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PNG logo files changed from 150dpi or lower to 300dpi ok for print?

New Here ,
Oct 10, 2020 Oct 10, 2020

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I need to print a large vinyl banner at 8' x 10'.

Some of the logo files provided to me were only provided in a small size (few inches large) at 150dpi or less.

In PS if I click Image > Image Size, then change the resolution to 300dpi and increase the size of each logo, will the image print clearly and not pixelated on the final banner?

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correct answers 1 Correct answer

LEGEND , Oct 11, 2020 Oct 11, 2020

No. You can't improv rubbish. Redraw or re-source.

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LEGEND ,
Oct 11, 2020 Oct 11, 2020

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No. You can't improv rubbish. Redraw or re-source.

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Community Expert ,
Oct 11, 2020 Oct 11, 2020

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...and here is that number 300 again. No, you don't need 300 ppi for print. You need 300 ppi for book and magazine print. At this size, 80-100 is more than enough.

 

Even so, those logos are useless and Test Screen Name is absolutely right.

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Community Expert ,
Oct 11, 2020 Oct 11, 2020

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A few years go, the brilliant @davescm posted the following formula and the math he used to create it.  You may want to apply it in order to determine the minimum ppi recommended based on viewing distance.  

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The ppi required can be calculated directly from the viewing distance.*

 

ppi required = 6878/Viewing distance           where viewing distance is in inches.

So for your example  5 feet away --    ppi = 6878/(12x5) = 6878/60 = 115ppi

 

 

* The math that arrives at the formula above is shown below:

A good human eye can resolve 60 line pairs per degree i.e. pairs of black and white lines.  So in half a degree we would have 60 single lines.

Screen Shot 2020-10-11 at 12.46.39 PM.png

 

Simple trigonometry means that the width containing those 60 lines can be calculated by:
Width of 60 lines = Viewing distance x tan(0.5°)

Once we know the width for 60 lines (or pixels) we can calculate how many lines per inch can be seen by using 60/Width of 60 lines.

 

So putting those together gives us 
ppi   =  60/(Viewing distance x tan(0.5°)) = 60/(Viewing distance x 0.0087269)  
         = 2/(Viewing distance x 0.000291)    

         = 1/(Viewing distance x .0001454)

         =  6878/Viewing distance

 

 

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