# Optical Forums > Ophthalmic Optics >  Cylinder Power in the 180-degree meridian

## Bob Price

Apparently, the formula "D=Dcyl sin^2 Ax" isn't the most accurate for finding the power of the cylinder in the 180-degree meridian. Is there a more accurate formula that can be used, without getting too technical?

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## Darryl Meister

> Is there a more accurate formula that can be used, without getting too technical?




It depends on what your application is. If you are just calculating thickness, your formula will suffice. If you are wanting to calculate prism at off-center points, you will have to get more technical.

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## HarryChiling

Will the equation gain more accuracy if measured in radians?

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## Bob Price

> It depends on what your application is. If you are just calculating thickness, your formula will suffice. If you are wanting to calculate prism at off-center points, you will have to get more technical.


My goal is to be able to figure out everything on a workticket without the use of our computer software. Right now I'm using a graphing calculator and hope to upgrade to an IPac. Hopefully this will allow me to have access to an Excel spreadsheet (that I can carry around) where I can combine all of the formula's into one organized spreadsheet. Just enter the Rx and all of the information that is needed to process a lens is calculated. This is the easiest way for me to learn and I have found it to be quite fun, but challenging. It seems that all of the optical calculators I have found come up with slightly different values. This is frustrating, but is it because of my question above, that there are different formulas to get the final result? On my question, it appears that by using the formula I mentioned for calculating decentration, then it loses accuracy and another formula needs to be used.

I don't know if you remember me calling you abut 5 years ago, when you helped me with the equation for figuring the tool needed. Somehow I erased the equation and now have renewed interest, since figuring out how to get it back in the calculator. Thanks for your help.

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## Darryl Meister

> Will the equation gain more accuracy if measured in radians?


If you mean using radians for _Ax_, you will actually get the same answer either way...?




> My goal is to be able to figure out everything on a workticket without the use of our computer software.


Then you will definitely need a more "technical" approach. Calculating prism accurately requires a couple of simultaneous equations that use quite a bit of trigonometry. Fortunately, you can just copy them from a textbook or something, but you would still need to verify the sign convention of your input and output and that sort of thing.




> This is frustrating, but is it because of my question above, that there are different formulas to get the final result?


Yes, but if they're correct, you'll get the same answer either way.

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## HarryChiling

I am not sure how else you could go about finding the power in that meridian without using the the 

D=Dsph+Dcyl*sin^2(180-axis)

What other method could you use to more accurately get this power? And where does this equation lose it's accuracy?

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## Darryl Meister

> What other method could you use to more accurately get this power? And where does this equation lose it's accuracy?


It's not so much the power he's worried about, but rather calculating the prism at points away from the optical center (or, conversely, calculating how to induce prism in order decenter the optical center).

Also, the _sine-squared_ formula just approximates the curvature of the surface through a particular meridian. It doesn't actually represent the "power," since a sphero-cylindrical lens technically _has no power_ in meridians other than the principal meridians (the rays of light refracted in these other meridians are _skew_ rays, and do not actually intersect at a focus). Also, recall from our earlier discussions of Keating's _Dioptric Power_ matrix that the power of a cylinder has both a _curvital_ component (i.e., the sine-squared component) and a _torsional_ component.

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## Bob Price

[QUOTE=Darryl Meister]Then you will definitely need a more "technical" approach. Calculating prism accurately requires a couple of simultaneous equations that use quite a bit of trigonometry. Fortunately, you can just copy them from a textbook or something, but you would still need to verify the sign convention of your input and output and that sort of thing.


Would you please lead me in the right direction for the equations. I might already have them under my nose and will look for them tonight. The books I am working with are "Understanding Lens Surfacing" and "Introduction to Ophthalmic Optics". Thanks again for your input.

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## Darryl Meister

> Would you please lead me in the right direction for the equations. I might already have them under my nose and will look for them tonight.


Just about any book on ophthalmic or geometrical optics should have them. _Clinical Optics_, _Principles of Ophthalmic Optics_, _Optics of Ophthalmic Lenses_, _Geometrical, Physical, & Visual Optics_, and so on.

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## Darryl Meister

For instance, you can find these equations to calculate the horizontal (Ph) and vertical (Pv) prism effects at a horizontal distance x and a vertical distance y from the optical center (compliments of _Clinical Optics_):

Ph = y * Fc * sin A * cos A + x (Fs + Fc * sin^2 A)
Pv = y (Fs + Fc * cos^2 A) + x * Fc * sin A * cos A

With some consideration, and proper adherence to a sign convention, these equations can also be used to solve the for the horizontal prism and vertical prism necessary to move the optical center to a desired horizontal or vertical location.

