# Optical Forums > Ophthalmic Optics >  What is an Aspheric???

## F.Bourreau

Hello,

who could give me scientific explanations of what is an aspheric /atoric surface???
Is it like a progressive surface a deal between aberation and esthetic?
or is it exact science with mathematical equations? I would guess it is because precision lenses are using 'aspheric' technology but is the precision asphericity and the ophtalmic one are the same????:hammer: 

Fred

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## chip anderson

By definition it is any surface or lens that is *not spherical*.  This includes progressive bifocals,  lenses that that have progressively longer or shorter curvatures on one or more surfaces (aspherics).  Toric surfaces are lenses that have cylinder (two surfaces on a single surface) Atorics are any surface that is *not toric.*   This includes spheres, and toric lenses that have surfaces that become progressively longer or shorter.

This can be precisely controlled surfaces to yield better optics, flatter lenses, progressive bifocals, or other effects.  It can also include poorly ground surfaces that are not toric or spherical due to plain old aberration.

Chip

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## Don Lee

> Hello,
> 
> who could give me scientific explanations of what is an aspheric /atoric surface??? 
> Fred


Fred,
Not to discount anything Chip has already said. He is right on.

An aspheric lens has more than one curve on one surface, most likely the front. Having several curves helps to promote good visual acuity throughout the entire lens instead of only through the optical center. These lenses are normally used for aphakic patients. Aspheric design also flattens the lens which makes it more aesthetically pleasing and allows for a wider frame selection.

Aspheric lens design became popular in the 1960's. It had already been used in telescopes.

Best regards,
Don

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## Steve Machol

Here is an excellent article on Aspheric Lenses by Darryl Meister:

http://www.optiboard.com/files/Asphe...plications.pdf

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## Steve Machol

And here's an article on Atoric leses that I wrote for VisionCare Product News:

http://www.visioncareproducts.com/19/lens_atoric.html

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## F.Bourreau

Thanks all of you for your answer but i have to be more precise in my question

What are the parameters to calculated the design of an Aspheric Single Vision lense?

is there a mathematic formula for creating a design for an aspheric SV lense? 

Each brend has aspheric Semi-finished blanks, are they equivalents? if not what are the differents?

Is it a complex compromise to design a SV Aspheric or simply a mathematic rule?

The precision optic hase also its "aspheric" lenses to eliminate aberations. are they equivalent design as ophtalmic?

Why suppliers now speak about Atoric SV lenses? (not the same asphericity horizontaly and verticaly)

do you know any good books where i cand find explanation of how i can calculate my own aspheric design?

Fred

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## Shutterbug

Fred - the curve can probably be calculated on an indivdual basis, but I doubt it will be helpful information for you.  It is unlikely that you can get a prescribed custom aspheric design, and if you can it will likely be pretty pricey.  What are you planning to use it for?  Eyes or instrumentation?

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## F.Bourreau

I can guarenty that it would be an helpfull information because i want to produce aspherical lenses from spherical blanks wich are realy cheaper.... if it is "just" a formula I just have to put in in a FreeForm Generator and the work is done...

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## Darryl Meister

Here is an article I wrote on the subject many years ago:

Aspherics Lenses: Optics and Applications

You cannot produce aspheric lenses using conventional toric surfacers (cylinder machines), since they are designed for fining and polishing lenses with _circular_ cross-sections and aspheric surfaces use non-circular cross-sections.

Aspheric lenses also require "optimization" of some form in order to calculate the surface, which generally requires a ray tracing program and a basic knowledge of polynomial and/or conicoid optical surfaces.

Best regards,
Darryl

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## Jacqui

It could be done using a freeform generator and a polisher with soft laps. It would take a lot of calculating and fancy footwork, but it is possible. "Why do you want to do it??" is the next question. There is an abundance of aspheric lenses available on the market.

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## Darryl Meister

But investing in a free-form process will run up to a million dollars or more. You would have to sell a _lot_ of aspheric lenses to recover such a cost.

Best regards,
Darryl

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## Jacqui

Lotsa lenses. Even with a small DAC freeform system ($300,000) it would take a very long time to recover the cost. My question is still "Why would you want to do it??" There are many good lenses on the market, check the prices and check the cost of doing your own.

