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Analysis of the symmetric Rhombic Drive
Figure 1 : Schematics of symmetric rhombic drivewith power
piston connected to the upper anddisplacer to the lower
section.
In Figure 1 we show theschematics of a rhombic drivewith power
piston connected
to the upper and displacer tothe lower section. The bars12,
1'2', 13, and 1'3' haveidentical length, L, and areconnected to the
cross-bars22' and 33' by pin/holeconnections. Joints 1 and 1'are
pin-hole connections aswell. The crank throws 01 and0'1' have
identical length, r,and the crank centers 0 and0' are at equal
distance, d+e,
from the piston axis. The sizeof the cross bars, 22' and 33',d,
is has no bearing on theanalysis of the variation of theheight of
the expansionspace,Ve, and thecompression space,Vc, as theangle
changes.In the configuration shown,the left crank will
turnclockwise and the right crankcounter clockwise in order
toachieve proper phase lagbetween expansion andcompression space.
They turnat the same angular velocitywhich can be accomplished
bytwo intermeshing counter-rotating gears.
General Geometric Relations
(1) Ve = Ce - Rsin() + L sin(&beta);
(2) Vc = Cc -2 L sin()
The angles and are related to each other by :
(3) e = Rcos() + L cos()
The constants Ce and Cc will be set such that the minima of Ve
and Vc are zero, respectively.Also note that the angle occurs
between the horizontal and each of the 4 bars, 12, 1'2', 13,1'3',
respectively. Finally, the crank can make complete revolutions only
if :
(4) L > R + e
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Determine Ce for Eq.(2) based on min{Ve}=0
It can be shown ( set derivative with respect to of Eq. (1) to
zero and taking Eq. (3) intoaccount ) that the minimum of Ve occurs
when in Fig. 1 the points 1 and 3 are exactly onopposite sides of
crank center 0, with point 1 above and to the left of point 0.
For this position the angles and can be determined :
(5) cos() = e/(L - R)
(6) = 180 -
Substituting this into Equation (1) and setting the constant Ce
such that the minimum of Vebecomes zero we get :
(7) Ve = -(L-R) sqrt [ 1 - e/(L-R) ] - Rsin() + L
sin(&beta);
Maximum of Ve
It can be shown ( set derivative with respect to of Eq. (1) to
zero and taking Eq. (3) intoaccount ) that the maximum of Ve occurs
when in Figure 1 the points 1 and 3 are exactly inline with crank
center 0 but this time both are below and to the right of point
0.For this position the angles and can be determined :
(8) cos() = e/(L + R)
(9) = 360 -
Substituting this into Equation (7) we get for the maximum value
of Ve :
(10) max{Ve} = sqrt [ (L+R) - e ] - sqrt [ (L-R) -e ]
Determine Cc for Eq.(2) based on min{Vc}=0
We have to distinguish between the case eRand e
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Eq. (13) has two solutions, one on either side of &alpha=0
with a local maximum for V c inbetween at &alpha=0. Again, we
demand min{Vc}=0 :
(14) Vc = 2 L ( 1 - sin(&beta) )
Maximum of Vc
Vc reaches its maximum when =180 and :
(15) cos() = (e+R)/L
(16) max{Vc} = 2 L ( sqrt[ 1 - ((e-R)/L)] - sqrt[ 1 - ((e+R)/L)]
) (eR )
= 2 L ( 1 - sqrt[ 1 - ((e+R)/L)] ) ( e
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= 180 - acos ( /(1-) ) ( for the minima )
In Eq. (20) we took =0 as the angle at which Vc reaches its
minimum regardless of thespecial circumstances for
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Last revised: 06/01/05
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