3. 1 Input situation: ir, ωI, MO, FCO
1.1 Condition
Input parameters: ir, ωI, MO, FCO
Unknowns: ωO, v0
1.2 Simplification
1.2.1 The simplification of the first Equation (2.214)
From 2.216 we can get
(LOin
− K) eµ∗
ΦO
=
MO
rO
+ (LOin − K) . (1)
For equation 2.214, we can substitute (LOin − K) eµ∗
ΦO
by (1) and then insert equation 2.218. So
we get
rIωI =
v0MO
EArO
+ rOωO . (2)
The equality of torques are
MO = rO(LOout
− LOin
) , (3)
MI = rI(LIout
− LIin
) . (4)
As LOin
= LIout
and LOout
= LIin
, we can get
MO = −MI
rO
rI
. (5)
With the help of Equation (5), (2) can be also wrote in the form of
rIωI = −
v0MI
EArI
+ rOωO. (6)
These two equations (2) and (6) build a bridge between ωI, ωO and MO, MI. From (2) (6), we can
get:
ωI =
rOωO
rI
−
v0MI
EArI
2
, (7)
ωI =
rOωO
rI
+
v0MO
EArIrO
. (8)
We also can get:
ωO =
rIωI
rO
+
v0MI
EArIrO
, (9)
ωO =
rIωI
rO
−
v0MO
EArO
2
. (10)
We have to use one of these four equations (7) (8) (9) and (10) to represent ωI or ωO in the
situation when it is not provided by the input conditions.
3
4. 1.2.2 The simplification of the second Equation (2.217)
From 2.216 we also can get
(LOin
− K) eµ∗
ΦO
− 1 =
MO
rO
(11)
From 2.218 and the formula of K we can get
(LOin
− K)
EA
=
rOωO
v0
−
EA
EA − m∗v0
2
(12)
For equation 2.217, we can substitute (LOin
− K) eµ∗
ΦO
− 1 by (11) and (LOin
− K) /EA by
(12), then insert 2.218. So we get
2FCO
tan(δ0) = EA − m∗
v0
2
2ϕ
rOωO
v0
− 1 +
MO
µ∗rOEA
−
rOωO
v0
−
EA
EA − m∗v0
2
ΦO −2m∗
v0
2
ϕ
(13)
1.2.3 Combine the equation (2) and the equation (13) into one equation
From (2), we can get:
MO =
(rIωI − rOωO)rOEA
v0
. (14)
Insert (12) and (14) in to 2.221, we can get
ΦO =
1
µ∗
ln
(rI ωI −rOωO)rOEA
v0
rOEA(rOωO
v0
− EA
EA−m∗v0
2 )
+ 1 =
1
µ∗
ln
rIωI(EA − m∗
v0
2
) − EAv0
rOωO(EA − m∗v0
2) − EAv0
(15)
Insert (14) and (15) in to (13), we can get
2FCO
tan(δ0) = EA − m∗
v0
2
2ϕ
rOωO
v0
− 1 +
rIωI − rOωO
µ∗v0
−
rOωO
v0
−
EA
EA − m∗v0
2
1
µ∗
ln
rIωI(EA − m∗
v0
2
) − EAv0
rOωO(EA − m∗v0
2) − EAv0
− 2m∗
v0
2
ϕ
(16)
We can use this equation (16) as the main function to be solved as soon as the value of FCO
is
provided in the input parameters.
In this situation, with the help of (10), (16) only depends on the unknown v0 and can be solved
with the starting value provided by 2.220.
1.3 How to calculate v0 when MO = 0
Condition: ΦI = ΦO, FCO
= 0, LIin
= LOin
= 0
Insert ΦO = 0 and 2.219 in to 2.217 and simplify it, we can get
rIωIϕm∗
v0
2
+ (FCO
tan(δ0) + EAϕ)v0 − EArIωIϕ = 0 (17)
And as v0 ≥ 0, solving the quadratic equation in v0 (17),we can get v0 described in 2.220.
4
5. 2 Input situation: ir, ωO, MI, FCI
2.1 Condition
Input parameters: ir, ωO, MI, FCI
Unknowns: ωI, v0
2.2 Basic formulas
From 2.214,we can get
LIin
=
EArIωI
v0
− EA . (18)
By 2.211 and the relationship of LIout
and LIin
we can get
LIout − LIin = [LIin − K]e−µ∗
ΦI
+ K − LIin
= (LIin
− K)(e−µ∗
ΦI
− 1).
(19)
As the longitudinal force decreases from LIin
from LIout
, we choose the negative sign when using
the formula 2.211.
The equality of torque MI is
MI = rI(LIout − LIin )
= rI(LIin
− K)(e−µ∗
ΦI
− 1) .
