8: astigmatic decomposition & obliquely crossed cylinders

obliquely crossed cylinders

power of a cylindrical lens along an axis other than the principal meridians

plano/+6.00 × 90

= power along the axis of the cylinder is plano

power along the meridian at 90 degrees to the axis is +6.00D

how is the power along some other meridian worked out?

plano/+6.00 × 90

F feta is oriented at an angle feta from the axis of the cylinder

F feta = F x sin feta ²

( sin feta, and then square this)

what is the F feta for plano/+6.00 × 90

= +6.00 x sin ² 45

= +6.00 × 0.5

= +3.00D

plano/ +6.00 × 90

1. feta = 10 degrees

= +6.00 x sin² 10

= +0.18D

2. sin feta is 80 degrees

= +6.00 x sin²80

= +5.82D

obliqquely crossed cylinders- graphical representation

plots the power of the whole lens as it is moved around

the angle is with respect to the axis of the cylinder not the trial frame notation

+4.50 / -1.75 × 165 ; graphical representation

shows change in meridional power ( ie power along a given axis) of the lens according to standard trial frame notation

combining cylinders

if the axes are the same:

-1.00/-0.50 × 175 + -1.25 / -1.00 × 175 = -2.25 / -1.50 × 175

if axes are not the same;

eg -1.00/-0.50 × 175 + -1.25/-1.00 × 165 : harder to combine

axis 5 and 175 is 5 degrees apart

what methods of calculations are used to calculate resultant power from the combination of prescriptions where the cylinder axes are obliqye with respect to eachother

graphical method: storkes construction

astigmatic decomposition

graphical solution for the combination of 2 lenses

-1.00 × 20 + -2.00 × 160

1 line at 20, other at 160 , an and the length of the lines represent the power of the cyl

graphical solution steps ( not needed)

1. transpose the lenses so that theyre in positive sphero cylindrical form

2. choose the lens with smaller numerical axis ( 180 is 0 )

3. draw a line to scale - length and axis direction which represents the first cylinder power F1 (OA)

4. draw second line F2 to scale at an angle to the first line of twice the difference in axis; AR

5. draw third line to complete triangle. the length in this lie represents the power of the resultant cyl

6. bisect the angle between the 1st line drawn and resultant.

7. new sphere power will be ( f1 + f2 - C) / 2

astigmatic decomposition

dividing up a prescription so that it can be represented by 3 terms

M = mean spherical power : S + ½ C

J₁₈₀ = a cross cylinder at 90/180

J₄₅ = a crossed cylinder at 45/135

work out M from +4.25 / -1.50 × 20

M = S + ½ C

M = +4.25 + -0.75 = +3.50

work out J₁₈₀ from +4.25 / -1.50 - x 20

= - 0.5C x cos(2A) A being the axis

take the axis of cl, and double it, then take cosiine of this

multiply it by minus half the cyl

= -0.5 x (-1.50) x cos(40) = 0.75 × 0.7660 = 0.575D

usualy cross cyl is around a half of the cyl power

work out J₄₅ from +4.25 / -1.50 × 20

-0.5C x sin (2A)

-0.5 x -1.50 . sin (40)

= 0.75 × 0.6428 = +0.482D

the J180 cross cyl component is bigger than 45; 180 is closer to 20 than 45

for both J180 and J45 to be exactly the same axes would need to be 22.5: midpoint

how is the resultant sphere worked out

take the sum of all the J180 terms and square it, do same for J45, then add the two together

take square root of this and add the sum of the M terms

how is the resultant cylinder worked out

take the sum of all J180 terms and square it, do the same for J45, add the two together, then square root this

multiply this by -2.00

how is resultant axis worked out

divide the sum of all the j45 terms by the sum of j180 terms. calc 0.5 the arctan of this

express the combination of the lenses in standard clinical notation

+4.25/-1.50 × 20 & -2.50/-1.00 × 10 (1)

• +4.25 / -1.50 × 20

M = +4.25 + - 1.50/2 = +3.50D

J180= -0.5 x -1.50 x cos(40) = +0.575D

J45= -0.5 x -1.50 x sin(40) = 0.482 D

• -2.50 / -1.00 × 10

M = -3.00

J180 = 0.470 D

J45= 0.171 D

sum of all components

M= +3.50 + - 3.00 = +0.50

J180 = +1.045

J45 = +0.653

resultant sphere power (2)

resultant cylinder power (3)

resultant axis (4)

= leaves the prescription as +1.73 / -2.46 × 16

(1)

(2) resultant

if have an axis that isnt near either of the 2 axes in question , needs adjusting by 90 degrees

so resultant axis would be 54

(1)

axis is different- not between the 2 axes

so adjust 90 degrees

so 24 + 90 = 114 is the new axis

(1)

axis isnt between 160 and 170

so add 180 to it to get it in between the 2 axis

= new axis is 164

what are the fixes for cylinder axis

for resultant axes less than 45 : use calculated value

for resultant axes greater than 45 and less than 135, add 90 to calculated value

for resultant axes greater than 135, add 180 to calculated value