chipnyman, thanks for the feedback. I had allready mentioned this solution in #10, but it didn't work either. Meanwhile I found a simple remedy: I installed CS2 from disk, and because it's an academic (teacher) version it doesn't need activation... Meanwhile other dark clouds are darkening the sun: I cannot install (additionally) CS6 because ' the installer could not be initialized'. This is a very common error and I think I've read all these cute suggestions how to overcome the problem, especially this one: 1. Uninstall CS6 2. Apply the Adobe Clean Tool (something like that) 3. Re-Install CS6 There is not even a log-file. Without having installed CS2 the situation is just the same. Best regards --Gernot Hoffmann ... View more

In Windows 7 (32bit) it's not possible to extract a self-extracting  EXE  file into a folder explicitly (without executing), neither by left/right/single or double clicking. I've tried Universal Extractor v.1.61.1 and I could extract the three EXE-files into one folder. I've uninstalled the already working Photoshop CS2 and started the whole process once again. It works until 'Insert Disk2' and then everything fails. Here is another recommendation: https://forums.adobe.com/thread/1337558?tstart=0 This is based on Adobe's enigmatic remark on filename/path confusion (as quoted by me previously in #6). http://helpx.adobe.com/x-productkb/policy-pricing/creative-suite-2-activation-end-life.html Photoshop CS2 is working perfectly, despite the much newerOS. It's really a pity that the installer seems to be somewhat sick or crazy Best regards --Gernot Hoffmann ... View more

Thanks – sorry – I don't understand.  In Windows 7 I can't extract an EXE file explicitly into a destination. Such a self-extracting EXE file is just executed, obviously based on a hidden extraction. Best regards --Gernot Hoffmann ... View more

How can we extract EXE files? And why is it necessary for self-extracting(?) EXE files? http://helpx.adobe.com/x-productkb/policy-pricing/creative-suite-2-activation-end-life.html Some 'extracted' information: 1. Uninstall existing CS2 applications If you have CS2, CS2 applications, Acrobat 7, or Audition 3 installed on your computer, uninstall them. Note: If you installed CS2, make sure to uninstall all applications in the suite. Windows Click Start, click Control Panel, and then double-click Add or Remove Programs. Select the CS2 applications you want to uninstall and click  Remove. Mac OS... 2. Download and install CS2 applications You can use the serial numbers provided as a part of the download only if you legitimately purchased CS2, CS2 applications, Acrobat 7, or Audition 3. Make sure that your computer meets these requirements: ... Select the language in which you bought the application: ... Sign in using your Adobe ID (usually your email address) and password. If you don't have an Adobe ID, click Get an Adobe ID... 3... 4... 5... 6... 7. Extract or double-click the downloaded file and follow the onscreen prompts to install the application. Problems: I'm having problems downloading or extracting –Try downloading again with a different web browser. –Try extracting the files again. I'm having problems installing – Invalid serial number error... – Insert disc two alert 1. Copy the contents of the C:\Creative Suite\Adobe Creative Suite 2.0\ folder to the C:\Creative Suite CS2\Adobe Creative Suite 2.0\ folder. 2. Reinstall the application. (End of info by Adobe, bold by me) Best regards --Gernot Hoffmann ... View more

I had downloaded the CS2 system from here Error: Activation Server Unavailable | CS2, Acrobat 7, Audition 3 and saved all files in the same folder. The PC is new, with Windows7 (32bit). For reasons of backward compatibility I have to use such a legacy equipment (I own as well CS6). The installation starts by clicking on CreativeSuiteCS2Disc1.exe and ends with a prompt Insert Disc2. Of course the disc is not accepted. How can I continue? This issue has been discussed here a couple of times, but I didn't find a solution. At this time Photoshop, Bridge and some fragments of InDesign are installed. Photoshop works fine. InDesign and Illustrator are missing. Clicking on CreativeSuiteCS2Disc2.exe starts the installer again, but without usable results. Best regards --Gernot Hoffmann ... View more

About MS Visio: This seems (still?) to work entirely in RGB, like all MS programs. CMYK is not available. Therefore it's the wrong tool for editing diagrams, flow charts etc. for offset printing and toner printers. RGB black for lines and small text should be printed by K-only. Colored lines by one or two inks instead of four inks in order to avoid visible registration errorrs. These diagrams should be vector graphics, then  a conversion by Photoshop (Vector  —> Vector) is impossible. Best regards --Gernot Hoffmann ... View more

