Help Calculating Compensated Power

[optimite]

If I have a customer that is prescribed:
Re: -9.00
Le: -8.50/2.00 x 45
and the BVD trial frame is 11mm

What would the required compensated power be if the customer chooses a frame which has a BVD of 14.5mm?
If you could show working out for this, that would be greatly appreciated

[uncut]

Welcome to the forum, optimite!

This patient would optically be better served with a frame that fitted at a closer vertex...I would change the frame that created that difference.

If, this is a homework question, I am sure Optiboard members will help you arrive at the answer. What have you worked out so far?

[optimite]

Thanks!

This is indeed a homework question, unfortunately. Haven't worked out anything new, as of yet.

[gmc]

Use the approximate vertex distance compensation formula.

power squared x millimeters moved
1000

OD
81 x 3.5 =0.2835 rounded to 0.25 diopter The fitting vertex is increased, moving the focal point in front of the retina.
1000

Minus power is required. The compensated power becomes -9.25

Follow the same steps for each of the primary meridian powers in the OS.

[MakeOptics]

Right eye:

-9.00 has a focal length of 1/-9.00 = -0.1111m or -111.1mm, if you are going to pull that lens out further from 11mm to 14.5mm that's moving the lens 14.5-11 = 3.5mm further out, so we adjust the focal length by -111.1 + 3.5 = -107.6mm or -0.1076m which has a dioptric value of 1/-0.1076 = -9.29D is the compensated power which is what you would order from the lab.

Left Eye same procedure but you need to compensate both meridians powers separately then compare them to get you new cylinder value:

Power@045 = -8.50D
Power@135 = -6.50D

@45/ 1/-8.50 = -0.1176m or -117.6mm
-117.6 + 3.5 = -114.1mm or -0.1141m
1/-0.1141 = -8.76D

@135/ 1/-6.50 = -0.1538m or -153.8mm
-153.8 + 3.5 = -150.3mm or -0.1503m
1/-0.1503 = -6.65D

-8.76/2.11 x 045

Step through the problem using simple procedures instead of long formulas.

[optimite]

Thank you so much, both of you!!!

[Rafael]

HSABE.XLS

Sorry no time for translation.

Regards
Rafael

[MakeOptics]

HSABE.XLS

Sorry no time for translation.

Regards
Rafael

Rafael, your on fire. Great calculator again. Nice to see other programmers out there. +1

[tmorse]

Right eye:

-9.00 has a focal length of 1/-9.00 = -0.1111m or -111.1mm, if you are going to pull that lens out further from 11mm to 14.5mm that's moving the lens 14.5-11 = 3.5mm further out, so we adjust the focal length by -111.1 + 3.5 = -107.6mm or -0.1076m which has a dioptric value of 1/-0.1076 = -9.29D is the compensated power which is what you would order from the lab.

Left Eye same procedure but you need to compensate both meridians powers separately then compare them to get you new cylinder value:

Power@045 = -8.50D
Power@135 = -6.50D

@45/ 1/-8.50 = -0.1176m or -117.6mm
-117.6 + 3.5 = -114.1mm or -0.1141m
1/-0.1141 = -8.76D

@135/ 1/-6.50 = -0.1538m or -153.8mm
-153.8 + 3.5 = -150.3mm or -0.1503m
1/-0.1503 = -6.65D

-8.76/2.11 x 045

Step through the problem using simple procedures instead of long formulas.

All good work, except missing final step if ordering from a lab...round off compensated Rx to nearest 0.12D.
So order OD: -9.25sph
OS: -8.75 -2.12 X 045 (bad form: your original cyl had no sign so I assume it was in minus cyl form.

[MakeOptics]

2nd post mentioned it was homework, computation was in plus cyl form and given bsck in exact same format as given to avoid confusion.

The excel spreadsheet and opticampus caculators all show non rounded answers as do I since this is not patient oriented, but theoretical my answer stands.

For lab ordering in the present 0.25 is the accepted standard as most online ordering systems dont allow 1/8 D. Still a valid point that the computation would need further refinement.