Help with Smoothing Number Cycle

Help and Feedback Scripting Support
#1

Hello! I have a loop in a script that increases/decreases a number value between a minimum/maximum amount, and the value does this without any issue. I am trying to find a way to smooth the amount of change towards the minimum/maximum ends, does anyone know how I can do this?

This is what I currently have without any smoothing:


local val = 1
local min = 0
local max = 100
local goingUp = true

while true do
      wait()
      if val < max and val > min then
            if goingUp == true then 
                  val = val + 1
            else
                  val = val - 1
            end
      else
            if goingUp == true then
                  goingUp = false
                  val = val - 1
            else
                  goingUp = true
                  val = val + 1
            end
      end
end

I would appreciate any help possible.

2 Likes
#2

if by “smooth” you mean make the process longer and not instant, in your wait you can do wait(0.1)

1 Like
#3

This can all be achieved with a for loop

for number = start, end, increment do

end

for variable = 1, 100 do -- Default is 1
     print(number) -- will print 1 - 100
end

for i = 50, 25, -0.25 do
     --Count from 50 to 25 in increments of -0.25
end
1 Like
#4

Thank you so much for trying to help, I should have further defined what I meant by “smoothing” though. I am trying to have the number slow down towards the max and minimum but remain relatively fast in the middle.

#5

Is your value going up and down constantly? Or is it just going from 0 to 100…

If so, you can just use tweening!

1 Like
#6

This is far more complicated, out of curiosity what are you using this for? Roblox has functionality built into Tweens that allow for nice phasing of parts using cubic interpolation.

1 Like
#7 
local val = 1
local min = 0
local max = 100
local goingUp = true
local waitTime = 0.1

while true do
      wait(waitTime)
      if val >75 or val<25 then
         waitTime = 0.2
      else
         waitTime = 0.1
      end

      if val < max and val > min then
            if goingUp == true then 
                  val = val + 1
            else
                  val = val - 1
            end
      else
            if goingUp == true then
                  goingUp = false
                  val = val - 1
            else
                  goingUp = true
                  val = val + 1
            end
      end
end

You can add more if statements to make it even smoother, this is just to give you an idea

3 Likes
#8

This solution won’t look the best, as it just changes the increment time, not how much is actually incremented. Yes it achieves the goal, but it will get stutter-y near the beginning and end.

2 Likes
#9

Agreed. Other than tweening this is the only thing that came to mind right now

2 Likes
#10

imma genius. This code will make it hella smooth. Again just giving an idea, you ll need to make some adjustments

3 Likes
#11

It could be possible to generate a linear interpolation function to overlay onto something like a cubic curve…

function PhaseValue(startv, endv, increment, number)
     for value = startv, endv, increment do
          local t = -math.sqrt(4-math.pow(value, 2))+2
          number = (1 - t) * startv + t * endv
          task.wait()
     end
end

local number = 0
PhaseValue(0, 100, 1, number)

EDIT: Realized I was using the end keyword!

1 Like
#12

Update on this solution, I did some research on using BĂ©zier curves in Lua, and came up with the following solution thanks to the following article:

function cubicBezier(t, p0, p1, p2, p3)
	return (1 - t)^3*p0 + 3*(1 - t)^2*t*p1 + 3*(1 - t)*t^2*p2 + t^3*p3
end

function interpolateValue(startv, endv, steps, value)
	for i = 0, 1, 1/steps do
		value = cubicBezier(
			i,
			Vector2.new(0, startv),
			Vector2.new(0.5, startv),
			Vector2.new(0.5, endv),
			Vector2.new(1, endv)
		).Y
		task.wait()
	end
	value = endv
end

local number = 0
interpolateValue(0, 100, 100, number)
1 Like
#13

You can play around with those two 0.5 values to increase the smoothness of the curve, so long as they add to 1 (ie change to 0, 0.25, 0.75, 1 in the Vector 2s for a smoother curve)

1 Like
#14

I’m using it to change the Y value constantly on a Vector3Value, do tweens work with that?

#15

Thank you so much for the help, I will try to use this and see what it does.

Edit: Your method seems to work well with what I am trying to do, thank you again.

1 Like