All About Vertex Distance with Calculations
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- čas přidán 22. 07. 2024
- Does vertex distance affect lens power or perceived lens power, or is it something entirely different? Either way, how do we calculate the effect it has on someone's vision?
Videos referenced:
What Is Compensated Lens Design?
• What Is A "Compensated...
Lens Tilt and Perceived Power
• Lens Tilt And Perceive...
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To derive that formula, just look at it in terms of focal lengths which means you take the reciprocals of the powers: 1/De=1/D -d , De=1/(1/D-d) , and De=D/(1-d*D)) and that is your formula BUT there is a minus sign in the bottom instead of a plus sign as you have it. The reason for that is you are looking at perceived power and this formula deals with the corrected(required) power. I teach physics and have appreciated your videos to learn more about optometry to provide additional material for my students. Your videos are great!
Whew! I always worry when I see a formula within a comment section. Thanks for the "Your videos are great!" comment. Keith is the behind the scenes guy here and he is also a real physics wizard with (unlike me) a big brain and everything so our stuff is carefully vetted before being posted. Please let us know if you need anything else brought to the big screen since we are always looking for new subjects. John
Hello Laramy, very interesting...thank you...
Awesome video again!
:)
Thankyou so much. Im becoming optometrist and these calculations just eat up my head every time i try solving them.
The trick is finally realizing that you are NOT trying to get them back to the original and all will be well.
@@LaramyKOptical my father is ophthalmologist but even he couldn't help me with these calculations, you did.
(ฅ´ω`ฅ)
Thanks, helped me a lot
Glad to hear!
Thank you! I wished more people understand about those with high powered eyeglasses. Most doctors tends to give you slightly weaker prescription and that creates an even bigger problem.
Never thought about that but yes I think the tendency would be to underpowered when at times a little over would actually be better.
@@LaramyKOptical Thank you again. In ideal world, eyeglases stay put in ideal position. For me, 90% of the time, it never is. I got high minus, so it's weaker than I'd like. The fact remains if it is a bit strong, I can always adjust outward for more comfortable viewing, but I can never get more power when I really needs it for distance.
Hey John! So you often day(it’s instilled in my brain by now), that we are not trying to bring the patient back to the written Rx power. But wouldn’t we want to try and adjust the frame so that we do get the lenses back to the power that was prescribed by the doctor for that particular patient? Ex: open the nose pads up to bring the frame closer (from 13mm vertex back to 10mm vertex).
If the lenses were made at the power of the as refracted position but not sitting in that position then YES by all means adjust the frame to the as refracted position. If you need 13 are at 15 and can get to 13 problem solved. If you moved a compensated lens from the as worn position you would be reversing the point of doing compensation. John
The equation of De in terms of DL and vertex distance is very useful. I derived the same result from the optical power of two thin lenses with separation distance d. However the sign of d does not seem right. I think your final result of the exam is correct. One could get -9.33 Diopter by simply using the equation of DL/(1+d*DL), while d is positive when vertex distance becomes larger.
Xi, I haven't worked a lens in series problem in 30+ years and wouldn't even know where to begin. I'm sure there are other ways of approaching the same problem. See: czcams.com/video/xi_fCBQgvo4/video.html
Again - you may find a site geared towards engineering far more helpful. I'm working from an entirely different level of understanding and application. What you (engineers) take for granted and understand is way beyond me. John
Question on the sphero-cylinder example (+18.00 -2.00 x 045) used in the lesson on Opticianworks site; when multiplying for 'd' that equals 0.003 x 18 = 0.054, the lesson is written as 0.05 without the '4'. I'm not understanding why we left off the '4' on this particular exercise? Can you explain? Was it an error?
Please email me and include a screen shot so I can see what you are asking about.
oh man. i know these will make sense over time but this makes my head hurt.
See it don't think it grasshopper... ;-) John
How i feel about everything related to optics ever:
Hi John, sorry for some dumb questions. I am an optical engineer who will do the ophthalmis lens design very soon. I know very little about the terminologies. What do PD and OC stand for? Thanks a lot!
