spreadsheet. Just enter the Rx and all of the information that is needed to process a lens is calculated. This is the easiest way for me to learn and I have found it to be quite fun, but challenging. It seems that all of the optical calculators I have found come up with slightly different values. This is frustrating, but is it because of my question above, that there are different formulas to get the final result? On my question, it appears that by using the formula I mentioned for calculating decentration, then it loses accuracy and another formula needs to be used.

I don't know if you remember me calling you abut 5 years ago, when you helped me with the equation for figuring the tool needed. Somehow I erased the equation and now have renewed interest, since figuring out how to get it back in the calculator. Thanks for your help.

**Darryl Meister** · 01-10-2006, 01:15 PM

Originally Posted by Harry

Will the equation gain more accuracy if measured in radians?

If you mean using radians for Ax, you will actually get the same answer either way...?

Originally Posted by Bob

My goal is to be able to figure out everything on a workticket without the use of our computer software.

Then you will definitely need a more "technical" approach. Calculating prism accurately requires a couple of simultaneous equations that use quite a bit of trigonometry. Fortunately, you can just copy them from a textbook or something, but you would still need to verify the sign convention of your input and output and that sort of thing.

Originally Posted by Bob

This is frustrating, but is it because of my question above, that there are different formulas to get the final result?

Yes, but if they're correct, you'll get the same answer either way.

**HarryChiling** · 01-10-2006, 01:31 PM

I am not sure how else you could go about finding the power in that meridian without using the the

D=Dsph+Dcyl*sin^2(180-axis)

What other method could you use to more accurately get this power? And where does this equation lose it's accuracy?

**Darryl Meister** · 01-10-2006, 01:39 PM

Originally Posted by Harry

What other method could you use to more accurately get this power? And where does this equation lose it's accuracy?

It's not so much the power he's worried about, but rather calculating the prism at points away from the optical center (or, conversely, calculating how to induce prism in order decenter the optical center).

Also, the sine-squared formula just approximates the curvature of the surface through a particular meridian. It doesn't actually represent the "power," since a sphero-cylindrical lens technically has no power in meridians other than the principal meridians (the rays of light refracted in these other meridians are skew rays, and do not actually intersect at a focus). Also, recall from our earlier discussions of Keating's Dioptric Power matrix that the power of a cylinder has both a curvital component (i.e., the sine-squared component) and a torsional component.

**Bob Price** · 01-10-2006, 08:53 PM

[QUOTE=Darryl Meister]Then you will definitely need a more "technical" approach. Calculating prism accurately requires a couple of simultaneous equations that use quite a bit of trigonometry. Fortunately, you can just copy them from a textbook or something, but you would still need to verify the sign convention of your input and output and that sort of thing.

Would you please lead me in the right direction for the equations. I might already have them under my nose and will look for them tonight. The books I am working with are "Understanding Lens Surfacing" and "Introduction to Ophthalmic Optics". Thanks again for your input.

**Darryl Meister** · 01-11-2006, 02:08 AM

Would you please lead me in the right direction for the equations. I might already have them under my nose and will look for them tonight.

Just about any book on ophthalmic or geometrical optics should have them. Clinical Optics, Principles of Ophthalmic Optics, Optics of Ophthalmic Lenses, Geometrical, Physical, & Visual Optics, and so on.