Biomedical Engineering Reference
In-Depth Information
Ta b l e 6 . 1
Constants used in
the numerical example
Constant
Value
Unit
Reference
s
−
1
k
2
9
.
76
Murtada et al. (
2012
)
s
−
1
k
3
4
.
00
Murtada et al. (
2012
)
s
−
1
k
4
0
.
05
Murtada et al. (
2012
)
s
−
1
k
5
9
.
76
Murtada et al. (
2012
)
s
−
1
k
7
0
.
002
Murtada et al. (
2012
)
s
−
1
η
21
.
55
Murtada et al. (
2012
)
q
1
/
2
0
.
37
µM
Murtada et al. (
2012
)
c
0
0
.
84
kPa
Murtada et al. (
2012
)
c
1
3
.
15
kPa
Murtada et al. (
2012
)
c
2
0
.
035
Murtada et al. (
2012
)
E
5300
kPa
Murtada et al. (
2012
)
σ
0
.
25
see text
μ
6800
kPa
see text
s
−
1
ν
0
.
03
Murtada et al. (
2012
)
s
−
1
b
5
.
0
see text
The free energy for the calcium ion concentration is take to be the simplest
form which allows for direct control of the intracellular calcium ion concentration,
namely,
1
2
q
2
.
ψ
q
=
(6.23)
Note that the factor 1
/
2 is not dimensionless but takes on the appropriate unit for
ψ
q
to be in Joule. The overlap function
f
must also be defined. It is generally reported
to be bell-shaped (Rachev and Hayashi,
1999
) or parabolic (Murtada et al.,
2012
)
and, for simplicity, we take it to be the normal distribution
exp
−
ε
ft
,
0
)
2
/
2
σ
2
,
f
=
(ε
ft
−
(6.24)
where
ε
ft
,
0
is the strain at optimal force generation and
σ>
0 is the width of the nor-
mal distribution taken to be 0.25. Finally, the constants
ν
and
μ
and the function
g
in Eq. (
6.17
) remains to be specified. To that end, assume
ν
to be the maximal short-
ening velocity for the smooth muscle and take it to be 0.03, see Fig. 9 in Murtada et
al. (
2012
). This assumption is motivated by Eq. (
6.16
) since
ν
is the maximal
contraction velocity for which the friction clutch generates a contraction force. To
determine
μ
, note that
ε
ft
=−
˙
0 in stationary isometric experiments and Eq. (
6.16
)
1
becomes
n
C
μνf
. By comparing this result with Eq. (20) in Murtada et al. (
2012
)
and substituting
ν
, the value for
μ
can be computed to be 6800 kPa. Finally, take
the function
g
to be
ε
ft
=
˙
∂ψ
cb
∂ε
cb
,
f
b
g
=
(6.25)