For clean illustrations you'll need a vector graphics program. Adobe Illustrator comes into mind, but until some time ago a line with arrow appeared longer by arrow length than the line without arrow....a really idiotic behaviour. Better use InDesign. This programm is probably not intended for these graphics, but for me it's really very pleasant (if you redefine a plain line as a line with arrow, then the line keeps it's length). There are all necessary features, lines with different arrows, stroked or/and filled boxes, typographical text in vector quality, grouping etc. etc. Here's an example: http://docs-hoffmann.de/cmsflow03072013.pdf My suggestion is based on a quiet workflow, doing everything manually, besides group, copy and paste. Best regards --Gernot Hoffmann  ... View more

»station_two« , I don't think we disagree (any more). Some years ago, CMYK printers used Adobe PostScript and were generally high quality devices, expensive because of both reasons. I'm surprised that such a low-cost printer is equipped with a PostScript clone (emulation) which is probably not bad, concerning the language implementation. I wouldn't use either such a printer for quality prints. da ich unter meinem wahren Namen schon mehrmals aus diesen Foren 'rausgeschmießen worden bin. Would you tell me your true name by private E-Mail? Best regards  --Gernot Hoffmann ... View more

station_two, please excuse me, if I should have misunderstood your post.  We are talking about true CMYK printers which can be operated as RGB printers. Their gamut is entirely defined by the CMYK inks and the paper or medium. How should it be possible to enlarge this gamut by an intermediate RGB work space? It is almost the same situation for offset printing, but the offset gamut is considerably small than that of monitors. Therefore we can be happy that Photoshop offers soft proofing. On the other hand, the gamut of inkjets is large enough for proof printing. Now let us compare some gamuts. Common 3D visualizations are not really suitable, therefore I'm using slices in Lab for a set of L values, here on p.19: http://docs-hoffmann.de/gamshow15052009.pdf AdobeRGB (red curve), ISOCoated-v2-eci (black curve), Inkjet Mutoh 6100 (gray curve). Curves for sRGB are found on other pages. No doubt, that RGB operated inkjets are widely used by Photographers for 'fine art printing'. That's not only a matter of gamut (which is not increased by RGB) but also a matter of black generation by multiple gray inks, by excellent dithering modes and knowledge how to avoid bronzing and how to reduce illuminant metamerism. Perhaps these methods are already optimized, therefore it's not necessary to offer to the photographer full control over all parameters. Best regards --Gernot Hoffmann ... View more

station_two, thanks for the feedback. I'm grateful for all corrections. Can we say RGB printers are optimal for all photographers who are not familiar with CMYK ? In my opinion RGB printing cannot be better than CMYK printing, but for many tasks one needs definitely the full access to the single inks. I remember the famous book by appreciated Dr.Hunt: The Reproduction of Colour http://www.amazon.de/Reproduction-Colour-Imaging-Science-Technology/dp/0470024259/ref=sr_1_1?ie=UTF8&qid=1418894029&sr=8-1&keywords=R.W.G.Hunt+The+reproduction+of+colour All diagrams were derived from RGB versions. Simple black lines were printed by CMYK with considerable registration errors. Better knowledge of prepress techniques could have preserved the work of this excellent color scientist. Best regards --Gernot Hoffmann ... View more

Post #16 does not appear in the list. Bruce Frazer (not Fracer), of course. G.H. ... View more