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## F.Bourreau

Unfortunately we can discuss a long time on the subject 'how long is the pay-back of generating aspherical lenses' but this is not my question... i just want to understand what is a SV aspherical topography (equations and so on)

.....

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## Darryl Meister

Generally, an aspheric surface is a rotationally-symmetrical surface of the form:

*z = ax^2 + bx^4 + cx^6 + dx^8...*

where z is the height of the surface, x is the distance from the center (or pole) of the surface, the coefficient a is related to the central curvature of the lens surface, and the remaining coefficients (b, c, d, etc.) are determined through optical optimization using lens design and/or ray tracing software.

Best regards,
Darryl

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## F.Bourreau

thank you darryl

i baught a good publication on the subject (which give some samples of coefficient) and i am now ready to make some tests (i already have the free form technology)
but i still intrested to get documentation on "how to calculate aspheric coefficients" if someone know good books....

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## Mr. Smith

Curved surfaces follow the geometric formula for conic sections:

Y(squared) = 2(R)(X) + (P)X(squared)

The shape of the conic section depends on the value of P, where R is the radius of curvature of the surface at the vertex. 
Values of P
 a) P > 1 you obtain an oblate ellipse
b) P = 1 you obtain a circle
c) 0<P<1 you obtain an prolate ellispe
d) P = 0 you obtain a parabola
e) P < 0 you obtain a hyperbola
If you take this formula and rotate through the Z axis you will obtain a surface that falls away from the apex at a constant rate. The easiest
surface to generate was P = 1 a circle, which when rotated through
Z gives a simple spherical surface the simplest to manufacture. No
all aspheric surfaces are not the same - It depends on what value of P
each manufacturer picks.

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## HarryChiling

Any good Calculus book will give you what you need to know. I will show you an example from mine (attached). If you turn your front base curve into radius insted of diopter it will give you the numbers that you would need in order to plug into the equations. The directrix would be the equvalent of the sagitta of your lens times 2. The F or focus would be the table your lens was sitting on. Your eccentricity would be:

e=|PF|/|Pl|

|PF|=r and |Pl|=d-r cos Theta

so e=r/d-r cos Theta or r=e(d-r cos Theta)

if we square both sides of the equation we would get

r^2=e^2(d-r cos Theta)^2

now we will turn this from polar coordiates to cartesian coordinates

r^2 becomes x^2 + y^2 (it's a triangle a^2 + b^2 = c^2 or in our case x^2 + y^2 = r^2)

the other side of the equation becomes 

e^2(d - x)^2 (the whole triangle thing again)

put it together and you get

x^2 + y^2 = e^2(d - x)^2

you could keep going but this equation in most cases will give me the answers I need

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## QDO1

A better question would be why use an aspheric design lens. or why does the designer chose one curve over another.  The original question is interesting - as we know we are trying to do 2 things with aspheric lenses

reduce off axis distortions there are many of thosereduce lens thicknessor bothHere is food for thought... if the front surface of a minus lens is flatter, and aspheric.. why is the whole lens thinner?

would a minus lens optimised for off axis performance would be thicker?

I would say to the poster.. there are a number of reasons for using an aspheric design lens, and lens thickness is one of them.. but any old curve wont do... thats why the big boys have huge R+D bugets, and many scientists that really understand the subject well. If you are starting out and not knowing the equations as a matter of course, then you probrably need to hire a scientist who does.  The starting point for any lens design is to solve a specific problem..  what problem are you trying to solve?

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## HarryChiling

An equation for aspheric lenses in a cartesian coordinate system is (ellipse):

[(x-h)^2/a^2]+[(y-k)^2/b^2)]=1

x = x coordinate (user defined)
y = y coordinate (the point you would be looking for)
h = x coordinate at the center of the ellipse
k = y coordinate at the center of the ellipse
a = 1/2 the diameter of the major axis in the ellipse (oblong axis)
b = 1/2 the diameter of the minor axis in the ellipse (shortest axis)

To simplify our equation we will place our elliptical lenses center at the center of our cartesian graph, this will make {h,k} = {0,0} reducing our equation down to:

(x^2/a^2)+(y^2/b^2)=1

we already know the b measure wich is equal to the radius of your lens so for instance if you have a 4 base lens in a 1.50 material  with the eccentricity equal to .5 we would get the radius of this lens:

