Factoring Torsional prismatic effect for Slab Off?

**Aarlan** · 06-22-2007, 11:42 AM

Hello all,

I have been wondering since I started playing around with torsional prismatic effects of lenses, would it be prudent to alter the horizontal component of a the near segment (prism seg, or decentering by grinding Dist OC differently and playing with Dec) in some cases.

Example

OD -4.00 SPH
OS +2.00 -4.00 x 45
Add +3.00

the imbalance between the two lenses is 4D @90, so you can calculate the Slab off easily.

My question stems from the HORIZONTAL prismatic effect of the OS as you move down into the Seg. Since the power @ 90 using sin^2 method is 'plano', there is no Vertical power/prism to speak of. BUt there is a substantial Horizontal displacement (or torsional component) which is noticable. Theoretically shouldn't we also adjust the spectacles (either adjust placement of the seg, or use a horiz prism seg )to minimize the effect of the Horizontal effect as well?

Just wondering.

Thanks

AA

**msparks7** · 07-06-2007, 09:57 AM

I'm only a pseudo expert, but in my experience when the eyes converge for reading the brain can tolerate a lot more horizontal image displacement than vertical. If it were otherwise Executive bifocals would not work in large frames (I still have bad memories from the 80's because of bad glasses like that). That said when you reach a certain point somewhere around 7 to 12 diopters you do need to address the induced horizontal prism with either a prism seg or a Franklin seg. Although I've never tried adjusting the location of the seg to compensate, it sounds like a workable idea, depending on how much you have to move the seg.
The variables affecting the 7 to 12diopters range are

Age. A younger patient can tolerate more the an older patient
A myopic patient can tolerate more than someone who is hyperopic
The individuals general ability to visually adapt to new images

:cheers:

I'll drink to anything

**HarryChiling** · 07-11-2007, 05:28 PM

$ Power = \left[ \begin{array}{ c c } S+C*\sin^2 \theta & -C*\cos \theta * \sin \theta \\ -C*\cos \theta * \sin \theta & S+C*\cos^2 \theta \end{array} \right] $

This is known as the power matrix, if you are looking for the torsional (xy)component look no further:

$Power_{xy} = -C*\cos \theta * \sin \theta$
$Power_x = S+C*\sin^2 \theta$
$Power_y = S+C*\cos^2 \theta$

This is the torsional component, if you wanted to find the prism using the power matrix you can multiply it by the 2x2 decentration matrix:

$ Decentration = \left[ \begin{array}{ c } X \\ Y \end{array} \right] $

The prism equation is power (D) times the decentration (d) is equal to the prism (P) or P=D*d in the matrix form this is:

$ Prism = \left[ \begin{array}{ c c } S+C*\sin^2 \theta & -C*\cos \theta * \sin \theta \\ -C*\cos \theta * \sin \theta & S+C*\cos^2 \theta \end{array} \right] * \left[ \begin{array}{ c } X \\ Y \end{array} \right] $

Now why not use this as your formula to figure slab, you can even get a more accurtae measure of prism by factoring the inset in the decentration instead of just using the vertical measure.

**Darryl Meister** · 07-14-2007, 03:48 AM

Honestly, the difference between using the power through the vertical meridian and using exact calculations that consider both torsional prism and near inset is less than 20% for even the worst cases (cyl at axis 45). For 2.0 prism diopters of vertical imbalance, for instance, the difference is 0.4 prism diopters, which is pretty close to the tolerance on prism imbalance anyway and certainly well within the vertical fusional reserves of the oculomotor system.

**HarryChiling** · 07-14-2007, 11:27 AM

Your absolutely right Darryl, probably not much need for the exact formula, but I just wanted to practice my $\LaTeX$ formatting. :D

**rolandclaur** · 10-13-2007, 10:59 PM

Hey Harry. What is the name of the equation that you listed above. Also, could you explain to me exactly what torsional component means in relation to optics. Thanks

**HarryChiling** · 10-14-2007, 12:48 AM

Originally Posted by rolandclaur

Hey Harry. What is the name of the equation that you listed above. Also, could you explain to me exactly what torsional component means in relation to optics. Thanks

Torsion is the representation of the amount of twist a sheet of light exhibits passing throgh a lens which is at a degree off from the [x,y] axis. The equation above is prentices rule, just not as we recognize it. The

$Power = \left[ \begin{array}{ c c } S+C*\sin^2 \theta & -C*\cos \theta * \sin \theta \\ -C*\cos \theta * \sin \theta & S+C*\cos^2 \theta \end{array} \right]$

is the dioptric power matrix which if you look at the P_1,1 you will notice is the equation for the power on the 180 degree meridian or x axis and P_2,2 represents the equation for the power along the 90 degree meridian or y axis.

