Optician Training: How To Use The 30 45 60 Rule To Determine Lens Powers In Oblique Meridians

[@Randomanimations19]

Thanks man you explain this well Ive been at my job for two weeks and decided im gonna study 30 minutes a day to become an optician a licensed one in Illinois just because I wanna know more about my job right now im a tech and I do digital optometrics and while I have an idea of how this stuff works I need to actually learn. Thanks for this free resource im gonna drill this 30 45 60 thing tonight for homework.

[@lynnettegarcia8125]

I started teaching a Dispensing Optician course with at my local Adult school and regularly refer my students to your video's as added references. I appreciate the breakdown and simplicity your video's provide, excellent resource and I joined your Opticianworks program as a refresher for myself, FABULOUS.

[@LaramyKOptical]

Thanks. We plan on starting the dispensing series the end of this month. Those should take us through 2022 and then it will be time to hang up my dry erase markers and wander off into retirement.

[@rickyachin]

I can't say this enough, The way you explain this makes it so easy to understand, I really appreciate the time and effort placed in doing these videos. It helps a great deal when explaining this to my staff. And a great refresher for myself as well.😀

[@RaquelHi]

I appreciate the fact that you go twice or more on explaining.

[@luvlessmess09]

I just started opticianry school. I was so lost but You! Thank you for these videos

[@Alebso]

Just who would be searching for a better lecture and explanation other than this?!! Super really super and cool with lovely humor 👍👍👍

[@LaramyKOptical]

Thanks! John

[@luisgomez1211]

Hello Laramy, I'm here to listen your interesting classes of Optical...now the 30 45 60 rule...Thank you...

[@marthaecarrillohernandez9564]

He is an outstanding teacher!! It explains everything soooo clearly! Thank you!!!!

[@LaramyKOptical]

And he thanks you for saying that. John

[@Singularity7425]

Thank you sir! Great lessons all around. They have helped so much!

[@LaramyKOptical]

Thanks! Love hearing from you. John

[@AceStrife]

These videos are slowly helping me, as a consumer, understand optometry in order to provide myself the ability to help prescribe a proper prescription, though I'm obviously missing all practical experience and it's difficult to piece together from what I've watched so far. I'm just slightly off in one of my eyes and over the last 3 years I've not had any contact lens prescription that completely fixed it. It's been bothering me so much and probably contributing to my eye getting significantly worse over a rapid period of time (due to brain trying to ignore it) that I started to look up things explaining prescriptions. If anyone sees this comment and can help push me in the right direction as to what's wrong, my issue is this: If I rotate the contact lens very slightly clockwise (or counterclockwise if looking AT my eye), my vision becomes "crystal clear" instead of "crystal blurry". My optician says I'm at an extreme value with my astigmatism and it's impossible to get a prescription because I'm "in-between" a value. My current Rx is -3.25 -2.25 x180. Other eye is -3.25 -1.75 x180. The first number has gone up more over time than the second number. Maybe things will make sense after I finish watching all the videos on this channel, but as of right now, I wish I actually knew an optician and could chat about things.

[@LaramyKOptical]

In contact lenses it is all about the middle number not the first number. You want to watch this: kzbin.info/www/bejne/omXJh5JtfpJrq8k Yes, as the middle number goes up a contact lens gets harder and harder to fit. Thing is most come in 5 degree steps. You must be between 5 degrees (hence when you rotate it things become clear) but if no lens comes in the exact degree that you need for perfect vision it ain't gonna happen. I think there are customs contact lenses available. I'd look for a true contact lens specialist over a doc or an optician.

[@jimhaggard7436]

It sounds like you need a contact lens with a slightly different axis of astigmatism. You might need a 170 axis instead of 180. Some contact lens brands come in five degree increments (175 might be perfect), but others are every 10 degrees. Another option is to lower the cylinder by a half and increase the sphere by a quarter. Your new Rx would be -3.50 - 1.75 x 180 and the other eye (if the same initial complaint) would be -3.50 - 1.25 x 180.

