# Optical Forums > Ophthalmic Optics >  Calculating minimum lens blank size

## Just Me

Just for curiosity's sake, how would one go about calculating Min. Lens Blank size of various lens type very accurately?


Thanks,
Just me

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## Diane

Just guessing that you need to calculate minimun blank size for single vision lenses, first, let me address that.

Determine the frame PD, which is (usually the addition of the eye size and the bridge size, if that is the measurement at the 180 cutting line.  Take the patient's PD that will be used for the eyewear, and subtract that amount from the Frame PD.  Say FPD is 70 and patient's PD is 60.  The difference is 10.  Measure the effective diameter of the eyewire.  (Should use the boxing system to get this measurement, it could be the longest horizontal measurement, the longest vertical measurement or the longest diagonal measurement.)  Say you get 56 mm.  Add the 10 to the 56 and include 1 millimeter for edging and you come up with a minimum of 67 mm minimum blank size. 

This is the really simple way of determining minimum blank size.   :)

Diane

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## Jeff Trail

Diane hit the SV solution on the button.. BUT when it comes to multi-focals it's a different kettle of fish (four eyed that is)
  The main problem is that even if you find out the effective cut out in diameter the location of the seg will throw off the math.. you can't hardly figure it unless you know the exact loyout of the seg location on the actual blank, which changes from design to design as well as manufacturer.
   Since the majority of labs stock RD22 in 74 mm blanks thats not a worry..and the over all majority of PAL's are 80 mm blanks also it's not a worry (effective cut out of 78 mm)
   I usually tell the opticians that use my lab that any frame with a ED over 63 charge for over size blanks.. then they do not have to worry...
    About the only time you REALLY need to know something about effective cut out diameter is if you are dealing with a aspherical lens.. in an aspherical lens you can not decenter the optical center via prism rings it HAS to be cut on center.. also since the optical center SHOULD be fit (even in SV's) in aspherical lens then the other formula is hard to figure (dbl+a)(pd-(a)+ED)
    But that formula still is a quickie for gtting a general idea on how to get an answer.. for asphericals it's pretty handy to keep the cutout charts that you can get from most of the lens makers and you can just dot the demo in the frame pop it over the chart and get an answer in two seconds  :)
    Those charts are real handy checking PAL's as well to see if you can get that low fitting with a certian design..

Jeff

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## shanbaum

> Originally posted by Just Me:
> _Just for curiosity's sake, how would one go about calculating Min. Lens Blank size of various lens type very accurately?
> _


Given that you used the term "very accurately", I must comment that the methods described so far (using the ED) are grossly approximate.

To get an accurate minimum blank size, one has to know a number of radii in each quadrant of the frame shape to be used, and the distance and direction that the blank to be used will have to be moved to achieve the correct layout geometry (Layout or Prism Reference Point location).

The starting point of each radius is then moved according to the movement of the blank and the distance to the end point recalculated.

The longest of these is the minimum blank size required, to which some amount should be added to allow for errors which can and will occur in all of the measurements involved.

This is a pretty simple process for some lenses, such as finished (or aspheric) single vision.  The most complex are semi-finished single vision blanks (on which the PRP can be located anywhere) and round seg multifocals, which can be rotated, locating the seg in innumerable positions, each with a different resulting blank movement.  And in the case of multifocals available in various configurations from different manufacturers (like flat-tops), the blank required depends on the blank selected.

If you visualize this process correctly, you should be able to see how the minimum blank requirement might be less than the ED - an assertion which when made has frequently drawn howls of objection from opticians.

Still curious?

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## HenryB

Yes it is possible for the minimum blank size to be smaller than the ED. I go round and round with the opticians here about uncuts and the so called ED that they supply to the lab. They complain that the blank size is too small when in fact they just plopped a ruler down on the frame and measured the widest point which in almost all cases will give you a fairly short number for the ED which is only exaggerated more the further away you get from a round type shape.

Correct me if I am wrong but the ED or "effective diamter" is a measurement taken from the geometric center of the frame out to the farthest point and multiplied x2. This cannot be measured by hand unless a frame is pretty much perfectly round or real close.

