Find percentage between 2 numbers that are between 0 and 1

Question

Say I have a timeline beginning at 0 and ending at 1. I have several points along the timeline and I want to lerp between them over time. e.g. pt1 to pt2, pt2 to pt3, pt3 to pt1. Once the value reachs 1, loop back to 0. But, the percentage from 1 back to 0 must be maintained.

So think of variable t as the current time on a clock between 0 and 1 and a and b are points on the timeline.

e.g.

a = 0.33
b = 0.5
t = 0.42

Finding what percentage t is between a and b?

"given number x in range between a and b, percentage = x - a  /  b - a."
0.42 is 53% between 0.33 and 0.5

Once t == b, the lerp between the two numbers should be 100%. Once this happens, a becomes 0.5, b becomes 0.33, and t continues as it was.

a = 0.5
b = 0.33
t = 0.72

The formula would not work the same because once it loops back passed 1 and becomes something like 0.3, it breaks. The distance between a and b is 0.83((1 - 0.5) + 0.33), so 0.3 should be somewhere like 80-90%. How would I find this?

Answer 1

$$ \frac{t-a}{(1+b)-a}$$ The $1+b$ allows the wrap around.

Answer 2

Your question is confusingly worded, but it seems like what you're saying is that there's some total time T, and then you have a variable t that keeps track of how much time since the beginning of the current loop. So in the first loop, the two variables are the same, but in the second loop, T is ahead of t by 1. If you've looped once, but haven't gotten to a yet your interval is going from b to a, but it's going from b from the first loop to a from the second loop. So the value of T at b is just b, but the value of T at a is a+1, and the value of T at t is t+1. The formula you originally had can be written as:

(current- beginning)/(end-beginning)

Now your current = t+1, beginning = b, end = a+1. So the formula comes out as:

(t+1-b)/(a+1-b)

If you plug in t = 3, a = .33, b = .5, you get

(.3+1-.5)/(.33+1-.5) = .8/.83 ~ .9639

In other word, your remaining time of .03 is ~3.61% of .83.