1
A
RCHIMEDES
’ S
CREW
: M
OMENT
A
RM
AND
P
OCKET
V
OLUME
John F. Hall
Professor of Civil Engineering, Emeritus
Caltech
Caltech Civil Engineering Report No. 2024
-
0
2
June 2024
2
T
able
of
contents
Page
Section 1. Introduction
4
Section 2.
Length of m
oment arm for the weight of
water in a
5
pocket
Section 3.
Pocket formation
9
Section 4.
Volume of water in a pocket
of the screw
11
Section 5.
Comparison with
other calculation method
s
for
pocket
13
volume
5.1
Rorres
13
5.2
Weisbach
1
4
5.3
Lubitz et
al.
15
Section 6.
Some other features of the pocket geometry of a
blade
17
screw
Section 7. Conclusions
1
8
Section 8. References
18
Section 9
.
Figures
2
1
Appendix 1
.
Daniel Bernoulli’s derivation for moment of
water
2
6
weight
Appendix 2. Antoine Parent’s
derivation for moment of
water
2
9
weight
3
Appendix 3. Computer code for finding the
water volume in the
3
3
pocket
of a blade screw
A3.1 Explanation of variables
3
3
A3.2 Notes
3
5
A3.3 Program
listing
3
6
The figure on the front page
shows an Archimedes’ screw of the blade type with
part of the casing cut away to reveal water
-
filled pockets
(
3,4
).
4
Section
1
.
Introduction
Archimedes
’
screw, which may or may not have been invented by Archimedes
(1)
,
is an ancient water lifting device whose use continues into modern times
, not only
as a pump but in hydroelectric generation
.
In the
pump
application
, w
ater
(or other
material)
is carried in pockets of a helical tube or channel that are filled at the
lower end of the screw, elevated and then discharged at the upper end as the screw
is turned
about its
longitudinal
axis
.
Shown in Figure
1
(
taken
from reference 2)
is a tube design in
which the tube
or
tubes
are wrapped around a
cylindrical
shaft.
When more than one tube
is
used,
they are placed in a nested arrangement.
Pockets are formed when the screw is
inclin
ed sufficiently. In another design, a
blade
or nested blades are
constructed
around
the
shaft
protrud
ing outward
,
which
are
encased in a
n
oute
r
cylinder
(Figure
2
, taken from references 3 and 4
)
.
The shaft and helical
blade
s
can either
be
free from
the
casing
and
turn
ed while the
casing
is held fixed, or the whole
arrangement
is
connected and
turn
ed as a unit.
In all these forms, the water
in a
pocket
is lifted
essentially
intact
,
as a unit in each pocket.
In his book
s
on machines, Julius Weisbach
(
3,
4
)
uses the term Archimedes
’
screw
for only the tube design
,
and
he
also refers to it as a water snail. The blade design
is referred to
just as
a water screw. In th
e present
paper, the term screw will be
used for both types, with tube and blade used to differentiate the two types.
S
crew
includes the shaft
and
helical
tubes or
the
shaft and
helical
blades with casing
.
Two operational aspects of the screw are considered in this paper: the vo
lume
of
water that can be carried in
a
pocket and the torque needed to
turn the screw and
lift
the water. The first is a problem of geometry
,
and
the second
involves
mechanics.
Knowing the weight of the water, the torque
can
easily
be
found by
work
-
energy
method
s
, as explained
below
in Section 2.
The earliest of these is the principle of
5
virtual velocities, which was in use
prior to 1700
,
before the concepts of work and
energy were established.
In 1726, Henri Pitot published a paper
(5)
dealing with a
tube screw whose radius is small compared to the radius of the shaft. Pitot used an
integrated form of the principle of virtual velocities (displacements instead of
velocities) to derive an expression for the torque, which actually applies
to the
general case and is equivalent to what is presented in Section 2. Pitot also
developed a procedure to find the pocket volume that applies only to t
he
case
of a
tube screw
with small radius.
Two other early attempts to find the torque, one by Antoine Parent (6) in 1705 and
the other by Daniel Bernoulli (7
,8
) in 1738, utilized moment equilibrium about the
screw’s axis. However, this method
is hampered by
having to know
the
length of
the
moment arm of the water
’s
weight vector
,
which would appear to be difficult to
deduce directly from the complicated geometry of the pocket of a screw.
For this
reason,
neither
Parent
nor
Bernoulli
solved the
general case; in fact, Bernoulli dealt
with the same special case as Pinot. Even so, a
s shown in this report for the first
time,
Bernoulli’s result
provides the key to obtaining an
easy
solution
for the
torque
in the general case
distinct from
work
-
energy methods
, and it also makes
possible
a simple and natural way to compute the volume of water carried by a
screw.
