Subjecting a JPEG Image to Successive Compression Steps
Greetings,
I've seen a few discussions on JPEG Quality, but can't find (here, or elsewhere on the interwebs - at least not without analyzing the compression algorithm directly) an answer to the question implied on the Subject line. To wit:
Let's say we have an uncompressed image, I, that we save as a JPEG with a Quality Factor of 60%. We'll call that compressed image, I60.
Now, let's say that we take the same, original, uncompressed image and save it as a JPEG with a Q of 70%, and we call that image, I70. Then, let's say we take that 70% JPEG and save it, again, only now with a Q of 60%... We'll call it, I7060.
Here's the question: Is the perceived Quality of I60 the same as the perceived Quality of I7060?
Experimentally, I can say that the sizes of the resulting files are... similar. But, at least in the experiment that I performed, the file size of I7060 was actually slightly larger than the file size of I60.
However, subjectively, I already have a hard time distinguishing between I and I60, so it's not likely that I'll be able to make a judgment about I60 vs. I7060.
Hopefully, someone with direct familiarity with the compression algorithm can make a pronouncement on this from the theoretical perspective.
So, what's the point? Well, I have a raft of images that were each assembled from a couple of uncompressed components, and then saved at 70% compression. They should have been saved at 60% compression. Unfortunately, although I have the original uncompressed constituents of each final product, I didn't save the intermediate, uncompressed, assemblies.
Hence, my question. I just want to know whether I can get the same effective result by re-saving those JPEGs at the higher compression ratio, or am I actually compressing the original image to "60% of the 70%" (i.e. a Q factor of 42% - Hey... whaddya know, I finally have an answer of 42, and I barely know what the question is... Douglas Adams would be proud.)
Thanks!
- s.west
p.s. Sorry for the long-winded post (I'm sure you're used to it from me by now), but I was trying to leave no aspect of the question open to confusion.