Keratron Scout Support

Dr. Richard Anderson's Extrapolation Primer

 

This message is for Wave geeks. If you don’t have much of an idea yet as to how to design Wave lenses it is a safe bet to ignore me for now. At this time I have no recommendations as to how anything I say here should affect how you design lenses.

 

Extrapolation of the Scout topography prior to Wave lens design matters, but how and why is not obvious. Extrapolation, by definition, would seem to imply that Scout uses the process to “fill in” the corneal curves between the last ring you have managed to obtain in any meridian and the limbus in the same meridian. A pure extrapolation would take a radial that is changing at some rate, or eccentricity, and continue that rate of change outward. This does not happen in Scout, nor should it necessarily.

 

What is very important to understand is that extrapolation tells Scout the corneal diameter. It knows, for instance, that the 28 rings of a full image might be spherical with a power of 43.3 diopters, such as your calibration ball. What it doesn’t know is what percent of the cornea it has captured. If you have a steep cornea it will have captured a smaller percent than a flat cornea. Flat corneas may take the 28th ring to the limbus while a steep cornea will show blank space

between the last ring and the limbus. Scout can’t see that relationship until you tell it.  After you extrapolate, Scout has to decide what the missing area might be. As an example, Scout might see that the topo ring in one meridian extends 4.5 mm from center, but the extrapolated circle at the same meridian is 6.0 mm from center. All of the blank space can only be caused by a steep cornea, so Scout will fill in the extra space with relatively steeper curves.

 

If on the other hand, Scout sees only 1.0 mm from the outer topo ring to the extrapolated circle, it will assume a flatter cornea and fill in with flatter curves.

 

Thus you control the peripheral curves where Ortho-K lenses are so critical in their fit partly by telling Scout to make the curves in question flatter or steeper by choosing your extrapolation points.

 

There are assumptions that Scout makes and they are not perfect. You can prove to yourself that it doesn’t know how to determine that the calibration ball is spherical to the limbus. It gets the closest to the actual curves when you make the corneal diameter over 16.0 mm, almost off the corners when placing the extrapolation points.

 

The way to measure how much this all affects a lens is to look at the sag. Just for review, think of the sag of a lens as the “rise” of the lens sitting on a flat surface. A larger sag (rise) means that the lens apex stands off the flat surface further, it is a steeper lens. The sag is measured at any diameter so that as the diameter gets smaller, the sag decreases. Sag is critical because it is the relation between the peripheral curve location and the apex of the cornea.

 

When measuring the sag of a Wave lens I suggest you pretty much ignore the “Lens sag = XXXX” that is displayed when your cursor is not held down. This, I think, is the extreme edge of the lens which is outside the display area where we do not see the actual curves of the last .5 mm or so. Since this is normally edge lift, go back towards the center a millimeter or so.

 

How much does this change the ultimate fit of a lens? Take a topo and measure the power on the color map at about 9.5mm along the 180. Extrapolate as you normally do and make a GSym lens with 0 apical clearance and bevel lift in the Autodesign Custom settings. Measure the curve that Wave is telling you fits in alignment at the same 9.5 mm location. My finding is that the curve is much flatter than the topo would indicate. Then go back to the topo and extrapolate very

large. Go back to Wave and redesign the same default lens. My finding is that the lens will now show alignment curves that are actually much closer to the topo curves at that location.

 

The difference in sag between the two lenses at that 9.5 location is about 5 microns while much larger at the displayed value for the full lens which is more on the order of perhaps 15 microns.

 

Food for thought. As I said up front, don’t change what you are doing based on any ideas here unless you already know what you are doing. :)

Richard

Richard L. Anderson, O.D.

Camarillo, CA