Does anyone have the "rule" and an example of ESTIMATING the cylinder value or power between the two principle meridians?
Not the power in oblique meridian formula but the 1/2 the amount rule-of-thumb thing?
I think I have it but need to be sure.
Thanks
Are you talking about the 30, 45, 60, 90 degree estimation or Thompson's Formula ( obliquely crossed cyl's....for over refracting)?
Originally Posted by John@OWDC
are you looking for this one ?
http://books.google.ca/books?id=yq5_...20lens&f=false
I guess I must mean the 30 45 60 90 degree estimation.
Something about 1/2 the cylinder value being present at ____.
25%, 50%, 75% and 100 %.
Sphere equivalent? Half of the cylinder power algebraically add to the sphere power.
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.
Thanks all.
I teach the axis as a percentage method, so if the axis is 20 degrees away add 20% of the cyl. Of course this method is more accurate for the middle and ends then it is for say the 1/4 or 3/4 since the cylinder is more a sin wave then a straight line, but it holds true enough for a new hire to grasp the concept. Once we've moved past that I teach the sin^2 for horizontal or cos^2 for vertical and go over why the cos^2 since I still have a hard time getting that concept across to seasoned opticians. It's more a trig thing than an optical thing.
cos(ang) = sin(ang - 90) = sin(ang + 90)
Since horizontal is the 180 or 0, then a lens at 45 degrees would have a formula that looks like SPH + CYL * sin(45 - 0) you can see that the 180 becomes redundant in a horizontal formula and the same holds true for a vertical. Only oblique meridian measures need the difference in angle, which is again another lesson in and of itself. That way an optician can go from beginner - intermediate - advanced with different topics. I try and use a three step method in most of the things I like to teach with the simplest rule of thumb taught first and then building upon that topic. Unfortunately I don't have too many pupils nowadays by choice.
dittos to optical24/7
Last edited by mshimp; 10-11-2014 at 08:03 PM. Reason: incomplete name
±45 degrees away from the principal meridians.
Last edited by Robert Martellaro; 10-14-2014 at 09:12 AM. Reason: Spell checker boogaloo!
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.
0% of cyl power is present on axis,
25% of cyl power is present 30 deg from axis,
50% of cyl power is present 45 deg from axis
75% of cyl power is present 60 deg from axis
100% of cyl power is present 90 deg from axis
Close enough to pass any test, cylinder as percentage is quick, dirty , and simple. When necessary for a more accurate version the formula is as simple as it gets I am not sure a rule of thumb will be worth learning.
Glad you said it, Harry! It's what I used for a couple of years to get through lab inspections a little quicker (where my calculator was free, jicky, and lacking trig functions.)Close enough to pass any test, cylinder as percentage is quick, dirty , and simple. When necessary for a more accurate version the formula is as simple as it gets I am not sure a rule of thumb will be worth learning.
Love the graph--except I think the blue line is sloping a little too gently? Intersects early and misses the 90. Easy tweak?
That's the error, since the axis 90 is the equivalent of 90% of the axis and not 100% you get a 10% error, that's also why the slope intersects the accurate formula earlier. I used that trick to pass the ABO way back when, I just needed to eliminate the bad answers not nail it exactly. Plus when we talk about power in an oblique meridian by definition we are wrong in assuming any one focal power is an accurate representation of focus..
You know, I misread you! I used theta/90 to give me a percentage multiplier for the cyl power:
(I'm in Green)
I like yours better for lower arcs off the axis, mine's a little less forgiving closer to the 90.
(Sorry my graph chopped off the arc values)
For test taking--yours is a fast easy slam dunk. Using mine for guaging unwanted prism is seemingly a far weaker case--(and I should say I erred on the side of caution when using this shortcut for quick math--close calls got reviewed by the doc.)
***
On a tangental subject (pun intended,) care to share more on oblique fuzziness?
Last edited by Hayde; 10-16-2014 at 11:05 AM. Reason: correcting terminology & further yammering
Theta/90 is actually better on the top end and not too difficult to remember, that's a great shortcut kudos.You know, I misread you! I used theta/90 to give me a percentage multiplier for the cyl power:
(I'm in Green)
I like yours better for lower arcs off the axis, mine's a little less forgiving closer to the 90.
(Sorry my graph chopped off the arc values)
For test taking--yours is a fast easy slam dunk. Using mine for guaging unwanted prism is seemingly a far weaker case--(and I should say I erred on the side of caution when using this shortcut for quick math--close calls got reviewed by the doc.)
***
On a tangental subject (pun intended,) care to share more on oblique fuzziness?
Thanks! Although I think you bring up a good point about oblique meridian powers. Since we're talking about approximations anyway, a more useful graph might be showing us "violin-shaped" lines (covering a broader margin of error at the obliques) instead of the skinny lines we have running all the way from 1 to 90. The overlaps of those 'hips' in the graph might refute any claim of a significant difference between any of these equations for any given cylinder power that was otherwise implied by our skinny lines.
There are currently 1 users browsing this thread. (0 members and 1 guests)
Posting Permissions
Reply With Quote
Bookmarks