Lets get back to the original subject.....the way you would have a smaller min. blank than the ED is this-

Let's say the measurement from the geo center out to the farthest point is 30mm....that would give you a 60 ED. But, lets also say this is some funky shape. The measurement from geo center to the opposite side of longest radii (the one used to calculate the ED) is only 20mm......if you add those two measurements together you will get the minimum blank needed as 50mm.....that of course does not assume some "fudge factor" which ya gotta throw in there. Now this also assumes that you are using a semi-finished SV blank and are grinding the OC accordingly to make it cut at that diameter. 

I believe that is a rough example of how this might work.........

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## shanbaum

> Originally posted by HenryB:
> _
> Correct me if I am wrong but the ED or "effective diamter" is a measurement taken from the geometric center of the frame out to the farthest point and multiplied x2. This cannot be measured by hand unless a frame is pretty much perfectly round or real close.
> 
> Let's say the measurement from the geo center out to the farthest point is 30mm....that would give you a 60 ED. But, lets also say this is some funky shape. The measurement from geo center to the opposite side of longest radii (the one used to calculate the ED) is only 20mm......if you add those two measurements together you will get the minimum blank needed as 50mm.....that of course does not assume some "fudge factor" which ya gotta throw in there. Now this also assumes that you are using a semi-finished SV blank and are grinding the OC accordingly to make it cut at that diameter. 
> _


Yup, the ED is twice the longest radius from the geometric center of the shape to the edge, and it can be _real_ hard to measure - probably not possible "by hand" without a special tool, like a box-o-graph.

Your description of how the minimum blank can be smaller than the ED is very close - until the very end.  It can happen with _any_ type of lens.  It's really just a matter of the blank movement in the frame, which can be in any direction, depending on the configuration of the blank and the layout geometry.  For this to happen does generally require that there be longer radii in the direction in which the blank is moving than those in the opposite direction - but not necessarily just the one opposite the radius that constitutes the ED.

The simplest case to try to visualize is the old Porsche Design-shape sunglass that had a much longer radius in the upper nasal corner than in the rest of the shape.  As you increase blank movement in and/or up towards that corner, the minimum blank size decreases, up to a fairly significant amount of decentration.

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## HenryB

Yea I see your point about it being possible with any lens. I was aware of that however, it is alot less likely that it will happen with, let's say a multifocal. There are obviously alot more factors involved including seg inset, drop, seg height, dec., etc......

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## Just Me

Ok... thanks so far, but keep going. :)  I'm curious about the fancy math that would be involved here.

Has anyone written any short program that could calculate min. blank size?

There's no special reason why I'm looking for the answers to this subject.  It's just that it's easy to find accurate powers of lenses with the right (as opposed to the left  :)) equations, but one never sees anything but general equations for finding MBS and I'm curious why.


Thanks again.  This is getting interesting.

Just Me

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## Steve Machol

I think I have a small program written in Basic that calculates minimum blank size.  Unfortunately it'll have to wait until I'm back in the office on Monday.  I'll look, and if I have it I'll post it here for you. 

------------------
Steve
OptiBoard Administrator

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## Darryl Meister

It's also important to distinguish between the minimum blank size for cut-out and the minimum blank size for thickness. Determining minimum blank size for cut-out involves deciding whether or not a desired lens blank will work for a given job (including prescription, frame, and measurements). Progressive lens cut-out and centration charts often have handy guides to quickly determine this kind of cut-out (or minimum blank size). You can also create your own for single vision and flat-top lens blanks -- if you know the manufacturer's segment inset and drop. Further, once you have determined the diameter of a finished stock lens, this is also the diameter to use for calculating thickness.

Determining minimum blank size for thickness involves computing the smallest possible diameter for the finished lens that will work for a given job. (This only applies to semi-finished lenses.) This is important for plus lenses, since the center thickness of the lens will increase as the lens diameter increases. Consequently, it behooves the lab to use the smallest diameter possible to minimize lens thickness. (In reality, the actual diameter of the finished lens may not change, but the center thickness while surfacing is reduced so that the minimum edge thickness of the lens is maintained once edged for the frame.) As Robert's posts have alluded to, accurately calculating this minimum blank size is a very complex process -- particularly for plus lenses. Another complication occurs when using plus lenses with cylinder power. These finished lenses, when produced with a constant minimum edge thickness around the entire blank, are elliptical in shape -- not round. This can sometimes allow for a smaller minimum blank size, if the ellipse happens to be optimally oriented (which is controlled by the axis of the Rx) with the dimensions of the frame. In general, a computer is the most accurate way to determine this blank size.

