Sizing photos exactly the same

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I saw another thread that was similar however my issue (challenge) is a bit different.

I have several photos, some from a digital camera, some from scanned prints that I need to be exactly the same size, dpi, everything. It seems that when I change things around in image properties (canvas size, image, etc) things get a bit different.

How can I batch or individually change these photos to all be 200 dpi, same dimentions, and the border I create be exactly the same? Is this enough information?

Thanks!
 
First off I'm going to say CROP tool, 200dpi and specify the size (i.e. 4x5).

Proportions may be different from your source material and you probably will not be able to keep the same length and width without cropping/losing some of the orignal image.

For example 4x5, 8x10, 16x20 all have the same proportions; 3.5x5, 5x7, 11x14 have the same proportions. But you can't enlarge a 5x7 into an 8x10 without losing some of the original image OR without distoring the original image.

Hope this answers your question.
 
Here's a workflow that I commonly use when sizing something like a 35mm negative for a 13x19 print.

1) Take a look at the canvas size of my original 3000dpi negative.
2) Divide 13 by 19 to get the canvas size ratio
3) Create a blank canvas based on that ratio, with a 3000dpi resolution, about the same size as my original negative. (At this point I may lose a little bit of the original image due to ratio diferences if want to go borderless because I'm too lazy to trim the finished print).
4) Print with preview, stretch to fit.
 
Hi,

You could create a new document to the size and dpi you want. Then create a layer with your border on it. Save this file as photo master etc as a psd (layered file). Then reopen it and call it whatever you want to call one of your pictures. Cut and paste one of your original images into the document and put it on a layer beneath the border. You can then resize your original image for best fit. Do the same process with your other pics.

I hope this makes sense:)
 

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