This is the same image as the previous post. My husband got a new monitor so the other scanner was down for a day while he installed new stuff in order to get more gb for his new monitor. It's hard for me to tell which of my images will scan better on my older scanner. I just know that when it doesn't look as good as the image I should rescan it. I didn't have this problem with all my oils. The background on this image looked so grainy. Since I haven't used burnt sienna I was wondering if the sedimentary quality of the paint was creating the problem or if was the scanner. I think it looks so much better rescanned. Which brings up the question of substituting an orange or maybe the light red for the burnt sienna. Any ideas?
Wednesday, May 28, 2008
Scanner
This is the same image as the previous post. My husband got a new monitor so the other scanner was down for a day while he installed new stuff in order to get more gb for his new monitor. It's hard for me to tell which of my images will scan better on my older scanner. I just know that when it doesn't look as good as the image I should rescan it. I didn't have this problem with all my oils. The background on this image looked so grainy. Since I haven't used burnt sienna I was wondering if the sedimentary quality of the paint was creating the problem or if was the scanner. I think it looks so much better rescanned. Which brings up the question of substituting an orange or maybe the light red for the burnt sienna. Any ideas?
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2 comments:
A new trick I've been using lately is using and Adjustment Levels Layer (in the layers pallete) and correcting the white balance using the eye-dropper options at the bottom right of the palette. The first eye dropper is your blacks, the middle is your grays, and the last are your whites. For my watercolors, I use the eye-dropper for the whites and then select the white of the paper in the image. Photoshop does the rest with very little fiddling involved. And using the Adjustment Layer, you don't affect the original image.
Hope this makes sense.
This explains it better: http://www.ofzenandcomputing.com/zanswers/1147
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