D=(n-1)/r
r=(n-1)/D
r=(1.5-1)/4
r=0.5/4=0.125 or 125mm

now we have the b measure we can compute the a measure with the eccentricity formula:

e= squareroot((a^2-b^2)/a)

if we square both sides we get

e^2=(a^2-b^2)/a

now we rework the equation to give us:

a^2=b^2/(1-e^2)

a^2=(125)^2/(1-(0.5)^2)
a^2=15625/(1-0.25)
a^2=15625/0.75
a^2=20833.33

our b^2 for the equation is:

b^2=(125)^2
b^2=15625

now lets say we would like to know what the base curve is 25mm from the center of the lens then we would set the x in our equation to the 25 and work it to find the y

[(25)^2/(20833.33)]+[y^2/(15625)]=1
[(625)/(20833.33)]+[y^2/(15625)]=1
625+[(20833.33)*y^2/(15625)]=20833.33
(20833.33)*y^2/(15625)=20833.33-625
y^2/(15625)=20208.33/20833.33
y^2/(15625)=0.97
y^2=0.97*15625
y^2=15156.25

now that we have the the x^2=625 and the y^2=15156.25 we can use the Pythagorean Theorem to solve for the radius at this point

r^2=x^2+y^2
r^2=625+15156.25
r^2=15781.25
r=125.62

now if we were to convert this to the equivalent dioptric power we would have our base

D=(n-1)/r
D=(1.5-1)/.12562
D=3.98 Diopters

This is a very simplified approach to aspherics and does not neccesarily apply to lenses most lenses are not true elliptical surfaces but have a uniform base curve in the middle of the lens and only use aspherics as after a certain distant from the center of the lens.  These equations were derived from keplars law of planetary motion if you would like to do more research on this.  If any part of my equatio is wrong please post it.

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## Darryl Meister

Keep in mind that the major (a) and minor (b) axes of the ellipse are not actually equal to the radius of curvature of the ellipse at the corresponding vertex. Nor could you use the Pythagorean theorem to determine the radius at any point using the X and Y coordinates (you actually use calculus). Also, the eccentricity (e) would have to be determined either analytically or numerically for a given front curve (radius) and lens power.

For optical lenses, a more useful description of a conic surface is:

*Sag = (x^2 * R) / [1 + SQRT(1 - p * x^2 * R^2)]*

where (x) is the horizontal distance from the center of the surface, (R) is the vertex curvature of the surface (or 1 / Radius), and (p) is a value that describes the eccentricity of the conic surface.

You would first determine the nominal base curve of the lens for a given prescription power and then compute the vertex curvature. From there, you would generally ray trace the lens while adjusting the p value until you eliminated or reduced lens aberrations of interest (e.g., oblique astigmatism).

Many modern aspheric lens designs use polynomial equations of the form:

*Sag = a * x^2 + b * x^4 + c * x^6...*

where (x) is again the horizontal distance from the center of the surface and (a), (b), (c)... are coefficients that must be determined. This formula can also be thought of as the expansion of a sphere, which is then deformed by manipulating the coefficients.

In this case, you would first determine (a), which is equal to one-half the vertex curvature (R). From there, you would ray trace the lens while adjusting each proceeding coefficient (i.e., b, c...) to reduce or eliminate lens aberrations. You could also play around with higher order terms to reduce thickness and such, as well.

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## fjpod

I could be wrong, but I am of the opinion that aspheric curves are chosen by manufacturers to improve the profile of a lens by making it flatter and thinner.  This in no way necessarily improves visual acuity, chromatic abberation, and the like.  

It does improve peripheral visual field through the lens in a high prescription, thereby reducing apparent image jump. 

Peripheral visual acuity is 20/200, and manufacturers capitalize on this with aspheric thinning techniques.  So, actual VA through slightly peripheral areas of an aspheric lens is reduced...and lots of patients have trouble with this if they are trying to see fine detail when looking through an off axis point.  Their spatial awareness may be better, but their VA is down.

IMHO

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## Darryl Meister

> Peripheral visual acuity is 20/200, and manufacturers capitalize on this with aspheric thinning techniques.


I think you might be confusing vision in the periphery of the retina with vision through the periphery of a spectacle lens (which is the principal concern of lens design). Much of the reduction in thickness afforded by non-cataract aspheric lenses comes from flattening the base curve. However, flattening the base curve introduces peripheral aberrations (chiefly, oblique astigmatism and power error). Aspheric surfaces allow the lens designer to flatten the base curve while maintaining good peripheral optics.

