Powers in Oblique Meridians Formula

This formula is more accurate that then percentage estimates in the 30-45-60 rule, and it also helps us solve the power at every point from 0 to 180. The formula says that;

“The total dioptic power (Dt) is equal to the sin of the axis from the merdian you are trying to solve for (either 90 or 180) squared, times the CYL power (Dc) plus the Sphere power (Ds). It is written as

Dt = (sin a)^2 x Dc + Ds

For example if you have a lens at -8.00 -2.00 112 and need to know the power at 180 because the PD is off center then you need to follow a few steps. First identify all the variables that you have.

Dt = ?

a = 180 - 112 = 68

Dc = -2.00

Ds = -8.00

So, Dt = (sin 68)^2 x -2.00 + -8.00 =

(0.9272)^2 x -2.00 + -8.00 =

0..8597 x -2.00 + -8.00 =

- 1.72 + -8.00 = -9.72 @180

You can check this by working from the transposed RX just as in the oblique formulas. -10.00 +2.00 022

Dt = (sin 22)^2 x +2.00 + - 10.00 = +0.28 + -10.00 = -9.72 @180

Practice Problems)

1) -4.00 -0.75 048 what is the power at 180?

2) What is the power of lens in the horizontal meridian with a power of -2.00 @037 and -3.00 @127?

3) What is the power of a lens in the vertical meridian with a power of -2.50 -1.25 076?

Answers)

1) -4.41 @180

2) First find the RX and transpose to double check. Then calculate the power at 180.

-2.36 @180

3) -2.57 @090