• Adobe After Effects Expressions

  Right angled Labels in expressions

  Posted by Adam Zygadlo on October 15, 2019 at 2:06 pm

  Hey there,

  Is it possible to use expressions to make right angle graphics? I will explain in detail below.

  Recently, I have decided to update the information labels that I use for my client projects and animations.

  Here’s the layout to the label:
  

  In previous builds, I built a basic rig with three points or nulls that I use to layout the label: A) Starting Point, B) Elbow for Arm, and C) Key information of the text.
  Then I used the beam effect to connect between the three nulls to form the arm for the label. Using null A and C, I can position them on stage to highlight key information on the screen. Once I am happy with the position of A and C, then I make sure that B is right-angled at 45-degrees. I used the pick whip to connect the height of null B to the height of null A and then manually adjust the weight of null B to align a 45-degree angle between null A and C.

  Is there a way to use expressions to produce a right angle on Null B where it calculates the angle different the other nulls, without manually adjusting it myself?

  I really appreciate your time and thank you in advance.

  Cheers,
  Addz

  Adam Jozef Zygadlo
  Motion Graphic Designer

  07708 124 915
  http://www.jarbrain.com
  ‘Filled with brainy goodness.’

  Adam Zygadlo replied 6 years, 7 months ago 2 Members · 2 Replies
• 2 Replies

• 

  Dan Ebberts

  October 15, 2019 at 4:25 pm

  Something like this maybe:

  a = thisComp.layer(“A”).position;
  c = thisComp.layer(“C”).position;
  d = a – c;
  x = Math.abs(d[1])*Math.tan(Math.PI/4);
  [a[0]-x,c[1]]

  Dan
• 

  Adam Zygadlo

  October 16, 2019 at 9:32 am

  Hey Dan,

  Thank you very much, that did the trick. I had to switch A and B in the last line of code, around in order to bend the elbow in the correct way to the label.

  a = thisComp.layer("Joint A").position;
c = thisComp.layer("Joint C (Point)").position;
d = a - c;
x = Math.abs(d[1])*Math.tan(Math.PI/4);
[c[0]-x,a[1]]


  Apart from that, it’s perfect and it will help save a small amount of time when producing 100’s of labels. Thank you Dan, you are a genius with code.

  All the best until next time,
  Cheers,
  Addz

  Adam Jozef Zygadlo
  Motion Graphic Designer

  07708 124 915
  http://www.jarbrain.com
  ‘Filled with brainy goodness.’

Reply to this Discussion! Login or Sign Up