You can learn more about lens design at Ophthalmic Lens Design.

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## HarryChiling

My equation was flawed but this one is correct to determine the curve at a given point on the lens if you knew the eccentricity

[(x-h)^2/a^2]+(y^2/b^2)=1

where

h = -(e^2*d)/(1-e^2)

a^2 = (e^2*d^2)/[(1-e^2)]^2

b^2 = (e^2*d^2)/(1-e^2)

if we were to take the same example given above and plug in the numbers 4 base lens with index 1.5 and eccentricity value of .5  and we wanted to find the radius of the lens 25 mm from the center of the lens.

first we need to convert the base to its radius

125mm from example given above

the we need to determine the directrix

d=r+(r/e)
d=125+(125/0.5)
d=125+250
d=375

now we can figure out the rest of our variables

h= -(0.64)(375)/(1-0.64)
h= -240/0.36
h= -666.66

a^2= (0.64)(140625)/(1-0.64)^2
a^2= 90000/0.1296
a^2= 694444.44

b^2= (.64)(140625)/(1-0.64)
b^2= 90000/0.36
b^2= 250000

with the variables solved then all that needs to be done is solve for the y in the equation

[(25+666.66)^2/694444.44]+(y^2/250000)=1
[(691.66)^2/694444.44]+(y^2/250000)=1
(478393.55/694444.44)+(y^2/250000)=1
(0.6889)+(y^2/250000)=1
y^2/250000=1-0.6889
y^2/250000=0.3111
y^2=0.3111*250000
y^2=77775

Now we can use the pythagorean theorem to solve for the radius of a point on the lens 25 mm from the center

r^2=y^2+x^2
r^2=77775+625
r^2=78400
r=280

Now if we were to convert back to dioptric value we would get 

D=(n-1)/r
D=(1.5-1)/.280
D=1.79 Diopters

This is the equation that is correct, I found the previous equation online and had a feeling it was not correct.  Darryl I have seen the equation in the form you have given as well but I cannot figure out how to get all the coefficients that you have explained about.  The only time I use the equation given above is if I wanted to find the base at the ED for thickness calculation purposes.  I would love to see a problem worked out with the equation you used.




> *Sag = (x^2 * R) / [1 + SQRT(1 - p * x^2 * R^2)]*

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## fjpod

No, I'm not confusing the two, but I do not think that peripheral asphericity improves visual acuity if the patient happens to move their line of sight and looks through a point off the optical center.  Their spatial awareness might be better than in a non-aspheric, but their VA is reduced.  If the wearer keeps his visual axis through the optical center of the lens, where there is no asphericity, he has good VA.  When a wearer keeps his visual axis through the optical center, his peripheral vision which is coming through off axis points through the lens, strikes parts of the peripheral retina which are incapable of 20/20 VA anyway.  This is what aspheric lens makers capitalize on.  Good optics in the center because wearers need 20/20 resolution, but you can compromise optics peripherally to get a thinner lens because the peripheral retina/brain is incapable of resolving 20/20.

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## Darryl Meister

> When a wearer keeps his visual axis through the optical center, his peripheral vision which is coming through off axis points through the lens, strikes parts of the peripheral retina which are incapable of 20/20 VA anyway.


Yes, but we're talking about what happens when a wearer moves his or her lines of sight _away_ from the optical center. If the wearer constantly looked through the optical center, we wouldn't have to worry about best form or aspheric design at all (spherical aberration generally isn't a big concern for most spectacle lenses).




> Darryl I have seen the equation in the form you have given as well but I cannot figure out how to get all the coefficients that you have explained about... I would love to see a problem worked out with the equation you used.


You would generally determine them numerically using a ray-tracing program of some sort. If you download my old OpticsLite spreadsheet from the file archives, it has a page that allows you to determine the p value while minimizing the aberrations of your choice. You do need MS Excel to run it though (I just haven't had the time to write a stand-alone VB program).

