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1
How would I incorporate elapsed time into linear interpolation?
I'm updating a value 60 percent closer to it's target value every frame using the formula a+(b-a)*alpha. This works, but if someone has low frames it'll take longer to reach it's target.
local target = 4
local value = 0
local last = tick()
while rs:wait() do
local time = tick()
local elaps = time - last
value = value + (target - value) * .6
end
How can I modify this formula to use the elapsed time?
0
You could probably just multiply by elaps*30 nicemike40 486 — 8y
1 answer
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2
Oh that's simple, but completely silly.
The issue is that in this equation, you're missing a fundamental scale of when the tween is supposed to finish. Are we supposed to go there in 2 seconds? 2 minutes? It's all about that end goal.
local target = 4
local value = 0
local last = tick()
while rs:wait() do
local time = tick()
local elaps = time - last
last = time
value = value + (target - value) * (1-0.6^elaps)
end
0
I probably should've included a note on this; I know it will never finish. Target is constantly changing, I'm making a smooth transition between value changes YellowoTide 1992 — 8y
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Then the time is irrelevant. If you have no target, your time is completely irrelevant. PlsNoDiscrimination 0 — 8y
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I have a target, it is dynamic. Currently I'm moving value 60% closer to target every frame. If someone has 30 fps, their value will be far behind someone's who has 60 fps (Assuming they start at the same value and have the same target). YellowoTide 1992 — 8y
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Right that's what you want. You'll need the delta, which is magic stuff might I add. Check my revised code. PlsNoDiscrimination 0 — 8y