Formula

[tyancy]

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!

[Maksymilian]

Hello,
I suppose that this link will help You
http://www.opticampus.com/tools/calculators.php
All the best:)

[Fezz]

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:

[tyancy]

Thank you so much for taking the time to reply! I promise to pass on a favor to a stranger today.

[OPTIDONN]

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!;)

[Fezz]

Thats Ok OPTIDONN. We still LOVE YOU :p .

[Jamesye]

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

[slackware]

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

[tmorse]

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

You do mean sin48 squared x (-5.00cyl), + (-2.00), right?;)

[slackware]

haha yes. i was low on caffein and the brain wasnt functioning, but thanks for the correction ;p