I am trying to understand how the math unfolds through this code, easingDoc
function easeInOutQuad (t, b, c, d) {
if ((t/=d/2) < 1) return c/2*t*t + b;
return -c/2 * ((--t)*(t-2) - 1) + b;
}
console.log(easeInOutQuad(0.3,50,150,1));
console.log(easeInOutQuad(0.95,50,150,1));I am trying to decipher the equation for both the conditions. While I understand, the condition for if ((t/=d/2) from this post describes what t/=d means and how it affects the calculation.
For example, (t/=d/2)<1 the code can be rewritten as following and it would produce the identical result. Such as,
function easeInOutQuad (t, b, c, d) {
if ((t/=d/2) < 1) return c/2*t*t + b;
return -c/2 * ((--t)*(t-2) - 1) + b;
}
function alternate (t,b,c,d){
if (2*t/d < 1) return c/2*(2*t/d)*(2*t/d)+b;
return 999999;};
console.log(easeInOutQuad(0.3,50,150,1));
console.log(alternate(0.3,50,150,1)); //(150/2)*(2*0.3/1)*(2*0.3/1)+50
console.log(easeInOutQuad(0.45,50,150,1));
console.log(alternate(0.45,50,150,1));//(150/2)*(2*0.45/1)*(2*0.45/1)+50I was wondering how can I express return -c/2 * ((--t)*(t-2) - 1) + b explicitly in equation like console.log(alternate(0.3,50,150,1)); //(150/2)*(2*0.3/1)*(2*0.3/1)+50 so that I can undersatnd what the math is for that part.
Probably the following 2 expressions could look tricky, here are the alternatives:
c = -1 * c; // -c
t = t - 1; // --t
And an example for -c/2 * ((--t)*(t-2) - 1) + b;
// -c/2 * ((--t)*(t-2) - 1) + b;
function alternate(t, b, c, d) {
c = -1 * c; // -c
t = t - 1; // --t
return (c / 2) * (t * (t - 2) - 1) + b;
};
console.log(alternate(0.3, 50, 150, 1));
console.log(((-1 * 150) / 2) * ((0.3 - 1) * ((0.3 - 1) - 2) - 1) + 50);