diff --git a/mrdna/model/spring_from_lp.py b/mrdna/model/spring_from_lp.py
index 5702c963ed87c9e1f72f95d2873e11410c6093c1..7cc237525423bb9cc1a6ce0b74c26f6cb5d08a8c 100644
--- a/mrdna/model/spring_from_lp.py
+++ b/mrdna/model/spring_from_lp.py
@@ -14,3 +14,44 @@ assert( (np.diff(_integral) <= 0).sum() == 0 )
def k_angle(sep,Lp):
val = np.exp(-sep/Lp)
return np.interp(val,_integral,_k) * 0.00030461742 # convert to degree^2
+
+""" Twist! """
+from scipy.special import erf
+_k_twist = np.logspace(-8,3,10000) # in units of kT
+with np.errstate(divide='ignore',invalid='ignore'):
+ _integral_twist = np.real(erf( (4*np.pi*_k_twist + 1j)/(2*np.sqrt(_k_twist)) )) * np.exp(-1/(4*_k_twist)) / erf(2*np.sqrt(_k_twist)*np.pi)
+def k_twist(sep,Lp,temperature=295):
+ kT = temperature * 0.0019872065 # kcal/mol
+ val = np.exp(-sep/Lp)
+ return np.interp(val,_integral_twist,_k_twist) * 2 * kT * 0.00030461742 # convert to degree^2
+
+def _TEST_get_spring_constant():
+
+ def __get_twist_spring_constant(sep, temperature=295):
+ import scipy.optimize as opt
+
+ """ sep in nt """
+ kT = temperature * 0.0019872065 # kcal/mol
+ twist_persistence_length = 90 # set semi-arbitrarily as there is a large spread in literature
+ ## <cos(q)> = exp(-s/Lp) = integrate( cos[x] exp(-A x^2), {x, 0, pi} ) / integrate( exp(-A x^2), {x, 0, pi} )
+ ## Assume A is small
+ ## int[B_] := Normal[Integrate[ Series[Cos[x] Exp[-B x^2], {B, 0, 1}], {x, 0, \[Pi]}]/
+ ## Integrate[Series[Exp[-B x^2], {B, 0, 1}], {x, 0, \[Pi]}]]
+
+ ## Actually, without assumptions I get fitFun below
+ ## From http://www.annualreviews.org/doi/pdf/10.1146/annurev.bb.17.060188.001405
+ ## units "3e-19 erg cm/ 295 k K" "nm" =~ 73
+ Lp = twist_persistence_length/0.34
+
+ fitFun = lambda x: np.real(erf( (4*np.pi*x + 1j)/(2*np.sqrt(x)) )) * np.exp(-1/(4*x)) / erf(2*np.sqrt(x)*np.pi) - np.exp(-sep/Lp)
+ k = opt.leastsq( fitFun, x0=np.exp(-sep/Lp) )
+ return k[0][0] * 2*kT*0.00030461742
+
+ seps = np.linspace(0.5,25,1000)
+ vals1 = np.array([__get_twist_spring_constant(s) for s in seps])
+ vals2 = k_twist(seps, 90/0.34 )
+
+ abs_diff = np.abs(vals1-vals2)
+ max_diff = np.max(abs_diff)
+ idx = abs_diff==max_diff
+ print("Max diff ({}) at {}, where sep = {}, vals1 = {}, vals2 = {}".format(max_diff, np.where(idx)[0], seps[idx], vals1[idx], vals2[idx]))
diff --git a/mrdna/segmentmodel.py b/mrdna/segmentmodel.py
index 5ccbe9311a21dcf7070d9c1fcda25ddd8e31a764..a2e410925a03be443a9b5a68b05fa5fcba1304de 100644
--- a/mrdna/segmentmodel.py
+++ b/mrdna/segmentmodel.py
@@ -16,6 +16,7 @@ from .model.CanonicalNucleotideAtoms import canonicalNtFwd, canonicalNtRev, seqC
from .model.CanonicalNucleotideAtoms import enmTemplateHC, enmTemplateSQ, enmCorrectionsHC
from .model.spring_from_lp import k_angle as angle_spring_from_lp
+from .model.spring_from_lp import k_twist as twist_spring_from_lp
import csv
@@ -1697,21 +1698,11 @@ class SegmentModel(ArbdModel):
def _get_twist_spring_constant(self, sep):
""" sep in nt """
- kT = self.temperature * 0.0019872065 # kcal/mol
twist_persistence_length = 90 # set semi-arbitrarily as there is a large spread in literature
- ## <cos(q)> = exp(-s/Lp) = integrate( cos[x] exp(-A x^2), {x, 0, pi} ) / integrate( exp(-A x^2), {x, 0, pi} )
- ## Assume A is small
- ## int[B_] := Normal[Integrate[ Series[Cos[x] Exp[-B x^2], {B, 0, 1}], {x, 0, \[Pi]}]/
- ## Integrate[Series[Exp[-B x^2], {B, 0, 1}], {x, 0, \[Pi]}]]
-
- ## Actually, without assumptions I get fitFun below
- ## From http://www.annualreviews.org/doi/pdf/10.1146/annurev.bb.17.060188.001405
## units "3e-19 erg cm/ 295 k K" "nm" =~ 73
Lp = twist_persistence_length/0.34
- fitFun = lambda x: np.real(erf( (4*np.pi*x + 1j)/(2*np.sqrt(x)) )) * np.exp(-1/(4*x)) / erf(2*np.sqrt(x)*np.pi) - np.exp(-sep/Lp)
- k = opt.leastsq( fitFun, x0=np.exp(-sep/Lp) )
- return k[0][0] * 2*kT*0.00030461742
+ return twist_spring_from_lp(sep, Lp, self.temperature)
def extend(self, other, copy=True, include_strands=False):
assert( isinstance(other, SegmentModel) )