javascript - Ease time between firing specific number of timeouts in a specific period of time -

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it's kind of math problem. want fire specific number of settimeout (the number based on array length) in specific period of time (say, 5 seconds).

the first settimeout should start @ 0 sec. , last @ 5 sec.. timeouts between should start ease-in effect, each timeout starts faster.

there's example ilustrates want achieve exactly.
i'm struggling around line:

next += timeperiod/3.52/(i+1);

which works perfect in demo example (for timeperiod), doesn't work different letters.length have used static number 3.52.

how calculate next?

var letters = [     'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t'  ];    var div        = $('#container');  var timeperiod = 5000; // 5 seconds;  var perletter  = timeperiod/(letters.length-1); // gives equal time between letters  var next       = 0;    for(var i=0; i<letters.length; i++){    settimeout(function(letter){      //div.append('<span class="letter">' + letter + '</span>');      // used "|" instead of letter, better redability:      div.append('<span class="letter">|</span>');    }, next, letters[i]);        // can't find logic here:     next += timeperiod/3.52/(i+1);      };      ///////////////// demo: ///////////////    var sec = timeperiod/1000;  var secondsinterval = setinterval(seconds, 1000);    var demointerval = setinterval(function(){    sec >= 0 || clearinterval(demointerval);    div.append('\'');  }, 30);    function seconds(){    sec || clearinterval(secondsinterval);    $('#sec').text(sec-- || 'done');  }  seconds();
.letter{    color : red;  }
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>  <span id=container></span>  <span id=sec class=letter></span>

var steps = letters.length; var target = timeperiod;  function easeoutquad(t, b, c, d) {   t /= d;   return -c * t*(t-2) + b; };  var arrayoftimeouts = new array(steps); var n; var prev = 0; for(var = 1; <= steps; i++){   n = easeoutquad(i, 0.0, target, steps);   arrayoftimeouts[i-1] = n-prev;      prev = n; }

this 1 should work input value.

fiddle

note graph appears fast believe discrepancy product of timing imperfections, sum of array equals timeperiod exactly.

more on easing equations


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