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Dear All, Sept 18, 2014
I have wondered for years how trees can whip around in the wind and =
normally spring back to their original position when the wind eases and =
I broached this on Naturens in 2003 (pasted below). Meanwhile another =
piece of this puzzle has come to my attention.
START OF PASTE\\\\\\\\\\\\\\\\\\\\\\\
Dear All, Feb 14, 2003
<snip>
Spruce trees normally=20
have a remarkable ability to bend with the wind and then spring back.=20
<SNIP>
Is anyone on naturens familiar with the mechanisms that enable=20
elasticity (molecular or cellular) in living trees ? There will=20
presumably be some reversible water movement, between living cells,=20
walls and intercellular spaces but there must also be some give and take =
at the molecular level, e.g. distortion and recovery of normally folded =
or coiled molecules in the cell wall or changes in water of hydration;=20
sort of an intramolecular shock absorber. Is lignin involved ?
END OF PASTE\\\\\\\\\\\\\\\\\\\\\\\\\\\
NOTICE: The following contains numerous calculations. A copy editor =
prize goes to the reader who locates the greatest number of =
calculating/transcription errors.
Driwater, a commercial product, is 2% Cellulose, 98% water and is =
solid at room temperature. I can not directly understand how this is =
possible but one must assume that water in the vicinity of Cellulose is =
no longer a free liquid and water sufficiently far away is a liquid. So =
there is apparently a gradual change in state with distance from =
Cellulose, as opposed to the sharp transitions portrayed in =
temperature/pressure state diagrams.=20
Also, while browsing through old copies of American Scientist, I =
came across an article on Colloidal Macromolecular Phenomena (Battista, =
O. A, 1965(20):151-173). Purified mcrofibrils of Cellulose (as in wood =
pulp) are schematically represented as very long chains of Cellulose =
variously associated with short portions of three to seven other long =
chains to form polymer microcrystals which are connected to other =
microcrystals by hinges of free (unassociated) lengths of Cellulose =
(chain length variable within hinges). With this kind of structure, =
Cellulose microfibrils can assume a range of longitudinal or lateral =
positions without rupture of chemical bonds.=20
In a vague way this 'explains' how living or steamed wood can bend. =
It is possible that the tendency of living trees to return to their =
initial shape when wind force abates is driven by elasticity of cell =
walls variously distorted into intercellular spaces (when compressed) or =
withdrawn from intercellular spaces when extended.
Returning to the example of Driwater (2% Cellulose in water being a =
solid at room temperature) brings to mind the behavior of water in a =
porous medium such as soil.=20
With decrease in effective pore size, water in soil becomes bound in the =
soil pores to an increasing degree. At tensions greater than pF 2 (200 =
cm of water) water in most soil profiles does not flow downward under =
the influence of gravity (i.e. it acts as though it were part of the =
solid phase). And as soil water content approaches zero all soils =
approach tensions of pF 7 (10^7 cm of water; or expressed another way =
100,000 metres of water tension).
Back of envelope approximations:
If total dissociation of Cellulose is assumed (no microcrystals) one =
can approximate the spacing between polymer chains as follows:
Assume the 2% solution implies 2 g Cellulose per 100 mL aqueous =
solution=3D=20
20 g Cellulose/dm^3.=20
Glucose (C6H12O6) has a molecular weight of 180.16; call this 180 =
and water is 18.
So the C6H10O5 units of Cellulose would have a molecular weight of =
~180-18=3D 162
According to=20
http://books.google.ca/books?id=3DNhwSMryGzY0C&pg=3DPA55&dq=3Ddimensions+=
of+cellulose+molecule+units&hl=3Den&sa=3DX&ei=3DoEsbVNGCEoOxyASTsYCgBQ&ve=
d=3D0CC0Q6AEwAQ#v=3Donepage&q=3Ddimensions%20of%20cellulose%20molecule%20=
units&f=3Dfalse
Fig. 3.4 the length of two units, each being C6H10O5), is 1.038 x 10^-9 =
M
or unit length of 0.519 x 10^-9 M
20/162 (Moles of units) X 6 x 10^23 Avogadro's number (units/mole) =
x 0.519 x 10^-9 M (length per unit)=3D 0.384 x 10^14 M or 3.84 x 10^14 =
dm (total polymer length).