Best regards,
Darryl

[This message has been edited by Darryl Meister (edited 06-24-2000).]

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## shanbaum

> Originally posted by Darryl Meister:
> _
> Determining minimum blank size for thickness involves computing the smallest possible diameter for the finished lens that will work for a given job._


I don't think the notion of a "minimum blank diameter for thickness" is terribly useful in surfacing.

As you mention, in the case of lenses with cylinder, surfaced lenses may not be round.  And when they're not, how do you make sense of the notion of "diameter"?

Consider an extreme example, for instance, +3.00 -6.00.  When fabricting such an rx from a blank larger than required for cutout, the "diameter" of the lens across the +3.00 meridian may be much shorter than across the -3.00; across the latter, the "diameter" for thickness purposes is, I guess, infinite.  For lenses such as these, the shape is not even so regular as an ellipse; they may be hourglass-shaped.

I run into the occasional lab rat who is stuck on the notion that there's a direct relation between minimum blank size (for cutout) and thickness - and as you know, it ain't necessarily so.

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## Jeff Trail

> Originally posted by shanbaum:
> _ I don't think the notion of a "minimum blank diameter for thickness" is terribly useful in surfacing.
> 
> As you mention, in the case of lenses with cylinder, surfaced lenses may not be round.  And when they're not, how do you make sense of the notion of "diameter"?
> 
> I run into the occasional lab rat who is stuck on the notion that there's a direct relation between minimum blank size (for cutout) and thickness - and as you know, it ain't necessarily so.
> 
> _


  Robert

  His is another of those "lab rats" who tend to sway towards the ones you run over, oops I mean into  :) ...I can name a few examples where lens thickness is related to cutout...

  1) Prism- you have 7^ BO or worse a high compound prism .. do you cut it on a "stock" lens? nope.. especially if it's a high plus power.. you would run out of lens "blank" before you got the total power ground in UNLESS you used an xtra thick blank..

  2) Execs- you tend to run out of lens for high pluses in that lower outside corner  :)

  3) Aspherical lens- where you need more thickness to give you a larger diameter of finished lens , since you can not, well SHOULD not grind it using prism rings for decentration.

  I can name more if you want? The thing is , in my opinion, optics is a little more complicted then to say it black or white with no shades inbetween .... I try to take it a job at a time   :Rolleyes: 

Jeff " Lab rat and PROUD of it" T.

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## Darryl Meister

Hi Robert,

All very good points, as usual. I agree with you in theory, to some extent, but...

I think we can both agree that the goal of determing the center thickness of a surfaced (plus) lens, more often than not, is to compute the thinnest possible center thickness that will allow the job to cut-out for the frame with a minimum edge thickness. 

Further, if any of its meridians contain plus power the center thickness of the lens is very likely going to be controlled by the "diameter" [sic] (or length) of the lens through that meridian. My point is this: If a spherocylindrical lens is surfaced to a specified center thickness, the shape of that lens will either be an ellipse or a circle at the points along the perimeter of the blank that meet the minimum edge thickness requirement. Of course, the minimum (center) thickness of minus lenses is not controlled by blank size. Now, while a frame and lens combination may require a finished lens of some arbitrary and unusual shape, that finished lens is generally going to start out as a circular or elliptical uncut lens blank (in the absence of cribbing). I'm ignoring progressives and other lenses with complex surfaces for the purposes of this discussion...

This is where our discussions diverged a bit, I think. Your contention is that these meridians may not necessarily have the same length (or radius) from the center of the lens. In which case, the lens would not have a single minimum blank size, and the lens would probably not even have a "diameter" as such -- since it doesn't have to be round. I agree completely with this.

I don't know that the notion of a "diameter" need only be applied to circles, however. Ellipses have major and minor axes, which one could think of as the major and minor "diameters" of the ellipse. These lines both pass through the center of the ellipse and intersect their two opposite curves -- two prerequisites for a "diameter." Using this argument, one could argue that a circle is a special case of an ellipse in which the major and minor diameters (or axes) are equal. Moreover, a sperocylindrical lens with plus power in both principal meridians will produce an ellipse if produced to the same minimum edge thickness around the perimeter.