[@AceStrife]

​@@jimhaggard7436 Unfortunately, I gambled on the axis thing and it resulted in actually worse fit, to the point of being permanently out of focus (unusable). There didn't seem to be a difference between 180 and 0, so choosing 170 or 10 would've been the same, if I understood things right. I also tested some old prescription with a cylinder and the vision was also worse.. but also in strength too. Since writing this comment 2 years ago, I've discovered that my eyes change during the day of wearing my contacts. My left eye gets more in focus, and my right eye goes out of focus. Eventually it gets bad enough I just have to take them out because of the annoyance. Have yet to go back to an optometrist to talk about it.. honestly haven't really left the house since Covid started. And yeah, I did look for lenses with adjustments down to 5, but never found any for astigmatism (toric). Custom lenses cost about 5x the price of stock lenses; completely unaffordable. Currently using -4.25/-2.25/180 and -3.50/-2.25/180, just to update the old comment.

[@jimhaggard7436]

@@AceStrife thanks for the response. Yes, 180 is equivalent to 0. When you look at the patient’s eyes and the hash mark on the Toric contact lens isn’t sitting snugly at 6:00 like it’s supposed to, you look at the rotation of the lens and follow the “LARS” rule. If the rotation is to the left, order a new trial toric lens that is ten degrees (or more depending on the amount of rotation) higher in number (010 degrees if the original was 0 or 180). If it rotates right, subtract ten degrees from the original 0 or 180, which would be 170.

[@kirtanpatel9518]

Very clear explanation

[@IdekWatMyUserShuldB]

I really needed this description! Thank you for spending the time to share this!

[@rafaelamaya1978]

Thank you very much for your videos. could we do the same with a progressive? And if we could, would it be the same for one conventional and one free form?

[@LaramyKOptical]

LOL Never even thought about that and certainly never got this question before - but "no." I can't even wrap my head around exactly why, but, no. I guess the bigger question would be, "why would you want to or need to?" The 30/45/60 rule falls under the Optical Parlor Tricks category and has little use in the real world especially with PIOMF calculators everywhere.

[@johnkaramouzis6996]

Hey! Great Videos! I think this technique need a little bit of refinement. A more precision approach to this is to divide your Cyl Power / 90 deegrees. Then you have how much power changed per degree. So in your last example its: 1.75 / 90 = 0,019 dpt / degree so at 180 degrees its +6,50 - (60 degree * 0,019 dpt/degree) = 5,36 dpt

[@sanjayghorai8521]

Great lessons,,, thanks sir...

[@LaramyKOptical]

Thank you!

[@ahmid464]

is it 0- 180 or 180-0 around the 10:40 mark

[@Nintendo_Gamer1889]

Very helpful thanks.

[@AkumaGaming22]

I'm a tad confused on the first example because the solution was -7.43 at 180 degrees but I thought you said it can't be beyond -7.00 or less than -5.00. sorry if it's a dumb question

[@LaramyKOptical]

Different example different powers.

[@Shraddhak98]

Will the powers at 90 and 180 always be the same then?

[@LaramyKOptical]

OpticianWorks.com

[@rohanbanerjee9852]

Sir if a person power is -6.50/ -3.50 * 30 then here we will multiply 0.25 with -3.50 directly?

[@venkatramb620]

Excellent sir.

[@greatestever9663]

Hello! First time viewers here and quick question? Is there a particular order where to watch these videos or just as they were uploaded? Thank you and look forward to studying with you!

[@LaramyKOptical]

Click Sort By and Date Added that would get you as close as you can. The perfect order appears on the OpticianWorks website.

[@krishanlal5680]

doyou have any video on how to convert positive cylinder to negative clinder and vice versa ?

[@LaramyKOptical]

YES! Same KZbin Channel.

[@krishanlal5680]

@@LaramyKOptical what is title of video for this ?

[@LaramyKOptical]

@@krishanlal5680 kzbin.info/www/bejne/qIewiGiXa9Z9bKs

[@krishanlal5680]

@@LaramyKOptical Thanks. Can you suggest some book (suitable for self study) for learning these basics which you teach ?

