Calculating Cylinder Power in an Oblique meridian

**musicvirtuoso** · 08-10-2010, 12:16 PM

Does anyone know a more exact way of calculating the power of cylinder in an oblique meridian other than the sine-squared approximation method: Fa = Fcyl [sin(a)]^2 ? Or does the torsional effect prevent from getting an exact power?

**Darryl Meister** · 08-11-2010, 12:31 AM

There has been a lot of debate over this topic over the years. Some argue that the oblique meridians of a cylinder do not produce a true "power" because rays of light refracted through these oblique meridians do not intersect; instead, they are "skew" rays. This is due to the torsional component. Others have argued that the cross-section of oblique meridians still has a curvature component, so you can think of the meridian as producing surface "power" of some sort.

In either case, any "power" produced through the oblique meridians would be due to the surface curvature through the meridians. And the "sine-squared" formula is an approximation of that curvature. More advanced optical applications, however, like vision science research or ophthalmic lens design, will generally consider some form of the "torsional" component in addition to the curvature through the principal meridians of the lens, referred to as the "curvital" component.

You can do a search on OptiBoard to pull up some old threads about this. We used to banter back and forth about this very topic every now and then in the Ophthalmic Optics forum. Michael Keating's Geometric, Physical, and Visual Optics also provides a lengthy discussion of this stuff. You can probably find a new or used copy online. I'll warn you, however, that at least a rudimentary knowledge of algebra and trigonometry is helpful when researching this stuff.

Best regards,
Darryl

**One science** · 08-11-2010, 04:15 PM

When a lens clock is used to measure the front surface powers of a plus power piano-cylinder will be found that time surface has a meridian of maximum power and a meridian of minimum power, the principal meridians. In other meridians the lens clock will read some intermediate power. Hope this will help.

**MikeAurelius** · 08-11-2010, 08:15 PM

There's certainly power there...and it varies by 1.1111 times the diopter power of the cylinder per degree of angle. Whether the eye "uses" it is another question altogether.

**Darryl Meister** · 08-12-2010, 12:08 AM

The issue of debate is that the oblique meridian of a cylinder has surface power, since oblique meridians have curvatures, but it does not have focal power, since oblique meridians do not actually produce a focus.

Best regards,
Darryl

**MikeAurelius** · 08-12-2010, 07:48 AM

Originally Posted by MikeAurelius

There's certainly power there...and it varies by 1.1111 times the diopter power of the cylinder per degree of angle. Whether the eye "uses" it is another question altogether.

oh...wow...I definately should *NOT* post when I am brain dead...

surfacing calculations use oblique power all the time to calculate the meridinal power to use prism for decentration. Each angle has 1.1111 PERCENT of the total cylinder power (from the hypothetical 0 to 90 degrees).

**MikeAurelius** · 08-12-2010, 07:50 AM

Originally Posted by Darryl Meister

The issue of debate is that the oblique meridian of a cylinder has surface power, since oblique meridians have curvatures, but it does not have focal power, since oblique meridians do not actually produce a focus.

Best regards,
Darryl

I find that to be a very interesting post, Darryl, and I can certainly see the point of the debate. We've always known that we can focus at the hypothetical 0 and 90 degrees of the cylinder, but extrapolating further, there MUST be a focus at all of the angle between, otherwise the dot of light that is focused on the back of the eye would be a cross instead of a (corrected) circle.

What do you think?

**Darryl Meister** · 08-12-2010, 10:02 PM

The "skew" rays refracted through a single oblique meridian of a cylinder lens do not come to a focus or intersect each other. Only rays through the principal meridians intersect each other in the same meridian, since there is none of the "torsional" effect discussed above. Skew rays do, however, contribute to image formation by intersecting rays refracted by the other meridians of the lens.

The rays refracted through the meridians of a cylinder lens generally intersect each other along two lines, corresponding to the focal lines of the lens. This bundle of rays, and the two focal lines that they produce, form the "Conoid" or "Interval of Sturm." Each ray refracted through the lens, whether through an oblique meridian or a principal meridian, contributes to the formation of both focal lines.

One way to think about the "torsional" component and the principal meridians of a cylinder lens is to use an analogy to a soda can, lying on its side. Now rest a ruler on the side of the overturned can, and press down on one end of the ruler. When the ruler is laid across the side of a can at a perfect right angle to the length of the can, and then pressed at one end, the ruler will move straight up and down, like a seesaw. Now turn the ruler so that it forms an angle between the side of the can and the length of the can, and press down again. This time, the ruler will "twist" as it pressed down, clockwise on one side and counter-clockwise on the other.

A cylinder lens does not actually produce a point focus on the retina. In reality, the cylinder lens creates an astigmatic focus, with two focal lines, that serves as the object for the eye. Assuming that the eye has an equal but opposite refractive error, the astigmatic focus of the cylinder cancels out or neutralizes the astigmatic focus of the eye, resulting in a single point focus. Just as a plano plus-cylinder lens set into a plano minus-cylinder lens of the same magnitude will result in a plano lens.

Best regards,
Darryl

**MikeAurelius** · 08-13-2010, 08:10 AM

Yes, ok, that I can understand (I've had two cups of coffee this morning).