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## HarryChiling

So let me see if this is correct if I would only want to move a lens horizontally then the equation would be

Ph = y * Fc * sin A * cos A + x(Fs + Fc * sin^2 A)
Ph = "0" * Fc * sin A * cos A + x(Fs + Fc * sin^2 A)
Ph = x(Fs + Fc * sin^2 A)

And similar would be used for the vertical prism.

Pv = y(Fs + Fc * cos^2 A)

So if this is true then what we were missing before was the decentration in the formula. Is that correct, and what measure does the x,y variable use mm or m?

-or-

Is the x,y component using a cartesian co-ordinate, where the Optical Center of the lens is equivalent to (0,0) or the center of the grid?

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## Darryl Meister

The variable x would represent the horizontal decentration (in cm, if I recall correctly) and y would represent the vertical decentration (again, in cm).

The basic sine-squared formula is a just an approximation of curvature, not prism. It will give you a good ballpark estimate of the prism along a particular meridian, but it fails to compute the prism that occurs perpendicularly to that meridian. For instance, consider the prescription +1.00 -2.00 x 045. Decentering this lens _horizontally_ will actually introduce a significant _vertical_ prism component.

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## Bob Price

> It will give you a good ballpark estimate of the prism along a particular meridian, but it fails to compute the prism that occurs perpendicularly to that meridian. For instance, consider the prescription +1.00 -2.00 x 045. Decentering this lens _horizontally_ will actually introduce a significant _vertical_ prism component.


 
Then going back to my opening question, is there a workable formula that will calculate this induced prism caused by an oblique axis? I would think that it isn't very criticle when you are blocking SV lenses using on center blocking, unless you have a large eye size and cut out is a problem. However,for multifocals using on center blocking, this must be more criticle when you need to move the OC horizontally and vertically to a desired location. 

Going back to another question I have,  I still don't know how to figure the prism axis needed for grinding slab off prism in the surface room (on center blocking). The whole goal is to get the prism in the exact location to achieve a strait line to your segment. This might be an easy calculation, but I must admit, I am still in the learning stages and have a lot of questions. Thanks.

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## Darryl Meister

> Then going back to my opening question, is there a workable formula that will calculate this induced prism caused by an oblique axis?


Yes, I posted them earlier in this thread.

Regarding your slab-off question, you don't position the slab line using prism. It is controlled by lens thickness.

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## Bob Price

Sorry to be so persistent, but I might not be asking the right questions. I think I need to be asking how to compensate for the prism angle, when trying to counter the OC moving vertically.

Here is a right lens example.

Poly Sv
-4.75 -3.75 Axis60
4.5 mm Decentration

x(Fs+Fc*Sin(Axis)^2)
.45(-4.75+-3.75*Sin(60)^2)=3.40  (Is this correct?)

Now, I wouldn't just grind 3.4 diopters @ 180, because the OC will move vertically. So, to keep the OC from moving perpendicular to the 180, I would have to alter the prism axis (and possibly the prism, I'm not sure). Is this the right way of thinking, and if so, how would I calulate this?

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## lensgrinder

You would grind 3.4 D of prism in the horiziontal meridian to move the OC 4.5mm, because that is where you want to move the OC.  In this case you want the base of the prism @ 180, because it is a minus lens.  Of course it is SV so it really does not matter.  If you ground the lens at an oblique axis it would move the OC in the horizontal and vertical meridian.  The question does arise, do you really want to grind prism with only 4.5mm of decentration?  That would depend on what blank size you need.

Hope this helps.

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## Bob Price

> The question does arise, do you really want to grind prism with only 4.5mm of decentration? That would depend on what blank size you need.
> 
> Hope this helps.


I realize that for SV this isn't extremely important, but would it be for Bi's and Tri's? When I try to duplicate our workticket with these formulas - SQRT(V^2+H^2)=Prism and Tan(V/H)=Axis - I don't get the same results. When your trying to move the OC to a desired location, there should be only one right answer. This is why I'm so persistent on trying to figure out what affect an oblique axis has when doing the calculations. I'm probably way off base and just didn't do the figuring correctly, using the formulas mentioned above.

Also, does the index of refraction have any affect on figuring for prism? Or, is it always Power*Decentration=Prism?

Thanks for all of the input.

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## lensgrinder

Yes refractive index does have an effect on prism in lower powers.  I would recommend that you read Don Whitney's paper on prism found in the file download area.  The name is Prentice's Rule - It's Applications and Limitations by Don Whitney

Could you give an example of what the  software computes versus what you get, using a bifocal?

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## Darryl Meister

Grinding base in or out prism, only, when trying to decenter the optical center of an oblique axis cylinder will move the optical center up or down, as well.

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## Bob Price

> Yes refractive index does have an effect on prism in lower powers. I would recommend that you read Don Whitney's paper on prism found in the file download area. The name is Prentice's Rule - It's Applications and Limitations by Don Whitney.


Thanks for leading me to this article. I'm not having much luck yet, but will continue to work on it.