In this
report
, friction
and leakage
are
neglected as
they
do not affect the ma
jor
ideas to be conveyed.
Thus
, the torque needed to
lift
the water at a
slow
steady
rate
is
equal to the torque needed to keep the screw from turning under the
sitting
weight of the water
.
Also, some of the notation used in original works discussed in this report has been
altered in order to maintain consistency.
Section
2
.
Length of m
oment arm for the weight of water in a pocket
6
When the screw turns through one revolution
, the torque does work equal to
2
휋푇
,
and
a
weight
푊
of water equivalent to that
contained in the pockets of the screw
rise
s
vertically
an amount
퐿
sin
(
훽
)
, where
푇
is the applied torque about the axis of
the
s
crew
,
퐿
is the pitch of the helix, and
훽
is the inclination angle of the screw
measured from horizontal (Figure 3). The external work equals the gain in
potential energy of the water; thus,
2
휋푇
=
푊퐿
sin
(
훽
)
,
(
1
)
from which the torque can be computed.
Note that in the coordinate system
employed,
푍
is vertical;
푋
and
푌
are in the horizontal plane; and the
inclination
of
the screw is accomplished by a rotation about the
−
푌
axis.
The axis of the screw
lies in the
푋푍
plane.
The
helix
angle
훼
(Figure 3)
can be specified in lieu of the pitch
퐿
:
tan
(
훼
)
=
퐿
2
휋
푅
퐶푇
(
tube
screw
)
(
2
)
or
tan
(
훼
)
=
퐿
2
휋
푅
푂
(
blade
screw
)
,
(
3
)
w
h
ich uses the radius
푅
퐶푇
of the centerline of the tube
with respect to the axis of
the s
crew
for the tube screw, or the radius
푅
푂
of the casing for the blade screw.
The angle
훼
is measured with respect to a plane perpendicular to the axis of the
screw
.
Note that the helix angle adjacent to the shaft
has a different value:
tan
(
훼
퐼
)
=
퐿
2
휋
푅
퐼
,
(
4
)
which applies to both the tube and blade screws
, and where
푅
퐼
is the radius of the
shaft.
Equation 1
can be
expressed
as follows:
푇
=
푊
cos
(
훽
)
·
퐿
2
휋
tan
(
훽
)
.
(
5
)
This applied torque must be equal and opposite to the moment
푀
that the weight
vector
of
the water carried by the screw makes with the axis of the s
crew
.
7
Referring
again
to Figure 3, the component of th
is
weight
vector
that makes a
moment about the axis of the s
crew
is
푊
cos
(
훽
)
.
Thus,
from Equation
5
,
퐿
2
휋
tan
(
훽
)
is the length of
th
e
moment arm. Evidently, this
expression
for the
moment arm
applies for both tube and blade screws, and the length of the arm does
not depend on any of the radial dimensions of the screw: radius
푅
퐼
of the shaft
and
the
tube radius
푅
푇
or
blade width
푅
푂
−
푅
퐼
, as long as
퐿
is constant.
Furthermore,
t
he moment arm is the same for a single helix or multiple helices in a nested
arrangement, for pockets filled with water or only partially filled
, for a single
pocket or multiple pockets
. T
he water weight
푊
depends on these parameters, but
the length of the moment arm
does not
.
The deduction
s
from the previous paragraph
about the length of the moment arm
are
surprising considering
that
the
geometry of a pocket in a tube or blade screw
is
compl
ex
. For example, Figure 4 shows
two
horizontal cross
-
sections through the
water
-
filled pocket of a
n inclin
ed blade screw
for the following parameters:
tan
(
훽
)
=
0.75
,
푅
푂
=
2
푅
퐼
,
퐿
=
2
휋
푅
퐼
in arbitrary length
units
,
and
푁
퐻
=
8, where
푁
퐻
is the number of
blades
(
푁
퐻
=
1
for a single continuous blade
,
>
1
for nested
blades
)
. The
outlines
in the figure
are the
boundaries of the water sections
and
are
comprised
of four types as marked: shaft,
casing
, helical
blade
above
and helical
blade
below
. The
solid
outline is for the water free surface at the maximum water
elevation
; the spill point is marked
. The
dashed
outline is at an intermediate
elevation, low enough so
that the
blade
above does not impinge
.
