To recap the formulas are:

F = 259(C + 255) / 255(259 – C) for the contrast correction factor. C is the contrast (range = -255 to 255).

and

R’ = F(R-128) + 128 for the actual contrast adjustment. R is the red colour component (range = 0 to 255).

Let’s assume that C is zero (i.e. no change in contrast). Using the first formula we work out that F is 1. Using the second equation we see that R’ is the same as R. I’ll illustrate this with some values:

R / R’
0 = 0
64 = 64
128 = 128
196 = 196
255 = 255

Let’s now assume that C is -255. F now becomes zero. The R’ values will then become:

R / R’
0 = 128
64 = 128
128 = 128
196 = 128
255 = 128

As you can see with C turned all the way down to -255 the result is a mid-level grey no matter what the R value is.

Let’s now assume that C is 255. F becomes 129.5. The R’ values will now become:

R / R’
0 = -16448
64 = -8160
128 = 128
196 = 8934
255 = 16575

And because the R’ values are outside of the acceptable range these will be truncated so in reality the values are:

R / R’
0 = 0
64 = 0
128 = 128
196 = 255
255 = 255

Now to explain the reason why the range of the contrast correction factor is 0 to 129.5. Yes it’s true that the range should be 0 to 128, but the value of 259 used in the contrast correction factor formula has been rounded for simplicity. To get a range closer to 0 to 128 you would need to use a value of 259.047619047619 instead. However, as you can see from the R’ values you get at the higher contrast levels they end up being quite extreme and end up being truncated anyway so that fact that the range is 0 to 129.5 instead of 0 to 128 is not very significant.

I hope that explains things more clearly for you. If you still have any questions then please do not hesitate to get back in touch with me.

Because I’m using a range of -255 to 255 here to adjust the contrast I have had to use this correction factor so that it will work with the second formula for the contrast adjustment. All the correction factor does is convert the range I’m using to a range of 0 to 129.5. If you are using a different range, e.g. -100 to 100, then that would obviously require a different correction factor to work properly.

I’ve had a look back through my notes, but I couldn’t find where I originally got this from. I can’t remember if I had calculated this myself or if I’ve read it in a book or online.

Kind regards,

Francis

Thanks

Thanks for the feedback.

I have placed another page on the site with a youtube video tutorial of the process. You’ll see all the steps required in Photoshop to achieve the result.

The downloadable application works in a slighty different way from this process but the results are similar.

http://sites.google.com/site/colinord/about-the-stereogram—3d-object-viewer-project/stereogram—see-the-3d-objects-using-photoshop

Regards,
Colin

Interesting approach, and different from the way that I had envisaged doing it. I like the fact that the original pattern is retained.

Nice work!

Kind regards,

Francis

The Stereogram – 3D Object Viewer application is now available to download. I will be putting a video up as well describing the process soon.

http://sites.google.com/site/colinord/stereogram-3d-object-viewer

Regards,
Colin Ord
http://www.colinord.com

The Stereogram – 3D Object Viewer pages are finally up at:
http://sites.google.com/site/colinord/stereogram-3d-object-viewer

The application will be available for download soon.

Regards,
Colin Ord