Member Site › Forums › PyRosetta › PyRosetta – General › Quadratic Function for Distance Constraints?
- This topic has 4 replies, 2 voices, and was last updated 11 years, 9 months ago by
Anonymous.
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January 27, 2014 at 2:21 pm #1814
Anonymous
I’m trying to develop a modified energy minimizer where, in addition to enforcing typical energetic interactions, I want to constrain distances between particular atoms. So I think I need to have a function associated with each constraint and I would like to use a parabolic/quadratic function to describe the behavior of the constraints. Is there a quadratic function in the constraints package of PyRosetta? I was just using the tab-autocomplete feature in iPython to look for something like this but I can’t seem to find what I am looking for. A Gaussian function might work but I would rather use a quadratic function instead if it exists. Thank you for your consideration.
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January 27, 2014 at 3:36 pm #9718
Anonymous
The harmonic constraint function has an x^2 dependence on the quantity of interest. (e.g energy integration of Hooke’s law)
If you want a more complex quadratic, with linear terms and offsets, you should be able to add those on with other constraint terms, although typical use case only requires a zero-centered quadratic.
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January 28, 2014 at 1:49 pm #9723
Anonymous
Okay, yeah that will work perfectly! It’s been a while since I took a physics class. Thanks for your help. Actually I have one more quick question. What is the name of the constraint that one would use to constrain distances? I thought it would be called “DistanceConstraint” but I only found “DistancePairConstraint.” Do I need AtomPairConstraint?
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January 28, 2014 at 3:27 pm #9725
Anonymous
Right – For AtomPairConstraint, the number that gets passed to the constraint function is the Cartesian distance between the pair of atoms you specify.
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January 28, 2014 at 4:09 pm #9727
Anonymous
Okay, that’s actually simpler than I thought it would be because now I don’t actually have to calculate any distances myself. Thanks again for your help!
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