Meanwhile I found some further informations: – The installation of the CP1515n driver does not add ICC profiles (one message). – This printer comes with HP Postscript (level) 3 emulation.    It is Postscript printer, but it's not original Adobe Postscript. – Two PPDs (Postscript printer page description programs). In one we have    CMYK as default color space, in the other RGB. For Postscript printers or Postscript RIPs I would use under all circumstances CMYK calibration (as I always did). This guarantees full control of the black generation: Black Max, CMYK Max, GCR mode or UCR mode, Black start. These settings depend strongly on the medium. The ICC profile depends on the settings. Furtheron one can use Rich black (images) or K-only (text, lines...). Printing ink limit swatches and linearization targets requires definitely direct access to inks. If the printer is treated as an RGB printer, then we have just one pure black: R=G=B=0 (very small values mean the same). No chance to distinguish the two basically very different blacks. The RGB values have to be converted into printable CMYK values (if we exclude true RGB printers, which work with lasers on film). Which conversion is used? It could be a generic method, or a more sophisticated one, which depends on the medium. The black generation parameters are inside the printer, probably not accessible (readable, writeable) during the profile generation. I would be greatful for explanations or hints how this works, because I want to improve and complete my document about Color management signal flows: http://docs-hoffmann.de/cmsflow03072013.pdf These questions are not answered in our well known books: Abhay Sharma, Understanding Color Management Bruce Fracer et al, Real World Color Management Best regards --Gernot Hoffmann ... View more

How to check the manufacturer's preferences RGB versus CMYK: Download the program ICC Profile Inspector: http://www.color.org/ICCProfileInspector2_4.zip Open one of a couple of available ICC-Profiles for the printer. Check by Profile Inspector whether it is an RGB–Lab profile or a CMYK–Lab profile.   If these profiles are RGB-profiles, then the manufacturer considers the printer normally as an RGB device. But stop! I'm using since many years 1. large format inkjet printers Mutoh 6100 and Epson 7890,     operated by RIP Colorgate Productionserver, earlier by RIPs Onyx and Best-Efi. 2. toner printers OKI 9600,    PostScript devices, operated by Photoshop and Acrobat. For the calibration I'm using in all cases large or small CMYK targets, printed without color management by Photoshop (older version, which has this mode) for the toner printer and by RIP for the inkjets. Profiles by GretagMacbeth ProfileMaker. The manufacturer's profiles for the Epson are RGB-profiles! All my profiles are of course CMYK profiles. The inkjets are often used for proof printing for offset. A remark to post #4: it seems that the mentioned test was done without any profile. Then it is of no relevance. Best regards  --Gernot Hoffmann ... View more

Hello Peter, thanks for the feedback. It's now 9 years ago that I had tested printing by OKI C9600, a PostScript Toner printer, at the beginning by PageMaker and later by InDesign CS2: http://docs-hoffmann.de/oki-ps-28082005.pdf P.17 says, "printing by Printer PostScript Color Management fails terribly – chaotic colors". Today I had repeated the test by InDesign CS6 – it's still the same desaster. This doesn't say, that InDesign is wrong per se, it's perhaps just a problem with this printer, which works (on the other hand) fine for PDFs. Meanwhile I had tried to explain color management by signal flow diagrams: http://docs-hoffmann.de/cmsflow03072013.pdf I don't have special insider knowledge regarding Adobe programs. Therefore this doc is not an explanation how Adobe programs definitely work. Best regards --Gernot Hoffmann ... View more

