Problem with contact lens correction

[sevalav]

Teacher give us this problem:
Patient's correction (Rx) is 0,00=-3,00 axis (visus 1,0 (20/20)) from right eye. We put on the patient's right eye contact lens the same power but it was rotate for 20 degree nasal-up.
Which mathematics method I must using to get value of dioptric error and axis of cylinder which this state inducing? What power of cylinder and axis I must put in trial frame in over-refraction to get clear vision for my patient?

Please, can someone help me with this?

[Darryl Meister]

Which mathematics method I must using to get value of dioptric error and axis of cylinder which this state inducing?

This problem relies on the mathematical combination of "crossed cylinders." Any standard textbook on ophthalmic optics should provide the necessary calculations. If you cannot locate them, we can post a step-by-step guide.

Essentially, you want to add the power of the rotated contact lens with the excess refractive power of the eye, which is equal to the opposite power of the specified prescription. The result of combining these two powers is the residual refractive error that you would measure in a trial frame (ignoring the additional vertex distance of the trial frame lenses).

If the prescription calls for a -3.00 D Sphere, for instance, the eye has an excess refractive power of +3.00 D.

Patient's correction (Rx) is 0,00=-3,00 axis (visus 1,0 (20/20)) from right eye

Unfortunately, your prescription is difficult for me to interpret, so I cannot help you work out the results of your particular problem. Can you rewrite it in the following format:

+/-0.00 Sph +/-0.00 Cyl 000 Axis

You can also verify the results of your calculations online at OptiCampus.com Crossed Cylinders Calculator.

[sevalav]

Patient's Rx (eyeglasses) is
Right eye: 0,00 Sph = - 3,00 Dcyl ax0
Contact lens (on right eye):
0,00 Sph = -3,00 Dcyl ax20

[Darryl Meister]

You can verify on OptiCampus.com that, after combining a power of +0.00 +3.00 x 180 (representing the eye) with a power of +0.00 -3.00 x 020 (representing the contact lens), the result would be a residual prescription of +1.03 DS -2.05 DC × 055.

[Dave Nelson]

You normally elicit the residual refractive value with manifest over-refraction, which is simply the same procedure for simple refraction. Predicting mathematically the probable residual error makes for an interesting exercise, but is not necessary. I also assume you are aware of the LARS princible for compensating for an axis shift while the lens is in place, left add, right subtract.

[sevalav]

David, this is my homework.



Darryl, thanks. Now I figure, in general.
But, for me, there is another problem. I must calculate this with graphical/mathematical method, with polar (circle) with 360 degrees. What are the rules in this method? I have a few books, but I cant find this approach.

[YrahG]

Originally Posted by sevalav

David, this is my homework.



Darryl, thanks. Now I figure, in general.
But, for me, there is another problem. I must calculate this with graphical/mathematical method, with polar (circle) with 360 degrees. What are the rules in this method? I have a few books, but I cant find this approach.

Graphical Method - Uses vectors to graph the powers on a polar coordinate.

1. convert cyls to same sign.
2. combine the spherical equivalent of both lenses, the spherical equivalent of the two lenses will be the same as the combined total.
3. Graph the vectors of the cylinder values.
4. measre the resultant vector (cylinder) and the axis.

(tips: we use 0 to 180 in prescriptions and a graph uses 0 to 360 so double the axis for each cylinder when graphing the vectors. The resultant vector is your cylinder power and the angle divided by two is your axis. Once the cylinder is determined then the spherical equivalent can be decontructed to find the sphere power. sph equiv = sph + cyl/2 -or- sph = sph equiv - cyl/2 ) (System for ophthalmic dispensing, pg318)

Formula Method - this method is the same as the above except you use the formulas instead of graphing the vectors and measureing.

S₁=sphere from the Rx with the lower axis
C₁=cylinder from the Rx with the lower axis
a₁=axis from the Rx with the lower axis
S₂=sphere from the Rx with the higher axis
C₂=cylinder from the Rx with the higher axis
a₂=axis from the Rx with the higher axis
S=new sphere
C=new cylinder
a=new axis

y=a₂-a₁
C²=C₁²+C₂²+2*C₁*C₂*cos(2y)
S=S₁+S₂+[(C₁+C₂-C)/2]
tan(2*theta)=[C₂*sin(2y)]/[C₁+C₂*cos(2y)]
a=a₁+theta

New power is S C x a

There is another formula method that is to me a simpler version than what's commonly known as Thompsons Formula from above. It's called astigmatic decomposition. (System for ophthalmic dispensing, pg320)

Astigmatic Decomposition - breaking down two cylinders into 3 components which can be directly added together, the mean refractive error, the cylinder effect at 180 and the cylinder effect at 045.