The line of action of the
cos
(
훽
)
component of the water weight
vector
passes
through the center of gravity of the water body
(marked c.g.
in Figure 4
)
, and it
lies
in
the
vertical plane that contains the
푋
-
parallel
horizontal d
ott
ed line
drawn in the
figure.
T
he moment arm for the water body about the axis of the screw is parallel
to the
푌
axis
,
which
is shown with length
퐴
.
Furthermore, since
퐴
is independent
of water depth, the bottom point of the pocket (marked
푐
in Figure 4) lies in th
e
same vertical plane
as the dotted line
.
It will now be
demonstrated
that
퐴
=
퐿
2
휋
tan
(
훽
)
without invoking the work
-
energy equation.
8
In his book
Hyd
rodynamica
,
published in 1738
(7,8)
,
Daniel Bernoulli deal
s
with a
single
helical tube wrapped around a
cylindrical
shaft, and he assume
s
the cross
-
sectional dimension
s
of the
tube
ar
e
small compared to the radius of the shaft.
Thus
,
each
cross
section of
th
is
thin
tube
i
s either empty or full
, i.e.,
the length
s
of
the transitions
from empty to full
and
from full to empty c
an
be ignored
.
One of
Bernoulli’s figure
s
is shown in Figure 5 where a pocket is denoted by
length
표푝푞
of the tube
:
표
being the spill point,
푝
the point with minimum elevation, and
푞
being the same elevation as
표
.
Using
calculus
,
Bernoulli
develops an
expression
for
the
moment
that
the weight of water in the pocket
makes about the
axis of the
s
crew
. From this expression, it can be
deduc
ed that the moment arm is
equal to
퐿
2
휋
tan
(
훽
)
,
and
this is true whether or not the
thin
tube’s
pocket
i
s f
ull
.
The only
condition is that the water at each end of the pocket ha
ve
the same elevation, which
must
be the case for static equilibrium. Bernoulli’s proof
is
not repeated here
in its
original form
, but
a
modifi
ed version
(for clarity)
appears in Appendix
1
.
T
he more general
case is
a
helical tube or channel
whose
cross
-
section
al
dimensions
are not small compared to the
radius of the
shaft
.
Consider that such a
tube or channel is
partitioned
into
푁
tubes of
relatively small
cross
-
section
so that
each one
i
s
similar to the
tube
Bernoulli analyzed.
Figure 6 shows the channel of a
helical blade screw
containing
such
a partition
(part a)
and
a plan view of the
푖
th
thin tube extending through a 360°
angle
of the helix (part b
, angle
휃
varying from
⎼
180°
at
퐴
to 180°
at
퐸
)
. Both these views are
for the screw in the vertical
position
, and the thin tube
shown in part b
gains elevation as
휃
increases.
Next, c
onsider the screw to be
inclin
ed
so that some of the thin
tube
s
form
their
own pocket
s
as shown in part c of Figure 6
,
which is a
n oblique
view
of
the
푖
th
thin
tube
. T
he
water in
all such
pockets
is
taken
to
have a common free
-
surface
elevation
푍
푠푢푟
that is
at or below the spill level for the
screw
pocket
, although
s
ome pockets
of thin tubes
that are too high
may be empty.
For the thin tube in
part c,
퐵퐶퐷
defines the pocket:
point
퐵
is the
spill point
for the thin tube
;
퐶
is at
the lowest elevation
;
and
퐷
is at the same elevation as
퐵
.
This tube is
shown
in a
9
partially filled state
where the water extends from point
1
through
퐶
to point
2
,
both
points
1
and
2
being
at the same elevation. The points
퐵
,
1
,
퐶
,
2
and
퐷
are
reproduced in part b of Figure 6 for reference.
As
is clear
from
Bernoulli’s result, the weight
vector
of
the
water
contained in each
pocket of
the
thin tube
s
has a moment arm about the axis of the screw of
length
퐿
2
휋
tan
(
훽
)
. Therefore,
the combined mass of
water
will
also
.
The reason is that
every thin tube shares the same values for
퐿
and
훽
.
Since this
combined mass of
water has a common free surface elevation,
it is in static equilibrium
and would
hold the same shape without the
partitioning into
thin tube
s
being present.
Thus,
for
the
screw
pocket,
any
amount of
water
with
a common free
-
surface elevation
that is at or below the spill
level of the pocket
will
have
퐴
=
퐿
2
휋
tan
(
훽
)
.
The same
argument applies equally well to
a
screw of the tube type.