A set of three numbers in xyY or XYZ defines a color.  How does it look? How can it be reproduced? A good occasion to go back to the roots. The CIE color system is essentially based on the work of Hermann Graßmann, one of the greatest scientists in the 19th century [1]. Mostly his four laws are quoted [2], but it boils down to this [3]: "The whole set of color stimuli constitutes a linear vector space, named tristimulus space." Colors behave like vectors in a three-dimensional space. A color is characterized by three numbers. A fourth number would be redundant (linearly dependent). A color is not characterized by just one spectrum. For one color exists an arbitrary number of different spectra, called metamers. The explanation of Graßmann's laws via spectra [4] is wrong. Nobody knows what a color 'really' is. Acccording to the philosopher Kant, we don't know what any object in the space 'really' is. We have just an impression by our senses, eventually enhanced by instruments. Colors are described by comparison with reference colors. For length and weight we need references as well: the meter, the kilogramm. Reference colors are (for instance) the three CIE primaries, spectral colors with well defined wavelengths (it's not important, that the first experiments were executed with a different set). Let's say R,G,B. According to Graßmann one needs exactly three 'primaries'. This should have ended a long dispute: three or four?, but it didn't. Now we can use these spectral colors as base vectors of a coordinate system. According to Kant, humans have an a-priori idea of space (without any proof), where we can position objects. By intuition the axes are orthogonal. Any other color can be described by a linear combination of the three primaries, according to Graßmann exactly in a vector space. Color matching means: find the three weights for the primaries to match a given (numerically unknown) color. Unfortunately, one needs for the matching of some colors negative weight factors, which is possible by the concept of vector space, but somewhat disturbing for real color mixing. In RGB we can describe a new vector space by base vectors X,Y,Z, which form a non-orthogonal coordinate system. All colors have non-negative coordinates X,Y,Z and this construct has the funny feature, that luminance is identical with Y, whereas the 'imaginary primaries' X and Z don't have luminance. Mathematically this is not a problem. The relation between the two spaces is given by a matrix Cxr and its Inverse: X=(X,Y,Z)'  (column vectors) R=(R,G,B)' X=Cxr R R=Crx X =Cxr^-1 X Conventionally we draw XYZ as a cartesian coordinate system (with orthogonal axes), and R,G,B is the set of non-orthogonal base vectors – the arrangement has been swapped. There is no natural law, which of the coordinate systems has to be shown as a cartesian. XYZ is universal. Other RGB-systems with new primaries can be added, either as working spaces like sRGB, aRGB or pRGB (ProPhotoRGB, which uses two non-physical primaries, which doesn't surprise, because we got used to entirely non-physical primaries X,Y,Z),  or device RGB systems like monitor spaces. Each RGB system is related to XYZ using a matrix, thus we can transform as well from one RGB-system to another. Now it's possible to render a threedimensional visualization of a color space by appropriate colors, for instance aRGB. But for a monitor these colors would be clipped almost at the sRGB boundary. Of course it's not possible to render pRGB correctly, because the blue and green primary are outside the human gamut, and the red primary would be too dark. Therefore it's wise, to use pRGB only for regions or real world colors. There might be still an objection: the reference system contains only spectral colors. How are the little saturated colors looking? This had been clarified by Graßmann: the vector addition of any two colors is valid mathematically and by appearance. These explanations may also help to end the dispute about the 'basic colors', especially for painting. Many artists have invented their own system. The answer is simple: basic colors are like primaries. Their location in the XYZ-space defines, which colors can be created with positive weight factors or mixing ratios. By the way: one shouldn't worry here about the difference between additive and subtractive color mixing. One may have serious doubts, whether Graßmann's laws and the whole CIE colorimetry is really valid, especially considering numerous optical illusions [5] and all these special effects in color appearance, like Helmholtz-Kohlrausch and other nonlinearities. It's perhaps surprising, that CIE colorimetry works so very well for image processing and device calibration for monitors, printers and cameras. Finally I would like to mention another scientist, which is not as popular as others, probably because his work is mathematically difficult: Jozef B.Cohen [6].    His work has been continued by William Thornton, Michael Brill and James Worthey (for a further search). Best regards --Gernot Hoffmann [1] Hermann Günther Graßmann (Grassmann) http://en.wikipedia.org/wiki/Hermann_Grassmann [2] http://de.wikipedia.org/wiki/Gra%C3%9Fmannsche_Gesetze [3] Claudio Oleari http://www.slidefinder.net/o/oleari10/32197498/p3 (6th slide). [4] http://en.wikipedia.org/wiki/Grassmann%27s_law_%28optics%29 . [5] http://www.michaelbach.de/ot/index.html [6] Jozef B.Cohen Visual Color and Color Mixture http://books.google.de/books?id=W8QeI5di7t4C&pg=PR13&lpg=PR13&dq=cohen+color&source=bl&ots=aYGoI0I99g&sig=IJEgdJjk3yDhi-1NvTD4GdyUIXk&hl=de&sa=X&ei=614hVM_dFsrMyAPbzoDwCA&ved=0CGsQ6AEwCA#v=onepage&q=cohen%20color&f=false ... View more

