Originally Posted by

**EdSheridan**
Thank you to everyone who has helped with this. At least I was able to use the math provided by Kwill212 to replicate for myself what I get from the OptiCampus Calculator.

From a different thread, Kwill212 wrote...

*Vertical prism is induced when horizontally decentering a cylinder. You can calculate it using Prentice's Rule and some Trig. The sphere power is irrelevant for this situation. Find the cylinder power at 90°. ***sin²**(135-90)*-4.00D=-2.00D. Then calculate the vertical decentration of the cylinder solving a right triangle. Assume you decenter horizontally 5mm. *Tan(135)=X/5) which means X=-5 where X = vertical decentration. Then simply use Prentice's Rule to determine the induced prism. **.5cm*2.00D=1D vertical prism. Obviously you would need to correctly input the signs to get the correct direction up or down, or draw it out. Your example doesn't really require trig if you know the basics, but this same method works for cylinder at any axis and power.*

For example.

-5.00 -3.25 x 062 decentered 6mm horizontally. Cylinder power at 90° is **sin²**(62-90)*-3.25D=-0.72D. Vertical decentration is *Tan(62)=X/6mm. X=11.28mm. Prentice's Rule is **1.128cm***.72D=.81D vertical prism. *

What I'm really having a hard time wrapping my head around is where the vertical decentration of 5mm and 11.28mm in the examples above come from? In the second example, we only moved the lens horizontally 6mm, yet we use 11.28mm of vertical decentration in prentice's rule for to calculate vertical prism. I know that is correct, but it's very hard for me to visualize.

Sometimes I start to see it when I think of a +5.00-5.00 x 45 lens as 2 different lenses placed on top of one another. One +5.00 sph and one plano - 5.00 x 45. As the -5.00 cylinder is moved horizontally at 45 degrees, it pushes the center down and out at equal rates.

I'm going to ponder some more. Thanks for all the help!!

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