I need help in finding the formula for vertex distance conversion for both glasses and contacts. I am not a math person so it needs to be simple. I do have the Optical Formula book but still having a hard time understanding. I have one that seems to work great for plus lenses but do I just subtract the same amount on minus?? Please help... It will be much appreciated.
misread your post
Your formula tells you how much power change you get when you move a lens of a certain power a certain number of millimeters. This is the same whether the lens is plus or minus.
But which direction?
When you move a lens away from the eye (say from vertex 10 to vertex 15) you get a power shift towards plus: A plus lens gets stronger (which is why you need to use a lens that was weaker in the first place to break even) and a minus lens gets weaker (so you use a higher initial power so the reduction is compensated for).
If you fit to a shorter vertex than the refractionist used, then you get a shift towards minus. Take the last paragraph and reverse everything.
For contacts, the fitting vertex will be zero, so you're always going to be dealing with reduced vertex: powers for plus contacts will be higher than the eyeglass Rx, and minus contacts will be lower power than the eyeglass Rx, all to maintain the same effective power.
Think of it this way. Regardless of whether the lens is plus or minus, when you move it away from the eye, the focal point is moved in front of the retina so more minus power (or less plus) is needed. When you move it closer to the eye, the focal point is moved behind the retina, so more plus power (or less minus) is needed.
Last edited by gmc; 03-06-2012 at 05:00 PM.
Here is a simple animation, the formula I have in there can be further worked to come to the point that is common in the text books, but I think simple is better and this animation shows how the formula was derived. (scroll over image to enlarge)
If you look in the back of almost any contact lens fitting manuel, catalogue, etc. It will give you the conversions done. You can also check Eyedock, they have computer programs for this. Also almost any authoritatve book on dispensing will have at least a chapter with formulas and charts on the subject.
Pretty sure Darryl Meister's Opticampus has a course or two on this. The CLSA university page will list several courses on this.
Chip
CAP...closer add plus. Shortcut (approximate) Power2 * vertex distance in meters. Power is sphere equivalent if there's cylinder. Handy for CLs. Follow up with an overrefraction.
Otherwise use power / (1 + change in vertex distance * power). Calculate for both principle meridians if there's cylinder, with the difference becoming the new cylinder. Axis stays the same.
Last edited by Robert Martellaro; 03-08-2012 at 05:11 PM.
Science is a way of trying not to fool yourself. - Richard P. Feynman
Experience is the hardest teacher. She gives the test before the lesson.
That is the formula, but at face value intuition in it's derivation is lost. The form I like to use is a bit more rudimentary, if we were to factor out the power in the equation we get a better idea of what's going on. For example:Originally Posted by Robert Martellaro
If we were to factor out the Power:
Since P/P=1 we can get rid of this, and since 1/P=focal length we can replace it with f
The added benefit is that we are now working with the same units, meters or mm if we so choose and it is not a difficult equation to put to memory or recreate. That is the key for me, I like modular pieces of a formula that I can further manipulate when needed.
Someone also mentioned in another thread to look to the Europeans for their way of learning optics and this is where I believe I once came to an understanding of this formula.
I enjoy your posts and always read with delight I look forward to a response or thoughts you may have.
Thanks.
The OP did emphasize 'easy'...
This was a test question when I took the ABO exam about 30 years ago. I nailed it, but I new (sic) I would.
Science is a way of trying not to fool yourself. - Richard P. Feynman
Experience is the hardest teacher. She gives the test before the lesson.
Vertex compensation formula was so confusing, that I used the rule of thumb:
If you move a 10D lens 10 mm, power will be altered by 1 diopter. Therefore, a 5D lens moved 10mm will alter the power by 0.50D.
Plus lens, power is decreased when moved closer (as with contact lens), so increase power to compensate. Minus lens, power will be increased when moved closer so decrease power to compensate.
For contact lens, do spherical equivalent for soft lens or drop the cylinder for RGPs.
It's not a linear relationship.
Science is a way of trying not to fool yourself. - Richard P. Feynman
Experience is the hardest teacher. She gives the test before the lesson.
All of the above is helpful, thank you. What I need at this time is the formula that will help me determine the Rx for a patient's needing a vertex distance other than the standard distance using the focal length formula approach.
try this
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