Stark Houdini It's Just Data!
In this overview I try to show beginners that are intimidated with Houdini, that all you are doing is playing around with Data. I create a simple gradient based off of all the colors in an image and line them up from darkest to lightest - all using simple concepts.
COMMENTS
3D Electronic Dreams 5 years, 4 months ago |
Nice ! Hoping to find more similar technical learning about Houdini
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fiddybux3d 5 years, 3 months ago |
Hi,
At around 11m 45s into the clip (see link):
https://youtu.be/dq5B3TeFeL8?t=707
...you mention using:
@ptnum == 55
Is the "55" an arbitrary figure used simply to identify if the code section is functional?
Michael Stark 5 years, 3 months ago |
Yep! I just chose a random number to show that it will select it. Nothing more than that.
redmirror 5 years, 3 months ago |
Haha, I'm not sure this video makes Houdini any less intimidating! Thanks for trying however.
Michael Stark 5 years, 3 months ago |
Just keep at it. We all started somewhere.
fiddybux3d 5 years, 3 months ago |
I've spent quite a few hours going over this, and I noticed overall there is a large swathe of white towards one side. This skews the average colour.
I tracked the issue down to the grid1 node, which in the video is set to Polygon for the Primitive Type.
If you change this to Point instead, then you can remove / bypass attribwrangle1 (because we don't need to 'addpoints' or 'removeprim' now), and it removes the large swathe of white going up to the outlier point 19800 (sort1 node).
I'm slowly getting the hang of this, but I agree with redmirror, it is daunting.
Michael Stark 5 years, 3 months ago |
Yep. It just depends on what you want to do. I used the polygon setting for the grid because it’s easy to show the comparison to resolution. The daunting part for you guys being new is just know what nodes there are and what to do with them. Once it starts to click more and more the daunting part will go away.
If I wanted to find the TRUE average color, I could take any matching colors and put them into an array or group so that there is only unique values. I promise, just stick with it and it will start to be second nature. I started off by making a shape and clicking a shelf tool. Then rebuilding it node by node to understand what was happening. A year from now you’ll rewatch this and be like “wow, this really isn’t difficult at all.”
fiddybux3d 5 years, 3 months ago |
Thanks Michael, for your responsiveness to these comments, and for your support. I'm already hooked!
Michael Stark 5 years, 3 months ago |
No problem! I think part of the reason you see these such complex and abstract Houdini shots is because once you learn all the main things...fluids, rbd, pyro...etc. you start being able to read between the lines at the data and think “wow, this one parameter could cause this thing to do a neat thing!” Then you just start doing them out of curiosity.
Like this video. It’s not particularly difficult. But a color is a vector - so just having that one attribute I was able to sort the colors. I guess what I’m trying to say is that from artists I’ve seen and known; it’s just a natural progression when using it.
fiddybux3d 5 years, 3 months ago |
I'm resisting the temptation of the sexy stuff as much as possible, so I'm avoiding Pyro, Fluids etc right now. I really just want to be able to box model and texture in H at the moment, and then chuck a bit of VEX in. I'm mindful that jumping straight into the sexy stuff will leave gaps in my basic knowledge.
Coming back to the tutorial for a moment, another thing I noticed when reading through the 17.5 docs today on my lunch, was that the 'color' variable is now reserved.
I didn't particularly bump into this problem, because I spelled it 'colour', but others might.
Michael Stark 5 years, 3 months ago |
Color is reserved as a vector type - v@Cd. If you create an attribute called color, it will just create an attribute called color. Even if you make v@color and set each vector, you're going to get 3 floats. So for example:
v@color = set(1,0,0); will get you color[0], color[1], color[2] with color[0] being equal to 1.
It won't change anything to the color red unless you write @Cd = v@color; But (you don't have to specify v before @Cd, it knows it's a vector type)
@Cd=set(1,0,0); On the other hand will effect the actual color.
In my video, I just had a float attribute named color. So if I did:
@Cd=@color; then each point would have equal RGB values equal to the @color attribute I made.
If I called it @myCoolAttrib, the results would've been the same.
But! Keep that in mind. The type of variables integer, float, vector, string...etc.
fiddybux3d 5 years, 3 months ago |
Thanks for explaining that to me Stark. *thumbsup*
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