Modeling Contractile Mechanisms:
Huxley 1957 Model " Courtesy"
With Slides
Stuart Campbell, U Kentucky and "
J. Jeremy Rice"
IBM T.J. Watson Research Center, P.O. Box 218,
Yorktown Heights, NY 10598 "
[email protected]"
914-945-3728"
"
"
Need to consider both how many
XBs are recruited and what is the
extension of attached XBs "
Thin filament - actin
Thick filament - myosin
* This requires a modeling approach that simultaneously
considers spatial and temporal aspects of XB cycling
Alternative theory proposed opposite
charges exist in thick and thin filament "
• Contraction from electrostatic attraction "
• Hard to reconcile with maximum velocity being
independent of sarcomere length "
1
Derivation of Huxley '57
Model"
Review key concepts from basic probability:
1.  Probability Density Functions
r
proton
Example:
electron
Consider an electron of a hydrogen atom. One could
draw an approximate probability density function (p.d.f.)
for its location in terms of distance from the nucleus = r.
Let h(r) be the function below that defines this p.d.f.
h(r)
r
r=0
The way to interpret a p.d.f. is that a probability can be
computed as an area under the curve. For example
the probability of the electron being closer than r0 is
computed as:
r0
Pr{r < r0 } = ∫ h(r )dr
0
h(r)
r=0
r0 r1
r
Likewise, the probability of the electron being between r0 and
r1
r1 is computed as:
Pr{r0 < r < r1} = ∫ h(r )dr
r0
2
By definition, the p.d.f. does not directly report the probability of
finding an electron at a given fixed distance, r0:
Pr{r = r0 } ≠ h(r )
In fact, the probability for any given distance, r0, is 0.
See that:
Pr{r = r0 } =
∫
r0
r0
h(r )dr = 0
More formally, we define the the p.d.f. as a limit:
⎡ u + Δu h(r )dr ⎤
⎡ Pr{u < r < u + Δu}⎤
⎢ ∫u
⎥
h(u ) = lim ⎢
=
lim
⎥ Δu →0 ⎢
Δu →0
⎥
Δu
Δu
⎣
⎦
⎣⎢
⎦⎥
A probability is always between 0 and 1 so if we
assume the electron is always found at some distance
from the nucleus then:
∞
Pr{0 < r < ∞} = ∫ h(r )dr = 1
0
General rule: The summation of p.d.f. over all possible values
must be = 1.
2. Conditional Probability
We define a conditional probability as probability of an
event A given that an event B has occurred. This is
written as:
Pr{A B} =
Pr{A & B}
Pr{B}
This relation may make more intuitive sense when
rearranged as:
Pr{A & B} = Pr{A B}Pr{B}
3
3. State Variables to Represent Probabilities of
Stochastic Events
When considering a random process like a channel
opening and closing, each channel looks like this:
f"
O"
C"
g"
For this channel, if we assume f = 1 s-1 and g = 2 s-1, then
steady-state probability of the channel being open is
computed as:
Pr{ channel = open} =
f
1
=
f +g 3
Of course, this does not mean the channel is 1/3 open
because only states "closed" and "open" exist.
However, if we were to average a large number of
channel responses together, we would get something
like this:
Run 1"
1"
0"
Sample many runs
}and average to get
Run 2
better estimate of
probability."
..
.
.
.
.
Run n
1"
1/3"
0"
time"
So instead of tracking a each channel, we can define a
state variable based system to capture the behavior of
the whole population. When implemented this way,
average behavior can calculated without averaging (and
associated noise and computational cost).
f"
O"
C"
g"
Value of state C = Pr{channel = closed} =
g
2
=
f +g 3
Value of state O = Pr{channel = open} =
f
1
=
f +g 3
4
Derivation of Huxley 57 Model
Assumptions:
1. Contractile machinery only
2. Plateau region of length-tension
3. Muscle fully activated
4. Constant velocity (parameter in model)
5. Crossbridges (XBs) always completes full cycle to
detach and uses 1 ATP in the process
6.  Single myosin near every A site and interaction
between this pair is independent of all other pairs of A
sites and myosins
Setup for Huxley '57 model"
equilibrium
myosin position"
X"
X"
Thick filament"
Thin filament"
A site"
A site"
We consider model on right to be equivalent to model on left.
