Die Kettenlinie (oder Katenoide) ist die geometrische Form, welche eine an ihren beiden Enden abgehängte Kette oder ein entsprechendes Seil unter der Belastung durch ihr Eigengewicht ausbildet. Das Eigengewicht sei dabei gleichmäßig über die Gesamtlänge der Kette (des Seiles) verteilt.
q
=
m
g
L
{\displaystyle q={\frac {mg}{L}}}
Gleichgewicht am Kettenelement
Bearbeiten
∑
F
x
=
0
:
−
H
+
H
+
d
H
=
0
{\displaystyle \sum F_{x}=0:\ -H+H+dH=0}
⇒
{\displaystyle \Rightarrow }
H
=
konst.
{\displaystyle H\;=\;{\mbox{konst.}}}
(Gl.1)
∑
F
y
=
0
:
−
V
+
V
+
d
V
−
q
d
s
=
0
{\displaystyle \sum F_{y}=0:\ -V+V+dV-q\ ds=0}
⇒
{\displaystyle \Rightarrow }
d
V
d
s
=
q
{\displaystyle {\frac {dV}{ds}}=q}
(Gl.2)
∑
M
P
=
0
:
H
d
y
−
V
d
x
−
q
d
s
d
x
2
⏟
≪→
0
=
0
{\displaystyle \sum M_{P}=0:\ H\ dy-V\ dx-q\underbrace {\frac {ds\ dx}{2}} _{\ll \rightarrow 0}=0}
⇒
{\displaystyle \Rightarrow }
d
y
d
x
=
V
H
{\displaystyle {\frac {dy}{dx}}={\frac {V}{H}}}
(Gl.3)
Die Differentialgleichung der Kettenlinie
Bearbeiten
d
s
2
=
d
x
2
+
d
y
2
{\displaystyle ds^{2}\;=\;dx^{2}+dy^{2}}
d
s
d
x
=
1
+
(
d
y
d
x
)
2
{\displaystyle {\frac {ds}{dx}}={\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}}
aus (Gl.2)
d
s
=
d
V
q
{\displaystyle ds={\frac {dV}{q}}}
1
q
d
V
d
x
=
1
+
(
d
y
d
x
)
2
{\displaystyle {\frac {1}{q}}{\frac {dV}{dx}}={\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}}
aus (Gl.3)
d
V
d
x
=
H
d
2
y
d
x
2
{\displaystyle {\frac {dV}{dx}}=H{\frac {d^{2}y}{dx^{2}}}}
d
2
y
d
x
2
=
q
H
1
+
(
d
y
d
x
)
2
{\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {q}{H}}{\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}}
mit
a
=
H
q
=
konst.
{\displaystyle a={\frac {H}{q}}={\mbox{konst.}}}
d
2
y
d
x
2
=
1
a
1
+
(
d
y
d
x
)
2
{\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {1}{a}}{\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}}
(Gl.4)
Die geometrische Form der Kettenlinie
Bearbeiten
(Gl.4):
y
″
=
1
a
1
+
y
′
2
{\displaystyle y''={\frac {1}{a}}{\sqrt {1+y'^{2}}}}
d
y
′
d
x
=
1
a
1
+
y
′
2
{\displaystyle {\frac {dy'}{dx}}={\frac {1}{a}}{\sqrt {1+y'^{2}}}}
d
y
′
1
+
y
′
2
=
d
x
a
{\displaystyle {\frac {dy'}{\sqrt {1+y'^{2}}}}={\frac {dx}{a}}}
y
′
=
sinh
u
{\displaystyle y'\;=\;\sinh u}
d
y
′
=
cosh
u
d
u
{\displaystyle dy'\;=\;\cosh u\;du}
cosh
u
1
+
sinh
2
u
d
u
=
d
x
a
{\displaystyle {\frac {\cosh u}{\sqrt {1+\sinh ^{2}u}}}du={\frac {dx}{a}}}
∫
d
u
=
∫
d
x
a
{\displaystyle \int du=\int {\frac {dx}{a}}}
u
=
x
a
+
C
1
{\displaystyle u={\frac {x}{a}}+C_{1}}
arsinh
y
′
=
x
a
+
C
1
{\displaystyle {\mbox{arsinh}}\;y'={\frac {x}{a}}+C_{1}}
y
′
=
sinh
(
x
a
+
C
1
)
{\displaystyle y'=\sinh \left({\frac {x}{a}}+C_{1}\right)}
Die Position des Koordinatensystems kann frei gewählt werden. Zwecks Vereinfachung wird hier das Koordinatensystem so positioniert, dass die horizontale Tangente an die Kettenlinie genau bei der Koordinate (0, a) anliegt.
