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Rotating Text arounda sphere in Invigorator Pro
Start this off by saying I found a similar post from a long time ago, but no resolution aside from purchasing another program was given. This is not an option for me.’
I am using Invigorator Pro v4.0 in After Effects 7.
I need to wrap 3D text around a sphere, like it is in orbit around the sphere, not mapped to the sphere itself. I will then rotate it sound the sphere. The rotation part I have figured out. It is making all the characters of the text equidistant from the center of the sphere that I cannot seem to figure out.
As I said earlier, buying ProAnimator is not an option, as much as I would like it.
Thanks!
Geno A. Palazzari
Creative Services Commerical Producer
KAAL-TV
Rochester, MN -
3 Replies
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Hi Geno.
In Invigorator, you’ll have to manually move each letter to the outside of the sphere. There is no auto-distribute in the Invigorator.
To do this in the Invigorator interface:
1. Create the sphere. Make it the size you want. Position it. Put it in Set 0.
2. Create your text block – let’s say you’re using “GENO”
3. Put the text block in Set 1.
4. Select the letter G. Use the Object Move and Tumble tools to put the G in position at a point outside the sphere.
5. Do step 3 for E, N, and O, placing and rotating the text letters in the appropriate places.You can then use the Set parameters in the ECW to rotate Set 1 about the Y axis. This will rotate each letter around the sphere. If the center of rotation isn’t perfect, you may need to adjust the pivot to be the center of the sphere.
Alternately, you can use the Stats dialog in the Object tab to manually set each object’s position and rotation. This may be easier than eyeballing it, but it does require some math.
Let us know if you have any questions.
Matt Rhodes
Zaxwerks, Inc. -
Matt,
Thanks for the response. I had that much figured out already. I should have been more specific in my original post. I have an logo rotating around the sphere on one side. Now I am trying to rotate an 18 character word on the back side, basically so when one is going to the dark side, the other is coming on.
I would like the characters in the word to be equidistant from the center of the sphere, so that each one is the same distance from the surface of the sphere. That way it looks like it is wrapped in orbit. Does this make sense.
The radius of my sphere is 295. Is this 295 pixels? Can I assume that the outside of my sphere is at z=295 (provided that I am looking straight at it from the Front Camera View).
Here is what I am attempting right now. Going under the assumption that the edge of my sphere is at 295, and the center of my text block is approx. 363, I am attempting to calculate… I think I am trying to calculate the points of the arc directly behind my text. That way I will be able to extrapolate those points outward to where my text block is and use the Stats dialogue to adjust each character individually.
Am I on the right track? Does anyone know how to calculate the points on an arc? It has been a long time since I had geometry. Time to find my old text books.
Seriously, any help would be appreciated.
Geno A. Palazzari
Creative Services Commerical Producer
KAAL-TV
Rochester, MN -
Hi Geno.
> The radius of my sphere is 295. Is this 295 pixels? Can I assume that the outside of my sphere is at z=295 (provided that I am looking straight at it from the Front Camera View).
– This is 295 world units (not quite pixels). Yes, if Z=295, it will be just on the outside of the sphere.> Am I on the right track?
– Yes. You have the right idea.> Does anyone know how to calculate the points on an arc?
– You can probably use the following:
Assumption: Your sphere is centered at (0,0,363).
Assumption: We’re distributing the phrase “*Eighteen Letters*” – which has 18 letters – along the back side of the sphere
We’ll start numbering our letters at 0 instead of 1 to make the math easier,
so Letter 0 is “*”, Letter 1 is “E”, etc. — we’ll use the symbol L to represent the letter number
Position X = 295*cos(PI*(L/18.0))
Position Y = 0
Position Z = 363 – 295*sin(PI*(L/18.0))
Rotation X = 0
Rotation Y = 90 + 180*(L/18.0)
Rotation Z = 0– That should get you close. Do the math for each character and write down the numbers somewhere.
Let us know if you have any questions.
Matt Rhodes
Zaxwerks, Inc.
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