Model is built around actin binding sites called A sites on thin
filament. Myosin heads from thick filament can bind to one
and only one nearby A site."
EM micrographs shows evidence
of crossbridges but no real detail"
5
Setup for Huxley '57 model"
X3 = 0"
X2 > 0"
X1 < 0"
equilibrium
myosin positions"
A site 1"
A site 2"
A site 3"
Forces are only considered in left-right directions parallel to thick
and thin filaments. When an A site is bound to myosin, force is
generated with respect to the distortion of myosin from its
equilibrium position. When A site is bound exactly at the
equilibrium position (i.e. X3), no force is generated. "
Setup for Huxley '57 model"
X3 = 0"
X2 > 0"
X1 < 0"
k
equilibrium
myosin positions"
A site 1"
k
k
A site 2"
A site 3"
When an A site is bound to myosin, force is generated with
respect to the distortion of myosin from its equilibrium position.
Linkage is assumed to be a simple spring so that T = kX. For
each A site with myosin bound, T = kX. Therefore, T1 < T3=0 <
T2. "
Setup for Huxley '57 model"
X1"
X2"
X4"
X3"
equilibrium
myosin positions"
Distance between A sites = l
A sites"
Model shows distance between A sites and equilibrium myosin
positions. A whole population of A sites is assumed to sample
equally all X values because A sites and equilibrium myosin
positions are unequally spaced. (p.d.f is constant)"
6
Position of myosin heads on thick
filament"
•  Pairs of heads emanate 180
degree apart in radial direction at
each step
axial direction
•  Radial direction of heads rotate
~60 degrees at next step in axial
direction (distance = ~14.3 nm)
•  a "pseudo-repeat" happens on
the 3th steps as heads will be
emanate in same radial direction
(distance = ~43 nm)
radial direction
Thin filament is a two-stranded helix of
actin monomers"
"Pseudo-repeat" 37
nm
"Pseudo-repeat" = 13 units
True repeat = 26 units
5.54 nm
2.77 nm
From http://www.kent.ac.uk/bio/geeves/Research/home.htm
Setup for Huxley '57 model"
X1"
X2"
X4"
X3"
equilibrium
myosin positions"
Distance between A sites = l (small L)"
A sites"
Model assumes distance l between A sites is large and
interactions are with only one nearby myosin. Hence, each
myosin can interact with only one A site at a time. Therefore,
the cases shown above (for X2 and X3) cannot happen."
7
Setup for Huxley '57 model"
X3 "
X2"
X1"
equilibrium
myosin positions"
Sliding in V > 0 direction
(if thick filament fixed) "
Sliding in V < 0 direction
(if thick filament fixed) "
"
and
The thick
thin filaments slide past each other at a constant
velocity V. We assume that the motion results from combined
action of many force generators acting across the whole muscle,
so the sliding velocity is not affected by the local attachment or
detachment events. Note: velocity is a parameter in the model."
Setup for Huxley '57 model"
X3 "
X2"
X1"
equilibrium
myosin positions"
Sliding in V > 0 direction
(if thick filament fixed) "
Sliding in V < 0 direction
(if thick filament fixed) "
" thin filaments slide past each other at a constant
As thick and
velocity V, the relative position of A sites compared to
equilibrium myosin positions changes. Therefore, when V>0,
X1, X2 and X3 all get smaller (less positive or more negative)
with time."
Attachment rates as function of X"
g2
4
3
x
2
f1 1
g1
f
0
h
g
XB attach only in this range"
XB can detach at any distortion"
8
Attachment rates as function of X"
X = 0"
X < 0"
X = h/2"
Attachment rate between A site and myosin
are a function of the relative distance
between the A site and equilibrium myosin
position. In the model, f(X) is the function
defined to control attachment. Function f(X)
increases linearly from X=0 to X=h."