Es gilt somit
y
′
=
0
{\displaystyle y'\;=\;0}
x
=
0
{\displaystyle x\;=\;0}
0
=
sinh
(
C
1
)
⇒
C
1
=
0
{\displaystyle 0=\sinh \left(C_{1}\right)\ \Rightarrow \ C_{1}=0}
y
′
=
sinh
x
a
{\displaystyle y'=\sinh {\frac {x}{a}}}
(Gl.5)
∫
d
y
=
∫
sinh
x
a
d
x
{\displaystyle \int dy=\int \sinh {\frac {x}{a}}dx}
y
=
a
cosh
x
a
+
C
2
{\displaystyle y=a\cosh {\frac {x}{a}}+C_{2}}
y
=
a
{\displaystyle y\;=\;a}
x
=
0
{\displaystyle x\;=\;0}
a
=
a
⋅
1
+
C
2
⇒
C
2
=
0
{\displaystyle a=a\cdot 1+C_{2}\ \Rightarrow \ C_{2}=0}
y
=
a
cosh
x
a
{\displaystyle y=a\cosh {\frac {x}{a}}}
(Gl.6)
d
s
2
=
d
x
2
+
d
y
2
{\displaystyle ds^{2}\;=\;dx^{2}+dy^{2}}
(
d
s
d
x
)
2
=
1
+
sinh
2
x
a
=
cosh
2
x
a
{\displaystyle \left({\frac {ds}{dx}}\right)^{2}\;=\;1+\sinh ^{2}{\frac {x}{a}}=\cosh ^{2}{\frac {x}{a}}}
∫
d
s
=
∫
cosh
x
a
d
x
{\displaystyle \int ds=\int \cosh {\frac {x}{a}}\;dx}
s
=
a
sinh
x
a
+
C
3
{\displaystyle s=a\sinh {\frac {x}{a}}+C_{3}}
Randbedingung:
s
=
0
{\displaystyle s\;=\;0}
x
=
0
⇒
C
3
=
0
{\displaystyle x\;=\;0\quad \Rightarrow \quad C_{3}=0}
s
=
a
sinh
x
a
{\displaystyle s=a\sinh {\frac {x}{a}}}
(Gl.7)
cosh
2
x
a
−
sinh
2
x
a
=
1
=
(
y
a
)
2
−
(
s
a
)
2
{\displaystyle \cosh ^{2}{\frac {x}{a}}-\sinh ^{2}{\frac {x}{a}}=1=\left({\frac {y}{a}}\right)^{2}-\left({\frac {s}{a}}\right)^{2}}
y
2
=
a
2
+
s
2
{\displaystyle y^{2}\;=\;a^{2}+s^{2}}
(Gl.8)
aus (Gl.2):
∫
d
V
=
∫
q
d
s
{\displaystyle \int dV=\int q\;ds}
V
=
q
s
+
C
4
{\displaystyle V\;=\;qs+C_{4}}
Randbedingung: für
s
=
0
{\displaystyle s\;=\;0}
V
=
0
⇒
C
4
=
0
{\displaystyle V=0\quad \Rightarrow \quad C_{4}=0}
V
=
q
s
{\displaystyle V\;=\;qs}
(Gl.9)
sin
φ
=
V
S
=
s
y
{\displaystyle \sin \varphi ={\frac {V}{S}}={\frac {s}{y}}}
S
=
q
y
{\displaystyle S\;=\;qy}
(Gl.10)
H
=
S
2
−
V
2
=
q
y
2
−
s
2
{\displaystyle H={\sqrt {S^{2}-V^{2}}}=q{\sqrt {y^{2}-s^{2}}}}
H
=
q
a
=
konst.
