Lens Tilt And Perceived Power: Why It's Important And How It's Calculated

[@SF-fb6lv]

This tilt parameter GREATLY improves my vision.

[@luisgomez1211]

Hello Laramy...very interesting...Lens Tilt And Perceived Power...

[@erodas8899]

Always enjoy watching your videos first and foremost, thank you! A thought came to mind when watching this, well a question. :) I am under the understanding that when minus lenses move closer to the eye they gain in minus power thus, when plus lenses move closer to the eye they loose plus power. So when adding panto I thought the lens power would become less plus not more plus. If you can help clear up my confusion that'd be great.

[@LaramyKOptical]

E Rodas, Sorry I missed this post: The lens isn't adding or loosing: the lens is creating a change in PERCEIVED power in relation to the position of wear. That change will remain in the power of the cylinder (minus) which is what we work in or in the actual sphere power. Watch the video on Compensated Lens Design it may help a lot. John

[@erodas8899]

Thank you very much 🙂

[@iokuok2]

Thank you for a really cool and in depth video. Along those lines I have a question that bothers me. Do you then take the OC height not only at a location where the patient is sitting eye to eye but also adjusting for the tilt in such a way that the line of sight comes in at a 90 degree angle either by dropping the height a millimeter for every 2 degrees of tilt or by having the patient tilt their head so that the face of their frame is perpendicular to the ground? Its seems to be real difficult to get a straight answer on this one...Thanks again!

[@LaramyKOptical]

When fitting a free-form lens and providing the lab with POW measurements you would mark the fitting height at the center of the pupil. The lens design software would then calculate any changes that would be made to account for tilt. You wouldn't concern yourself with it. I know you will see this: “Lower the OC 1mm for every 2 degrees of pantoscopic tilt.” And to be honest I actually have no idea what it means. In almost 30 years I have never met an optician that mentioned doing anything like it! You might toss it up as a question on Opticians On Facebook or the The Optical guild on FB? John

[@blairtaylor9552]

@@LaramyKOptical I'm so glad you admitted that! I'm sure MANY have read through that formula for tilt, and were still baffled why the need to lower the OC by a mm. for every 2 degrees? In fact, I'm still a little baffled by THIS video- IF the OC of the lens is directly in front of the pupil, and then you tilt the lens, keeping your horizontal axis exactly bisecting the pupil, then wouldn't an increase in cylinder below that axis, mean a decrease in cylinder must happen in the lens above that axis? The lower part of the lens is tilting TOWARD the eye; the top part of the lens is tilting AWAY from the eye. It stands to reason that the effects would be exactly opposite each other, so although distortion would surely increase, it's seems that part of the power increasing and part decreasing, would be a wash. Trying to visualize...……. ?

[@doc-vs-eng9678]

will it be the same for minus lens??

[@LaramyKOptical]

Math will be the same.

[@LaramyKOptical]

If you look around online I think you can find a calculator for these formulas. Maybe OptiCampus through OptiBoard - maybe? My math skills aren't good enough but someone with strong math and spreadsheet skills could create one.

[@doc-vs-eng9678]

@@LaramyKOptical thank you sir

[@milescityvisionclinic7693]

I work as an Optician in Dr Lunde's office and watch your videos, as a refresher coarse. I have noticed you do not have on on Abbe values. Can you post on it would be a great help for those just starting out. thanks.

[@LaramyKOptical]

Head on over to www.OpticianWorks.com. I cover it in detail there. I'll take a peek at what I have and see if I have enough for a video lesson. John

[@reenabhatia5564]

Can we provide power glasses to a person whose eyes lense are tilted age 67.....

[@arthurshalgian2641]

Thank you for the whole information you presented its for us. My question is : which of the lenses is the alternative for the prescription: OD:-1,25 sph OS::1.00sph .i suppose i don’t have -1.00 on my stock and It’s an emergency glasses .which of the alternative on those tree lenses are the best choice as an emergency glasses ? Using -.075 -.025 x90 which the true power is -.100 or -1.00 -0.25 x180 which the true power is -1.00 or using -0.75 sph ???

[@LaramyKOptical]

Rule of thumb we were taught, over minus and under plus. I'm not sure I would ever see -1.00 + or - even 0.75 as being an "emergency" but in my opinion none of those choices would be any better or worse than the other. 0.25 in any "direction" isn't going to make much of a difference. John

[@jordancochran7675]

As someone who learns audio-visual i wish the math descriptions were less "this times that plus this" instead using double the index of refraction or sin of tilt squared....

