Martin's formulas as used here
New sphere = F × (1 + sin²θ ÷ 3)
Induced cylinder = F × tan²θ
- F
- The sphere power in dioptres.
- θ
- Tilt angle in degrees: pantoscopic tilt for the sphere and the cylinder at axis 180, wrap for the cylinder at axis 90.
- 3
- Twice the refractive index, taken as 1.5.
Worked examples
| Lens | Tilt | Wrap | Sphere | Tilt cylinder | Wrap cylinder |
|---|---|---|---|---|---|
| −4.00 | 10° | 0° | -4.04 D | -0.12 D | +0.00 D |
| −4.00 | 0° | 20° | -4.00 D | +0.00 D | -0.53 D |
| +6.00 | 12° | 0° | +6.09 D | +0.27 D | +0.00 D |
Read the first row as −4.04 / −0.12 × 180 and the second as −4.00 / −0.53 × 90. Twenty degrees of wrap puts half a dioptre of unwanted cylinder in front of a moderate myope, which is why wrapped frames in higher powers need lenses compensated for the position of wear.
What the calculator leaves out
- The CYL field is not used in the calculation. Results are for the sphere only.
- The sphere includes the effect of pantoscopic tilt only; wrap changes the sphere too (by hand, −4.00 D at 20° of wrap: −4.00 × (1 + sin²20° ÷ 3) ≈ −4.16 D).
- Tilt and wrap are treated separately, and the index is fixed at 1.5.
- Freeform labs compensate with the full position-of-wear data: tilt, wrap, vertex distance and the frame shape. See the medical disclaimer.
Tilt and the height of the optical centre
Pantoscopic tilt causes these errors when the line of sight passes through the lens at an angle. Lowering the optical centre by about half a millimetre per degree of tilt, roughly 5 mm for 10°, makes the optical axis pass through the eye's centre of rotation and removes most of the effect for straight-ahead gaze.
Questions
How much pantoscopic tilt is normal?
Most dress frames are fitted with about 8° to 12°. At 10° a −4.00 D lens changes by only 0.04 D of sphere and 0.12 D of cylinder, so compensation matters mainly in wrapped frames with 15° to 25° of face-form wrap, and in higher powers.