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## JRS

Yes, the prism has a tendency to follow (or track) along the cylinder axis. So to get 2 diopters in on a lens with a cylinder axis of 45, would probably need a prism axis of about -174 (354) degrees. Obviously eye dependent, etc. I got the math around here somewhere Bob. I'll try to dig it up for you. And I tried to leave you a bit of info on EXCEL functions in your other post.

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## HarryChiling

Their is an article that would be relevant to your question


Analysis of Clinical Approximation in Applying Prentice's Rule to Decentration of Spherocylinder Lenses - Ophthalmic Physiological Optics 1984;4(3) pages 265-273

It should have relevant information, if I can get the article I will let you know and send you off a copy.

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## Bezza

Jalie's Principals of Ophthalmic Lenses covers this topic very well, theres trigonometrical solutions and some neat graphical constructions that you can use also. I had the pleasure of spending a week with Mo Jalie on a revision course for my SMC(tech), what that guy doesnt know about ophthalmic lenses isnt worth knowing

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## Darryl Meister

I agree.

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## HarryChiling

It can be done by using prentice's rule in the form of a power matrix.  This will account for the torsional component of the spherocylindrical lens.
for the example -4.75 -3.75 x 060 with a decentration of 4.5mm

The horizontal prism is 3.4 and the vertical prism is 0.73

It is a little complicated to explain but if you pm me I will send you a scan of the equations that I used or if you look in Geometric, Physical, and Visual Optics by Keating you will find it.

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## HarryChiling

Darryl I believe that the formula that you posted is the prentice in matrix form however you forgot the (-) sign to factor in




> Ph = y * Fc * sin A * cos A + x (Fs + Fc * sin^2 A)
> Pv = y (Fs + Fc * cos^2 A) + x * Fc * sin A * cos A


it should read

Ph = y * Fc * sin A * cos A - x (Fs + Fc * sin^2 A)
Pv = x * Fc * sin A * cos A - y (Fs + Fc cos^2 A)
note: x and y are decentration values in cm

or simplifyed to

Z=-Ph

Z = 2x1 prism matrix
P= 2x2 power matrix
h= 2x1 decentration matrix (in cm) 

Double check my work because as you know I always make stupid mistakes when it's late night.

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## Darryl Meister

> It can be done by using prentice's rule in the form of a power matrix.


Yes, and you'll notice that the solution to Keating's matrices results in a form of the equations I provided earlier (post #10). Bob has had the only equations he needs to calculate the horizontal and vertical prism components. Of course, he'll need to compound them into a single resultant prism, but that's not difficult.




> Darryl I believe that the formula that you posted is the prentice in matrix form however you forgot the (-) sign to factor in


It's just a difference in sign convention.

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## Bob Price

Thank you to everyone for the response. Although I've been in the lab since I was a kid, this is the first time I am really focused on studying optics. J.R. gave me one of my first real eye openers to the technical side of the lab when he visted us back in the late 90's. Then, I was more interested in the calibration and maintenance side of things. Now, it is almost to the point where a person needs to know nothing about optics to work in a lab. So, thank you J.R. and I would be interested in whatever info you can find.(Thanks for recognizing my promotion and my Mom's doing great)

I would also be interested in the articles that Harry is talking about. This will probably explain how green I am, but I don't know what "pm me" means or anything about Matrices.

Bezza, thanks for the tip on Jalie's Principals of Ophthalmic lenses. I'll put it on my list of books I will be getting.

Darryl, with what time I've had this week, I have been working with your formula's and also on the tip that someone gave regarding Whitney's thoughts for Prentice's Rule and low powered lenses. Although I don't have it figured out yet, I will keep working on it as time allows. There is still something I'm missing in Whitney's and your formulas that aren't allowing me to match our workticket. Hey, it's probably my crooked math.

I'll give an example of a right eye from a workticket and see if someone can duplicate it with what we've discussed so far.

-3.75 +1.00 AX 5

Material = Poly
On-center blocking
Bifocal = 6.5 in 6.5 down
Inset = 2.5
OC = 23
Seg Ht = 18

We need to move the OC In(or towards 0) 4mm and down 1.5mm.
The workticket shows the prism @1.24 and the prism axis @166.

Thanks

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## Darryl Meister

> There is still something I'm missing in Whitney's and your formulas that aren't allowing me to match our workticket


Whitney's article doesn't really apply to your question. What is your work ticket telling you?

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## Bob Price

> Whitney's article doesn't really apply to your question. What is your work ticket telling you?


I wasn't able to duplicate the prism and prism axis that our system came up with, so I thought maybe that the Index of Refraction might play a part in figuring the amount of prism needed.

Here is my example again, with what the workticket came up with.