Now without the torsion present in the equation how do we knwo which lens this matrix represnts? For instance:

-3.00 -2.00 x 045

Power_x=S+C*sin²(theta)
Power_x=(-3.00)+(-2.00)*sin²(45)
Power_x=(-3.00)+(-2.00)*0.5
Power_x=(-3.00)+(-1.00)
Power_x=(-4.00)

Power_y=S+C*cos²(theta)
Power_x=(-3.00)+(-2.00)*cos²(45)
Power_x=(-3.00)+(-2.00)*0.5
Power_x=(-3.00)+(-1.00)
Power_x=(-4.00)

Whoa if we were to ignore torsion and I was to propose to you what lens has a power of -4.00 at 090 and -4.00 at 180? Looking over my math above you would easily say well -3.00 -2.00 x 045, but what about -4.00 sph? If we were to compute the torsional component of this Rx

Power_torsion=(-C)*sin(theta)*cos(theta)
Power_torsion=(2.00)*sin(45)*cos(45)
Power_torsion=2*0.5*0.5
Power_torsion=0.50

Now the matix can only represnt the power -3.00 -2.00 x 045, this in effect adds a third demension to our curvature by not just representing the curve along these two meridians but also the sharpness in which these curves twist. I hoep that kind of explains it and if not Darryl seems to be more eloquent than me and he originally shined the light on this component to me.

**Darryl Meister** · 10-14-2007, 01:21 PM

Darryl seems to be more eloquent than me and he originally shined the light on this component to me

I would try to visualize placing a 2x4 board directly across a log, and then trying to balance on the log, while standing with your feet at each end of the 2x4 board. If the 2x4 straddles the log at a 90-degree angle, you will have a tendency to rock sideways as you attempt to balance the board across the log. If the 2x4 is placed along the length of the log, however, you will have a tendency to rock forwards and backwards. Although the log is now supporting the length of the 2x4, you will feel your ankles bending back and forth as you try to stabilize the width of the board on the rounded surface of the log.

Now, what happens when the 2x4 is placed at an angle across the log, no longer directly across it or directly along it? Now, you will have a tendency to to rock both sideways as well as forwards and backwards. The 2x4 will actually rotate slightly as you distribute your weight to either leg, causing your body to twist or contort slightly. This log represents a really big cylinder, and the tendency to "twist" is the torsional component of the cylinder. You can demonstrate this effect for yourself with a soda can and a popsicle stick or some other flat object, like a credit card, by placing your fingers at each end of the stick or card and then rocking it at various angles around the can.

[ADVANCED]

In terms of differential calculus, a surface can be described by a function z = f(x,y). The curvatures of this surface at the center of the coordinate system are described by the "second partial derivatives" of the function. The horizontal curvature of this surface is equal to the second partial derivative with respect to x,

$\frac{d^2f}{dx^2}$

This is analogous to the sine-squared component of the formulas described earlier. And the vertical curvature is equal to the second partial derivative with respect to y,

$\frac{d^2f}{dy^2}$

This is analogous to the cosine-squared component of the formulas described earlier. Finally, the torsional-like curvature of the surface is given by the second mixed partial derivative (the derivative with respect to x, then y, or vice versa),

$\frac{d^2f}{dxdy}$

This is analogous to the sine-cosine component of the formulas described earlier, and is also associated with the amount of cylinder power or astigmatism at axis 045.

[/ADVANCED]

**HarryChiling** · 10-14-2007, 02:37 PM

Originally Posted by Darryl Meister

I would try to visualize placing a 2x4 board directly across a log, and then trying to balance on the log, while standing with your feet at each end of the 2x4 board. If the 2x4 straddles the log at a 90-degree angle, you will have a tendency to rock sideways as you attempt to balance the board across the log. If the 2x4 is placed along the length of the log, however, you will have a tendency to rock forwards and backwards. Although the log is now supporting the length of the 2x4, you will feel your ankles bending back and forth as you try to stabilize the width of the board on the rounded surface of the log.

Now, what happens when the 2x4 is placed at an angle across the log, no longer directly across it or directly along it? Now, you will have a tendency to to rock both sideways as well as forwards and backwards. The 2x4 will actually rotate slightly as you distribute your weight to either leg, causing your body to twist or contort slightly. This log represents a really big cylinder, and the tendency to "twist" is the torsional component of the cylinder. You can demonstrate this effect for yourself with a soda can and a popsicle stick or some other flat object, like a credit card, by placing your fingers at each end of the stick or card and then rocking it at various angles around the can.

[ADVANCED]

In terms of differential calculus, a surface can be described by a function z = f(x,y). The curvatures of this surface at the center of the coordinate system are described by the "second partial derivatives" of the function. The horizontal curvature of this surface is equal to the second partial derivative with respect to x,

$\frac{d^2f}{dx^2}$

This is analogous to the sine-squared component of the formulas described earlier. And the vertical curvature is equal to the second partial derivative with respect to y,

$\frac{d^2f}{dy^2}$

This is analogous to the cosine-squared component of the formulas described earlier. Finally, the torsional-like curvature of the surface is given by the second mixed partial derivative (the derivative with respect to x, then y, or vice versa),

$\frac{d^2f}{dxdy}$

This is analogous to the sine-cosine component of the formulas described earlier, and is also associated with the amount of cylinder power or astigmatism at axis 045.

[/ADVANCED]

Thanks that was a better explanation of it than I could come up with. I like the analogy of the soda can and the popsicle stick.