To your point, the finished lens shape need not be circular or elliptical at all... It could be the exact shape of the frame, for that matter. In practice, though, wouldn't you say that your generator will start with a circular semi-finished lens blank and then spit out an elliptical or circular finished lens, in most cases (again, ignoring cribbing)?

Also, in your example of a spherocylindrical lens with a +3.00 DS -6.00 DC power, -3.00 D diameter wouldn't be infinite since the limiting thickness would occur through the +3.00 D meridian. The diameter of the -3.00 D meridian would be no longer than the initial semi-finished blank size of the lens, while the +3.00 D meridian could be shorter (if the lab chose to).

Although the relationship between minimum blank size and thickness may not be direct, there is certainly a relationship, nonetheless. For most jobs (frame and lens combinations), an 80-mm minimum blank size is going to be thicker somewhere than a 60-mm blank size.

For the most part, though, I think that we're just talking about differences in semantics. You're simply referring to a more general approach. And I do agree that "minimum blank size" might be a misnomer in this context or, at the very least, misleading. Although the concept of a minimum blank size or diameter might be an oversimplification to help lab personnel -- particularly before the advent of high-speed computing -- I believe that there is still a bit of method to that madness...

However, in all honesty, you deal with this kind of stuff more than I do -- and may very well have a better perspective on it!

Best regards,
Darryl

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## shanbaum

> Originally posted by Darryl Meister:
> _
> &lt;snip&gt;
> My point is this: If a spherocylindrical lens is surfaced to a specified center thickness, the shape of that lens will either be an ellipse or a circle at the points along the perimeter of the blank that meet the minimum edge thickness requirement. 
> &lt;snip&gt;
> To your point, the finished lens shape need not be circular or elliptical at all... It could be the exact shape of the frame, for that matter. In practice, though, wouldn't you say that your generator will start with a circular semi-finished lens blank and then spit out an elliptical or circular finished lens, in most cases (again, ignoring cribbing)?
> 
> Also, in your example of a spherocylindrical lens with a +3.00 DS -6.00 DC power, -3.00 D diameter wouldn't be infinite since the limiting thickness would occur through the +3.00 D meridian. The diameter of the -3.00 D meridian would be no longer than the initial semi-finished blank size of the lens, while the +3.00 D meridian could be shorter (if the lab chose to).
> _


Of course, I was thinking of the infinitely expandable semi-finished blank you told me you guys were going to start making... 

Maybe I'm visualizing this badly (memory has to be refreshed from time to time), but it seems to me that were the putative +3 -6 to have a center thickness such that the plus meridian has a smaller diameter than the original blank size, the shape of the lens (as surfaced) would not be elliptical, would it?  I seem to recall that these turn out sort-of hourglass-shaped.  I can see that for lenses of plus power in both major axes, yes, these would be circular or elliptical.

You're right, this topic is a semantic issue.  But inasmuch as you've helped clear up some other semantic issues that made opticianry harder than it has to be, I think you would agree that while semantics are infrequently illuminating, they often serve to obfuscate.

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## shanbaum

> Originally posted by Jeff Trail:
> _
> 
>   1) Prism- you have 7^ BO or worse a high compound prism .. do you cut it on a "stock" lens? nope.. especially if it's a high plus power.. you would run out of lens "blank" before you got the total power ground in UNLESS you used an xtra thick blank..
> 
> _


1) Has to do with minimum blank thickness, nothing to do with size (other than: larger blanks _may_ be thicker, and in the case of single vision, reduce the need for prism for decentration).

Now that you mention it, this is further evidence of the decoupling of MBS and thickness.  For a given Rx, changing the amount of decentration you grind on an SFSV blank changes the MBS, but has no effect whatsoever on the finished lens thickness.




> Originally posted by Jeff Trail:
> _
> 
>   2) Execs- you tend to run out of lens for high pluses in that lower outside corner   :)
> 
>   3) Aspherical lens- where you need more thickness to give you a larger diameter of finished lens , since you can not, well SHOULD not grind it using prism rings for decentration.
> _


2) Yes, you do; 

3) No, you shouldn't.