Finally,
it will be mentioned that a correct expression for the
length of the
moment
arm
for the water weight vector
in a blade screw
, obtained by geometric analysis,
was published by Antoine Parent in 1705
(
56
)
, more than 30 years before Daniel
Bernoulli
’s
solution for a
thin tube
.
However,
in his largely
unknown
paper,
Parent makes a faulty assumption
, but which
d
oes
not affect
the
result.
He
assume
s
that the
weight vector
of the water in a pocket
passes through
the bottom
point for the pocket. Although this is true
only
when
there is
a small amount of
water in the pocket, it leads to the correct expression for the
length of the
moment
arm for any
water level since th
is
length is independent of water level, as explained
above. A summary of Parent’s solution
, which is probably the first attempt to
analyze Archimedes’ screw mathematically,
is presented in Appendix 2.
Section 3
. Pocket formation
10
Before continuing, some comments will be made about the conditions under which
a pocket forms. These comments apply to a blade screw
; some generalization
would be
needed for
a tube screw.
Figure 7
presents
a bade screw in the vertical position
and
in
two inclined positions
showing the outer edge
of the blade
along the casing and the inner edge along the
shaft.
The angles of these two helices are
훼
and
훼
퐼
, respectively
,
measured from
a
plane perpendicular to the axis of the screw.
As the screw is
inclined
, a pocket
begins to form adjacent to the casing when the slope of
the outer edge at
휃
=
−
휋
2
becomes
zero
(part b of figure)
.
This
location
is initially the spill point, but as the
inclin
ation increases, this spill point moves inward, reaching
the shaft when the
slope of the inner edge becomes zero at
휃
=
−
휋
2
(part c of figure).
With further
inclin
ation, the spill point remains adjacent to the shaft and moves in the direction
of
decreasing
휃
.
Thus, the condition for the spill point to be adjacent to the
shaft
is
휋
2
−
훽
≥
훼
퐼
.
(
6
)
Using
Equation 4
, this can be written as a condition on
푅
퐼
as
푅
퐼
≥
퐿
2
휋
tan
(
훽
)
,
(
7
)
where the right side happens to be the length of the moment arm.
There appears to
be no reason Equation 7 should not be satisfied
for a practical design
, so it will be
assumed to hold.
It also means that in a partition into
푁
thin tubes, all
푁
tubes will
have
form
ed
their own pockets.
The maximum
possible
water surface elevation in
the
pocket
of a screw
is that of
the spill point, and the
lowest
elevation
in the pocket is its
bott
om point. The
bottom point separates off from
the spill
point after the pocket
first
forms
(Figure
7, part b),
and then
it
moves in the direction of
increasing
휃
as the screw is
inclined
further
. It remains adjacent to the casing.
11
Section
4
.
Volume of water in a pocket
of the screw
The thin
-
tube
partition
described in
S
ection
2
represent
s
a natural subdivision of
the volume inside the
tube of a tube screw or the
channel of a blade screw. The
thin tubes
form a convenient basis for computing the volume of water contained in
a pocket
for
a
given free surface elevation
푍
푠푢푟
of the water
that is at or below the
spill level.
With reference to Figure 6b
with the screw in the vertical position
, the location of
a
point
along
the
푖
th
thin tube in terms of
its
radi
al coordinate
푅
푖
and
the
angle
휃
is
푋
=
푅
푖
·
cos
(
휃
)
푌
=
푅
푖
·
sin
(
휃
)
푍
=
퐿
휃
2
휋
+
풵
푖
,
}
(
8
)
where
풵
푖
is the elevation of the
푖
th
thin tube at
휃
=
0
.
After
the screw is
rotated
about the
−
푌
axis
to an angle
훽
with respect to
horizontal
, the
푍
coordinate of this
point become
s
푍
=
−
푅
푖
·
cos
(
휃
)
cos
(
훽
)
+
[
퐿
휃
2
휋
+
풵
푖
]
sin
(
훽
)
.
(
9
)
The
corresponding
expression for
푋
with the screw inclined
also becomes a
function of
훽
, but it will not be needed
here
.
A thin tube such as shown in Figure
6c will either be entirely empty, or it will dip
into and then
emerge
out of the water at points 1 and 2, respectively
, at
angles
휃
푖
1
and
휃
푖
2
.
The volume
of
water in the pocket is
given
by
summation
over the
푁
thin
tubes:
푉
=
∑
푅
푖
(
휃
푖
2
−
휃
푖
1
)
퐴
푖
푁
푖
=
1
,
(
10
)
where
퐴
푖
is the area of the
푖
th
thin
tube
as determined from a cross
-
section
parallel
to the axis of the screw
,
such as shown in Figure 6a
.