1. break both rx's down into the mean refractive error, MRE = sph + cyl/2
2. get the cylinder effect at 180 for both rx's, C₁₈₀ = C*cos(2*axis)
3. get the cylinder effect at 045 for both rx's, C₀₄₅ = C*sin(2*axis)
4. Add the two MRE's, C₁₈₀, and C₀₄₅ together.

To break the results down into sphero cylinderical form:

C_r = (C₁₈₀² + C₀₄₅²)^1/2 Either the plus or minus sign can be used for this value to get the Rx's in either form.

tan(2*axis_r) = C₀₄₅/C₁₈₀

S_r = MRE - C_r/2

If you look carefully you can see the similarities between the two formulas, but the decompositions is a much more elegant solution you'll also notice that a logic step is taken out of the equation which computes the y variable in the Thompson Formula you no longer have to keep track of the highest value axis. (Ophthalmic lenses and dispensing, pg25)

[YrahG]

Originally Posted by Darryl Meister

You can verify on OptiCampus.com that, after combining a power of +0.00 +3.00 x 180 (representing the eye) with a power of +0.00 -3.00 x 020 (representing the contact lens), the result would be a residual prescription of +1.03 DS -2.05 DC × 055.

With the risidual error to neutralize it you would change the signs while leaving the axis the same so it would net out to "0":

+1.03 -2.05 x 055
-1.03 +2.05 x 055 +
0.00 0.00 x 055

[Darryl Meister]

Sorry for the delayed response; I've been traveling.

With the risidual error to neutralize it you would change the signs

Yes, I should have been clearer about that point.

Graphical Method...Formula Method...

Fortunately, the problem can be simplified considerably, since the cylinder powers are equal in magnitude and opposite in sign. In this case, the equation for the resultant cylinder power simplifies to:



where Ang is the angle between the two cylinder axes. This is also the equation for a Stokes lens. The sign of this resultant cylinder can be set to the desired cylinder convention (either plus or minus).

The new sphere power is given simply by:



The new cylinder axis is simply midway between the original cylinder axes, once both have been transposed to the same cylinder convention.

In our example, we have:

Plano +3.00 x 180 (eye)
Plano -3.00 x 020 (lens)

The original angle between these two cylinders is 20 degrees. Using our Stokes lens equation, we have for the resultant cylinder power:




And the resultant sphere power is:




Finally, after transposing the two original cylinder powers into minus cylinder form (i.e., +3.00 -3.00 x 090 and Plano -3.00 x 020), we arrive at the new cylinder axis midway between the two: (90 + 20) / 2 = 55.

[YrahG]

Originally Posted by Darryl Meister

Sorry for the delayed response; I've been traveling.


Yes, I should have been clearer about that point.


Fortunately, the problem can be simplified considerably, since the cylinder powers are equal in magnitude and opposite in sign. In this case, the equation for the resultant cylinder power simplifies to:



where Ang is the angle between the two cylinder axes. This is also the equation for a Stokes lens. The sign of this resultant cylinder can be set to the desired cylinder convention (either plus or minus).

The new sphere power is given simply by:



The new cylinder axis is simply midway between the original cylinder axes, once both have been transposed to the same cylinder convention.

In our example, we have:

Plano +3.00 x 180 (eye)
Plano -3.00 x 020 (lens)

The original angle between these two cylinders is 20 degrees. Using our Stokes lens equation, we have for the resultant cylinder power:




And the resultant sphere power is:




Finally, after transposing the two original cylinder powers into minus cylinder form (i.e., +3.00 -3.00 x 090 and Plano -3.00 x 020), we arrive at the new cylinder axis midway between the two: (90 + 20) / 2 = 55.

If memory serves me correct you mentioned the stokes lens and used the equation in your paper on cylinder axis tolerance in the file directory. Not only applicable in this case but even more elegant of a solution kudos to that, one note it will only work in this scenario as long as the cylinders of the two Rx's are equal.

[Darryl Meister]

one note it will only work in this scenario as long as the cylinders of the two Rx's are equal

Yes, that is correct. Fortunately, that will typically be the case for rotation of contact lenses and other "off axis" errors.

[sevalav]

Thanks for all your suggestions, finally, here is my homework. Now, if there is some mistake, please, tell me.

[YrahG]

Originally Posted by sevalav

Thanks for all your suggestions, finally, here is my homework. Now, if there is some mistake, please, tell me.

Looks good our graphics are very nice looking what program did you make them in?

[sevalav]

I am using OpenOffice.org. It is free, and here is link:
http://download.openoffice.org/