But neither comment speaks to the relationship between the blank size required for cutout, and thickness.

Imagine - exec, +2.00 -4.00 45, droopy frame shape (long lower temporal radius). Lens would need to be relatively thick (whatever the add) so as not to "run out of lens in that lower outside corner" as you wrote, right?  Now, change the Rx axis to 135.  Different, isn't it?  

But the minimum blank size - for cutout - is exactly the same in both cases.

And I didn't mean "Lab Rat" in a disparaging way.  Just folks who work in labs.  Been one myself.  Term of endearment.

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## Steve Machol

All of the points made above are valid.  Calculating minimum blank size (MBS) _accurately_ can be a complicated procedure.  Nonetheless, here's the simplified formula for calculating MBS:

=============================================

MBS = 2*ERad + 2*[SqrRoot(HDec^2 + VDec^2)] + EA

where,

MBS = Minimum blank size

ERad - Effective radius (the measurement from the geometric center of the frame to it's farthest point)

HDec = Horizontal decentration (the distance the lens will move horizontally for cut-out)

VDec = Vertical decentration (the distance the lens will move vertically for cut-out)

EA = Error adjustment (this is your 'fudge' factor to account for measurement and processing errors)

Note: All measurements are in millimeters.

=============================================

This formula should give you a pretty good approximation.  However, as pointed out by Robert and others, it will not be exact.  Furthermore, to accurately calculate the HDec and VDec on multifocal lenses, you'll also need to know: a) seg placement, b) optical center placment, c) seg heights, d) monocular PDs, etc.  

------------------
Steve
OptiBoard Administrator


[This message has been edited by Steve Machol (edited 06-27-2000).]

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## Darryl Meister

_ Of course, I was thinking of the infinitely expandable semi-finished blank you told me you guys were going to start making..._ 

Hey, don't start giving out our new product ideas!

_ ...it seems to me that were the putative +3 -6 to have a center thickness such that the plus meridian has a smaller diameter than the original blank size, the shape of the lens (as surfaced) would not be elliptical, would it?  I seem to recall that these turn out sort-of hourglass-shaped..._

Yes, in this case, the finished lens would look like an hourglass or bow-tie. I was referring to the plus-on-plus case. I haven't actually seen anyone grind the hourglass kind before (either they crib it first or keep some edge thickness at the diameter of the original blank).

_ You're right, this topic is a semantic issue.  But inasmuch as you've helped clear up some other semantic issues that made opticianry harder than it has to be, I think you would agree that while semantics are infrequently illuminating, they often serve to obfuscate._

Point well taken. And more accurate language would certainly be useful. But I can't take all the credit for that kind of stuff! Let us not forget who brought us the "LRP." I still think the R is actually supposed to stand for "Robert!"   ;)

Best regards,
Darryl

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## Matthew

I know its probably already answered but here is the formula.

ED + Decentration X 2

Effective diameter plus the decentration multiplied time 2

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## AM

im a student Optom

R: +2.50DS / +2.50DC x 180 with 2 prisms up and 3 prisms out

If the lens size is 48 mm what is the minimum blank size??

can someone please explain and show how this can be done because all the equations are confusing.

thanks

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## Judy Canty

You need frame information and patient's PD to calculate this.

MBS = ED + (2)(per-lens decentration) + 2 mm

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## AM

im a student optom,

Frame:           Eyesize                                  46 ú 40
Distance between Lenses 20


Mono 31
Mono 35

R: +2.50DS / +2.50DC x 180 with 2 prisms up and 3 prisms out

If the lens size is 48 mm what is the minimum blank size??

can someone please explain and show how this can be done because all the equations are confusing.

thanks

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## AM

frame has 66mm bcd

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## Judy Canty

What is the ED (Effective Diameter) of the frame?

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## AM

the above was all i have been given,

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## Judy Canty

Ok, I'm going to assume that 48 is the ED.

Decentration for the OD is 2 in.

48 + 4 + 2 = 54

Decentration for the OS is 2 out.

48 + 4 + 2 = 54

MBS = 54

Somebody check me here.  It's been a while.

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## AM

can you just tell me what OS and OD stand for ? and any chance u can just explain the calculations and steps so i can do it myself in the future

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## drk

Dude, you cannot be an optometry student.