-3.75 +1.00 AX 5

Material = Poly
On-center blocking
Bifocal = 6.5 in 6.5 down
Inset = 2.5
OC = 23
Seg Ht = 18

*The workticket shows that 1.24 diopters of prism needs to be ground at axis 166 to move the OC to the desired location.
*

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## lensgrinder

Bob,
I calculated the work ticket using the formula's that Darryl provided:

Ph=1.48
Pv=0.08375

Resultant prism = 1.48 @ Axis 176

Using the sine squared method I got:

Ph=1.5
Pv=.56

Resultant prism = 1.6 @ 159

Well I guess my calculations got you no where. Sorry.  I do notice that on Innovations software that it does not compute prism correctly, so I am going to grind some on Monday using Innovations calculations and the formula's that Darryl provided and see which is better.  I have been  wanting to do this since he posted the formula's anyway.  I will post my results when finish.

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## HarryChiling

Bob the solution to your problem lies in the fact that you are using prism to decenter the OC and you are blocking the lens on center.  To determine how much prism you will need you have to first find the seg inset

inset=(farpd-nearpd)/2

the the seg drop

drop=segheight-(B/2)

now this is how far off the geometric center that you want to place the lens  the drop would be comparable to the y decentration in the formula and the inset would be equal to the x decentration.  The Prism is a way of moving the OC from the geometric center of the lens to the correct position.  (geometric center being 6.5 in and 6.5 down)

Given your parameters we would not be able to determine the correct prism.  We need the far pd, the near pd, and the frame b at least to figure it out.  If you provide these parameters I will walk you through the formula, as for the matrix equation and the formula for prism that was posted by Darryl it came from Keatings book that Darryl had suggested to me.  Keating uses matrix's extensively throughout the book to solve common optical problems and once you gain a understanding of these matrix's these equations start to make more sense in my mind.  Like for example this equation is prentice's rule 

Prism=Power*decentration

but it is expresses in the form

prism matrix=power matrix*decentration matrix

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## lensgrinder

> To determine how much prism you will need you have to first find the seg inset
> 
> 
> inset=(farpd-nearpd)/2
> 
> the the seg drop
> 
> drop=segheight-(B/2)


He did provide the inset.  Which is 2.5mm.  Also he wants the OC 5mm above the seg(OC ht 23 Seg ht 18).  He blocked his lens on center at 6.5 in and 6.5 down.  So he needs to move the OC 4mm in and 1.5mm down.

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## Bob Price

I never really gave Darryl's formula's a chance with a bifocal lens. At the time, I was concerned with a single vision lens and determining the affect of an oblique axis.

Here is what I come up with for the example I posted.

Ph = y * Fc * sin A * cos A + x (Fs + Fc * sin^2 A)
Ph= .15*-1.00*sin(95)*cos(95)+.40(-2.75+-1.00*sin(95)^2)=-1.48

Pv = y (Fs + Fc * cos^2 A) + x * Fc * sin A * cos A
Pv = .15(-2.75+-1.00*cos(95)^2)+.40*-1.00*sin(95)*cos(95)=-.38

SQRT(1.48+.38)=1.36
Tan-1(.38/1.48)=14.4
180-14.4=165.6

So, with this formula, I come up with 1.36 diopters at axis 166. This is within .12 diopters of the workticket for prism and right on for axis. Getting closer. Did I do my calculations correct? 

Now, how to figure the affects of an oblique axis when figuring the prism axis to grind on a SV. This isn't crucial unless cut out is an issue, but I still am curious how to figure it.

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## HarryChiling

nice!

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## HarryChiling

nice!

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## lensgrinder

Greetings Bob,

I got the same thing you got in the horizontal meridian, but in the vertical meridian it is:

Pv=0.030456816

This is different than what I posted earlier.  The prism axis will be:

tan-1(a)=0.030456816/1.48
a=1.18 degrees

Your prism axis will be 180-1.18=178 degrees
Prism amount will be:
0.03456816 squared + 1.48 squared = 2.19
square root 2.19 = 1.47

1.47 @ 178 degrees


As far as for SV use the same formula's.

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## JRS

Bob, is the difference you are seeing - your calc versus the work ticket... related to the "prism index" (or ring index, or by whatever name) stored within the software? I think you find it in params.vis on the DVI platform.

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## Bob Price

> Bob, is the difference you are seeing - your calc versus the work ticket... related to the "prism index" (or ring index, or by whatever name) stored within the software? I think you find it in params.vis on the DVI platform.


I'm going to be out of town thru Wednesday, but will take a look on Thursday. The only reason I am doing all of this is to try and learn. I feel that if I can duplicate our workticket with an Excel spreadsheet, I will have accomplished something. The trouble I'm having today seems to be trying to get the prism axis to come up in the right quadrant. This is more of an Excel issue, and a definite challenge, and I haven't even started entering prescribed prism into the mix.

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## Darryl Meister

> My goal is to be able to figure out everything on a workticket without the use of our computer software... The only reason I am doing all of this is to try and learn.