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## HarryChiling

Darryl,

If I understand it correctly your program is in a sense a lens design tool.  This is a great program for designing lenses.  My interest was peaked a while ago when someone posted a question involving an aspheric lens and the true power.  Someone posed a question about the aspheric surface and off my mind went.  I used Keplers Law of Planetary motion to come up with my answers.  I have only so far looked at the ellipse, I don't really see a need for the hyperbola or the parabola, its the only one that I would think would be used for the design of a lens.  I have an do enjoy your opticslite program it's genius.  I have a hard time looking at the answer if I have not had any research or understanding into how it was derived.  YOU CAN CALL ME HARD HEADED.  It's just the way I was raised.  My equation is usefull in determining properties of a lens when you have the parameters and you just need to know the power of the curve at a given point.  I would take and figure the base at the ED and use this to calculate my thickness.  It is not completely accurate due to the fact that most lenses are not aspheric all the way through, but it works.  I have some experience with visual basic if you need to farm out some of the code work let me know I would be more than willing to give you a hand.

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## Darryl Meister

> If I understand it correctly your program is in a sense a lens design tool


More or less.




> I used Keplers Law of Planetary motion to come up with my answers.


Kepler's laws do make much use of conic calculus, since orbits are generally elliptical with the sun at one focus, but that does seem like a roundabout way to figure this out. ;)




> I have only so far looked at the ellipse, I don't really see a need for the hyperbola or the parabola, its the only one that I would think would be used for the design of a lens.


As it turns out, an aspheric lens can use a variety of conic sections, including hyperbolic, parabolic, oblate elliptical, and prolate elliptical curves, depending upon how flat the lens is, whether the aspheric surface is on the front or back, and so on. That's the advantage to using the p value, which manipulates the curve through the entire conic family (p = 1 is a circle).




> I have some experience with visual basic if you need to farm out some of the code work let me know I would be more than willing to give you a hand.


At some point, I'll probably convert the entire spreadsheet into a Visual Basic program. But I'll have to burn through many a vacation day to do it.

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## HarryChiling

Yes I made mistakes in my calculations above again I used the eccentricity as .8 instead of .5 but you get the drift.  I am working on a calculator for this equation as well.  Post it soon

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## QDO1

> I can guarenty that it would be an helpfull information because i want to produce aspherical lenses from spherical blanks wich are realy cheaper.... if it is "just" a formula I just have to put in in a FreeForm Generator and the work is done...


why not employ the sevices of a optical scientist when you buy your free-form machine, that (in proportion) will be small fish.  If you dont know anything about Aspheric lenses, why are you trying to make them.  All the coeficents, in the equations are there for a purpose, get one wrong, and yes you will make an aspheric lens, but unfortunatally they will be of no use to a patient

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## HarryChiling

I have the calculator online at 

www.harrychiling.com/equations/index.html

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## Payman

> Here is an article I wrote on the subject many years ago:
> 
> Aspherics Lenses: Optics and Applications
> 
> You cannot produce aspheric lenses using conventional toric surfacers (cylinder machines), since they are designed for fining and polishing lenses with _circular_ cross-sections and aspheric surfaces use non-circular cross-sections.
> 
> Aspheric lenses also require "optimization" of some form in order to calculate the surface, which generally requires a ray tracing program and a basic knowledge of polynomial and/or conicoid optical surfaces.
> 
> Best regards,
> Darryl


Dear Darry,
Would you please send me a copy of the article you introduced in your message by email? it seems the link does not exist or has error to download.
my email address is paymanrajai@gmail.com

Regards
Payman

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## lensgrinder

Here is the link

It is found in the Optiboard file directory.

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## EyeFitWell

> No, I'm not confusing the two, but I do not think that peripheral asphericity improves visual acuity if the patient happens to move their line of sight and looks through a point off the optical center.


I thought that was exactly the point of aspheric lenses.
I found this site a while back looking for a picture of aspheric vs. nonaspheric lenses (B/C our VSP pts want to know why there is a specific charge on their reciept for aspheric when they upgrade from CR-39).  According to this the benefits are:
1. thinner, lighter.
2. reduced image minification/magnification visible when looking at the eye of the wearer 
3. *better vision on the periphery of the lenses*.  (in other words, away from the OC)
4. lenses fit better in the frame when working with strong plus
http://www.allaboutvision.com/lenses...ric-lenses.htm

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## Darryl Meister

> I thought that was exactly the point of aspheric lenses.


The point of an aspheric lens for low to moderate powers is to allow the use of flatter, non-standard Base curves to improve cosmetics and weight. Aspheric lenses cannot really "improve" optical performance compared to a standard or "best form" lens using spherical curves. They just allow similar optical performance to be obtained with thinner, flatter lenses.

