80
B. M˘ anescu et al.
• for the rigid body 2:
m ¨
x C 2 = −R B x + R A x − R C x , m ¨
y C 2 = −R B y − R C y + R A y − m 2 g,
(35)
J 2 ε 2 = −
x B − x C 2
R B y +
y B − y C 2
R B x +
x A − x C 2
R A y −
y A − y C 2
R A x
−
x C − x C 2
R C y +
y C − y C 2
R C x ;
(36)
• for the rigid body 1:
m ¨
x C 1 = −R A x + R O x , m ¨
y C 1 = −R A y + R O y − m 1 g,
(37)
J O ε 1 = −(x A − x O )R A y + (y A − y O )R A x +
x C 1 − x O
m 1 g − M e1 .
(38)
Assuming that the law of rotational motion of the crank OA is known,
var phi 1 (t) = ωt, (30)–(38) form a system of 14 linear equations with 14 unknowns
(the reactions R O x , R O y , R A x , R A y , R B x , R B y , R C x , R C y , R D x , R D y , R E x , R E y , N , and
the equilibration moment M e1 ).
4 Numerical Example
For the numerical application, we consider the following values: AB = 0.043 m,
BC = 0.128 m, CA = 0.099 m, CE = 0.103 m, OA = 0.030 m, BD = 0.130 m, d =
0.086 m, e = 0 m, Y E ∈ [0.100, 0.120] m, ω 1 = 100π rad/s, ϕ 1 ∈ [0, 2π ] rad, the
diameter of the piston d p = 0.075 m, position of the cylinder head y ch = 0.2088 m,
the thickness of the elements thick = 0.008 m, and the density of the material
ρ = 7800 kg/m
3 . The iteration steps are: dY E = 0.00025 m, and dϕ 1 =
π
180
rad.
Some characteristic diagrams are captured in Fig. 3.
Analyzing the diagrams, we may state that:
• not all the reactions follow the same rule (increasing or decreasing) when the
parameter Y E increases. One may say nothing about the shape of these laws of
variation;
• when Y E increases, the axial force N (and consequently, the wear decreases), and
the equilibration moment M e 1 (and consequently, the necessary inertial moment
of the flywheel) decrease;
• the compression ratio decreases when the parameter Y E increases, the variation
being almost a linear one;
B. M˘ anescu et al.
• for the rigid body 2:
m ¨
x C 2 = −R B x + R A x − R C x , m ¨
y C 2 = −R B y − R C y + R A y − m 2 g,
(35)
J 2 ε 2 = −
x B − x C 2
R B y +
y B − y C 2
R B x +
x A − x C 2
R A y −
y A − y C 2
R A x
−
x C − x C 2
R C y +
y C − y C 2
R C x ;
(36)
• for the rigid body 1:
m ¨
x C 1 = −R A x + R O x , m ¨
y C 1 = −R A y + R O y − m 1 g,
(37)
J O ε 1 = −(x A − x O )R A y + (y A − y O )R A x +
x C 1 − x O
m 1 g − M e1 .
(38)
Assuming that the law of rotational motion of the crank OA is known,
var phi 1 (t) = ωt, (30)–(38) form a system of 14 linear equations with 14 unknowns
(the reactions R O x , R O y , R A x , R A y , R B x , R B y , R C x , R C y , R D x , R D y , R E x , R E y , N , and
the equilibration moment M e1 ).
4 Numerical Example
For the numerical application, we consider the following values: AB = 0.043 m,
BC = 0.128 m, CA = 0.099 m, CE = 0.103 m, OA = 0.030 m, BD = 0.130 m, d =
0.086 m, e = 0 m, Y E ∈ [0.100, 0.120] m, ω 1 = 100π rad/s, ϕ 1 ∈ [0, 2π ] rad, the
diameter of the piston d p = 0.075 m, position of the cylinder head y ch = 0.2088 m,
the thickness of the elements thick = 0.008 m, and the density of the material
ρ = 7800 kg/m
3 . The iteration steps are: dY E = 0.00025 m, and dϕ 1 =
π
180
rad.
Some characteristic diagrams are captured in Fig. 3.
Analyzing the diagrams, we may state that:
• not all the reactions follow the same rule (increasing or decreasing) when the
parameter Y E increases. One may say nothing about the shape of these laws of
variation;
• when Y E increases, the axial force N (and consequently, the wear decreases), and
the equilibration moment M e 1 (and consequently, the necessary inertial moment
of the flywheel) decrease;
• the compression ratio decreases when the parameter Y E increases, the variation
being almost a linear one;