[@LaramyKOptical]

@@krishanlal5680 www.OpticianWorks.com

[@brendagomez8311]

Hi I’m trying to watch your videos in order but you keep talking about a previous video what video is that please help ASAP

[@LaramyKOptical]

Click on Sort By and you can have them show up in the order they were posted.

[@brendagomez8311]

@@LaramyKOptical will do thank you! I will be taking the ABO with a friend of mine and we love your videos ❤️

[@LaramyKOptical]

@@brendagomez8311 Of course the videos are in perfect order and augmented by great content and additional examples on the OpticianWorks website!

[@wetworkwolf]

Could you please do a video on how to calculate prism based on lenses that do not follow the 30-45-60 rule? (Rx at x020, x010, x015?)

[@LaramyKOptical]

They are in video library. Power in oblique meridians.

[@ebrahimmahmoud2654]

Thanx alot sir, but Q1) what about 15 & 75 degrees?? Q2) why you calculate plus cyl for 90 axis & calculate minus cyl for 180 axis??

[@liliadelima352]

That’s awesome! Thanks so much!

[@LaramyKOptical]

Thank YOU, Lilia!

[@romm1112]

Thank you. I just found you. I have 12 years working in a lab and now I want to get my ABO license. Can you please tell me what of your videos should I start to have an order?

[@LaramyKOptical]

Rudy, You will want to sign up for the OpticianWorks program. The videos appear in order as you work through the rest of the material. John

[@holls33mm]

I feel like all I've learned so far is math and formulas. Does it ever end? I mean, don't get me wrong, I seriously appreciate all the visualizations, because without them i'd be lost. You're the best. But It seems so math-based!

[@LaramyKOptical]

Yes, yes it does. You may find the full OpticianWorks program a little easier to navigate. The foundation you need to build on is all math and science yet the actual building is all general theory and then the good stuff - frames, lenses, sales, making glasses, etc. John

[@tariqshahtalib1726]

Sir please give a proper numbering to your videos that we can watch it and understand Thanks

[@LaramyKOptical]

Click on the Oldest button and that will get you about right. Better join OpticianWorks.com.

[@sarahmason5818]

Hello! Just a quick question about your notation - you use '@' but the actual axis of the power is then perpendicular to this? Is that correct? Thank you!

[@LaramyKOptical]

When I use "at" then I am indicating the power at a given axis. If I say -4.00 at 137 I mean you would find -4.00 at 137. We can keep running with this if you like. Just email me through the OpticianWorks website. John

[@zhane8018]

Thank you, very helpful

[@bpark10001]

The explanation you get into on the surface about the origin of the standards (where 0 degrees is, etc.) is good, but you need to prepare a video that addresses the prescription standards in totality. Nowhere have I seen such a reference explained in either a course or a textbook, collected in 1 place. What reference do the lensometer and phoropter manufacturers consult to determine how to build these instruments? (I was faced with this problem when I repaired B&L lensometer with a broken belt between the target and the indicator dial.) Apparently there is a standard for "near IPD", measured at 40mm, used for the setting of bifocals and progressive lenses. True? Imagine you are making a pair of glasses for a martian, to be sent by rocket-mail. You need to explain the standards completely so the martians can make the phoropter and determine the proper Rx to send to you. Those standards start with the meter as the length standard, and the diopter and testing distances derived from that for "near" and "far" vision. Then comes the angle standards for cylinder and prism, and the BO/BI/BU/BD prism standards, which in my opinion should be replaced with prism magnitude/base angle (using same angle convention as that used for cylinder, except going full circle).

[@johnseegers9448]

Brian, I just did a refurb on an old B&L myself. The standard for power and alignment on that was a Younger GaugeMaster. It sounds like you might be looking for something more in depth? If that is the case you are really getting into optical engineering over opticianry. If you have the time, interest and expertise I'll be happy to work with you on creating a video or video series. I know a little and JA Optronics and Marco may be willing to help a little too.

