• Adobe After Effects Expressions

  How do I avoid a small jump in the first line when using a text animator?

  Posted by David Cabestany on February 11, 2020 at 11:00 pm

  I have a text layer in which the text on the 1st line writes on automatically each time a marker is passed via an expression, the text on the 2nd appears as a single line by moving from below into its final position. That is also controlled by an expression and the animation repeats itself each time a marker is passed.

  My problem is that depending on the length of the first line a small jump is noticeable in some of the lines, you can watch an example of this here:

  https://www.dropbox.com/s/nhfp7xgfp1y52o4/bumpSample.mp4?dl=0

  The 3rd text, at :04 seconds moves slightly upwards.

  The problem there is a bunch of small texts in this layer and the issue happens on some of them not all.

  So my question is is there something I can add to the expression that will compensate and prevent that slight jump?

  Below are the expressions I’m using, I think the problem lies in the offset, which because I only have two lines it stands around 50%

  Thanks in advance for your help,
  D.

  For the text source:
txt = value;
n = 0;
if (thisComp.marker.numKeys > 0){n = thisComp.marker.nearestKey(time).index;
if (thisComp.marker.key(n).time > time) n--;
if (n > 0) txt = thisComp.marker.key(n).comment;
}
txt;

  1st animator, write on the 1st line of text: spd = 65; // characters per second
  n = 0;
  if (thisComp.marker.numKeys > 0){
  n = thisComp.marker.nearestKey(time).index;
  if (thisComp.marker.key(n).time > time) n--;
  }
  if (n > 0){
  t = time - thisComp.marker.key(n).time;
  dur = text.sourceText.length/spd;
  linear(t,0,dur,0,100);
  }else
  0

  2nd animator, 2nd line moving up:

  spd = 65; // characters per second
n = 0;
if (thisComp.marker.numKeys > 0){
n = thisComp.marker.nearestKey(time).index;
if (thisComp.marker.key(n).time > time) n--;
}
if (n > 0){
t = time - thisComp.marker.key(n).time;
dur = text.sourceText.length/spd;
linear(t,0,dur,100,0);
}else
0


  David Cabestany replied 6 years, 3 months ago 1 Member · 1 Reply
• 1 Reply

• 

  David Cabestany

  February 12, 2020 at 2:54 pm

  No one?

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