are nearly satisfied. Approximate values of
r
1
,
r
2
,
r
3
{\displaystyle r_{1},r_{2},r_{3}}
VII.
Test of the first hypothesis.
log
r
1
,
log
r
2
,
log
r
3
{\displaystyle \log r_{1},\log r_{2},\log r_{3}}
N
=
A
1
B
1
r
1
−
3
+
B
2
r
2
−
3
+
B
2
r
2
−
3
+
A
3
B
3
r
3
−
3
s
1
=
A
1
r
1
+
A
1
B
1
r
1
−
2
s
2
=
r
2
−
B
2
r
2
−
2
s
3
=
A
3
r
3
+
A
3
B
3
r
3
−
2
s
=
1
2
(
s
1
+
s
2
+
s
3
)
s
−
s
1
,
s
−
s
,
s
−
s
3
.
{\displaystyle {\begin{aligned}N&=A_{1}B_{1}r_{1}^{-3}+B_{2}r_{2}^{-3}+B_{2}r_{2}^{-3}+A_{3}B_{3}r_{3}^{-3}\\s_{1}&=A_{1}r_{1}+A_{1}B_{1}r_{1}^{-2}\\s_{2}&=r_{2}-B_{2}r_{2}^{-2}\\s_{3}&=A_{3}r_{3}+A_{3}B_{3}r_{3}^{-2}\\s={\tfrac {1}{2}}(s_{1}+s_{2}+s_{3})\\s&-s_{1},s-s_{},s-s_{3}.\end{aligned}}}
The value of
s
−
s
2
{\displaystyle s-s_{2}}
R
{\displaystyle _{\text{R}}}
=
(
s
−
s
1
)
(
s
−
s
2
)
(
s
−
s
3
)
s
{\displaystyle ={\sqrt {\frac {(s-s_{1})(s-s_{2})(s-s_{3})}{s}}}}
tan
1
2
(
v
2
−
v
1
)
{\displaystyle \tan {\tfrac {1}{2}}(v_{2}-v_{1})}
=
R
s
−
s
2
{\displaystyle {\frac {_{\text{R}}}{s-s_{2}}}}
p
{\displaystyle p}
=
2
(
s
−
s
2
)
N
{\displaystyle ={\frac {2(s-s_{2})}{N}}}
tan
1
2
(
v
3
−
v
2
)
{\displaystyle \tan {\tfrac {1}{2}}(v_{3}-v_{2})}
=
R
s
−
s
1
{\displaystyle ={\frac {_{\text{R}}}{s-s_{1}}}}
tan
1
2
(
v
3
−
v
1
)
{\displaystyle \tan {\tfrac {1}{2}}(v_{3}-v_{1})}
=
s
−
s
2
R
{\displaystyle ={\frac {s-s_{2}}{_{\text{R}}}}}
For adjustment of values:
1
2
(
v
3
−
v
1
)
=
1
2
(
v
2
−
v
1
)
+
1
2
(
v
3
−
v
2
)
{\displaystyle {\tfrac {1}{2}}(v_{3}-v_{1})={\tfrac {1}{2}}(v_{2}-v_{1})+{\tfrac {1}{2}}(v_{3}-v_{2})}
e
sin
1
2
(
v
3
+
v
1
)
=
p
r
1
−
p
r
3
2
sin
1
2
(
v
3
−
v
1
)
e
cos
1
2
(
v
3
+
v
1
)
=
p
r
1
+
p
r
3
−
2
2
cos
1
2
(
v
3
−
v
1
)
tan
1
2
(
v
3
+
v
1
)
e
2
{\displaystyle {\begin{aligned}e\sin {\tfrac {1}{2}}(v_{3}+v_{1})&={\frac {{\frac {p}{r_{1}}}-{\frac {p}{r_{3}}}}{2\sin {\tfrac {1}{2}}(v_{3}-v_{1})}}\\e\cos {\tfrac {1}{2}}(v_{3}+v_{1})&={\frac {{\frac {p}{r_{1}}}+{\frac {p}{r_{3}}}-2}{2\cos {\tfrac {1}{2}}(v_{3}-v_{1})}}\\\tan {\tfrac {1}{2}}(v_{3}+v_{1})&e^{2}\end{aligned}}}
ϵ
=
1
−
e
1
+
e
{\displaystyle \epsilon ={\sqrt {\frac {1-e}{1+e}}}}
a
=
p
1
−
e
2
{\displaystyle a={\frac {p}{1-e^{2}}}}
tan
1
2
E
1
=
ϵ
tan
1
2
v
1
{\displaystyle \tan {\tfrac {1}{2}}E_{1}=\epsilon \tan {\tfrac {1}{2}}v_{1}}
tan
1
2
E
2
=
ϵ
tan
1
2
v
2
{\displaystyle \tan {\tfrac {1}{2}}E_{2}=\epsilon \tan {\tfrac {1}{2}}v_{2}}
tan
1
2
E
3
=
ϵ
tan
1
2
v
3
{\displaystyle \tan {\tfrac {1}{2}}E_{3}=\epsilon \tan {\tfrac {1}{2}}v_{3}}
τ
1
calc.
=
a
3
2
(
E
3
−
E
2
)
+
e
a
3
2
sin
E
2
−
e
a
3
2
sin
E
3
τ
2
calc.
=
a
3
2
(
E
2
−
E
1
)
+
e
a
3
2
sin
E
1
−
e
a
3
2
sin
E
2
{\displaystyle {\begin{aligned}\tau _{1{\text{ calc.}}}&=a^{\frac {3}{2}}(E_{3}-E_{2})+ea^{\frac {3}{2}}\sin E_{2}-ea^{\frac {3}{2}}\sin E_{3}\\\tau _{2{\text{ calc.}}}&=a^{\frac {3}{2}}(E_{2}-E_{1})+ea^{\frac {3}{2}}\sin E_{1}-ea^{\frac {3}{2}}\sin E_{2}\end{aligned}}}
