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1
Tweening Color3 values in sequence?
So I'm trying to create a smooth transition between Color3 values, in a rainbow-like effect. I've come up this conclusion:
01 | local function TweenColor(gui,new,prop,time) |
02 | local v = gui[prop] |
03 | time = time * 30 |
04 | local saved = { |
05 | r=new.r-v.r; |
06 | g=new.g-v.g; |
07 | b=new.b-v.b; |
08 | } |
09 | for i = 1,time do |
10 | gui[prop] = Color3.new( |
11 | gui[prop].r+saved.r/time, |
12 | gui[prop].g+saved.g/time, |
13 | gui[prop].b+saved.b/time |
14 | ) |
15 | wait() |
16 | end |
17 | end |
This will tween color A, to color B very smoothly. However, I still have yet to figure out a compact way to make a rainbow effect and have the function tween through all it's colors.
If anyone knows a good way to accomplish this, I'd very much appreciate it. Thanks.
0
well a rainbow starts at red and goes to blue so start increasing red then increasing green and decreasing red and then the same with blue maybe ProfessorSev 220 — 9y
1 answer
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4
ProfessorSev's comment isn't far off. For the effect you're looking for, you could cycle the hue value and convert to RGB, as follows:
01 | local hue = 0; -- this will cycle from 0-359 |
02 | local SAT = 1; -- satutation; constant |
03 | local LUM = 1; -- luminance; constant |
04 |
05 | local SPEED = 1; -- hue degree/frame |
06 |
07 | function hsvToRgb(h, s, v) -- from https://github.com/EmmanuelOga/columns/blob/master/utils/color.lua - essentially ramps different values depending on which sixth 'h' is in |
08 | local r, g, b |
09 |
10 | local i = math.floor(h * 6); |
11 | local f = h * 6 - i; |
12 | local p = v * (1 - s); |
13 | local q = v * (1 - f * s); |
14 | local t = v * (1 - (1 - f) * s); |
15 |
View all 35 lines...
That could be severely simplified, but the main loop is basically the logic.