Taking The Average Of A Dihedral Angle

Question

[dimkal]

I have a time series a dihedral angle measured from a Molecular dynamics simulation. This dihedral angle fluctuates around 0 degrees. How do i properly take an average of something that's either +1 and +359 degrees, and take the periodicity of the system in to account to get zero???

I guess i have to use polar coordinates, but entirely sure on how to do this.

Answer 1

[Lars Juhl Jensen]

I do not know of a way to do this with a single formula, but you can solve it as an optimization problem.

Let us call the unknown average x and the measured angles x_1, x_2, x_3, etc. For a given value of x, you can define a sum of squared errors, which is easy enough to calculate in polar coordinates:

error = sum_i (diff(x, x_i)**2), where diff(x, x_i) = min(abs(x-x_i), 360-abs(x-x_i))

You now simply find the value of x that minimizes this error function using whatever numeric optimization algorithm you like.

[dimkal]

yes that seems like it would work, and this is what i'm sort of doing at the moment. However I'm looking to see if there's a much simpler approach.

Answer 2

[Ahill]

Editing this answer, I'll suggest a solution to the optimization problem posed by Lars. The "average" angle phi that minimizes the cartesian distance to the observed angle unit vectors on the unit circle would be:

phi = atan(sum_i(sin(theta_i))/sum_i(cos(theta_i)))    [if sum_i(cos(theta_i))>=0]
      atan(sum_i(sin(theta_i))/sum_i(cos(theta_i)))+pi [otherwise]

where theta_i are the observed angles (in radians).

In R:

mean.angle <- function(theta.i) {
  sum.sin <- sum(sin(theta.i))
  sum.cos <- sum(cos(theta.i))
  mean.angle <- atan(sum.sin/sum.cos)
  phi <- ifelse(sum.cos>=0, mean.angle, mean.angle+pi)
}

[Lars Juhl Jensen]

I don't think that will work. If you tage the average of 1 and 359 degrees you will get 180, but the real answer would be 0.

[dimkal]

Oops i made a mistake in my question, it should of been +359 and not -359. in which case your algorithm would average it as ~180, but it should be zero.

[dimkal]

but the correct answer should be zero.

[Ahill]

I see I've misunderstood the question, agree original answer will not work. See edit to answer above.

Answer 3

[Misplacedisgrace]

If you convert to the interval [0,2pi], you should get a mean of pi. Mean-centering that data by subtracting pi should give you data on the interval [-pi,pi], where the mean of a uniform distribution is 0. You could also try that with +/-180, if the conversion is too time consuming. The difference in sign corresponds to mechanical interpretations of opposite forces. I would also try to convert to flory convention, depending on the polymer you're working with. Hope I didn't misunderstand your question, I'm interested a more thorough explanation.

[Lars Juhl Jensen]

Sorry, but I think you have misunderstood the problem. Suppose I start out with two angles of 1 and 359 degrees. After your transformations, the average becomes 0, again corresponding to 180 degrees whereas the right answer is 0 degrees.

Answer 4

[Kris]

What if you try the following transformation of your angles (in degrees):

[0,360) ---> (-180, 180]

newangle = oldangle, if oldangle is between 0 and 180 degrees

newangle = oldangle - 360, if oldangle is between 180 and 360 degrees

(In other words, if the point of the unit circle lies in quadrants III and IV (and only then), you choose a different angle to represent this point, an angle that differs from your angle by 360 degrees and is negative)

With this transformation, angle +1 stays angle +1, and angle +359 becomes angle -1, so the two angles average to 0. The uniform distribution on [0,360) gets mapped to the uniform distribution on (-180,180] and the new average is 0. Does that work?