[@LaramyKOptical]

As someone who struggles with math and has some serious math anxiety myself I don't think the math can be broken down anymore than it is. That is kind of the entire point of my video series. Some concepts can only be reduced so much. And I'm not being sarcastic but 100% honest - if you think you could a better job then make a video and I will gladly post it here and on the website - you'll get full credit for it. Do a good enough job I might even pay you for it. John

[@amitverma3764]

Could you please do an example on oblique axis please. eg. +3.00 -1.00 080 (keep all other variables same). I have talked to lots of faculties, but never got a definite answer, and different answer from every faculty, which made me more confused than clear. Some say to use Cross Cyl when the axis are not the same. Ok makes senses but at what point. How do I start off. Even books only talk about 090 and 180 axis. Just wanted to see what you have to say about oblique axis. Even if you cannot make a video, then please reply to my comment and just show me how you get answer for each step. You don't have to break down and show answer for each expression like you show in this video, JUST the full expression and FINAL answer for EACH step. If it does require cross cyl at some point, then just give the two expressions for cross cyl and then the final answer, do not show detail. 1. Power at 090 Your Answer........................ 2. Calculate New Sphere Your answer.... 3. Calculate induced cyl Your answer................................... x . Last step . final resultant power patient is looking through Thanks Amit

[@LaramyKOptical]

I have been searching for years (quite literally) for someone to explain crossed cylinder and I have never found anyone. I recall working the formula in school (a full blackboard if I remember) but what it was all about I have no idea. Much like your request here I have asked someone to provide an actual working example and show the steps performed. More or less a case study in crossed cylinder. No luck at all... As to your step-by-step I don't mean to be "direct" but you have to do the math (as shown in the video) so to speak. It is a multi step problem involving two different formulas. You can't "distill" it down past the steps and formulas required. John

[@amitverma3764]

@@LaramyKOptical Thanks for your reply, so I am not the only one. I have worked out TWO ways to do this. One way I thought would be to calculate the power at 090 or 180 depending on if we doing TILT or WRAP, and use that to calculate the New Sphere and Induced Cyl. For Example: Lets assume rx is +3.00 -1.00 060. Using Oblique Formula we get +2.75 in 090 meridian which is our old sphere (S), keeping everything else same n = 1.586, theta = 015 pantoscopic tilt. New Sphere(S') = +2.808 D Induced Cyl(C') = +0.202 D Perceived Rx = +2.808 +0.202 180 Original Cyl = +1.000 150 (original cyl in + form) Sphere for original cyl would be PLANO. Thats the only way I can think to use cross cyl. Now we can use cross cyl here to get the final answer; unlike 090 axis since we are working on oblique axis, the final answer will also be in oblique. I haven't calculated that yet but wanted to see if I am on right track. It would be totally opposite for WRAP ------------------------------------------------------------------------------------------------------------ Second way would be using the same example above for TILT. Calculate powers at 090 and 180 using OM formula Rx(090) = +2.75 D Rx(180) = +2.25 D Lets write this in 090 format due to tilt so we don't have to transpose +2.75 -0.50 090 and now continue like a normal problem. What are your thoughts? Both seems valid methods. I am still working on it, if I come to conclusion, will get back to you.

[@LaramyKOptical]

@@amitverma3764 As I mention in that video and several others I'm not a math guy. I can "see" what is going on, follow the steps and get where I need to go but beyond that I'm afraid I'm not your guy. Bear of little brain here...

[@heatherharrington7274]

Will this be on the ABO? Or advanced test?

[@LaramyKOptical]

Heather, No one knows what will be on the exam. On the basic you may see a question about inducing cylinder by lens tilt but you sure WILL NOT need to calculate it! On the advanced I should hope that you would. John

[@francisfrancis4219]

In the compensated freeform lens design video you made mistakes implying tilt induces cylinder in the vertical meridian; and wrap induces power in the horizontal meridian. You mentioned these things almost as though they were mutually exclusive. Here you prove yourself (formerly) wrong where the change in cylinder is created due to tilt, and you show that perceived power is affected also when you tilt. So it seems like you made this video to account for the errors implied in the previous video, but you make no account for making up for any mistakes. Certainly the previous video does not make sense until you watch this one.