Honestly, Bob, if you really want to learn this, I think you're going about it the wrong way. You're basically trying to find a formula that will replicate your lab software's results, without actually learning how or why it's producing those results. The work ticket of a good lab calc program is the product of dozens of complex calculations; it will take you awhile to reproduce those results with this kind of trial-and error-method. As Bezza, Harry, and I have suggested, you might consider getting a good book on ophthalmic optics to help with some of the base principles in conjunction with any support you get here on OptiBoard.

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## Bob Price

> Honestly, Bob, if you really want to learn this, I think you're going about it the wrong way. You're basically trying to find a formula that will replicate your lab software's results, without actually learning how or why it's producing those results.


I agree with what you are saying. I have your book and one by Clifford Brooks that I am working with to help me along. I plan on purchasing the ones everyone has suggested, also. Unfortunaltely, I'm not the type to be able to sit down, read a book and retain the information. I need to be hands-on and work thru it. You are right about the computer software being complex, as I am finding that out more and more. Thanks for your help.

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## JRS

I can't answer directly for Bob, but I get the impression he is trying to figure out why his work order prints certain results. As you mentioned - the software system's results are a complex set of instructions (results) based on many parameters that mimic (or hope to mimic) his equipment and his methods.

However, I agree with you Darryl... if you understand the basic principles, then understanding why the ticket says "do this" becomes easier.

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## shanbaum

> Sorry. I do notice that on Innovations software that it does not compute prism correctly, so I am going to grind some on Monday using Innovations calculations and the formula's that Darryl provided and see which is better.


'Scuse me?  Want to provide an example?

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## JRS

(LOL)...Was wondering when you'd take notice of that statement Robert. I was going to ask the same thing, but figured you'd follow up.

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## Darryl Meister

> I can't answer directly for Bob, but I get the impression he is trying to figure out why his work order prints certain results


My impression was that he was trying to duplicate the entire work ticket.




> Unfortunaltely, I'm not the type to be able to sit down, read a book and retain the information. I need to be hands-on and work thru it.


Yes, but posting a question looking for a formula that produces the right results isn't working through it, it's simply asking for the answer from someone who _has_ sat down and read the book. ;) Of course, we're all certainly glad to help either way.

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## lensgrinder

> 'Scuse me?  Want to provide an example?


OK.  I spoke out of turn.  It was wrong of me to imply that the software engineers and any other engineers @ GC have made errors.  I am in no position to make a statement like that.  I have no where near the knowledge that the people @ GC have.  My experience comes solely from practical experience.  That being said I do have an example:

OD -0.25 -1.50 X 80
OC=19
Seg Ht.=14
PD 63.5/60.5

Innovations calls for 0.8 @ 169 the OC is roughly located 0.5 out  and 4 mm above seg.  Granted it does not make much of a difference it just seems consistent.
When I perform the calculations I come up with 0.4 @ 170.  Now the OC is roughly located 1.5 out and 5 mm above the seg.  
Again I want to state that I do not claim to know more than the Innovations team.  I have only tested this maybe a dozen times over the course of a couple of months.  I know GC has done a lot more testing than me.

Any way I hope I cleared up my stupid comment.

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## shanbaum

> OK. I spoke out of turn. It was wrong of me to imply that the software engineers and any other engineers @ GC have made errors. I am in no position to make a statement like that. I have no where near the knowledge that the people @ GC have. My experience comes solely from practical experience. That being said I do have an example:
> 
> OD -0.25 -1.50 X 80
> OC=19
> Seg Ht.=14
> PD 63.5/60.5
> 
> Innovations calls for 0.8 @ 169 the OC is roughly located 0.5 out and 4 mm above seg. Granted it does not make much of a difference it just seems consistent.
> When I perform the calculations I come up with 0.4 @ 170. Now the OC is roughly located 1.5 out and 5 mm above the seg. 
> ...


 
You don't have to apologize, but you do have to give me a little more information before I could tell you how it comes up with that number, namely, the surface blocking instructions, material index, prism index - or, better yet, if you send the XML file for that job to me (located in the Ticket folder), that contains all that and more, and I would certainly take a look at it.

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## Darryl Meister

> It was wrong of me to imply that the software engineers and any other engineers @ GC have made errors.


Especially in a forum moderated by Gerber-Coburn's Chief Software Architect. ;)

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## lensgrinder

> You don't have to apologize, but you do have to give me a little more information before I could tell you how it comes up with that number, namely, the surface blocking instructions, material index, prism index - or, better yet, if you send the XML file for that job to me (located in the Ticket folder), that contains all that and more, and I would certainly take a look at it.


Thanks, I will get that information to you the next chance I get.

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## Darryl Meister

Bob, Did you ever get your formulas sorted out?

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## Bob Price

> Bob, Did you ever get your formulas sorted out?