However, for cataract lens designs above +8.00 D, which represents the mathematical limit of the prescription range of "best form" lenses, aspheric lenses _can_ improve optical performance compared to spherical lenses.

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## Payman

Dear Fred,
Your question does not have an straight answer. It was a part of my M.Sc.Thesis 5 years ago. I can introduce you some crucial references as follow:
1- Bennett 1988, Aspheric and continuous curve contact lenses, optometry Today; 2 jan. pp 11-14, 27 Feb. pp 140-142; 23 April pp 238-242; 30 July pp 433-444; 5 Nov.pp 630-632
2- Bennett 1968, Aspherical Contact Lens Surfaces, The ophthalmic optician, part one ?, part two:30 Nov. pp 1297-1311; part three: 8:1037-1040 or 1969:9:222-224 & 229-230
3- Baker, Ray tracing through non-sphercial surfaces. Proc.Phys.Soc (Lond) or Proc.R.Soc. 1943:55:361-4
4- Hopkins, the wave theory of abberations, Clarendon Oxford 1950.

Payman

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## LxGoodies

Hello @Darryl Meister,

Thanks for your your sagitta formula for R-P. For a programmer (not an optometrist) like me, some terminology is difficult to grasp intuitively. Maybe I can add something usefull to this thread.. your remark..




> Nor could you use the Pythagorean theorem to determine the radius at any point using the X and Y coordinates (you actually use calculus).


Indeed, I use the following Pascal procedure for that.. It takes three subsequent x,y points from the result polynome, it yields the midpoint (xc,yc), maybe forum members can use this too.



```

function SOLVE_2r(a,b,c,d,e,f: real; var x,y: real) : boolean;var t: real;begin t:= (b*d - e*a); if ((t>-1e-14) and (t<1e-14)) then begin SOLVE_2r:=false; exit; end; x:= (b*f - e*c)/t; y:= (a*f - d*c)/(-t); SOLVE_2r:=true;end; function find_arc_center(x1,y1,x2,y2,x3,y3: real; var xc,yc: double): boolean;//  Given three points (x1,y1) (x2,y2) (x3,y3) determine arc center (xc,yc)var a,b,c,d,e,f,x,y: real;begin find_arc_center:=false; a := x3 - x1; b := y3 - y1; c := (x3*x3 + y3*y3 - x1*x1 - y1*y1)/2.0; d := x3 - x2; e := y3 - y2; f := (x3*x3 + y3*y3 - x2*x2 - y2*y2)/2.0; if SOLVE_2r(a,b,c,d,e,f,x,y) then begin   xc:=x; yc:=y;   find_arc_center:=true end;end; 


```

Kind regards,

:) 
Lex

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## Darryl Meister

Looks interesting. I might have to experiment with it a bit...

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## LxGoodies

> Looks interesting. I might have to experiment with it a bit...


It is always needed for visualisation things, in 3d it can also be done. By the way, above version is speed optimized, to support graphics. Double precision calculation (and slower) must be done by changing the code to



```

function SOLVE_2r(a,b,c,d,e,f: extended; var x,y: extended) : boolean;var t: extended;begin t:= (b*d - e*a); if ((t>-1e-14) and (t<1e-14)) then begin SOLVE_2r:=false; exit; end; x:= (b*f - e*c)/t; y:= (a*f - d*c)/(-t); SOLVE_2r:=true;end; function find_arc_center(x1,y1,x2,y2,x3,y3: double; var xc,yc: double): boolean;//  Given three points (x1,y1) (x2,y2) (x3,y3) determine arc center (xc,yc)var a,b,c,d,e,f,x,y: extended;begin find_arc_center:=false; a := x3 - x1; b := y3 - y1; c := (x3*x3 + y3*y3 - x1*x1 - y1*y1)/2.0; d := x3 - x2; e := y3 - y2; f := (x3*x3 + y3*y3 - x2*x2 - y2*y2)/2.0; if SOLVE_2r(a,b,c,d,e,f,x,y) then begin   xc:=x; yc:=y;   find_arc_center:=true end;end; 


```


Lx

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## Darryl Meister

> Double precision calculation (and slower) must be done by changing the code to


I'll have to convert it all into either Visual Basic or C++ before I can do anything with it anyway. ;)

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