Issue with curriculum

[MX1]

I’m prepping for the ABO, and a question I’ve encountered in my textbook is what is the power at 180 for:

-2.00 -1.00 x 030

If I’m not mistaken, that asks me what % of the power is at 150?

How do I calculate that?

Thank you.

[Robert Martellaro]

I’m prepping for the ABO, and a question I’ve encountered in my textbook is what is the power at 180 for:

-2.00 -1.00 x 030

If I’m not mistaken, that asks me what % of the power is at 150?

How do I calculate that?

Thank you.

The basic ABO test will only ask you to solve for axes of 30/120 (x 0.25), 45/135 (x 0.50), and 60/150 (x 0.75). No calculators required.

We need to know how to solve for mostly the 90 and 180 meridians to determine the degree of prismatic error, vertical prism imbalance, estimating thickness, etc.

In your example, the power at 30 degrees is -2.00. The power at 180 is 0.25 x -1.00 + (-2.00) = -2.25

https://www.passyouropticalboards.co...-any-meridian/

Hope this helps,

Robert Martellaro

[tmorse]

I’m prepping for the ABO, and a question I’ve encountered in my textbook is what is the power at 180 for:
-2.00 -1.00 x 030
If I’m not mistaken, that asks me what % of the power is at 150? How do I calculate that?

You are indeed mistaken.
Look in your textbook for placing Rx powers on an optical or power cross. Think protractor, a 1/2 circle with 0 degrees at bottom right, 090 degrees at vertical and 180 degrees at bottom left), and you see that 0 degrees and 180 degrees are both located in the same horizontal meridian. If 0 axis is same as 180-0 axis, then 090 is same axis as 270, but we never write any axis beyond 180 degrees. But we never see written 0 degrees as an axis position, we Dr's always substitute 180 axis for 0 axis.
So when you are asked for power in the 180 meridian, it is the same as asking for power in the axis 0 degree meridian. And we see from Rx: -2.00-1.00 x 030 the change from cylinder axis 030 to axis 0 degrees is 30 degrees or 1/4 of cylinder (or 25%) of -1.00 cylinder being involved. This % cylinder -0.25 is added algebraically to -2.00- sphere component of Rx to give you a total horizontal power on the optical cross as -2.50 @180.

Or, you can transpose the Rx -2.00-1.00 x 030 to its +cylinder form... -3.00+1.00x120.

Now for power at 180 meridian, the % amount of total +1.00cylinder from axis meridian 120 to meridian in question 180, is 60 degrees or 75% of +1.00 cyl... +0.75cyl, which is added to -3.00 sphere component of transposed Rx algebraically to total the same -2.25 @ 180 as shown above. And of course it's the same, since 0 and (180) horizontal meridian is the same meridian on the optical cross.