The
angles
휃
푖
1
and
휃
푖
2
are
found
numerically
by solving Equation
9
with
푍
set to
푍
푠푢푟
.
A search that begins
12
at
휃
=
−
휋
and proceeds
by increasing
휃
will encounter point 1 first, then point 2.
The lat
t
er can occur for positive
휃
.
No solution means that the pocket is empty, its
bottom being above the water surface.
The total volume of water
carried by the
screw
should include the water in all the pockets.
To determine the maximum amount of water
a
screw
pocket can hold
, i.e.
,
the
pocket volume,
the
spill
elevation
should be used for
푍
푠푢푟
.
For a blade screw, the
peak of the
helical line made by the lower
blade
of the channel
adjacent to the shaft
is the spill point
,
and
it
is marked in Figure
4
; see the previous
discussion in
S
ection
3
.
This elevation is found from Equation 9 modified with
푅
퐼
,
휃
푠푝
and
풵
퐿
as follows:
푍
푠푝
=
−
푅
퐼
·
cos
(
휃
푠푝
)
cos
(
훽
)
+
[
퐿
휃
푠푝
2
휋
+
풵
퐿
]
sin
(
훽
)
,
(
11
)
where
휃
푠푝
=
−
휋
+
arcsin
(
퐿
2
휋푅
퐼
tan
(
훽
)
)
,
(
12
)
and
푠푝
denotes spill point.
풵
퐿
is the elevation of the
lower blade
at
휃
=
0
with the
screw in the vertical position.
Equation 12
comes from taking the derivative of
Equation 9
with respect to
휃
, setting it equal to zero, then solving for
휃
=
휃
푠푝
.
If the
water
volume at a lower water surface elevation is desired, the elevation of
the bottom of the pocket should be known for reference. This elevation
푍
푏표푡
is
given by Equation 11 with
푅
퐼
replaced by
푅
푂
and
휃
푠푝
replaced by
휃
푏표푡
:
푍
푏표푡
=
−
푅
푂
·
cos
(
휃
푏표푡
)
cos
(
훽
)
+
[
퐿
휃
푏표푡
2
휋
+
풵
퐿
]
sin
(
훽
)
,
(
13
)
where
휃
푏표푡
=
−
arcsin
(
퐿
2
휋
푅
푂
tan
(
훽
)
)
,
(
14
)
and
푏표푡
denotes bottom point.
The angle
휃
푏표푡
also comes from
taking
the
derivative of Equation 9
with respect to
휃
and setting it
to zero.
13
S
pill
and bottom
point
s
for a blade screw
occur in the
general area
s
indicated in
Figure 4.
For a tube
screw
, the
se
elevation
s
require a broader
search.
A computer code is presented in Appendix 3 for finding the
water
volume for a
blade screw using the above method
, except that
휃
푏표푡
is used as the starting angle
in the search for
휃
푖
1
and
휃
푖
2
.
Section
5
. Comparison with
other
calculation method
s
for pocket volume
5
.1
Rorres
(
9
)
An algorithm
to compute the
pocket
volume for a screw of the blade type
by
numerical integration
is described in the appendix of
Rorres’
2000
paper. The
procedure
determine
s
the areas of the water for cross
-
sections perpendicular to the
axis of the s
crew
and then integrate
s
these areas along th
is
axis over the extent of
the pocket. The geometries of these cross
-
sections are complex, comparable to
what is shown in Figure
4
for horizontal cross
-
sections. The author mentions the
need to be careful with bookkeeping and choosing various angles in their correct
ranges.
In the reference, n
umerical results are given for an example with
tan
(
훽
)
=
0.75,
푅
푂
=
2
푅
퐼
,
퐿
=
2
휋
푅
퐼
,
a
nd
푁
퐻
=
8
(same parameters as used in Figure 4)
.
Equation 7 is satisfied.
The
pocket volume
is stated
as equivalent
to
푉
=
0.1703
·
휋
푅
푂
2
퐿
푁
퐻
.
[This equation only applies for
푅
푂
,
퐿
and
푁
퐻
as given, not arbitrary
values of these parameters.]
Using the method described
in
S
ection
4
, the same
result
to all
4
significant digits
is obtained when the channel is divided into an
arrangement of
16
(high) by
20
(wide) thin tubes of
equal
rectangular cross
-
section
(
푁
=
320
). For a
n
8
by
10
arrangement
(
푁
=
80
),
0.1706
is obtained instead of
0
.1703, still very accurate.
Computation times on a modern laptop are trivial.