To put this in context, the equatorial radius is 6,378,388 metres =
which is equivalent to an equatorial circumference of about 4 x 10^8 dm. =
So 20 grams of Cellulose has a total polymer chain length sufficient to =
circle the earth at the equator 960,000 times. Holy smoke. [i.e. 38.4 x =
10^13/ 4 x 10^8 =3D 9.6 x 10^5=3D 960,000]
So the final step is to space a polymer length of 3.84 x 10^14 dm =
equally within a volume of one dm^3. This would be difficult to solve =
directly but iteration can quickly reach an approximate solution.
=20
And for ease of approximation I will consider one strand to lie =
against the side wall of the container so e.g. 100 strands would divide =
the distance of 1 dm into 100 equal parts of 1 mm. These equally spaced =
strands will run in the X, Y & Z directions i.e. from left to right, =
bottom to top and front to back directions. Consequently the number of =
strands available for any one direction is 3.84/3 x 10^14 i.e. 1.92 x =
10^14.
Considering one direction, say top to bottom: if strands were 1 =
micron (10^-5 dm) apart there would be 10^5 strands in each of 10^5 =
panels for a total of 10^10 strands; each strand being 1 dm long. This =
leaves a residue of 1.92 x 10^4 dm to be distributed so a spacing of one =
micron (10^-6 metres) is slightly too large.=20
To add another row and column would require 2 x 10^5 +1 strands so =
if the residue were distributed equally to this additional row and =
column then the strands would each be (1.92 x 10^4)/(2 x 10^5 +1)=3D =
19200/200001 =3D 0.096 dm long; too insignificant to consider..=20
=20
This requires a leap of faith but if a column of cubes, with edges =
defined by hydrophilic polymers, can be considered have to capillary =
properties then by calculating the 'capillary rise' the intensity with =
which water is bound by this column could be estimated.
A column of cubes that has edges of10^-6 metres=3D 10^-4 cm would =
have a cross-sectional area 10^-8 cm^2. A capillary tube of this =
cross-sectional area would have a radius of 0.00005642 cm.=20
Wetting angle of water on Cellulose is given in the above reference =
as ranging from 20-30 degrees. Plugging the above into a general =
equation for rise of liquids in capillary tubes: h (cm)=3D (2T Cos =
wetting angle)/rdg where T is surface tension of 72.8 (dynes/cm), =
wetting angle is 25 degrees, r is radius (cm), d is density g/cm^3 (1), =
g is gravitational accelleration (980 cm/sec^2) leads to
h=3D 2 x 72.8 x 0.906/(0.00005642 x 980)=3D 2385.8 cm; a very strong =
affinity for water.
Taking one atmosphere pressure to be 1020 cm water at room =
temperature (drawing on memory) then it would take a pressure of about =
2.3 atmospheres to force water from this column into a saturated ceramic =
plate that had very fine pores (to prevent air entry).
Yt, Dave Webster, Kentville
=20
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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD>
<META content=3D"text/html; charset=3Diso-8859-1" =
http-equiv=3DContent-Type>
<META name=3DGENERATOR content=3D"MSHTML 8.00.6001.23588">
<STYLE></STYLE>
</HEAD>
<BODY>
<DIV>Dear All, =20
=
Sept=20
18, 2014</DIV>
<DIV> I have wondered for years how trees can whip =
around in=20
the wind and normally spring back to their original position when =
the wind=20
eases and I broached this on Naturens in 2003 (pasted below).=20
Meanwhile another piece of this puzzle has come to my =
attention.</DIV>
<DIV> </DIV>
<DIV>START OF PASTE\\\\\\\\\\\\\\\\\\\\\\\</DIV>
<DIV>Dear=20
All, &nb=
sp; &nbs=
p; =20
Feb 14, 2003<BR><snip></DIV>
<DIV>Spruce trees normally <BR>have a remarkable ability to bend with =
the wind=20
and then spring back. <BR><SNIP></DIV>
<DIV>Is anyone on naturens familiar with the mechanisms that enable=20
<BR>elasticity (molecular or cellular) in living trees ? There will=20