4
3
2
f
1
x
g
h
0
f(X)"
Detachment rates as function of X"
X = 0"
X1 < 0"
X > h"
In the model, g(X) is the function defined to
control detachment as a function of the relative
distance between the A site and equilibrium
myosin position. In the positive range, g(X)
increases linearly from X=0. In the negative
range, g(X) is large so that the negative
distortion XBs ("draggers") detach quickly."
4
3
2
f
1
x
0
g
h
g(X)"
Let n(x,t) be a conditional probability describing the
likelihood that an XB is attached given that the A site is
at displacement x from the nearest XB equilibrium
position.
To be more rigorous:
⎡ ⎧
⎢ ⎪ A site has
n( x ,t ) = lim ⎢ Pr ⎨
Δx →0
XB attached
⎢⎣ ⎪
⎩
⎫⎤
A in range ⎪⎥
⎬
x to x + Δx ⎪⎥
⎭⎥⎦
Conditional probability vertical
bar means given that
9
Note that a more intuitive function describes when an A
site is attached and the A site is between x and x + Dx
n̂( x ,t ) = lim
Δx →0
⎡ ⎧⎛
1 ⎢ ⎪⎜ A site has
Pr ⎨⎜
Δx ⎢ ⎪⎜ XB attached
⎢⎣ ⎩⎝
⎞
⎛
⎞⎫⎤
⎟
⎜ A in range ⎟⎪⎥
⎟ & ⎜ x to x + Δx ⎟⎬⎥
⎟
⎜
⎟⎪⎥
⎠
⎝
⎠⎭⎦
We can use rule from basic probability
Pr{A & B} = Pr{A B}Pr{B}
Make a substitution for probability inside limit
⎡ ⎧
1 ⎢ ⎪ A site has
Pr ⎨
Δx →0 Δx ⎢
XB attached
⎢⎣ ⎪⎩
nˆ ( x, t ) = lim
⎫ ⎧
⎫⎤
A in range ⎪ ⎪ A in range ⎪⎥
⎬ Pr ⎨
⎬
x to x + Δx ⎪ ⎪ x to x + Δx ⎪⎥
⎭⎥⎦
⎭ ⎩
Substitute limit above with product of limits below
⎡ ⎧
⎢ ⎪ A site has
nˆ ( x, t ) = lim ⎢Pr ⎨
Δx →0
XB attached
⎢⎣ ⎪⎩
⎫⎤
⎡ ⎧
⎫⎤
A in range ⎪⎥
1 ⎢ ⎪ A in range ⎪⎥
Pr ⎨
⎬⎥ • Δlim
⎬⎥
⎢
x to x + Δx⎪ x→0 Δx
⎪ x to x + Δx⎪⎥
⎭⎦
⎭⎥⎦
⎣⎢ ⎩
ˆ ( x, t ) = n( x, t ) • hˆ( x, t )
n
Conditional
Probability
Probability Density
Function
ˆ ( x, t ) = n( x, t ) • hˆ( x, t )
n
ĥ is a probability density function describing the
positions of A sites with respect to equilibrium
XB positions
Model assumes ĥ is constant over all possible x
values between -l/2 and l/2 as shown below:
ĥ
-l/2
1/l
l/2
x
10
Continue with derivation
For steady-state response:
∂n dx ∂n dt
+
=0
∂x dt ∂t dt
Apply chain rule:
We know that:
dn(x, t )
=0
dt
dt
=1
dt
dx
=v
dt
Rearrange to get:
∂n
∂n
v+
=0
∂x
∂t
∂n ∂n
−v
=
∂x ∂t
∂n
= rate of attachment to A site ∂t
rate of detachment from A site
= f (x)[1 − n(x, t )]− g (x)n(x, t )
dn(x, t )
=0
dt
⇒ use n(x ) instead of n(x, t )
Consider steady state where
∂n(x )
= f (x )[1 − n(x )] − g (x )n(x )
∂t
Unattached A site fraction
Combine above results:
−v
−v
Attached A site fraction
∂n(x ) ∂n(x )
=
∂x
∂t
∂n(x )
= f (x )[1 − n(x )] − g (x )n(x )
∂x
Can find solution for specific
cases if we define f(x) and g(x)
with units of s-1
g2
4
x < 0 : f (x ) = 0 ; g (x ) = g 2
f1 x
gx
; g (x ) = 1
h
h
gx
x > h : f (x ) = 0 ; g (x ) = 1
h
3
2
0 < x ≤ h : f (x ) =
x
f1 1
g1
f
0
h
g
11
Apply constraint that if V ≠ 0 then solution is continuous at
x=0 and x=h. One can write:
n (0 − ) = n ( 0 + )
n( h − ) = n( h + )
Now solve for three region assuming V > 0:
x < 0 : f (x ) = 0 ; g (x ) = g 2
Region 1 -
∂n dn
dn(x, t )
=
in steady state where
=0
∂x dx
dt
−v
∫
dn
= − g2n
dx
dn
=
n
Integrate to get:
∫
ln n =
g2
dx
v
g2
dx + C0
v
n( x) = C1e
g2 x
v
where C1 ( = eCo ) is a constant T.B.D.