Please excuse me, your diagram for ProPhoto RGB isn't correct: http://www.google.de/imgres?imgurl=http://upload.wikimedia.org/wikipedia/commons/thumb/e/eb/CIExy1931_ProPhoto.svg/325px-CIExy1931_ProPhoto.svg.png&imgrefurl=http://en.wikipedia.org/wiki/ProPhoto_RGB_color_space&h=369&w=325&tbnid=JVEXfCNViLaYzM:&zoom=1&tbnh=90&tbnw=79&usg=__f59EVs8luFswOka_UXVcKMiQ77g=&docid=fFh0ao7j-Tt-6M&hl=de&sa=X&ei=EHceVJu8J5DcatujgcAI&ved=0CDwQ9QEwAw&dur=9029 Especially we have the red primay at the rightmost tip of the horseshoe. The blue and green primaries are mathematical constructs without any physical meaning, but this red is a spectral color, though it cannot be reproduced by technical devices like monitors or printers. It is even beyond the code range of CIELab in Photoshop, coordinate (a) ist clipped at 127. Your test seems to prove, that the linear interpolation from white to red should create a considerable magenta cast, because the trajectory hits the purple line of the sRGB triangle. My suspicion was, that the effect were caused by gamut clipping by the monitor. Therefore I've desaturated the monitor by 50%: In my opinion the new gradient bar has a uniform hue, but everywhere a weak magenta tint. One may ask: why desaturate as much as 50%? The answer is given on page 44 (top left) of this doc: http://docs-hoffmann.de/munsell15052009.pdf sRGB has red for L=62 at appoximately a=60, b=70. pRGB red has to be desaturated toward these limits. The graphic shows as well, that a pRGB red would be very far from the range of practical colors as represented by the Munsell system. Other considerations: Lines of constant hue are not straight lines in CIE xyY. The constant hue curve for white to red is bent toward the right side (somewhere shown in Hunt, Measuring Colour). In CIELab it's a little better, but for blue we have a well-known desaster: Linear interpolation between dark blue and gray in a plane L=const. generates halfways a magenta tint. A white point shift (as suggested) cannot be the solution, IMHO, because these tints are local phenomena. One could counteract by ICC-Profiles with correction tables or, more systematically, by elaborated color appearance models, like CIECAM02. Thanks for your always interesting questions. Best regards --Gernot Hoffmann ... View more

Thanks for the clarification - the bottom is flat, though it looks like a pyramide (but the triangle doesn't fit AdobeRGB. The green primary should be more inside). I'm not able to view Lindbloom's graphics because of a missing plug-in. The human gamut in CIELab is a cone-like volume, unlimited on top. For color spaces and for optimal object colors the volume is closed, with the white point on top. Wyszecki & Stiles, Fig. 2(3.3.9). The shape is not unique, it depends strongly on the illuminant. This excellent publication shows many graphics: Masakoa, Berns, Fairchild et al: Number of discernible object colors is a conundrum www.cis.rit.edu/fairchild/PDFs/PAP32.pdf Best regards --Gernot Hoffmann ... View more

Sorry for the delay, I've been travelling and pondering over your ColorThink diagram. Here we see a diagram 3D-xyY according to Adobe: http://dba.med.sc.edu/price/irf/Adobe_tg/models/ciexyz.html Note that the base of this object is flat, not like a pyramide in the ColorThink diagram (black lines). Here we see my diagram xyY in 2D-xy with contours Y=const: http://docs-hoffmann.de/ciegamut16012003.pdf p.2 + p.12 Both representations are equivalent. Now I'm trying to interprete the ColorThink graphics: The appearance of an RGB color cube in  XYZ, which is now called XYZ-cube, is shown here: http://docs-hoffmann.de/ciexyz29082000.pdf p.5 Three base vectors R,G,B represent the three 'dark edges' of the XYZ-cube. The other 9 edges are formed by linear combinations of the base vectors like C=rR+gG+bB with special factors r,g,b,  fixed 0 or 1 and in the range (0...1): For instance C=1R+(0...1)G is the edge from red to orange. The XYZ-cube is an affine transformed regular cube. Straight edges of the regular cube are mapped to straight edges of the XYZ-cube. What we see in the ColorThink graphics is a 3D-color space which seems to be a perspective transform of the XYZ-cube, 'erected' over the chromaticity plane xy with Y as third coordinate. But this object does not have straight edges, therefore it's not a perspectively mapped XYZ-cube, because perspective mapping (like affine mapping) would have preserved straight edges. What is it? 3D-CIELab over xy? Sorry, I don't know how this object had been generated.    About the other questions: A white point conversion is generated by varying the lengths of the base vectors or the directions or both. The base vectors appear in xy as 'primaries' (the nomenclature may vary). If only the lengths of the base vectors are changed (and not their direction), then the primaries in xy are the same. The new white point in XYZ is achieved by Wn=Rn+Gn+Bn. If we apply previously used factors r,g,b, the all colors are shifted automatically for the new white point. A color with flat spectrum (in the relevant  range) is called 'Equal Energy Stimulus'. Such a spectrum does not exist for any natural or technical light. The color in XYZ is the Equal Enery White point, which ist mapped to xy by a straight line through this point and the point X=Y=Z=0, intersecting the plane X+Y+Z=1 in xe=ye=1/3. The Planckian locus is a curve of chromaticities. It can be shown exactly and by approximation for different 'correlated color temperatures' in xy: http://docs-hoffmann.de/coltemp18102003.pdf Obviously any arbitrary light source or illuminant doesn't have an assigned white point in XYZ at Y=100 (Y is arbitrary). Therefore a diagramm Y over xy wouldn't make any sense here, in my opinion. A good argument for the introduction of the chromaticity diagram! Now I hope, these explanations are more or less correct. I don't know how to edit already posted contributions, therefore some errors might have been left. Best regards --Gernot Hoffmann  ... View more