(20)
And the axial equality of force on the input pulley is
FCI
= −
ϕI
S dθ
= −
ϕI
L(EA − m∗
v0
2
) − m∗
v0
2
EA
2 tan(δ0)EA
dθ
= −
EA − m∗
v0
2
2 tan(δ0)EA
π−ϕ−ΦI
−(π−ϕ)
Ldθ +
π−ϕ
π−ϕ−ΦI
Ldθ +
m∗
v0
2
(π − ϕ)
tan(δ0)
= −
EA − m∗
v0
2
2 tan(δ0)EA
LIin
(2π − 2ϕ − ΦI) +
π−ϕ
π−ϕ−ΦI
(LIin
− K)e−µ∗
(θ−(π−ϕ−ΦI ))
+ K dθ +
m∗
v0
2
(π − ϕ)
tan(δ0)
= −
EA − m∗
v0
2
2 tan(δ0)EA
LIin
(2π − 2ϕ − ΦI) + (LIin
− K)
1
−µ∗
eµ∗
(π−ϕ−ΦI −θ)
+ Kθ
π−ϕ
π−ϕ−ΦI
+
m∗
v0
2
(π − ϕ)
tan(δ0)
= −
EA − m∗
v0
2
2 tan(δ0)EA
2LIin
(π − ϕ) + (LIin
− K)
e−µ∗
ΦI
− 1
−µ∗
− ΦI +
m∗
v0
2
(π − ϕ)
tan(δ0)
.
(21)
2.3 Simplification
2.3.1 The simplification of the Equation (21)
Equation (21) can be simplified the same way as (13). So we can get
2FCI
tan(δ0) = − EA − m∗
v0
2
2(π − ϕ)
rIωI
v0
− 1 +
MI
(−µ∗)rIEA
−
rIωI
v0
−
EA
EA − m∗v0
2
ΦI +2m∗
v0
2
(π−ϕ)
(22)
5
6. 2.3.2 Combine the equation (6) and the equation (22) into one equation
If MO = 0, it is necessarily ΦI = ΦO and LOin
= K. So
ΦI = −
1
µ∗
ln
MI
rI(LIin
− K)
+ 1 (23)
is only defined for MI ≥ rI(K − LIin
).
From (6), we can get:
MI = −
(rIωI − rOωO)rIEA
v0
. (24)
From (18) and the formula of K we can get
(LOin − K) =
rIωI
v0
−
EA
EA − m∗v0
2
EA (25)
Insert (24) and (25) in to (23), we can get
ΦI =
1
−µ∗
ln −
(rI ωI −rOωO)rI EA
v0
rIEA(rI ωI
v0
− EA
EA−m∗v0
2 )
+ 1 =
1
−µ∗
ln
rOωO(EA − m∗
v0
2
) − EAv0
rIωI(EA − m∗v0
2) − EAv0
(26)
Insert (24) and 26 in to (22), we can get
2FCI
tan(δ0) = − EA − m∗
v0
2
2(π − ϕ)
rIωI
v0
− 1 +
rOωO − rIωI
−µ∗v0
−
rIωI
v0
−
EA
EA − m∗v0
2
1
−µ∗
ln
rOωO(EA − m∗
v0
2
) − EAv0
rIωI(EA − m∗v0
2) − EAv0
+ 2m∗
v0
2
(π − ϕ)
(27)
We can use this equation (27) as the main function to be solved as soon as the value of FCI
is
provided in the input parameters.
2.4 How to calculate v0 when MO = 0
Condition: ΦI = ΦO, FCI
= 0, LIin
= 0
Insert ΦO = 0 and 2.219 in to (21) and simplify it, we can get
rOωO(π − ϕ)m∗
v0
2
+ (−FCI
tan(δ0) + EA(π − ϕ))v0 − EArOωO(π − ϕ) = 0 (28)
And as v0 ≥ 0, solving the equation (28),we can get
v0 =
(−FCI
tan(θ0) + EA(π − ϕ))
2
+ 4EAm∗rO
2ωO
2(π − ϕ)
2
− EA(π − ϕ) + FCI
tan(δ0)
2m∗rOωO(π − ϕ)
(29)
2.5 How to calculate v0 when MO = 0
If MI = 0, it means that the output torque MO = 0. With the help of Equation (7), equation (27)
only depends on the unknown v0 and can be solved with the starting value provided by (29).
6
7. 3 Input situation: ir, ωI, ωO, FCO
3.1 Condition
Input parameters: ir, ωI, ωO, FCO
Unknowns: MO, v0
3.2 How to calculate v0 when MO = 0
Condition: ΦI = ΦO, FCO
= 0, LIin = LOin = 0
If ωO/ωI = rI/rO, it means that the output torque MO = 0. Then we can get v0 by 2.220.
3.3 How to calculate v0 when MO = 0
If ωO/ωI = rI/rO, it means that the output torque MO = 0. We can solve equation (16) in
unknown v0. We can choose 2.220 as the starting value of v0.
4 Input situation: ir, ωI, ωO, FCI
4.1 Condition
Input parameters: ir, ωI, ωO, FCI
Unknowns: MI, v0
4.2 How to calculate v0 when MO = 0
Condition: ΦI = ΦO, FCI
= 0, LIin = LOin = 0
If ωO/ωI = rI/rO, it means that the output torque MO = 0. Then we can get v0 by (29).