Region 2 -
0 < x ≤ h : f (x ) =
−v
f1 x
gx
; g (x ) = 1
h
h
dn f1
g
= x(1 − n) − 1 xn
dx h
h
− vh
dn
= f1 x(1 − n) − g1 xn
dx
= f1x − ( f1 + g1 ) xn
= x( f1 + g1 )(
f1
− n)
f1 + g1
Rearrange to get:
dn
f1
(−n +
)
f1 + g1
=
f1 + g1
xdx
vh
12
Integrate to get:
ln(−n +
−n+
n=
f1
f + g x2
)= 1 1
+ C2
f1 + g1
vh 2
⎧ f + g x2
⎫
f1
= exp⎨ 1 1
+ C2 ⎬
f1 + g1
⎩ vh 2
⎭
⎧ f + g x2 ⎫
f1
− C3 exp⎨ 1 1 ⎬
f1 + g1
⎩ vh 2 ⎭
where C3 = eC2
x > h : f (x ) = 0 ; g (x ) =
Region 3 -
g1 x
h
If we assume shortening then no crossbridges can be
attached at x>h. This is equivalent to n(x,t)=0 for x>h.
Now determine the constants using the continuity
conditions:
+
−
n ( h ) = n( h )
0=
⎧ f + g x2 ⎫
f1
− C3 exp⎨ 1 1 ⎬
f1 + g1
⎩ vh 2 ⎭
C3 =
f1
f + g h2
exp( − 1 1
)
f1 + g1
vh 2
Substitute constant back into original equation:
n( x ) =
f +g
− 1 1 ( h2 - x2 ) ⎤
f1 ⎡
2 vh
⎢1 − e
⎥ for 0 < x < h
f1 + g1 ⎣
⎦
Find C1 using the other continuity condition: n(0− ) = n(0+ )
n(0 − ) = C1e
−
0x
v
= C1
C1 =
n (0 + ) =
f +g
− 1 1 h2 ⎤
f1 ⎡
2 vh
⎢1 − e
⎥
f1 + g1 ⎣
⎦
f +g
− 1 1 h2 ⎤
f1 ⎡
2 vh
⎢1 − e
⎥
f1 + g1 ⎣
⎦
13
Substitute constant back into original equation:
n( x ) =
f +g
g x
− 1 1 h2 ⎤ − 2
f1 ⎡
2 vh
v
for x < 0
⎢1 − e
⎥e
f1 + g1 ⎣
⎦
Make a change of variables:
Define
φ=
h
( f1 + g1 )
S
v=
and
S
V
2
where S is a full sarcomere length (~ 2 mm), S/2 is a half
sarcomere length, and V is normalized velocity in half
sarcomere lengths per second.
The three regions can now be defined as:
g2 x
φ
− ⎤
f1 ⎡
V
SV
⎢1 − e ⎥e
f1 + g1 ⎣
⎦
x2
φ
( 2 −1) ⎤
f1 ⎡
V
h
0 < x < h : n( x ) =
⎢1 − e
⎥
f1 + g1 ⎢⎣
⎥⎦
x < h : n(x ) = 0
x < 0 : n( x ) =
Must define rates for XB cycling. Can use the ratios
below:
g
g
3
1
f1 + g1
=
2
f1 + g1
16
= 3.919
The following plots on following slides are the result.