[@AkashKumar-gn6sh]

Thank you , but I have confusion . In which type of problem I can use this formula.

[@LaramyKOptical]

You could use it on any lens that has cylinder. It is more of a short-cut for estimating powers so you can work a basic prism problem. It doesn't really have a specific use. People like it for taking timed multiple choice exams. John

[@rayfreeman3387]

Help me please I’m doing the online course and I can never pass the quiz.

[@LaramyKOptical]

Email me through the website please. John

[@ngtuckchun8459]

may i know why are the -5.00 and the -7.00 which correspond to the thickest and thinnest part of the lens be 90 degrees apart. and why at this angle the astigmatism is corrected. with respect, why not make the angle be say 30 degree/120 degree apart?

[@johnseegers9448]

Ng Tuck Chun: Good question! 1) Until just a few years ago surfacing equipment was all design around a diamond "cup." The cup was/is only capable of creating two curves which are 90 degrees apart. Modern free-form lenses are getting close to variable degree correction. 2) Think of the cornea which is round and has a convex shape. Quite simply a lens with curves ground 90 degrees apart on a round lens with a convex backside works great for correcting astigmatism. 3) Keep in mind that the two principle meridians may be 90 degrees apart but they are then rotated around to best match the cornea. 4) To create a lens with variable curvature to correct astigmatism it would require sophisticated corneal lens mapping which then matched equally sophisticated lens design. Short answer - because it works well and its simple! John

[@mariatheodoridou8263]

Hey, your videos are really helpful but I have one question... How does it come that 30 degrees away represent the 25% of the cylinder? Since 50%(2 times the 25% of the cylinder) is represented by 45 degrees which is not two times the 30 degrees.. Doesn't it have to be divided equally?

[@LaramyKOptical]

Everything I know is in the video... ;-)

[@mariatheodoridou8263]

@@LaramyKOptical I see now thanks !

[@walkingmonument]

Last time? I thought this was the first lesson in the series

[@mln180]

Love it

[@jonathanclingenpeel8104]

One Question, why do you round at the hundredths place and not the thousands place. I know all Rx's will be written to the hundredth but since the math gives us a number into the thousands place why do we not leave it there? I got -6.875 Deg at the 90 Deg. I know if I round it I end up with the same answer as you. I just find it a little odd why we are rounding.

[@LaramyKOptical]

LOL You are asking me a math question? Have you watched all my videos? Anyway - I would GUESS that we do it remain in our comfort zone of the common place holding for lens powers used in common optical formulas. And remember this is an estimated power not an absolute. The percent rule is a short-cut. So, close enough works quite nicely. Do I follow the rules of proper math? Probably not. I don't even know what they are anymore.. John

[@jonathanclingenpeel8104]

@@LaramyKOptical I do appreciate the response, not many KZbinrs get back to us anymore. I have watched many of your videos in prep for my ABO test coming up in the spring and I do have to say they are very informative and you explain some stuff better than my many mentors have. When it comes to proper math are there even proper rules lol

[@kenhong7732]

I just found an interesting fact. (sin30)^2=1/4，(sin45)^2=1/2，(sin60)^2=3/4. Is that correct? thank you for inspiring me.

[@LaramyKOptical]

Assuming that you included the power in there somewhere sure. That looks like a version of the power in oblique meridian formula. As I always say, if it works for you and you get the results you need, go for it! John

[@اعرفحقك-ه3ض]

thanks for you so mush

[@LaramyKOptical]

You are welcome.

