# Optical Forums > Ophthalmic Optics >  Decentration to Lens Tilt

## HarryChiling

So I have been racking my brain on this, how to convert decentration into lens tilt.

d = decentration (mm)
r = radius (mm)
t = tilt



The value comes out a little high though, I have found that it should be about 3/4 of that value.  I have put pencil to paper and ray traced it and come up with the formula above but in Spectacle Optics the value comes out consistently about 3/4 of the value I come up with.  What am I missing, is their a paper that I could source that would have an explanation and an example?  I need to cite something so it is important to have a source not just the correct answer, can anyone help?

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## HarryChiling

Anybody?

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## Darryl Meister

I use a slightly different formula. Taking the inverse tangent is not necessary, for instance. I also incorporate a "fudge" factor to allow for variability in edging configuration. These formulas are only accurate if the bevel if perfectly parallel to the frame eyewire. Although mini-bevels or "hide-a-bevels" come close to this relationship, the bevel of high-powered lenses will often be a little more centered on the thickest edge (temporal edge for minus and nasal edge for plus).

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## HarryChiling

> I use a slightly different formula. Taking the inverse tangent is not necessary, for instance. I also incorporate a "fudge" factor to allow for variability in edging configuration. These formulas are only accurate if the bevel if perfectly parallel to the frame eyewire. Although mini-bevels or "hide-a-bevels" come close to this relationship, the bevel of high-powered lenses will often be a little more centered on the thickest edge (temporal edge for minus and nasal edge for plus).


I see so you could even factor in a bevel arrangement to the formula. I don't get what you mean about not taking the inverse tangent do you mean:



Small angle approximation?

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## Darryl Meister

You really don't even need to think of it as a small angle approximation. If you consider the amount of decentration as the distance that occurs along the front surface of the lens, decentration essentially becomes a measurement of arc length. Consequently, the tilt of the lens in radians becomes equal to the ratio of the decentration arc length to the front surface radius, or tilt = d / r.

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## HarryChiling

> You really don't even need to think of it as a small angle approximation. If you consider the amount of decentration as the distance that occurs along the front surface of the lens, decentration essentially becomes a measurement of arc length. Consequently, the tilt of the lens in radians becomes equal to the ratio of the decentration arc length to the front surface radius, or tilt = d / r.


Just for arguements sake, we don't arc the PD stick when taking decentration measurements and when laying out we don't arc the decentration so wouldn't it be more acurate to use the tangent?  Even though at the amounts of decentration and radii that are feasable the approximation would be accurate to any degree measureable?

Dude thanks for all your help I don't know where I would be without you.

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## kcount

Just so I can understand, at what point does this equation become pertinent in clinic?

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## HarryChiling

Would I be more accurate in using:






> Just so I can understand, at what point does this equation become pertinent in clinic?


In the day to day operation it would be too tedious to perform many of these calculations to each and every patient walking into your shop, but many do find Darryl's Spectacle Optics application to be invaluable. Someone had to write the code for that program and since the computer is doing all of the day to day calculations as a programmer it is easier for me to be intricate in the programming stage to create a great program not just a good one. I am in the process of rewriting my compensation program so i would like to put as much effort into it's accuracy so that the next rewrite doesn't have to be performed for some time.

I am also considering includeing a decompensation function to allow someone to rework the formula and find out if the job was fabricated properly.  It requires work and I think Darryl is the undisputed expert when it comes to wrap compensation this is due to his attention to detail and if I hope to create a great program I need to tap the experts, that's the reason for my question and also the reason why i would think Darryl is one of the only people that would respond with a straight answer.  In reality I benefit very little from the question other than to have someone as knowledgable give me an answer.

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## Darryl Meister

> Just for arguements sake, we don't arc the PD stick when taking decentration measurements and when laying out we don't arc the decentration so wouldn't it be more acurate to use the tangent? Even though at the amounts of decentration and radii that are feasable the approximation would be accurate to any degree measureable?