Things have been really hectic at work and at home, however, I still work on it every chance I get. I don't have it figured out yet and still plan on purchasing some more books to help me out. This will be an ongoing process that I will have to stay committed to. I tend to get going in too many directions and need to focus on one task at a time.

By trying to use the formulas, that are in the books I have, to surface a lens in the lab, seems to be where the challenges are. I want to be able to put all of this to practice and believe me, I know how over my head I am right now. It seems that the formulas the lab software uses are more detailed and accurate and I just need some help in gathering that type of information. Although I am looking for the answers, I don't expect to have them given to me. 

Thanks for the follow up.

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## Bob Price

I am seeking some more advice, as I seem to be stalled in my search. Even though I would like to be able to duplicate our workticket, I realize it helps to know how to walk before you try and run, so that will have to come later.

This thread started with me wanting to know how to figure the prism axis on a SV, with an oblique axis, when I was moving the OC for decentration, and has gone in a little different direction, for which I'm glad. In trying to take things one step at a time, I'm going to focus on grinding prism in a lens (SV or Multifocal, but mainly Bi's & Tri's) to move the OC to a desired location. Using the formulas that Darryl provided - 
Ph = y * Fc * sin A * cos A + x (Fs + Fc * sin^2 A)
Pv = y (Fs + Fc * cos^2 A) + x * Fc * sin A * cos A - 
I feel I have been able to accomplish this, but know that the index of the lens affects prism and should be figured into these formulas somehow or the formulas in the Whitney article need to be used  somehow. I did run some tests from our system, in different indexes with the same power, and saw that the prism amount changed for each index. I thought that the Whitney article was the answer, but remember Darryl saying that this didn't pertain to what I was trying to accomplish. Even if it did, I haven't been successful in figuring out either formula in the article. I think my problem was in the difference between the Front Surface Power (1.53), which I thought was n-1/ti-1*true curve, and the True Front Surface Power. I also tried taking the prism result, from the above equation, and multiplying by ti-1/n-1, but that didn't seem to be the answer. 

Am I going down the right road or on a complete dead end? Any advice would be much appreciated.

P.S. Someone did give an example on the Whitney formula and I will find it and look it over.

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## Darryl Meister

> I feel I have been able to accomplish this, but know that the index of the lens affects prism and should be figured into these formulas... I did run some tests from our system, in different indexes with the same power, and saw that the prism amount changed for each index.


As JR noted earlier, the difference might very well be due to a "prism index" setting (though, unless you're blocking prism into the lens, I'm not entirely sure why this setting would matter). You could probably check this quite easily by running a sample work ticket with a significant amount of prescribed prism (e.g., 4.0 PD), and then adjusting the prism index from its highest to lowest setting in order to see whether it changes the values significantly. If you _are_ blocking for prism, you might also check to see whether the software rounds the calculations to match your prism ring steps.




> I thought that the Whitney article was the answer, but remember Darryl saying that this didn't pertain to what I was trying to accomplish.


Whitney's article deals with something akin to _ray traced_ prism. Prentice's rule is indeed valid for any power, at least until the effects of spherical aberration kick in. The difference is that Prentice's rule applies to a perfect horizontal translation of the lens. A focimeter, on the other hand, does not translate a lens across a flat horizontal plane when it is moved across the lens stop, but rather it translates the lens across an arc equivalent to the back surface of the lens (since the lens stop is always perpendicular to the back surface).

But I don't know that your computer software is even getting into this level of detail, though it could be. And, even if it were, the differences would be marginal in most prescriptions (while the linear distance may seem large in Whitney's examples, the prism is very small and would not fail ANSI tolerances). If it is getting into this level of detail, it may be doing these calculations at a level of sophistication or complexity that you will have a hard time reproducing.

Nevertheless, once you have eliminated the prism index as a potential culprit, you could verify the accuracy of your calculations by running a sample work ticket using a relatively flat, high-powered plus lens (e.g., a +3.00 sphere on a 4.00 Base). If the prism results are closer for this job than for a -0.25 sphere on a 6.00 Base, it could be due to calculating ray traced prism versus Prentice's prism.

Also, ensure that the software you're comparing your results against is producing dead-on accurate results for these prism calculations in the first place, or you may be spending a lot of time trying to reproduce small systematic errors.

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## Bob Price

[QUOTE = As JR noted earlier, the difference might very well be due to a "prism index" setting QUOTE]

I know that the software has a Prism Index setting listed for each material, but am not sure how or if it figures it into the equation. I will try and find this out. If the Prism Index is taken into account when figuring prism, then is it possible to factor it into your formulas to get similar results? All of this still baffles me, because there should only be one correct amount of prism and prism axis that correctly moves the OC to the proper location. 