[@walkingmonument]

@@LaramyKOptical ooh! She called your work mush

[@timashaq5908]

you are amazing! Thank you

[@sadeqajeena]

supper great sir

[@LaramyKOptical]

Thanks

[@jkourk23]

i think that the value "-0.93" is the decrease of the power of the cylinder so the cylinder on 180 is -3.75 - (-0.93)= -2.82 and the power at 180 is -6.50 + (-2.82)= -9.32 and not -7.43 which is a big decrease since the cylinder will be furthermore reduced at 45 from 150 by 50% and at 60 from 150 by 75%

[@hell00O0]

In a meridian divided every 90 degrees from 0 to 180, as per your calculation distribution it should be 400% with each 90 degree distribution as 100%. Thus, going from 0 degree to 90 degree (irrespective of actual number but a 90 degree separation), it indicates 100%. Therefore a 25% away from 0 degree will be 22.5 degrees and not 30 degrees. Accordingly, the 25 45 60 degrees will reflect about 27.78%, 50%, 66.67% respectively. To make it truly 30 45 60 degree rule the correct percentages should be 33.33%, 50%, 66.67% respectively. The calculation for the power at 180 - 0 or 90 degress is following the correct procedure though, there is a mistake in percent calculation of 90 degrees. for a 90 degrees from zero, 90 degrees = 100% thus, 45 degree = 50% is correct though, the 25% of 90 degree is 22.5 degrees and not 30 degrees. Likewise, 75% is 67.5 degrees and not 60 degrees. While dividing 90 degrees into four quarters it is important to divide 90 by 4 and not 180 by 4. 90/4 = 22.5 degrees for each 25% increment. 180/4 = 45.0 degrees for 25% increment though it is for 0 to 180. The cases / examples you are working on are only 90 degrees apart. Accordintly, the rule should be called 25 50 75 rule and the numbers should be 25% away from lower power meridian. Hope this helps clarify the error in calculations. Thanks for the videos!

[@LaramyKOptical]

Thanks for your comment. The relationship between the angle and power is not linear. Please see kzbin.info/www/bejne/oqe7aH6EhZapirs to understand more about how the value is calculated.

[@hell00O0]

@@LaramyKOptical Thanks for your reply, I agree with you that powers are not linear though, it does not add up to 30 45 60. I watched the video that you mentioned above in the response and it has nothing to do with clarifying as there is an error. This video has a conceptual error and must be corrected as it will instill wrong lesson and many ABO aspirants who are preparing for their exams will learn a wrong calculation. During this video in the early part, you do mention about how you have arrived at 30 45 60 and they corresponds to the 25% 50% 75% respectively when going from 0 to 90 degrees. That particular portion is based on wrong calculation. Again, in 0 to 90 degree meridian, a 30 degree will be 30/90*100=33.33% and not 25%, 45 degree is 50% but then again 60 degree is not 75% but 66.66%. If we assume the 0 to 180 degree meridian then also 30 45 60 do not represent 25% 50% 60%. While you say above that powers are not linear, your message in the video is conveying exactly that they are linear. How is 30 45 60 degrees equal to 25% 50% 75% in oblique meridians? Please help me undertand it. Thanks again.

[@LaramyKOptical]

@@hell00O0 You can't divide the circle up into equal bits and expect the power to follow proportionally because it's a trig function, not a linear one. Your best bet is to do the calculations for yourself (using the power in oblique meridians formula) to see that 30, 45, and 60 do in fact correspond to 25%, 50%, and 75%. Thanks!

[@hell00O0]

@@LaramyKOptical For using the oblique meridian formula you need the values for cylinder and sphere powers. In the video on 30 45 60 rule this formula was not used to arrive at the percentage values. Moreover, these values could not be arrived at by using OM formula as that video never mentioned about meridians being oblique. It only talks about rotating 90 degrees and how the power fluctuates between the two highest powers, example -7.00 to -5.50 in case 1 that was discussed. Will appreciate some help on this. Thanks.

[@meygaagquule5860]

thanks sir

[@LaramyKOptical]

You are welcome!

[@Supporthimanityandnotai247]

Isn't this transposition?

[@LaramyKOptical]

No.

[@madelomalacao6064]

Im 6 years optician here in Philippines i want to go abroad looking for job..can anyone help me to find job abroad as optician..

[@EvilBraTT]

Shame on you for making a mistake :) Great Series of Videos!!!