Results for Huxley '57 model"
X = h/2"
n(h/2) = 0.8
X = h-"
X < 0"
(a)"
(c)"
n(h-) = 0.8
(b)"
(b)"
(a)"
(c)"
In isometric conditions (V=0), for given distances between X=0
and X=h, there is a high probability of attachment of the A sites to
the myosin. Attached XBs with positive distortion are "pullers"."
14
Results for Huxley '57 model"
X = h/2"
X < 0"
(a)"
X = -h/2"
(b)"
(c)"
n(-h/2) ~ 0
n(h/2) ~ 0.4
(c)" (b)"
(a)"
As." velocity increases (V>0), probability of attachment of the A site
to the myosin can be above zero for X<0 because XBs may attach
between X=0 and X=h and be dragged to negative distortions."
Results for Huxley '57 model"
X = h/2"
X < 0"
X = -h/2"
(b)"
(a)"
(c)"
n(-h/2) ~ 0.1
n(h/2) ~ 0.15
(c)" (b)"
(a)"
As velocity increases (V>0), the average distortion of the
."
attached XBs decreases (less positive or more negative). At
Vmax the "pullers" (X>0, i.e. (a)) and "dragger" (X<0, i.e. (b))
cancel so that net tension T = 0. Detached XBs don't contribute. "
Now we want to generate a Force-Velocity relation.
Velocity is an input parameter, and force is computed.
Must define force per single XB as
forceXB = kx
Where x is the distortion (= distance from A site to
equilibrium position of attached crossbridge)
Now we want to compute total tension as a the sum of the
the contributions of all the XBs. To do this, we need to
compute an expected value that is analogous to
average value. First define an expected value as:
< u >= ∫ u pdf(u)du
expected value of u
value of u
probability density
function of finding u
15
In our case, we want expected value of tension for all A
sites, both attached and detached. We can write
expected value as:
Force of attached A site
Force of detached A site
[
]
⎧
⎫
< TXB >= ∫ ⎨kxnˆ ( x) + 0 • hˆ( x) − nˆ ( x) ⎬ dx
−∞
⎩
⎭
∞
Detached A sites
Attached A sites
[
]
∞ ⎧
⎫
1 = ∫ ⎨nˆ ( x) + hˆ( x) − nˆ ( x) ⎬ dx
−∞
⎩
⎭
True p.d.f. of all A site being at distance = x
Recall these features of the model:
nˆ ( x ) = n( x ) • hˆ( x )
Conditional probability of an A
site having an attached XB given
its distance is x from the nearest
equilibrium XB position
A p.d.f. describing likelihood
of A sites being distance x
from the nearest equilibrium
XB position
Model assumes ĥ is constant over all possible x
values between -l/2 and l/2 as shown below:
ĥ
1/l
l/2
-l/2
x
Multiplying the terms produces:
⎧n( x) / l ; − l / 2 < x < l / 2
ˆ ( x) = ⎨
n
;
otherwise
⎩ 0
Then substituting into expected value of force:
∞
< TXB >= ∫ kxnˆ ( x)dx
−∞
=∫
l/2
−l / 2
kx
n( x)dx
l
Substitute for n(x) and integrate to get:
< TXB >=
2
φ ⎡
−
f1 kh 2 ⎧
1 ⎛ f1 + g1 ⎞ V ⎤ ⎫
⎪ V
⎪
⎟⎟
⎥⎬
⎨1 − (1 − e V ) ⎢1 + ⎜⎜
f1 + g1 2l ⎪ φ
2 ⎝ g2 ⎠ φ ⎥⎪
⎢
⎣
⎦⎭
⎩
16
For isometric force, set v=0 to get:
max
< TXB
>=
f1 kh 2
f1 + g1 2l
Now normalize force by isometric force to get:
T / Tmax =