[QUOTE =  Whitney's article deals with something akin to _ray traced_ prism[/QUOTE]

This is totally new to me and have only heard of it since joining this site. From what I am gathering, it takes into account the power (and whatever else) at every position of the lens. Is Ray Tracing something that is found thru calculations or measuring? Don't worry, I won't go diving into this until I'm more seasoned.

[QUOTE = Also, ensure that the software you're comparing your results against is producing dead-on accurate results for these prism calculations in the first place, or you may be spending a lot of time trying to reproduce small systematic errors.[/QUOTE]

Our OC's spot really good, so this is all I have to go on to determine that the software produces spot on results. Of course, when we calibrate our generator, a lot of time goes into making sure there is as little prism as possible in a plano lens.

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## Darryl Meister

> I know that the software has a Prism Index setting listed for each material, but am not sure how or if it figures it into the equation.


It doesn't. The prism calculated by the equation tells you how much prism you need. The prism index setting tells you how much prism you need to surface into your n1 index material using an n2 index prism ring in order to produce the desired results. This will adjust the prism by an amount equal to (n2 - 1) / (n1 - 1).




> If the Prism Index is taken into account when figuring prism, then is it possible to factor it into your formulas to get similar results?


You would just reverse the process described above.




> All of this still baffles me, because there should only be one correct amount of prism and prism axis that correctly moves the OC to the proper location.


For a given lens configuration, There Can Be Only One. (Cue the Queen theme music from _Highlander_.)




> Is Ray Tracing something that is found thru calculations or measuring?


It is done through analytical calculations. Ray tracing in the traditional sense of the term is generally used in lens design, though good surfacing software should account for various optical effects, like the astigmatism induced by prism, the change in prism produced tilt angle in low powers, and so on, which can only be determined using certain ray tracing calculations. (I know that SOLARx -- SOLA's surface software -- does, since it is written by the same guys who write our ray tracing programs, but someone like Shanbaum or JR would have to speak to whether or not this is typically the case with commercial surfacing software.)

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## Bob Price

> It doesn't. The prism calculated by the equation tells you how much prism you need. The prism index setting tells you how much prism you need to surface into your n1 index material using an n2 index prism ring in order to produce the desired results. This will adjust the prism by an amount equal to (n2 - 1) / (n1 - 1).


In the information that you and J.R. have provided, I am finally seeing that I already have the correct amount of prism needed for any given RX, using the formulas that you provided. Now, it is a matter of knowing the equipment settings in the software and converting it to the amount on the workticket.

I'm not totally following your example above. From my previous example where I had a poly ST28, and I came up with a prism value of 1.36D and the workticket called for 1.24D. n1 would be the index of refraction of the lens and n2 would be the prism ring index. Is the prism ring index a value like 1.523 or is it the depth or diameter of the ring?

So, is this similar to what happens for laps cut in different indexes? A radius of curve is the same in 1.53 or 1.60, but is expressed differently depending on what index the lap is cut in.

Am I getting hotter or colder?

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## JRS

Bob, in your LMS System, the prism index - per material, may be the same indice, or may equal material index. You should look at the settings within the software.

If you take 3 materials and create 3 equal Rx's (plano powered SV with 3 diopters BI prescribed) and print tickets... you will likely see 3 different amounts printed for "grind this prism at generator", yet all 3 will produce 3 diopters of BI prism when viewed through an inspection device.

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## Bob Price

> Bob, in your LMS System, the prism index - per material, may be the same indice, or may equal material index. You should look at the settings within the software.
> 
> If you take 3 materials and create 3 equal Rx's (plano powered SV with 3 diopters BI prescribed) and print tickets... you will likely see 3 different amounts printed for "grind this prism at generator", yet all 3 will produce 3 diopters of BI prism when viewed through an inspection device.


J.R.,

   Yes, the prism indexes are different. As is the ring depth and Prism ring diameter. I did run tests with a plano with 3 diopters base in and the prism amount required at the generator was different for each material. The higher the index the less prism required.

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## Darryl Meister

> I did run tests with a plano with 3 diopters base in and the prism amount required at the generator was different for each material. The higher the index the less prism required.


Assuming your prism index (n2) is set to 1.530 and you are using polycarbonate (n1 = 1.586), you should be seeing this much prism relative to your prescribed prism:

*(n2 - 1) / (n1 - 1) = (1.530 - 1) / (1.586 - 1) = 0.904*

Or, roughly 1 - 0.904 = 9.6% less prism. This would change 3.0 prism diopters to 2.7 prism diopters.

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## Darryl Meister

To clarify, you need a weaker prism ring in a higher index lens material since the actual strength of the prism ring increases as the refractive index of the material increases. A 3.0 prism ring calibrated for a tooling index of 1.530 will produce exactly 3.0 prism diopters in a lens material with a refractive index of 1.530, but 3.31 prism diopters in a lens material with a refractive index of 1.586 -- and only 2.82 prism diopters in a lens material with a refractive index of 1.499.