{\displaystyle H\;=\;qa={\mbox{konst.}}}
(Gl.11)
Die Traktrix (Schleppkurve, Hundekurve) ist die Evolvente der Kettenlinie und die Kettenlinie ist die Evolute der Traktrix.
tan
η
=
d
y
t
r
d
x
t
r
=
y
t
r
x
−
x
t
r
{\displaystyle \tan \eta ={\frac {dy_{tr}}{dx_{tr}}}={\frac {y_{tr}}{x-x_{tr}}}}
y
t
r
2
+
(
x
−
x
t
r
)
2
=
a
2
{\displaystyle y_{tr}^{2}+\left(x-x_{tr}\right)^{2}=a^{2}}
d
y
t
r
d
x
t
r
=
y
t
r
a
2
−
y
t
r
2
{\displaystyle {\frac {dy_{tr}}{dx_{tr}}}={\frac {y_{tr}}{\sqrt {a^{2}-y_{tr}^{2}}}}}
∫
d
x
t
r
=
∫
a
2
−
y
t
r
2
y
t
r
d
y
t
r
{\displaystyle \int dx_{tr}=\int {\frac {\sqrt {a^{2}-y_{tr}^{2}}}{y_{tr}}}dy_{tr}}
x
t
r
=
a
2
−
y
t
r
2
−
a
ln
|
a
+
a
2
−
y
t
r
2
y
t
r
|
+
C
5
{\displaystyle x_{tr}={\sqrt {a^{2}-y_{tr}^{2}}}-a\ln \left|{\frac {a+{\sqrt {a^{2}-y_{tr}^{2}}}}{y_{tr}}}\right|+C_{5}}
Randbedingung: für
x
t
r
=
0
{\displaystyle x_{tr}\;=\;0}
y
t
r
=
a
⇒
C
5
=
0
{\displaystyle y_{tr}\;=\;a\quad \Rightarrow \quad C_{5}=0}
x
t
r
=
a
2
−
y
t
r
2
−
a
ln
|
a
+
a
2
−
y
t
r
2
y
t
r
|
{\displaystyle x_{tr}={\sqrt {a^{2}-y_{tr}^{2}}}-a\ln \left|{\frac {a+{\sqrt {a^{2}-y_{tr}^{2}}}}{y_{tr}}}\right|}
(Gl.12)
Eigentlich sind schon fast alle Formeln für die Kettenlinie bekannt. Ein kleines Problem gibt es noch: Wie soll der Wert für den Seilparameter
a
=
H
q
{\displaystyle a={\frac {H}{q}}}
Die Lösung für a muss mittels zusätzlicher Daten gefunden werden. Nachfolgend wird eine Lösung für die Vorgabe von Seillänge L und den Abständen der Aufhängepunkte, nämlich b und h gezeigt (Ketten- bzw. Seildreieck).
Gegeben sind L , b , h . Gesucht ist der Seilparameter a .
h
=
y
2
−
y
1
=
a
(
cosh
x
2
a
−
cosh
x
1
a
)
{\displaystyle h=y_{2}-y_{1}=a\left(\cosh {\frac {x_{2}}{a}}-\cosh {\frac {x_{1}}{a}}\right)}
(
h
a
)
2
=
cosh
2
x
2
a
+
cosh
2
x
1
a
−
2
cosh
x
2
a
cosh
x
1
a
{\displaystyle \left({\frac {h}{a}}\right)^{2}=\cosh ^{2}{\frac {x_{2}}{a}}+\cosh ^{2}{\frac {x_{1}}{a}}-2\cosh {\frac {x_{2}}{a}}\cosh {\frac {x_{1}}{a}}}
L
=
s
2
−
s
1
=
a
(
sinh
x
2
a
−
sinh
x
1
a
)
{\displaystyle L=s_{2}-s_{1}=a\left(\sinh {\frac {x_{2}}{a}}-\sinh {\frac {x_{1}}{a}}\right)}
(
L
a
)
2
=
sinh
2
x
2
a
+
sinh
2
x
1
a
−
2
sinh
x
2
a
sinh
x
1
a
