🚀 go-pugleaf

Dr Judy wrote:
 > "Dick DeadEye" <anonymouse@yahoo.com> wrote in message
 > news:431C9A7C.4000901@yahoo.com...
 >
 >> Dr Judy wrote:
                <snip>
 >>> The reason you can see (sort of) in the distance  with your old near
 >>> glasses is your large amount of astigmatism.  In one meridan your
 >>> current distance prescription for  is -2.00 and in the other
 >>> meridian it is +1.00, and the equivalent sphere is -0.50.   Your
 >>> current near presciption is -0.25 in one meridan and +2.75 in the
 >>> other, equivalent sphere is +1.25.  Your 1988 near glasses
 >>> prescription is -0.25 in one meridan and +2.50 in the other,
 >>> equivalent sphere is +1.00, virtually the same as required for near.
 >>> You can also see that the -0.25 meridian is not far off from the
 >>> +1.00 you need at distance, so by twisting an angling the glasses
 >>> you could see at distance, although a little blurry.

Okay, I found a page that seems to give a decent explanation and 
illustration:

           http://www.eyecarecontacts.com/optical_lens_prescriptions.html

So Cyl is the dioptic spread between meridian boundaries:  Sph equals 
the meridian's dioptic value at angle Axs 0° ("Meridian_0°"--or is it 
"Meridian_90°"?), and Sph + Cyl is the other meridian's value at angle 
Axs ("Meridian_Axs°")--why not just express it as "Mer_0,Mer_x"?
So, based on the "1976" glasses, my left eye's lens may be Meridian_0° = 
-1.50 and Meridian_80° = -1.50 + 3.25 = +1.75 (Meridian_20° may be 
something like -0.65, Meridian_40° may be +0.05 and Meridian_60° may be 
+0.90), whereas my current glasses correct for Meridian_0° = -2.00, 
Meridian_85° = +1.00 (distance) and Meridian_0° = -0.25, Meridian_85° = 
2.75 (close-up).

               <snip>

 >>#> (Sph = "Sphere", Cyl = "Cylinder", Axs = "Axis";
 >>#>  Adj.Sph, Adj.Cyl = Adjusted Sph, Cyl = Sph/Cyl + Add)
 >>#>
 >>#>      Year    Sph     Cyl    Axs
 >>#>      ====   =====   =====   ===
 >>#>  *** 1976?  -1.50   +3.25    80 (Right eye = -1.25  +3.75  105)
 >>#>      1988?  -0.25   +2.75    80
 >>#>      1990   -3.00   +4.00    85    Add    Adj.Sph   Adj.Cyl
 >>#>      2002   -2.00   +3.00    80   =====   ==============
 >>#>      2004   -2.00   +3.00    85   +1.25    -0.75     +4.25
 >>#>      2005   -2.00   +3.00    85   +1.75    -0.25     +4.75
 >>
 >>
 >>> Looking at the equivalent spheres (your prescription if the
 >>> stigmatism was not in the glasses and refraction was done using only
 >>> spherical lenses) tells the tale:  your distance ES is -0.50 and the
 >>> old glasses provide +1.00, so you would see about well in the
 >>> distance with those near glasses as a 1.00 to150 myope would:
 >>> somewhat blurry but tolerable.
 >>
 >> What is the actual equation for finding the equivalent spherical?
 >
 >
 > Sphere plus 1/2 the cyl, respecting signs.
 >
 > So your 1976 Rx  eqiv sphere is (-1.50 ) +  (+3.25/2)  = +0.12 which
 > is even closer to -0.50.  It also is about 0.50 undercorrecting your
 > myopia and thus it provides some improvement at near as well.

I see (no pun intended! P=), it's .5*(Meridian_0° + Meridian_Axs°) 
--though, would it be safe to say that that is just an approximation (it 
is actually the divided difference of a definite integral)?

-"Dick DeadEye"