φ=
2
φ ⎡
⎧
−
< TXB >
1 ⎛ f + g1 ⎞ V ⎤ ⎫
⎪ V
⎪
⎟⎟
⎥⎬
= ⎨1 − (1 − e V ) ⎢1 + ⎜⎜ 1
max
< TXB > ⎪ φ
2 ⎝ g2 ⎠ φ ⎥⎪
⎢
⎣
⎦⎭
⎩
h
( f1 + g1 )
S
These parameters give best fit
to experimental data at right
g1
3
=
f1 + g1 16
g2
= 3.919
f1 + g1
Successes of model - Good basic framework for cycling XB distribution
- Reproduces Force – Velocity relationship
- Reproduces energy use vs. tension relationship
- Superb first attempt given the knowledge of the
system at the time
Problems - XB cycle is simplistic
- Restrictive set of conditions
- isotonic, constant velocity
- full activation
- Cycling rate increases with lengthening causing increased
ATP usage in disagreement with experimental results
Excel-based simulation
package for Huxley '57
(developed by D. Yue and J. Rice)"
17
Huxley '57 assumes continuous time
and space - first must make discrete
for Excel simulation"
Solve model on a grid of points as shown below"
Dx"
Dt"
N(-15)"
t(0)"
N(-8)"
N(0)"
N(8)"
N(15)"
t(5)"
t(15)"
Attachment and detachment rates"
Rates from Huxley '57 model
4
3
2
f
1
x
As implemented in Excel model
0
g
h
Attachment and Detachment Rates as Function of Distortion
25
g2(x)
20
Rate (1/s)
15
g(x) = g1(x) + g2(x)
10
f(x)
5
g1(x)
0
-0.00002
-0.000015
-0.00001
-0.000005
0
0.000005
0.00001
0.000015
0.00002
distortion (nm)
How to run model"
Input velocity "
(-0.002-0.137)"
Model
parameters"
Force is
computed"
Normalized
velocity and
force are
computed"
18
This model uses a discrete time and
space approximation to continuous
values equations from Huxley '57 "
dn( x, t ) ∂n( x, t ) dx ∂n( x, t ) dt
=
+
dt
∂x dt
∂t dt
We know that:
dx
=v
dt
dt
=1
dt
dn ( x, t ) ∂n( x, t )
∂n( x, t )
=
v+
dt
∂x
∂t
Then use Euler integration to evolve forward in
time at every location"
n( x, t + Δt ) = n( x, t ) +
dn ( x, t )
Δt
dt
From previous page:"
dn ( x, t ) ∂n( x, t )
∂n( x, t )
=
v+
dt
∂x
∂t
Substitute to get:"
n( x, t + Δt ) = n( x, t ) + (
∂n( x, t )
∂n( x, t )
v+
)Δt
∂x
∂t
From class derivation:"
∂n
= rate of attachment to A site ∂t
rate of detachment from A site
= f (x)[1 − n(x, t )]− g (x)n(x, t )
Substitute to get:"
⎡ ∂n( x, t )
⎤
n( x, t + Δt ) = n( x, t ) + ⎢
v + f (x )[1 − n(x, t )] − g (x )n(x, t )⎥ Δt
⎣ ∂x
⎦
Now we just need to compute approximation
for ∂n( x, t ) using a difference equation"
∂x
19
One minor complication is approximation for
the spatial derivative - must use different
forms for positive and negative velocities"
Positive velocity:"
∂n( x, t ) n(x + Δx, t ) − n(x, t )
≈
∂x
Δx
⎡ n(x + Δx, t ) − n(x, t )
⎤
n( x, t + Δt ) = n( x, t ) + ⎢v
+ f (x )[1 − n(x, t )] − g (x )n(x, t )⎥ Δt
Δx
⎣
⎦
Negative velocity:"
∂n( x, t ) n(x, t ) − n(x − Δx, t )
≈
∂x
Δx
⎡ n(x, t ) − n(x − Δx, t )
⎤
n( x, t + Δt ) = n( x, t ) + ⎢v
+ f (x )[1 − n(x, t )] − g (x )n(x, t )⎥ Δt
Δx
⎣
⎦
Slightly rearrange previous results to