Without further explanations these graphics are rather dubious, in my humble opinion. http://docs-hoffmann.de/ciexyz29082000.pdf On p.8 we can see, how XYZ is mapped onto the plane X+Y+Z=1 as coordinate system xy by a perspective projection: The original Y-axis is not orthogonal to the plane xy, therefore the quoted diagrams are not correct. It's of course possible to construct a kind of substitution Y* orthogonally to xy. As long as we are talking about unrelated colors, like an arbitrary set of spectral light sources with arbitrary radiant powers, the human gamut in XYZ is indeed a cone, which has its vertex at X=Y=Z=0 and which intersects the plane X+Y+Z=1 by this horseshoe curve (plus purple line for mixtures of extreme spectral red and blue). About power: http://en.wikipedia.org/wiki/Radiant_flux "radiometric power", correctly: radiant power or radiant flux The situation becomes different under certain circumstances where the unlimited cone is replaced by a finite volume or by finite curves on this cone: 1) The radiant power of the spectral light sources is limited and equal, e.g. 1 Watt.     The curve on the cone shows the spectral loci. In Wyszecki&Stiles in chapter 3.3.3     in Fig. 4(3.3.3) exactly this situation is illustrated: we can see the cone, the horseshoe and     the spectral locus on the cone.     As already mentioned, this issue has been further evaluated by Jim Worthey ("amplitude     not left out" and illustrated by me and of course by himself  (see references):     http://docs-hoffmann.de/jimcolor12062004.pdf 2) The volume of Optimal surface colors, concept by Roesch, as already explained.     Fig. 5(3.7) in W&S shows Y orthogonally over xy, which is not correct, but we understand     the meaning. This 'wet sack' has on top Y=100, which is the luminance of the illuminant. 3) A color space is a subset of a finite gamut volume in XYZ. There is always a white point and    nothing can become brighter than the white point. An RGB space is represented in XYZ by an    affine distorted cube. Mapping to xy delivers a triangle, but here we can see, what happens    if the "amplitude is not left out":    http://docs-hoffmann.de/ciegamut16012003.pdf    The sections Y=const. with the dark sides of the cube appear still as triangles, but the sections    with the bright sides by other polygons (p.12 for sRGB).    A threedimensional graphic would be possible, Y over xy. Best regards --Gernot Hoffmann ... View more

Some guidelines for a clean workflow: 1. Don't use sRGB but a valid measured profile for the monitor. 2. Synchronize color settings manually or by Bridge. 3. Check whether these settings are valid as well in Acrobat Pro for viewing (they are     not automatically synchronized in Acrobat Pro 8, which I'm still using).     Use 'Output Intent overrides single settings', or so. 4. Define in InDesign an RGB and a CMYK workspace as document color spaces.     InDesign exports Grayscales correctly, but the monitor preview can be wrong. Grayscale     should not be used. 5. Place only images which are already in these RGB or CMYK color spaces. 6. For inkjet printing use only RGB images with embedded profile. 7. For CMYK offset printing use only CMYK images with embedded profile. 8. Don't redefine image color spaces in InDesign for individual images. 9. For any printing, especially by inkjet, use in Export as PDF    – Leave Colors Unchanged (European Prepress doesn't change colors)    – Embed all Profiles      or    – Use always the same profiles for inkjet RIP printing 10. For offset printing use Export as PDF/X with pure CMYK content.    Assign Output Intent (the valid CMYK ICC profile). Single image profiles will be stripped. In cases of color shifts: a) Import PDF pages in Photoshop in the assumed color space and check appearance. b) Analyze PDF by Acrobat Pro (which cannot be explained here, except for Acrobat Pro 8).     Eventually show us an example.  A PDF can be much more complex, but then the process will be even more prone to errors. Everything in my humble opinion... Best regards --Gernot Hoffmann ... View more

