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Adjusting Dan’s Squash and Stretch Expression
I am trying to make the expression only effect the X axis instead of the Y. I can’t quite get the expression right. I swapped out the X and Y in the expression, but it did not work. How can I get this to only squash and stretch the X axis?
maxDev = 10; // max deviation in pixels
spd = 20; //speed of oscillation
decay = 1.0; //how fast it slows downn = 0;
if (marker.numKeys > 0){
n = marker.nearestKey(time).index;
if (marker.key(n).time > time){
n--;
}
}if (n > 0){
t = time - marker.key(n).time;
y = scale[1] + maxDev*Math.sin(spd*t)/Math.exp(decay*t);
[value[0],y]
}else{
value
} -
4 Replies
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Try it this way:
maxDev = 10; // max deviation in pixels
spd = 20; //speed of oscillation
decay = 1.0; //how fast it slows downn = 0;
if (marker.numKeys > 0){
n = marker.nearestKey(time).index;
if (marker.key(n).time > time){
n–;
}
}if (n > 0){
t = time – marker.key(n).time;
x = scale[0] + maxDev*Math.sin(spd*t)/Math.exp(decay*t);
[x,value[1]]
}else{
value
}Dan
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Worked like a charm! Thanks Dan!
So, I tried to just interchange the X and Y values, which did not work, and I noticed you must also change the ‘value[1]’ portion.
Why does x=scale[0] buy y=scale[1]?
I would figure whichever active parameter would equal 1.
Still learning how these expressions work. -
Multi-dimensional properties like scale are represented by JavaScript arrays. So, x is always represented by the value at index 0, y is at index 1, (and z would be at index 2).
Dan
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