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Calculating Change In Position(Delta) Using Motion Tracking Data
Hi there, I am fairly new to using expressions in After Effects and I am in need of some guidance. I am working on a project where I am using the X,Y position of a Null Object (Generated via a motion track) to calculate the velocity of the tracked object.
This the expression I am using to pull the X,Y position and find it’s change in location every frame:
((effect("X")("Slider")-effect("X")("Slider").valueAtTime(time-(1/3)))^2)+((effect("X")("Slider")-effect("X")("Slider").valueAtTime(time-(1/3)))^2)
I am storing the X,Y position as two slider values, as you can see from the expression.My problem is that even though the X and Y values are changing as they should be, the expression output is remaining at 4.00 and for a specific frame it will jump up to 6.00 and then go back to 4.00. I am baffled as to why, I assume that I am doing something wrong with using the valueAtTime…
Here is a screenshot of all of my sliders and their values: https://i.imgur.com/soLIO.png
And here are all of my expressions: https://i.imgur.com/bUPTv.pngThank you in advance for your help!
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2 Replies
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Well, one thing that jumps out at me is that you’re using the ^ operator, which doesn’t do what you want it to. The ^ is a bitwise XOR operation:
https://en.wikipedia.org/wiki/Bitwise_operation#XOR
What you want to do is probably closer to:
x = effect(“X”)(“Slider”) – effect(“X”)(“Slider”).valueAtTime(time – framesToTime(1));
Math.pow(x, 2);Also I’m not sure that all of your expressions are necessary for your stated goal. If you want the rate of change of the null’s position have you tried:
thisComp.layer(“End of Shot”).transform.position.speed;
The result is in pixels/second. No need to get Pythagoras involved 🙂
Darby Edelen
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Thank you so much, the XOR operation explains the strange results.
[Darby Edelen] “Also I’m not sure that all of your expressions are necessary for your stated goal. If you want the rate of change of the null’s position have you tried:
thisComp.layer(“End of Shot”).transform.position.speed;
The result is in pixels/second…”
No I did not know about this but I will defiantly use it instead! 🙂
Thanks again for the help, I really appreciate it!
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