4.3 How to calculate v0 when MO = 0
If ωO/ωI = rI/rO, it means that the output torque MO = 0. we can solve equation (27) in
unknown v0. We can choose (29) as the starting value of v0.
5 Input situation: ir, ωI, MO, FCI
5.1 Condition
Input parameters: ir, ωI, MO, FCI
Unknowns: ωO, v0
5.2 How to calculate v0 when MO = 0
Condition: ΦI = ΦO, FCI
= 0, LIin = LOin = 0
If the output torque MO = 0 , then we can get v0 by inserting 2.219 into (29):
v0 =
(−FCI
tan(θ0) + EA(π − ϕ))
2
+ 4EAm∗rI
2ωI
2(π − ϕ)
2
− EA(π − ϕ) + FCI
tan(δ0)
2m∗rIωI(π − ϕ)
. (30)
7
8. 5.3 How to calculate v0 when MO = 0
With the help of Equation (10), equation (27) only depends on the unknown v0. We can choose
(30) as the starting value of v0.
6 Input situation: ir, ωI, MI, FCO
6.1 Condition
Input parameters: ir, ωI, MI, FCO
Unknowns: ωO, v0
6.2 How to calculate v0 when MO = 0
Condition: ΦI = ΦO, FCO
= 0, LIin = LOin = 0
If the input torque MI = 0 ,it means that the output torque MO = 0. Then we can get v0 by 2.220.
6.3 How to calculate v0 when MO = 0
If the input torque MI = 0. it means that the output torque MO = 0. With the help of Equation
(9), equation (16) only depends on the unknown v0 and can be solved with the starting value
provided by 2.220.
7 Input situation: ir, ωI, MI, FCI
7.1 Condition
Input parameters: ir, ωI, MI, FCI
Unknowns: ωO, v0
7.2 How to calculate v0 when MO = 0
Condition: ΦI = ΦO, FCI
= 0, LIin = LOin = 0
If the input torque MI = 0 , it means that the output torque MO = 0. Then we can get v0 by (30).
7.3 How to calculate v0 when MO = 0
If the input torque MI = 0. it means that the output torque MO = 0. With the help of Equation
(9), equation (27) only depends on the unknown v0 and can be solved with the starting value
provided by (30).
8 Input situation: ωI, ωO, FCI
, FCO
8.1 Condition
Input parameters: ωI, ωO, FCI
, FCO
Unknowns: rO, ϕ and v0.
8
9. 8.2 How to calculate v0 when MO = 0 and MO = 0
If ωO/ωI = rI/rO, it means that the output torque MO = 0. From 2.187, we can get
rI = rO + cos(ϕ)dA . (31)
Similar to 2.189, we can get the starting value of rO is
rOS
= (0.5 lR − dA
2
+ dalign
2
)/pi . (32)
Then with the help of (31), equations 2.188, (16), (27) only depend on unknowns rO, ϕ and v0. We
can choose equations (32), 2.190 and (29) as the starting value of rO, ϕ and v0.
9 Input situation: ωI, MI, FCI
, FCO
9.1 Condition
Input parameters: ωI, MI, FCI
, FCO
Unknowns: rO, ϕ and v0.
9.2 How to calculate v0 when MO = 0 and MO = 0
If the input torque MI = 0. it means that the output torque MO = 0. Then with the help of (9),
equations 2.188, (16), (27) only depend on unknowns rO, ϕ and v0. We can choose equations (32),
2.190 and (30) as the starting value of rO, ϕ and v0.
10 Input situation: ωI, MO, FCI
, FCO
10.1 Condition
Input parameters: ωI, MO, FCI
, FCO
Unknowns: rO, ϕ and v0.
10.2 How to calculate v0 when MO = 0 and MO = 0
If the input torque MO = 0, then with the help of (10), equations 2.188, (16), (27) only depend on
unknowns rO, ϕ and v0. We can choose equations (32), 2.190 and (30) as the starting value of rO,
ϕ and v0.
9
10. A For the situation 8
In matlab, the code can give a nearly correct solution.
• rI part = (rI ∗ omgI ∗ (E ∗ A − mstar ∗ v0
2
) − E ∗ A ∗ v0) =
2.727524235557020e + 04 − 5.313356954436232e − 11i
• rO part = (rO ∗ omgO ∗ (E ∗ A − mstar ∗ v0
2
) − E ∗ A ∗ v0) =
1.942531552161276e + 04 + 2.744090420739602e − 11i
• log(rI part/rO part) = 0.339402278119434 − 0.000000000000003i
• phiO = log(rI part/rO part)/mustar = 0.588736433786180 − 0.000000000000006i
• phiO degree = 180/π ∗ phiO = 0.588736433786180 − 0.000000000000006i
• phiO degree ref = 33.732112521095750
• v0 = 9.635874606404668 − 0.000000000000000i
• v0ref = 9.635874561660197
In the C++ code, when rI part/rO part < 0, it cannot further calculate the logarithm value.
10