While each lens material will need the same amount of prism, the actual amount of prism produced by the _wedge angle_ of the prism ring will vary from material to material. Consequently, a single prism ring will not produce a consistent amount of prism across different lens materials. Though you can convert back and forth using that expression, (n1 - 1) / (n2 - 1) or (n2 - 1) / (n1 - 1), depending on which way you need to go.

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## Bob Price

Darryl,

I brought a couple worktickets home tonight, to work on you're latest post, and stummbled across something I never realized (or maybe still don't). The Rx is a CR-39 ST35 +2.00 -100 Ax 70 OU. The OC needs to be moved in 3.5mm and down 2mm. Right eye calls for prism of .45 @322, the left eye .68 @ 226. If I use the formulas that we have been discussing in this thread, I would come up with the same result. In examining the workticket you can plainly see that the OC is traveling differently thru the cylinder of the right lens compared to the left lens. If the left eye axis was 110, I would imagine that the prism amount would be the same. 

Sorry to bring up something else when I don't even have a handle on our main topic, but I wanted to get yours and everyone elses opinion.

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## Bob Price

I must have blown it and you guys are scratching your heads thinking, "This guy is an idiot." In regards to my last post, does the formula I'm using take this into account and I'm not seeing it? Also, I think that as the OC height changes a different angle is created in relation to the cylinder and needs to be taken into account.

 Maybe this is what your saying about learning the basics first, because I don't know the inner workings of the formula yet, to know everything it is figuring.

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## Darryl Meister

Hi Bob, I won't have time to go through the math for a bit, but you are flipping the sign of the decentration between the right and left lenses, right?

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## Bob Price

I will be now. Thanks.

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## Bob Price

I spent some time yesterday going over my inconsistent results. It all boiled down to not having the formula plugged into my calculator correctly. I went thru the formulas piece by piece and got much better results. Another missing piece was the prism index, which I'm pretty sure is set to 1.498.

I did notice more inconsistencies in the higher powers and am wondering if this is because of a possible "prism rounding" in the software? Also, when dealing with a plano, my calculations call for no prism when there should be around .05. Any suggestions would be greatly appreciated.

Darryl mentioned using the proper sign convention for the decentration between the right and left eyes. Is this just for the vertical? For instance, if the bifocal is 6mm down and the difference between the seg ht and OC is 3mm, I would use a -.3 for the left eye but not for the right? A more in depth explanation of this would also be greatly appreciated.

Thanks.

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## Darryl Meister

> Darryl mentioned using the proper sign convention for the decentration between the right and left eyes. Is this just for the vertical?


No, it's just for horizontal.

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## lensgrinder

> Darryl mentioned using the proper sign convention for the decentration between the right and left eyes. Is this just for the vertical? For instance, if the bifocal is 6mm down and the difference between the seg ht and OC is 3mm, I would use a -.3 for the left eye but not for the right? A more in depth explanation of this would also be greatly appreciated.


Bob,
If you think of a coordinate system with 0 degrees to the right 180 degrees to the left 90 degrees at top and 270 degrees at the bottom. Now think of two of them side by side like you were looking at the person. On the left you have your right eye and on the right you have your left eye. When you move the OC in or towards 0 for the right eye you are moving in the positive direction. When you move out or towards 180 you are moving in the negative direction. The opposite applies to the left lens, except when you move toward 90 or up then you are moving in the positive direction and the same applies to the right eye. When you move a left lens in and down (towards 180 and towars 270) you are moving both in the negative direction. I hope this helps.

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## Bob Price

> Also, when dealing with a plano, my calculations call for no prism when there should be around .05.


I'm thinking that if I were to figure the power, in these formulas, using the radius of the front and back curves (combined with thickness and index of refraction), then when figuring for prism it would take into account that it is not a true plano and that prism needs to be ground for decentration. 

I've been messing around a little bit with this tonight, but would like to be shut down, if I'm way off base again, before I waste a bunch of time.

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## lensgrinder

> No, it's just for horizontal.


Why do you not use negative numbers in the vertical?

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## Darryl Meister

> Why do you not use negative numbers in the vertical?


You do use positive and negative numbers for the vertical meridian, but this convention doesn't change between the right and left eyes (as you noted in your earlier post). For horizontal decentration, on the other hand, the sign convention is flipped between the two eyes.

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## lensgrinder

> You do use positive and negative numbers for the vertical meridian,


OK, good. I just wanted to make sure I was not loosing my mind.  I misunderstood when you said it is just for horizontal.  Thanks for clearing that up.

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## Aarlan

This is on the ABO, right?


AA

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## Darryl Meister

> This is on the ABO, right?


Not this kind of stuff. You may need to know that a lens with cylinder power has no cylinder power along its axis and maximum cylinder power at 90 degrees away from its axis, but that's about it.

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