PD is pupillary distance (distance between the humans pupil center or visual axis). OC is the optical center of the lens (but for this purpose it would be vertical movement to meet the humans eye). There are 100+ other videos on the OpticianWorks CZcams Channel where you found that one that might help make things clearer. John
sir it will be kind of you if you could help us find the vortex distance for an astigmatic prescription in spectacle lens
-4.50-3.75*180 OD and -3.75-2.75*20 OS for contact lens
Sorry - not a contact lens person. You may find a chart online somewhere? It isn't just vertex you have to know to fit a topic CL you also have to have the ability to see its rotation and how to account for it. That is in the specialty range for a topic CL.
I actually got a 9.33 using a different method.
Given is 9 diopters with 10 mm vertex distance. First we calculate for the distance of the Retina to the back of the lens by getting the reciprocal of the lens power plus the vertex distance (i.e 1/9 Diopters = 0.11 meters ~111mm) :
111 mm + 10 mm = 121 mm (this is the value or focal length we want to achieve when adjusting the lens power to compensate for the new vertex distance)
solution:
let x be the new pair's power in diopters and we will have an equation like this:
1000(1/x) + 14 mm = 121 mm
We then solve for x:
x = 1000 / (14 mm + 121.11 mm)
x = 1000 / 107.11
x = 9.33 Diopters
an easy equation is: Optical Power = Lens1_Power+Lens2_Power-d*Len1_Power*Lens2_Power. With this simple equation, one could quickly calculate the correct lens power for vertex distance d.
In your example, you often use Freeform lens. What if customer orders normal lens? Will it affect results?
The lens will be ground to the "prescription" written to the 0.25 diopter.
@@LaramyKOptical 😘😘😘
DE=D*D/1000*dmm is the easiest way
Just be carful that they ask for compensated power or effective power?
And then you can add or subtract to the lens power
Compensated power
Plus lens away from eye = less plus power
Toward eye = more plus power
Minus lens away from eye = more minus power
Toward eye= less minus power
Foe effective power is opposite
Not necessarily any "easier" than any other approach. Just another way of looking at it/doing it. I believe that is the method Martin G uses in the other video. Whatever works for you...
sir how measure near pd
Distance PD - 4 if doing by hand.
ok thanks
@@LaramyKOptical hi John, when to -4 and when to -3 for near PD?
@@ngNudraconis Since we are talking about manual or by-hand measurements it really doesn't matter. Rule of thumb is -2 for intermediate -4 for near. Unless power is quite high a mm here or there isn't going to matter much. John
So...
1. A minus lens that moves further away from the tested vertex distance (lens becomes less minus). I would order a lens more minus in
prescription.
2. A minus lens that moves close from the tested vertex distance (lens becomes more minus). I would order a lens less minus in
prescription.
3. A plus lens that moves further away from the tested vertex distance (lens becomes more plus). I would order a lens less plus in prescription.
4. A plus lens that moves closer from the tested vertex distance (lens becomes less plus). I would order a lens more plus in prescription.
Correct me if I’m wrong.
Hussan, Everything I know, everything I understand about the subject of vertex is in one of the three videos, this one or the other two I mention or on the OpticianWorks website. John
This statement conflicts with lens design terminology. For an optical engineer, lens optical power is 1/EFL. So a positive lens power is larger when it has a shorter focal length. Therefore, when the separation between the two positive lenses is shorter, the optical power is larger since the overall focal length is shorter.
Um, thanks. But the video isn't for optical engineers... As I mention numerous times, "If I had an engineer size brain I wouldn't have become an optician."
@@LaramyKOptical hi John, I do enjoy your videos very much. It helps me to catch the ophthalmic lens language. There are a lot of differences which I need to learn.
Instead of 1+ ( )use 1-( ) skips the additional last step
To get the compensated lens power.
CAN SOMEONE FROM INDIA DO INTERNSHIP AT YOUR PLACE ?
Anyone is welcome to arrange a visit and spend some time at the lab. We don't offer any internships. But - if you could pay your own and wanted to spend a few weeks doing on the job training that is something we might be able to arrange. If you have basic knowledge and a desire (or better yet) an actual need to learn wholesale optical lab work then 7 to 10 actual working days might give you a good head start. That would let you work at least a half day at each station.
@@LaramyKOptical Thankyou for your kind information sir 😇