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## Judy Canty

> can you just tell me what OS and OD stand for ? and any chance u can just explain the calculations and steps so i can do it myself in the future


No, I think that's why you're in school.

Unless, of course, you're willing to forward your tuition to me.   :Cool:

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## kcount

if memory serves OD means Optometric Doctor and OS means Optician-Sinner

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## Darryl Meister

You cannot calculate the minimum blank size "very accurately" with the data provided. Chances are, they want you to calculate the minimum blank size using the traditional rule-of-thumb method in conjunction with a calculation for the decentration required to induce the specified prism (3.0 Base Out and 2.0 Base Up) with a finished stock lens.

Begin by calculating the normal horizontal decentration of the lens required to align the prism reference point (or optical center, in the absence of prism) with the pupil center (PD). As Judy indicated, it is 2 mm in for the right eye. You can find this by subtracting the monocular PD from one-half the BCD/DBC measurement, or 66 / 2 - 31 = 2 mm in.

Then you must calculate the additional decentration from this point required to induce the specified prism. They have simplified the problem considerably by keeping the cylinder axis at 180. The power of this lens through the horizontal meridian is +2.50 D and the power through the vertical meridian is +2.50 + 2.50 = +5.00 D. The decentration required to induce 3.0 pism diopters Base Out is given by:





Note that pllus powers require decentration in the direction of the prism base, so this represents 12 mm of decentration out.

Now, since this decentration is out, the original decentration of 2 mm in, should be subtracted from this value, since the prism decentration is in the opposite direction. The new horizontal decentration is 12 mm Out - 2 mm In = 10 mm out. The decentration required to induce 2.0 pism diopters Base Up is given by:





Again, the decentration is up for Base Up prism. Finally, we need to determine the total horizontal and vertical decentration using:





The rule of thumb for calculating minimum blank size (MBS) is:



The rule of thumb for calculating minimum blank size typically relies on the effective diameter (ED) of the frame. Since one was not provided in this particular case, you should just use the eyesize or A measurement of 46 mm, although this would result in an error in practice:





The _smallest_ possible minimum blank size is therefore 67.5 mm. Again, in reality, the minimum blank size would be larger than this value. In fact, the minimum blank size could reach up to 81.6 mm in this case, for a perfectly rectangular frame (not that these are made). Typically, a 1 mm edging allowance is also added to the final result.

I have attached some diagrams to illustrate these calculations.

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## OpticalMessiah

My head was spinning for a while there, but you put things right back into perspective, Darryl (pardon the pun!) My students would have been amazed! 
BTW, I always thought that OS stood for Optician Sinister! Imagine my embarrassment!

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## Darryl Meister

> My head was spinning for a while there, but you put things right back into perspective, Darryl (pardon the pun!) My students would have been amazed!


I'm glad problems like this appear on OptiBoard occasionally; it keeps me from forgetting some of this stuff after 20 years.

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## Lab King

PD-NB/2
ANSWER-FRAME SIZE X 2 
ADD 1 MM FITTING ALLOWANCE 

Example
frame size = 52
nose bridge = 19
PD = 62
 then
62-19/2
ans - frame size x 2

21.5-frame size x 2
30.5 x 2 = 61

add 1mm Fitting allowance 
61+1 = 62mm

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## Rafael

http://www.optiboard.com/forums/show...466#post347466

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## elaneo

> if memory serves OD means Optometric Doctor and OS means Optician-Sinner


This site is hilarious.

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## dgbradd1

the formula i have used for many years in measuring for single vision cout out is this 

eyesize of frame is 56mm you would then add the dbl which is 19mm this gives you frame pd of 75mm then you take away the patients pd say 58mm that will then give you 17mm you add that to the eyesize of 56mm this gives you 73mm then i add a 4mm buffer due to coating ring around the lens edge whether it be hard coated or ar coated this also includes the normal 1mm buffer for fitting into frame.giving you an effective diameter of 77mm therefore you would need to grind,unless it was such a low power you could move pd`s to within tolerance something i am not at all keen on.

56+19+75
75-58=17
17+56=73
73+4=77mm

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## pyite1979

Very helpful information.
Thank you

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