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  #71  
Old 06-10-2006, 07:55 AM
Steve Conway's Avatar
Steve Conway Steve Conway is offline
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Ro, I'll go by and take a closer look. Did notice it was there, but have not gone into it for info and stuff.

Appreciate the tip.

Steve

Quote:
Originally Posted by byRo
Steve, there is an active forum at their site.
There many people, as you, have criticised the usage of the window's "real estate".
The next version should be a little better.

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  #72  
Old 06-30-2006, 07:36 PM
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Cameraken Cameraken is offline
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Is it me or is there a bug?

Mike Blackney built a 5 pointed star but the code was too slow so I tried to speed it up however I can’t get the ‘maths’ to work.

The angles should be 72, 144, 216, 288 and 360 degrees for a 5 pointed star.

When blending two gradients the slider goes from 0 to 100 which equates to 0 to 90 degrees. So the slider must be set at 40 to equal 36 degrees and 80 for 72 degrees but when I use these settings the star is not correct. I can easily correct this but I would rather find what is wrong.

Clicking Linear or Smooth Gradients does not help although linear seems the best. Is my logic wrong or is there a bug?

Ken.
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File Type: zip 5-pointed Star Shorter BUG_.zip (1.3 KB, 4 views)
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  #73  
Old 07-01-2006, 01:39 PM
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Quote:
Originally Posted by Cameraken
When blending two gradients the slider goes from 0 to 100 which equates to 0 to 90 degrees. So the slider must be set at 40 to equal 36 degrees and 80 for 72 degrees but when I use these settings the star is not correct. I can easily correct this but I would rather find what is wrong.
Sorry, Ken. It's a bit more complicated than this. When you blend the orthogonal ( ) gradients @ 40%, that means you get 40% of one and 60% of the other.
Remember your "trig" classes? This is a right-angled tringle with the vertical side of 40 and the horizontal side of 60, so the opposite angle is arctan(40/60) - which is 33.69 degrees (and not 36).
Starting with 36 degrees and working back, you'll end up with a blend of 42.08%.
The other blend, instead of 80% is in fact 75.48%.

(the thresholds I just lined up by eye)

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File Type: zip 5-pointed_Star_Fixed_byRo.zip (1.3 KB, 4 views)
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  #74  
Old 07-02-2006, 01:19 PM
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Cameraken Cameraken is offline
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Hi Rô

Thank You. That solves the problem and answers my question perfectly.
I knew something was wrong, I just couldn’t see where. I have re-done my spreadsheet of angles and I have even made an angle calculator which I am attaching here. It’s made in VB and so requires VB run-time.

I have searched the net for info on gradients, offsets and curves but can’t find anything useful.
Now if I could only find such a good explanation of multiply, squared etc?

Re.
http://www.filterforge.com/forum/rea...=623#nav_start

I don’t think a Gamma Spreadsheet (or a zip file) will upload at FF. Maybe you could post them here. I would be interested in seeing them.
Is this any help?
http://www.rskey.org/gamma.htm

I am trying to keep up. What is “the square root trick using the Gamma curve”?

Thanks again.

Ken.
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File Type: zip FilterForge Angles.zip (3.8 KB, 2 views)
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  #75  
Old 07-02-2006, 07:01 PM
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byRo byRo is offline
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Quote:
Originally Posted by Cameraken
I have searched the net for info on gradients, offsets and curves but can’t find anything useful.
Now if I could only find such a good explanation of multiply, squared etc?
OK, I hear you.

Quote:
Originally Posted by Cameraken
.....“the square root trick using the Gamma curve”?
When making a more mathematically based filter, square and square root functions are very useful. There is an easy way to do the square, but the only way I know to do the square root is with the gamma function.

In the snippet attached the Gamma curve generates a sqr(c) (squared) output when we specify a gamma value of -30.10.
Feeding that into the gradients and adding (blend, normal, 50%) you get a gradient of sqr(x)+sqr(y). That's good, but for it to be linear we need the square root of that. Running it through the Gamma adjustment, value 30.10, we end up with a luvly linear radial gradient. r = sqrt(sqr(x) + sqr(y))

Problem is that the Gamma isn't exactly the square root - though only a nitpicker would worry about that. (but, seeing as I am one.........)

(Excel file attached)


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File Type: jpg FF-Circle-Gamma.jpg (96.9 KB, 9 views)
Attached Files
File Type: zip FF-Gamma.zip (2.6 KB, 3 views)
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  #76  
Old 07-04-2006, 01:00 PM
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Cameraken Cameraken is offline
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Hi Rô

Thanks, I have studied and (I think) understand that. I guess as more of these snippets are written then the need to understand them fully will diminish. (We can just use them)
However, I like to have some idea of the ‘how it works’.

One (last?) question if I may, My Sepia Filter which (I agree) I should not have used an impulse curve to select shadows, midtones and highlights.
This filter requires some fading or smoothing in the transitions areas. (Like Stroker wrote for Craig’s filter in FM). What would be the best way to do this?


Attached is an example of the 5 point star angles in action. Happy 4th July America.

Ken.
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File Type: jpg USA.jpg (72.0 KB, 10 views)
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  #77  
Old 07-05-2006, 11:00 AM
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byRo byRo is offline
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Quote:
Originally Posted by Cameraken
....I should not have used an impulse curve to select shadows, midtones and highlights.
No problem with the impulse curve.
What happens is that when you put in a value of 100% for the plateau, there's no room for manouver. Change the value to 0%, increase the width to 100% and use positions of -50, 0 and 50.
.......Nice smooth transitions.

The flag looks perfect to me! Great work!!

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  #78  
Old 07-06-2006, 03:05 PM
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Cameraken Cameraken is offline
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Thanks Rô, That’s much better

I was never happy with that filter, I will send in an update.

Ken.
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