<BR>presumably be some reversible water movement, between living cells,=20
<BR>walls and intercellular spaces but there must also be some give and =
take=20
<BR>at the molecular level, e.g. distortion and recovery of =
normally=20
folded <BR>or coiled molecules in the cell wall or changes in water of=20
hydration; <BR>sort of an intramolecular shock absorber. Is lignin =
involved=20
?</DIV>
<DIV>END OF PASTE\\\\\\\\\\\\\\\\\\\\\\\\\\\</DIV>
<DIV> NOTICE: The following contains =
numerous=20
calculations. A copy editor prize goes to the reader who=20
locates the greatest number=20
of calculating/transcription errors.</DIV>
<DIV> </DIV>
<DIV> Driwater, a commercial product, is 2% Cellulose, =
98%=20
water and is solid at room temperature. I can =
not directly understand=20
how this is possible but one must assume that water in the vicinity of=20
Cellulose is no longer a free liquid and water sufficiently far =
away is a=20
liquid. So there is apparently a gradual change in state with distance =
from=20
Cellulose, as opposed to the sharp transitions portrayed in =
temperature/pressure=20
state diagrams. </DIV>
<DIV> </DIV>
<DIV> Also, while browsing through old copies of =
American=20
Scientist, I came across an article on Colloidal Macromolecular =
Phenomena=20
(Battista, O. A, 1965(20):151-173). Purified mcrofibrils of Cellulose =
(as in=20
wood pulp) are schematically represented as very long chains of =
Cellulose=20
variously associated with short portions of three to seven other long =
chains to=20
form polymer microcrystals which are connected to other microcrystals=20
by hinges of free (unassociated) lengths of Cellulose =
(chain=20
length variable within hinges). With this kind of structure, Cellulose=20
microfibrils can assume a range of longitudinal or lateral=20
positions without rupture of chemical bonds. </DIV>
<DIV> </DIV>
<DIV> In a vague way this 'explains' how living or =
steamed=20
wood can bend. It is possible that the tendency of living =
trees to=20
return to their initial shape when wind force abates is driven by=20
elasticity of cell walls variously distorted into intercellular =
spaces=20
(when compressed) or withdrawn from intercellular spaces when =
extended.</DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV> Returning to the example of Driwater (2% =
Cellulose in=20
water being a solid at room temperature) brings to mind the behavior of =
water in=20
a porous medium such as soil. </DIV>
<DIV>With decrease in effective pore size, water in soil =
becomes=20
bound in the soil pores to an increasing degree. At =
tensions=20
greater than pF 2 (200 cm of water) water in most soil profiles =
does not=20
flow downward under the influence of gravity (i.e. it acts as though it =
were=20
part of the solid phase). And as soil water content approaches =
zero all=20
soils approach tensions of pF 7 (10^7 cm of water; or expressed another =
way=20
100,000 metres of water tension).</DIV>
<DIV> </DIV>
<DIV>Back of envelope approximations:</DIV>
<DIV> If total dissociation of Cellulose is assumed =
(no=20
microcrystals) one can approximate the spacing between polymer chains as =
follows:</DIV>
<DIV> Assume the 2% solution implies 2 g Cellulose per =
100 mL=20
aqueous solution=3D </DIV>
<DIV>20 g Cellulose/dm^3. </DIV>
<DIV> Glucose (C6H12O6) has a molecular weight of =
180.16;=20
call this 180 and water is 18.</DIV>
<DIV>So the C6H10O5 units of Cellulose would have a molecular weight of =
~180-18=3D=20
162</DIV>
<DIV> According to </DIV>
<DIV><A=20
href=3D"http://books.google.ca/books?id=3DNhwSMryGzY0C&pg=3DPA55&=
dq=3Ddimensions+of+cellulose+molecule+units&hl=3Den&sa=3DX&ei=
=3DoEsbVNGCEoOxyASTsYCgBQ&ved=3D0CC0Q6AEwAQ#v=3Donepage&q=3Ddimen=
sions%20of%20cellulose%20molecule%20units&f=3Dfalse">http://books.goo=
gle.ca/books?id=3DNhwSMryGzY0C&pg=3DPA55&dq=3Ddimensions+of+cellu=