[@markbeuligmann7363]

The 30 45 60 rule is only a rough way of estimating the power in the meridian of error. The rule is accurate only for the 45° position. While 45° is exactly analogous to 50% of the cylinder power, 30° and 60° are not exactly analogous to 25% and 75% respectively. Why? in a 25-50-75 scale, the adjacent points are equidistant from each other. Each position is 25 points from its nearest neighbors. That is not exactly analogous to 30-45-60. In that scale, the points are NOT equidistant from each other. In short, 30° is not exactly analogous to 25%, and 60° is not exactly analogous to 75%. For better accuracy, one would have to use a 22.5 45 67.5 rule. Even then, you would only be finding the true power in the meridian of error if the meridian of error was at one of those three precise positions. For real accuracy, you must use the powers in oblique meridians formula. Please correct me if I'm wrong.

[@LaramyKOptical]

I wonder if you plugged your numbers into the powers in oblique meridians formula to test your theory. 😉Hint: trigonometry

[@markbeuligmann7363]

@@LaramyKOptical I did as you suggested. I plugged the numbers into the powers in oblique meridians formula. It appears to confirm that the 30 45 60 rule produces accurate results. It still seems to me that it shouldn't, but I yield to the proof in the formula. I would like to understand how that is possible, but the logic eludes me. It seems to defy simple math. How can equidistant points be perfectly analogous to points that are not equidistant?

[@keithbenjamin6056]

@@markbeuligmann7363 it's not simple arithmetic, it's trigonometry. You have two component vectors that do not change at equal rates. Closer to 0 deg, the magnitude of your vertical component changes much faster than your horizontal component. Likewise, closer to 90 degrees the magnitude of your horizontal component changes much faster than your vertical. Try to envision the length of the sides of the triangle as the angle moves through the circle.

[@markbeuligmann7363]

@@keithbenjamin6056 Thank you! And thanks to LaramyK Optical as well. May I infer from your response that the power of the lens does not change in a simple linear fashion as one moves from the sphere power to the cylinder power?

[@markbeuligmann7363]

I think I have answered my own question. I created a spreadsheet for the lens powers in oblique meridians formula (and Prentices's Rule because they are used together). Starting with a given sphere and cylinder value, I entered different values for the axis in equal intervals and noted the difference in the lens power in the meridian of error. Indeed, the change in lens power was not the same between all the intervals. (If anyone is interested in receiving a copy of that spreadsheet, just reply to this message.)

[@TheGozalus]

I wonder why I am watching these videos when I am an aviation student.

[@LaramyKOptical]

Because they are, "out of this world" good? BadumBump... John

[@sajadmazhari]

where is the last time, this is your oldest video, on this channel...

[@LaramyKOptical]

This one: kzbin.info/www/bejne/imjInmWAhZitjas

[@blairtaylor9552]

So, there will be 100% power change, from the lowest power in the lens, to the highest power in the lens, yet only 90 degrees within which that 100% power change happens. Between high and low power meridians, at 45 degrees, would be exactly 50% change. HOWEVER... 30 degrees would come out to 33.33333% of the power, NOT 25%. It's a pretty lousy approximation formula, when it could have been much closer to actual: 30deg.=33% / 45deg.=50% / 60deg.=66%. NOT John's fault, but it's mathematically pretty far off for approximations at 30 & 60 degrees. Agreed? (100%/90degrees = 1.111, times the degrees; ie.: 30 deg. x 1.111 = 33.333%). It's a pretty flawed rule, if we're only going to round the result to 2 places anyway! Oh, I'm a grammar Nazi- "the way the cornea is shapened". No such word. "the way the cornea is shaped" or, "if the cornea is Ill-shapen" are gooder inglish, lol. :-) Still love your stuff!!!

[@keithbenjamin6056]

Hi Blair. It's not straight arithmetic it's based on trigonometry. sin^2(30) =.25, sin^2(45) = .5, and sin^2(60)=.75 Thanks for taking the time to keep us straight though!

[@اعرفحقك-ه3ض]

your friend in algeria

[@TheDejatube]

🦁🙋🏻‍♂️SHARE KO WHAT IS GOOD FRAME FOR FIRST TIME 👤#7PTSDANXIETY