**HarryChiling** · 08-13-2010, 10:50 PM

Originally Posted by musicvirtuoso

Does anyone know a more exact way of calculating the power of cylinder in an oblique meridian other than the sine-squared approximation method: Fa = Fcyl [sin(a)]^2 ? Or does the torsional effect prevent from getting an exact power?

If you have time to meet up I am in the area next week and can let you look over the book Darryl referenced above. It is definitely a great read and provides a lot of information.

**musicvirtuoso** · 08-14-2010, 11:40 AM

Originally Posted by HarryChiling

If you have time to meet up I am in the area next week and can let you look over the book Darryl referenced above. It is definitely a great read and provides a lot of information.

That sounds great! You're going to be in DC next week? Send me a message.

**Wlada** · 07-11-2015, 05:52 PM

http://tobiasbreitenstein.de/HFAK/apps/CurvedGlases/

**MakeOptics** · 07-11-2015, 10:39 PM

Originally Posted by Wlada

http://tobiasbreitenstein.de/HFAK/apps/CurvedGlases/

Nice app, I was reading through your blog and saw the arduino. I am an arduino enthusiast as well. I have created a whole bunch of little household trinkets. Nice to see a hardware and software programmer in the crowd.

**MakeOptics** · 07-13-2015, 05:43 PM

Originally Posted by Wlada

http://tobiasbreitenstein.de/HFAK/apps/CurvedGlases/

You could also improve accuracy on Keatings equation by substituting the geometric mean instead of the arithmetic mean when compensating the torsional component in the dioptric matrix. It won't make for a huge difference, especially since the whole equation is third order anyway but it does lend a slight advantage. Also the refraction even at the optical center needs compensation due to the ray path being offset by refraction through the thickness of the lens and emerging at a slightly off axial point, a simple prism compensation can help to bring that vector back inline.

Further a lens tilt is going to require a bit more information from the user and calculation on the part of the calculator. The frame tilt can be easily computed with a simple protractor, however most lenses are decentered in a real life scenario requiring additional compensation.

Another point of tilt compensation is the bevel placement, in sunglasses which represent the best possible scenario the lens tilt is easier to compute as the bevel is often placed at the front so the lens can be compensated for the decentration alone. Worst case scenario and the most likely scenario is a frame fit scenario where the tracer provides Z data to to place the bevel more accurately. In the preceding scenario the thickness as well as the bevel placement must be factored in and compensated for. In the case of a lens with positive horizontal and vertical decentration often the frame tilt leads to too high a compensation and often a non-adapt in moderate to higher Rx's. I have seen many anecdotal pre compensations from opticians to try and correct for these variances with little understanding as to why the effect is present. Blendowski glosses over a few of these points in his paper on the subject which provides an improvement to Keating's formula and mentions it as a side note in a discussion with Hoffman. With very little effort the formulas can be derived and factored in.

**Wlada** · 07-13-2015, 05:58 PM

Originally Posted by MakeOptics

Nice app, I was reading through your blog and saw the arduino.

Make Optics, I am sorry, but blog is not mine, and I am not an author of these apps. I just posted this link, it seemed to me appropriate.

**MakeOptics** · 07-14-2015, 05:22 AM

Originally Posted by Wlada

Make Optics, I am sorry, but blog is not mine, and I am not an author of these apps. I just posted this link, it seemed to me appropriate.

Welcome either way, a great tool. Thank you for sharing it.

**Wlada** · 07-21-2015, 05:32 PM

Originally Posted by MakeOptics

Blendowski glosses over a few of these points in his paper on the subject which provides an improvement to Keating's formula and mentions it as a side note in a discussion with Hoffman. With very little effort the formulas can be derived and factored in.

Can you tell me what is the name of paper from Blendowski?

**MakeOptics** · 07-21-2015, 07:27 PM

Originally Posted by Wlada

Can you tell me what is the name of paper from Blendowski?

http://www.fbmn.h-da.de/~blendowske/...S_jan_2002.pdf

**Wlada** · 07-22-2015, 02:59 AM

Thanks!

**iokuok2** · 01-19-2018, 07:17 PM

(7 and a half years later...That certainly nudges the comfort zone of my basic certification. So the "soda can" on the correcting lens and the "soda can" in the eye are on the same axis, not at 90 degrees apart , just one is concave, the other convex at the same but opposite strength?
I'm wondering just out of curiosity, given the skew rays and torsional effect is the impact of cylinder linear or would persons (under the bell curve) with high power sphere issues experience more impact by the same power of needed cylinder correction as mild sphere corrections with the same needed cylinder addition? --for example -1.00 -.25 X 180 or -8.00 -.25 x 180? Does this contribute then to more skewing and greater sensitivity across the thicker steeper correction even for a mild cyl if one already has need of a larger correction? Is one of these two wearers going to notice the impact more than the other if that small amount of cyl is off in their correction?

**Tallboy** · 01-20-2018, 01:32 PM

iokuok2 I've always thought it was best to view it as a percentage of total power in the strongest meridian. I think a -1.00 -0.25 @180 would notice the impact much more. I would be interested to know if I am wrong in thinking this, I'll probably do some reading today on it.