Hello, does someone know the formula for finding the specific power at a given meridian. I don't know where else to find this information.
Thanks!
Hello, does someone know the formula for finding the specific power at a given meridian. I don't know where else to find this information.
Thanks!
Hello,
I suppose that this link will help You
http://www.opticampus.com/tools/calculators.php
All the best:)
All the best Max :)
One of our very own Optibaord Authors; OPTIDONN, wrote a great article for the magazine Opticourier.
Check it out. I believe there is a mistake....that was corrected in a later edition! Good luck!
http://www.opticourier.com/1webmagaz...prep/index.asp
:cheers::cheers::cheers::cheers:
Thank you so much for taking the time to reply! I promise to pass on a favor to a stranger today.
Of course in the article there is an error. The axis for the right eye was supposed tp be 180 and the left needs to be 4. Hey what can I say I was all jacked up on cold medicine!;)
Thats Ok OPTIDONN. We still LOVE YOU :p .
the formula as follow:
F (t)= F(tc)X(sin g square)+ F(s)
F (t)--- total power
F(tc)---- Total cylinder
g--------angle of change in axis from axis in Rx to axis in new meridian
F (s)----- sph power
James
say you want to find the power for 90 degrees, but the rx is: -2.00 -5.00 x42 what i do is dafca = degrees away from current axis 90-42=48 on the calculator, i type in sin48squared x -2.00 + -2.00 i dont have a calculator infront of me but that's what i do the formula would be sin(dafca)squared x cylinder + sphere works like a charm
haha yes. i was low on caffein and the brain wasnt functioning, but thanks for the correction ;p
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