{\displaystyle \left({\frac {L}{a}}\right)^{2}=\sinh ^{2}{\frac {x_{2}}{a}}+\sinh ^{2}{\frac {x_{1}}{a}}-2\sinh {\frac {x_{2}}{a}}\sinh {\frac {x_{1}}{a}}}
(
L
a
)
2
−
(
h
a
)
2
=
−
2
+
2
(
cosh
x
2
a
cosh
x
1
a
−
sinh
x
2
a
sinh
x
1
a
)
{\displaystyle \left({\frac {L}{a}}\right)^{2}-\left({\frac {h}{a}}\right)^{2}=-2+2\left(\cosh {\frac {x_{2}}{a}}\cosh {\frac {x_{1}}{a}}-\sinh {\frac {x_{2}}{a}}\sinh {\frac {x_{1}}{a}}\right)}
Mit Anwendung der Summationsformel
cosh
(
α
±
β
)
=
cosh
α
cosh
β
±
sinh
α
sinh
β
{\displaystyle \cosh \left(\alpha \pm \beta \right)=\cosh \alpha \cosh \beta \pm \sinh \alpha \ \sinh \beta }
(
L
a
)
2
−
(
h
a
)
2
=
−
2
+
2
cosh
(
x
2
a
−
x
1
a
)
=
2
(
cosh
b
a
−
1
)
{\displaystyle \left({\frac {L}{a}}\right)^{2}-\left({\frac {h}{a}}\right)^{2}=-2+2\cosh \left({\frac {x_{2}}{a}}-{\frac {x_{1}}{a}}\right)=2\left(\cosh {\frac {b}{a}}-1\right)}
Weitere Vereinfachung durch
cosh
2
α
=
cosh
2
α
+
sinh
2
α
{\displaystyle \cosh 2\alpha \;=\;\cosh ^{2}\alpha +\sinh ^{2}\alpha }
2
α
=
b
a
{\displaystyle 2\alpha \;=\;{\frac {b}{a}}}
(
L
a
)
2
−
(
h
a
)
2
=
2
(
cosh
2
b
2
a
+
sinh
2
b
2
a
−
1
)
{\displaystyle \left({\frac {L}{a}}\right)^{2}-\left({\frac {h}{a}}\right)^{2}=2\left(\cosh ^{2}{\frac {b}{2a}}+\sinh ^{2}{\frac {b}{2a}}-1\right)}
(
L
a
)
2
−
(
h
a
)
2
=
2
(
1
+
sinh
2
b
2
a
+
sinh
2
b
2
a
−
1
)
{\displaystyle \left({\frac {L}{a}}\right)^{2}-\left({\frac {h}{a}}\right)^{2}=2\left(1+\sinh ^{2}{\frac {b}{2a}}+\sinh ^{2}{\frac {b}{2a}}-1\right)}
(
L
a
)
2
−
(
h
a
)
2
=
4
sinh
2
b
2
a
{\displaystyle \left({\frac {L}{a}}\right)^{2}-\left({\frac {h}{a}}\right)^{2}=4\sinh ^{2}{\frac {b}{2a}}}
L
2
−
h
2
=
2
a
sinh
b
2
a
{\displaystyle {\sqrt {L^{2}-h^{2}}}=2a\sinh {\frac {b}{2a}}}
(Gl.13)
oder auch in dieser Form
L
2
−
h
2
b
=
k
=
sinh
b
2
a
b
2
a
{\displaystyle {\frac {\sqrt {L^{2}-h^{2}}}{b}}=k={\frac {\sinh {\frac {b}{2a}}}{\frac {b}{2a}}}}
(Gl.14)
ξ
=
b
2
a
{\displaystyle \xi ={\frac {b}{2a}}}
f
(
ξ
)
=
k
ξ
−
sinh
ξ
=
0
{\displaystyle f(\xi )\;=\;k\xi -\sinh \xi \;=\;0}
ξ
{\displaystyle \xi \;}
a kann
grafisch
per Nullstellenberechnung (Newtonsches Näherungsverfahren, o.ä.)
oder per Näherung durch Reihenentwicklung
sinh
ξ
=
ξ
+
1
3
!
ξ
3
+
1
5
!
