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scotwllm
Inspiring
October 15, 2021
Frage

Does Dodge & Burn Avoid Extremes?

  • October 15, 2021
  • 1 Antwort
  • 393 Ansichten

I have a scan of a 50-year-old photo of someone's face. The subject was wearing dark blue, and the color has seeped into the chin and jaw area. Above it there is u-shaped blown-out highlights area. I'm trying to use dodge & burn to even out the area. The dodge tool works everywhere except where it's the darkest and the burn tool works everywhere that isn't the lightest. It's like the tools just skip over those areas. Does Photoshop protect these extreme areas for some reason?  I thought that might be the case and screengrabbed my monitor into a new file. It couldn't have any hidden layer information that way, right? 

 

The attached photo shows the original, after dodging and after burning. I cleared away distracting portions of the image. in the middle "dodge" illustration you can see PS didn't touch the darkest areas. In the rightmost "burn" image, you can see it didn't touch the brightest pixels.

 

Any idea why this is happening? 

 

Scott

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1 Antwort

lambiloon
Community Expert
Community Expert
October 15, 2021

Hi dodge and burn not works fine on scanned pictures you need a digital camera picture to work with them in perfect way check the given video which shows the use of dodge n burn...regards

 

https://youtu.be/rgxa7zq-LPA

Ali Sajjad / Graphic Design Trainer / Freelancer / Adobe Certified Professional
scotwllm
Inspiring
October 16, 2021

Hi Ali --

 

Thank you for the link to portrait hacks video. There's some great stuff in there.

 

The photo I'm working with was taken in 1967, before the days of digital cameras, but not long after the creation of the Lescaux cave paintings in France. We hadn't even landed on the moon, yet. 

 

Photoshop doesn't necessarily know where the pixel information came from, and you can dodge and burn scanned images.  I discovered that if I apply a Gaussian blur of .5 pixels to the area, D&B works fine again. See the attache jpg showing the problem area.

 

Scott