But you _do_ "arc" the decentration; that's the whole point of these formulas. Otherwise, I'm not sure what you're attempting to calculate when you refer to converting lens tilt from decentration. I had originally assumed that you were calculating the same thing I do: the tilt of the optical axis as you decenter a meniscus lens, which occurs along an arc.

And, yes, there is a difference in the projected decentration along a flat plane tangent to the front curve and the arc length decentration along the front curve, but this difference is negligible; for a 6.00 Base with 5 mm of decentration, for instance, the difference is only 0.005 mm. Nevertheless, you could still calculate the correct arc length decentration from the projected decentration, although this adds extra steps for very little improvement in accuracy.

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## Darryl Meister

> Just so I can understand, at what point does this equation become pertinent in clinic?


Because the formula shows that decentration can affect lens tilt, and therefore optical performance, it's pertinent to any pair of eyeglasses you fit that requires decentration. The differences in optical performance are clinically insignificant, however, for most prescriptions. High plus lenses will suffer the most.

That said, there really isn't a clinical _application_ of this formula, although some software tools that compensate prescriptions for the position of wear may rely on this principle when calculating lens tilt.

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## HarryChiling

> But you _do_ "arc" the decentration; that's the whole point of these formulas. Otherwise, I'm not sure what you're attempting to calculate when you refer to converting lens tilt from decentration. I had originally assumed that you were calculating the same thing I do: the tilt of the optical axis as you decenter a meniscus lens, which occurs along an arc.
> 
> And, yes, there is a difference in the projected decentration along a flat plane tangent to the front curve and the arc length decentration along the front curve, but this difference is negligible; for a 6.00 Base with 5 mm of decentration, for instance, the difference is only 0.005 mm. Nevertheless, you could still calculate the correct arc length decentration from the projected decentration, although this adds extra steps for very little improvement in accuracy.


Thanks just checking to make sure I am being more accurate, it makes sense that the axis does arc, but often the layout doesn't arc the decentration so would it be better to have an allowance for that or am I being crazy and overthinking this?

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## HarryChiling

So with thickness as a factor I would take the thickness of the temporal edge and the thickness of the nasal edge and depending on my bevel lets say 2/3, I would multiply the difference by 2/3 then that would be used as the numerator in my tangent and the denominator would be my minimum blank size, is that on the right track?

Example:

Lens with 2mm nasal edge and a 5mm temporal edge and a minimum blank size of 60 would be:

2-5 = -3 
-3 * 2/3 = -2
-2/60 = tan(t)
t = -1.9 degrees

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## Darryl Meister

That's probably a good approximation of the tilt due to differences in edge thickness, which you would then add or subtract from the tilt due to decentration.

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## HarryChiling

> That's probably a good approximation of the tilt due to differences in edge thickness, which you would then add or subtract from the tilt due to decentration.


Wow, amazing how intricate you get without even trying.  Kudos bro.

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## Wes

You two make my brain hurt.  Darryl, nice as always.  Harry, just turn in the paper.

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## braheem24

> equal to the ratio of the decentration arc length to the front surface radius, or tilt = d / r.





> 





> Lens with 2mm nasal edge and a 5mm temporal edge and a minimum blank size of 60 would be:
> 
> 2-5 = -3 
> -3 * 2/3 = -2
> -2/60 = tan(t)
> t = -1.9 degrees


 
You guys must have some killer pickup lines!

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## Darryl Meister

> You guys must have some killer pickup lines!


Well, I'm still single, so apparently "Hey Baby, I like your cosines" really doesn't work that well.

Best regards,
Darryl

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## braheem24

Unfortunatly math capabilities are inversely proportional to charm where

Charm= -social skills/math abilities cubed.

Edit... Here's a graph where x=math abilities and y=charm

ok, i give up. can someone nerdier graph y=-xcubed?



That was exhausting, no wonder there's no hot chicks in calc3.

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## braheem24

> Well, I'm still single, so apparently "Hey Baby, I like your cosines" really doesn't work that well.
> 
> Best regards,
> Darryl


You try softly whispering soh cah toa?

It drives them crazy.

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## Darryl Meister

I'll have to give it a try!

Best regards,
Darryl

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