lose+molecule+units&hl=3Den&sa=3DX&ei=3DoEsbVNGCEoOxyASTsYCgB=
Q&ved=3D0CC0Q6AEwAQ#v=3Donepage&q=3Ddimensions%20of%20cellulose%2=
0molecule%20units&f=3Dfalse</A></DIV>
<DIV>Fig. 3.4 the length of two units, each being C6H10O5), is =
1.038 x=20
10^-9 M</DIV>
<DIV> or unit length of 0.519 x 10^-9 M</DIV>
<DIV> 20/162 (Moles of units) X 6 x 10^23 =
Avogadro's=20
number (units/mole) x 0.519 x 10^-9 M (length per unit)=3D 0.384 x 10^14 =
M or 3.84=20
x 10^14 dm (total polymer length).</DIV>
<DIV> </DIV>
<DIV> To put this in context, the equatorial radius is =
6,378,388 metres which is equivalent to an equatorial circumference of =
about 4 x=20
10^8 dm. So 20 grams of Cellulose has a total polymer chain length =
sufficient to=20
circle the earth at the equator 960,000 times. Holy smoke. [i.e. 38.4 x =
10^13/ 4=20
x 10^8 =3D 9.6 x 10^5=3D 960,000]</DIV>
<DIV> </DIV>
<DIV> So the final step is to space a =
polymer length=20
of 3.84 x 10^14 dm equally within a volume of one dm^3. This would be =
difficult=20
to solve directly but iteration can quickly reach an approximate =
solution.</DIV>
<DIV> </DIV>
<DIV> And for ease of approximation I will consider =
one strand=20
to lie against the side wall of the container so e.g. 100 strands would =
divide=20
the distance of 1 dm into 100 equal parts of 1 mm. These equally =
spaced=20
strands will run in the <EM>X, Y </EM>& <EM>Z </EM>directions =
i.e. from=20
left to right, bottom to top and front to back directions. Consequently =
the=20
number of strands available for any one direction is 3.84/3 x 10^14 i.e. =
1.92 x=20
10^14.</DIV>
<DIV> </DIV>
<DIV> Considering one direction, say top to bottom: if =
strands=20
were 1 micron (10^-5 dm) apart there would be 10^5 strands in each of =
10^5=20
panels for a total of 10^10 strands; each strand being 1 dm =
long. This=20
leaves a residue of 1.92 x 10^4 dm to be distributed so a spacing of one =
micron=20
(10^-6 metres) is slightly too large. </DIV>
<DIV> </DIV>
<DIV> To add another row and column would require 2 x =
10^5 +1=20
strands so if the residue were distributed equally to this additional =
row and=20
column then the strands would each be (1.92 x 10^4)/(2 x 10^5 =
+1)=3D=20
19200/200001 =3D 0.096 dm long; too insignificant to =
consider.. </DIV>
<DIV> </DIV>
<DIV><EM> </EM>This requires a leap of faith but if a =
column=20
of cubes, with edges defined by hydrophilic polymers, can be considered =
have to=20
capillary properties then by calculating the 'capillary rise' the =
intensity with=20
which water is bound by this column could be estimated.</DIV>
<DIV> </DIV>
<DIV> A column of cubes that has edges of10^-6 =
metres=3D 10^-4=20
cm would have a cross-sectional area 10^-8 cm^2. A =
capillary=20
tube of this cross-sectional area would have a radius of 0.00005642 cm. =
</DIV>
<DIV> </DIV>
<DIV> Wetting angle of water on Cellulose is given in =
the=20
above reference as ranging from 20-30 degrees. Plugging the above into a =
general=20
equation for rise of liquids in capillary tubes: h (cm)=3D (2T Cos =
wetting=20
angle)/rdg where T is surface tension of 72.8 (dynes/cm), wetting angle =
is 25=20
degrees, r is radius (cm), d is density g/cm^3 (1), g is gravitational=20
accelleration (980 cm/sec^2) leads to</DIV>
<DIV>h=3D 2 x 72.8 x 0.906/(0.00005642 x 980)=3D 2385.8 cm; a very =
strong affinity=20
for water.</DIV>
<DIV> </DIV>
<DIV> Taking one atmosphere pressure to be 1020 cm =
water at=20
room temperature (drawing on memory) then it would take a pressure of =
about 2.3=20
atmospheres to force water from this column into a saturated ceramic =
plate that=20
had very fine pores (to prevent air entry).</DIV>
<DIV> </DIV>
<DIV>Yt, Dave Webster, Kentville</DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV> </DIV></BODY></HTML>
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