Thanks for the clarification and further information. This will hopefully enable the scripting and layer mask experts here to help you further, and as well me concerning the application of Illustrator. There is one issue which I'm still not understanding: how do you decide which parts of the graphic belong to one layer? Meanwhile you may have a look here, an application of Live Trace by Illustrator: http://docs-hoffmann.de/casamunsell24062012.pdf 3 MBytes. Explanation in German and English on the first pages. Best regards --Gernot Hoffmann ... View more

(1) http://en.wikipedia.org/wiki/Jigsaw_puzzle (2) http://en.wikipedia.org/wiki/Printed_circuit_board (3) http://www.lpkf.de/produkte/rapid-pcb-prototyping/software/circuitcam-pcb.htm ---- Thanks for providing the images. I thought you were planning to produce jigsaw puzzles (1), but now I see: printed circuit boards (PCBs), (2). In my opinion your workflow contains a logical bug: the layers are combined in one flat image and distinguishable only by colors and, concerning the holes only by topological analysis: a black circle inside a white eyelet is a drill hole. By eyelet I mean an annular or quadratic ring, sometimes called pad. If the design was made by adhesive tape (your design looks so) then one uses normally two plastic sheets, one for the top layer, another for the bottom layer. Drill holes have to be identified by human 'pattern recognition'. If the design is made by CAD, then the working piece is separated from the beginning by layers: top, bottom, holes, text, varnish (for a double sided connected-through PCB). Considering this, late separation of layers is obsolete, in my humble opinion. Before we, nevertheless, discuss the possibilities of Photoshop or Illustrator, please explain: What's the material? You're talking of paper, but normally we have copper on plastic. How are the conductor tracks separated? By lasering a 'stroke' around the contour of a contiguous area (tracks + eyelets) or by lasering a 'fill' anywhere outside these areas, leaving only the red parts? I think there are a couple of commercial solutions for CAD–CAM in this field already available, e.g. (3).  Best regards --Gernot Hoffmann ... View more

Better use Illustrator. There you can choose, whether a stroke is applied symmetrically on a path, outside only or inside only. Good old PostScript tradition. Best regards --Gernot Hoffmann ... View more

Can't yuo use Export > Paths to Illustrator? The image has 1000 x 1000 Pixels. The path starts and ends at 500,500. The numerical values are not integer pixel coordinates but somewhat rounded. The path in the image is not stroked (for clean information). It's a screenshot of the unstroked path as shown in Photoshop. Best regards --Gernot Hoffmann Text file : %!PS-Adobe-2.0 %%Creator: Adobe Photoshop(TM) Pen Path Export 7.0 %%Title: (Path.ai) %%DocumentProcSets: Adobe_Illustrator_1.1 0 0 %%ColorUsage: Black&White %%BoundingBox: 0 0 1000 1000 %%HiResBoundingBox: 0 0 1000 1000 %AI3_Cropmarks: 0 0 1000 1000 %%DocumentPreview: None %%EndComments %%EndProlog %%BeginSetup Adobe_Illustrator_1.1 begin n %%EndSetup 0.0 0.0 0.0 1.0 k 0 i 0 J 0 j 1 w 4 M []0 d %%Note: 500.7495 499.2504 m 500.7495 899.5502 l 800.5996 899.5502 l 800.5996 100.4497 l 500.7495 499.2504 l n 500.7495 499.2504 m 500.7495 899.5502 l 800.5996 899.5502 l 800.5996 100.4497 l 500.7495 499.2504 l n %%Trailer %%EOF ... View more