ξ
5
+
…
{\displaystyle \sinh \xi =\xi +{\frac {1}{3!}}\xi ^{3}+{\frac {1}{5!}}\xi ^{5}+\ldots }
erfolgen.
s1 und s2 aus den Geometriebeziehungen (siehe Abb. am Anfang dieses Kettendreieck-Abschnittes) ermitteln.
h
=
y
2
−
y
1
=
s
2
2
+
a
2
−
s
1
2
+
a
2
{\displaystyle h=y_{2}-y_{1}={\sqrt {s_{2}^{2}+a^{2}}}-{\sqrt {s_{1}^{2}+a^{2}}}}
h
=
(
L
+
s
1
)
2
+
a
2
−
s
1
2
+
a
2
{\displaystyle h={\sqrt {(L+s_{1})^{2}+a^{2}}}-{\sqrt {s_{1}^{2}+a^{2}}}}
Nach einigen Umformungsschritten resultiert daraus die quadratische Gleichung
s
1
2
+
s
1
L
+
(
L
2
−
h
2
4
−
a
2
h
2
L
2
−
h
2
)
=
0
{\displaystyle s_{1}^{2}+s_{1}L+\left({\frac {L^{2}-h^{2}}{4}}-{\frac {a^{2}h^{2}}{L^{2}-h^{2}}}\right)=0}
s
1
1
,
2
=
−
L
2
±
h
2
1
+
4
a
2
L
2
−
h
2
{\displaystyle s_{1_{1,2}}=-{\frac {L}{2}}\pm {\frac {h}{2}}{\sqrt {1+{\frac {4a^{2}}{L^{2}-h^{2}}}}}}
s
1
=
−
L
2
+
h
2
1
+
4
a
2
L
2
−
h
2
{\displaystyle s_{1}=-{\frac {L}{2}}+{\frac {h}{2}}{\sqrt {1+{\frac {4a^{2}}{L^{2}-h^{2}}}}}}
(Gl.15)
s
2
=
L
+
s
1
=
L
2
+
h
2
1
+
4
a
2
L
2
−
h
2
{\displaystyle s_{2}=L+s_{1}={\frac {L}{2}}+{\frac {h}{2}}{\sqrt {1+{\frac {4a^{2}}{L^{2}-h^{2}}}}}}
(Gl.16)
(Gl.13) und der mathematischen Beziehung
coth
α
=
1
+
1
sinh
2
α
{\displaystyle \coth \alpha ={\sqrt {1+{\frac {1}{\sinh ^{2}\alpha }}}}}
s
1
=
−
L
2
+
h
2
coth
b
2
a
{\displaystyle s_{1}=-{\frac {L}{2}}+{\frac {h}{2}}\coth {\frac {b}{2a}}}
(Gl.17)
s
2
=
L
2
+
h
2
coth
b
2
a
{\displaystyle s_{2}={\frac {L}{2}}+{\frac {h}{2}}\coth {\frac {b}{2a}}}
(Gl.18)
tan
γ
=
h
b
{\displaystyle \tan \gamma ={\frac {h}{b}}}
y
~
+
f
=
y
1
+
(
x
~
−
x
1
)
tan
γ
{\displaystyle {\tilde {y}}+f=y_{1}+({\tilde {x}}-x_{1})\tan \gamma }
y
′
(
x
~
)
=
tan
γ
=
h
b
=
sinh
x
~
a
=
s
~
a
{\displaystyle y'({\tilde {x}})=\tan \gamma ={\frac {h}{b}}=\sinh {\frac {\tilde {x}}{a}}={\frac {\tilde {s}}{a}}}
x
~
=
a
arsinh
s
~
a
=
a
arsinh
h
b
{\displaystyle {\tilde {x}}=a\;{\mbox{arsinh}}{\frac {\tilde {s}}{a}}=a\;{\mbox{arsinh}}{\frac {h}{b}}}
y
~
=
a
cosh
x
~
a
=
a
cosh
(
arsinh
h
b
)
=
a
cosh
(
arcosh
h
2
b
2
+
1
)
=
a
h
2
b
2
+
1
{\displaystyle {\tilde {y}}=a\cosh {\frac {\tilde {x}}{a}}=a\cosh \left({\mbox{arsinh}}{\frac {h}{b}}\right)=a\cosh \left({\mbox{arcosh}}{\sqrt {{\frac {h^{2}}{b^{2}}}+1}}\right)=a{\sqrt {{\frac {h^{2}}{b^{2}}}+1}}}
f
=
y
1
+
h
b
(
a
arsinh
h
b
−
x
1
)
−
a
h
2
b
2
+
1
{\displaystyle f=y_{1}+{\frac {h}{b}}\left(a\;{\mbox{arsinh}}{\frac {h}{b}}-x_{1}\right)-a{\sqrt {{\frac {h^{2}}{b^{2}}}+1}}}
(Gl.19)
