Matching Indy Multiply color mode

Tech

Well-known member
Is there an accurate way to match multiply mode used in Indy from Photoshop? I have a project with a grayscale image applied with 4C and then multiply onto a pattern (vector art) 4C background. The point of this exercise is that we want white areas not to show through and pickup the background pattern.

I tried using applying the same 4C and varies color adjustments within photoshop on original image, but I could only get it within the ballpark, just not the exact color matching what multiply mode created.

Ideally we don't want to rasterized the vector pattern in photoshop because the pattern has very fine lines and I don't want soft lines. This will be a last resort.
 
So what you are saying is that the Multiply mode is behaving different in Indy than it is in P'Shop? (just want to make sure I'm understanding the question)
 
Rasterized background color and pattern in Pshop > apply the 4C onto the grayscale > switch to multiply mode.... same steps used within InDy doesn't work in Pshop. I'm guessing Pshop applies 4C differently over the grayscale and that causes multiply to behave differently as well.

I'm out of ideas... some days, you just have to hate Adobe for adding all these options within Indy and allow designers to play with the wrong toys. ~.~
 
In PS are you working in CMYK or RGB multiply is very different in CMYK and RGB. Using RGB or CMYK as transparency blend space will also affect the result.
Also since the multiply is a mathematical operation between the channels, the level of GCR in your working CMYK does affect the colours you get in InD.
Hope this helps to understand what you are asking for.
BTW if you are multiplying vector and pixel you will get pixel art anyway.
 
Designer found an odd solution and appears fine on screen and print out... simply color the selection of the image again without coloring the art using the same background color then apply the previous grayscale art with multiply again in Indy. The idea is that multiply can only go through one layer or one object.

I'm skeptical of it but have no better suggestion. We'll find out soon enough from printer's proof.
 

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