Is there a way to always start at 0 using math.sin() Lua

Question

Edit: This question is about Roblox Lua.

I'm using math.sin(tick()) to get a variable number and would like for it to always start at 0. Is this possible using math.sin? Is there something else I can use other than tick() to make this work?

Example:

for i = 1, 10 do
    local a = math.sin(tick())+1
    print(a)
    wait()
end
wait(1)
for i = 1, 10 do
    local a = math.sin(tick())+1
    print(a)
    wait()
end

My goal is to have this number start at 0 every time and then increase from there. So, it would start at 0 then increase to 2 and then decrease back to zero and continue modulating between 0 and 2 for as long as I continue calling it. Using the example above the number starts at any arbitrary number between 0 and 2.

@Nicol It's a ROBLOX function, it returns the number of seconds since January 1st 1970 (I think). Has some decimal places as well. — warspyking, Jan 26 '16 at 00:23
If you save the value of `tick()` on the first iteration, you can pass `tick() - start_time` to `math.sin()`. That would start with `1.0` rather than `0.0`, but it should give you a starting point for a solution. — Keith Thompson, Jan 26 '16 at 02:13
@KeithThompson thx. I think I can make your suggestion work also. That's really all I needed is a constant starting point. And I think that may be more efficient than the solution I came up with. — JJJ, Jan 26 '16 at 02:35
@KeithThompson - `local a = 1 - math.cos(tick()-start_time)` starts with zero ;-) — Egor Skriptunoff, Jan 26 '16 at 20:24

JJJ · Answer 1 · 2016-01-26T03:04:33.820

2

I took a different approach and came up with this. It does exactly what I wanted to do with math.sin(tick()). If anyone knows other ways to accomplish this I would like to know.

    local n = 0
    local m = 0
    local Debounce = true

    local function SmoothStep(num)
        return num * num * (3 - 2 * num)
    end

    while Debounce do
        for i = 1, 100 do 
            wait()
            m = m+.01
            n = SmoothStep(m)
            print(n)
    if not Debounce then break end
        end

        for i = 1, 100 do
            wait() 
            m = m+.01
            n = SmoothStep(m)
            print(n)
    if not Debounce then break end
        end
    end

edited Jan 26 '16 at 03:04

answered Jan 26 '16 at 02:05

JJJ

83
5

1

It's not *exactly* the same. If you graph the printed values, you'll see a triangle wave rather than a sine wave. If that suits your purposes, that's fine. – Keith Thompson Jan 26 '16 at 02:12
1

I see. This does suit my purpose, but yeah, it's not exactly the same. I'm editing it and adding a smoothstep function to make it closer to a true sine wave. – JJJ Jan 26 '16 at 02:24

score 1 · Accepted Answer · answered Jan 27 '16 at 08:14

1

To non-Roblox users: tick() returns the local UNIX time. wait(t) yields the current thread for t seconds, the smallest possible interval being roughly 1/30th of a second.

Given that math.sin(0) equals 0, what you have to do is subtract the tick() inside the loop with the time the loop began at. This should make the expression inside math.sin start at roughly 0 at the beginning of the loop.

local loopstart = tick()
for i = 1, 10 do
    local a = math.sin(tick() - loopstart)+1
    print(a)
    wait()
end

answered Jan 27 '16 at 08:14

xaxa

26
3

Thank you, xaxa. This is a working solution using math.sin(tick()). @KeithThompson made the same suggestion and it works. It starts at 1 instead of 0, but the most important part is that it starts at the same number every time. Perfect. I'll except this as the answer. – JJJ Jan 27 '16 at 21:42

Is there a way to always start at 0 using math.sin() Lua

2 Answers2