better correspond to Excel formula"
n(x + Δx, t ) − n(x, t )⎤
⎡
n( x, t + Δt ) = Δt ⎢ f (x )[1 − n(x, t )] − g (x )n(x, t ) + v
⎥ + n ( x, t )
Δx
⎣
⎦
(V>0 case)"
Corresponds to n(x,t)"
Corresponds to (1-n(x,t))*f(x) - n(x,y)*g(x)"
Discrete time
step =Dt"
Corresponds to V*dn(x,t)/dx"
Model results for V = 0"
Shows h"
n(x) - Conditional Probability as Function of Distortion
9.00E-01
8.00E-01
Conditional Probability
7.00E-01
6.00E-01
5.00E-01
4.00E-01
3.00E-01
2.00E-01
1.00E-01
0.00E+00
-0.000015
-0.00001
-0.000005
0
0.000005
0.00001
0.000015
Distortion (nm)
20
Model results for V = ~.25Vmax"
Shows h"
n(x) - Conditional Probability as Function of Distortion
5.00E-01
4.50E-01
Conditional Probability
4.00E-01
3.50E-01
3.00E-01
2.50E-01
2.00E-01
1.50E-01
1.00E-01
5.00E-02
0.00E+00
-0.000015
-0.00001
-0.000005
0
0.000005
0.00001
0.000015
Distortion (nm)
Calculation of n(x) is n(x,t) after
waiting for "steady-state""
Calculation of sample n(X) values
(See arrows on n(x) plot for sample locations)
5.00E-01
4.50E-01
Conditional Probability
4.00E-01
3.50E-01
N(1)
3.00E-01
N(8)
2.50E-01
N(15)
2.00E-01
1.50E-01
1.00E-01
5.00E-02
0.00E+00
0
500
1000
1500
2000
2500
time (delta t)
3000
3500
>
"
Data points taken at "steady state"
Combine contributions of kx and h
"
kl
2
kx
-l/2
l/2
ˆ( x )
h
x
1/l
-l/2
l/2
x
k*x*h_hat as Function of Distortion
0.6
kx • hˆ( x)
0.4
Force / cm
0.2
0
-0.000015
-0.00001
-0.000005
0
0.000005
0.00001
0.000015
-0.2
-0.4
-0.6
distortion (nm)
21
Multiply to help compute <TXB>"
kx • hˆ( x)
n(x)
k*x*h_hat as Function of Distortion
n(x) - Conditional Probability as Function of Distortion
0.6
5.00E-01
4.50E-01
0.4
Conditional Probability
4.00E-01
3.50E-01
X
2.50E-01
2.00E-01
1.50E-01
Force / cm
0.2
3.00E-01
0
-0.000015
-0.00001
-0.000005
0
0.000005
0.00001
0.000015
-0.2
1.00E-01
-0.4
5.00E-02
0.00E+00
-0.000015
-0.00001
-0.000005
0
0.000005
0.00001
-0.6
0.000015
distortion (nm)
Distortion (nm)
k*x*h_hat*n(x) as Function of Distortion
9.50E-02
7.50E-02
Force / cm
5.50E-02
3.50E-02
1.50E-02
-0.000015
-0.00001
-0.000005
-5.00E-03
0
0.000005
0.00001
0.000015
-2.50E-02
Distortion (nm)
Compute <TXB> by summing
over all distortions"
k*x*h_hat*n(x) as Function of Distortion
9.50E-02
7.50E-02
Force / cm
5.50E-02
3.50E-02
1.50E-02
-0.000015
-0.00001
-0.000005
-5.00E-03
0
0.000005
0.00001
0.000015
-2.50E-02
Distortion (cm)
This step corresponds to this step in the continuous time
derivation:
∞
∞
l / 2 kx
< TXB >= ∫ kxnˆ ( x)dx = ∫ kxhˆ( x)n( x)dx = ∫
n( x)dx
−∞
−∞
−l / 2 l
Exercises"
1.  Generate a classical force-velocity curve
using the discretized Huxley 57 solver in
Excel. Plot the result."
2.  Choose one of the model parameters to
alter, and predict its effects on the forcevelocity curve."
3.  Use the model implementation to test
your hypothesis."
22