Varignon frame

At point ${\displaystyle \mathbf {v} }$ the sum of all forces is 0

If the holes have locations Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \mathbf x_1, \dots, \mathbf x_n} and the masses of the weights are ${\displaystyle m_{1},...,m_{n}}$ then the force acting at the i-th string has the magnitude ${\displaystyle m_{i}\cdot g}$ (${\displaystyle g=9.81{\text{m/sec}}}$: constant of gravity) and direction ${\displaystyle {\tfrac {\mathbf {x} _{i}-\mathbf {v} }{\|\mathbf {x} _{i}-\mathbf {v} \|}}}$ (unitvector). Summing up all forces and cancelling the common term ${\displaystyle g}$ one gets the equation

(2):${\displaystyle \ \mathbf {F} (\mathbf {v} )=\sum _{i=1}^{n}m_{i}{\frac {\mathbf {x} _{i}-\mathbf {v} }{\|\mathbf {x} _{i}-\mathbf {v} \|}}=\mathbf {0} }$.

(At the point of equilibrium the sum of all forces is zero !)

This is a non-linear system for the coordinates of point ${\displaystyle \mathbf {v} }$ which can be solved iteratively by the Weiszfeld-algorithm (see below)[2]

The connection between equation (1) and equation (2) is:

(3): ${\displaystyle \ \mathbf {F} (\mathbf {x} )=\nabla D(\mathbf {x} )={\begin{bmatrix}{\frac {\partial D}{\partial x}}\\{\frac {\partial D}{\partial y}}\end{bmatrix}}.}$

Hence Function ${\displaystyle D}$ has at point ${\displaystyle \mathbf {v} }$ a local extremum and the Varignon frame provides the optimal location experimentally.

Varignon frame: example

Level curves

For the following example the points are

${\displaystyle \mathbf {x} _{1}=(0,0),\ \mathbf {x} _{2}=(40,0),\ \mathbf {x} _{3}=(50,40),}$

${\displaystyle \mathbf {x} _{4}=(10,50),\ \mathbf {x} _{5}=(-10,30)}$

and the weights

${\displaystyle m_{1}=10,\;m_{2}=10,\;m_{3}=20,\;m_{4}=10,\;m_{5}=5}$.

The coordinates of the optimal solution (red) are ${\displaystyle (32.5,30.1)}$ and the optimal weighted sum of lengths is ${\displaystyle L_{\text{op}}=1679}$. The second picture shows level curves which consist of points of equal but not optimal sums. Level curves can be used for assigning areas, where the weighted sums do not exceed a fixed level. Geometrically they are implicit curves with equations

${\displaystyle \;D(\mathbf {x} )-c=0,\;c>L_{\text{op}}\;}$ (see equation (1)).

Case ${\displaystyle n=2,m_{1}=m_{2}=1}$, the level curves are confocal ellipses

• In case of ${\displaystyle n=1}$ one gets ${\displaystyle \mathbf {v} =\mathbf {x} _{1}}$.
• In case of ${\displaystyle n=2}$ and ${\displaystyle m_{2}>m_{1}}$ one gets ${\displaystyle \mathbf {v} =\mathbf {x} _{2}}$.
• In case of ${\displaystyle n=2}$ and ${\displaystyle m_{2}=m_{1}}$ point ${\displaystyle \mathbf {v} }$ can be any point of the line section ${\displaystyle {\overline {X_{1}X_{2}}}}$ (see diagram). In this case the level curves (points with the same not-optimal sum) are confocal ellipses with the points ${\displaystyle \mathbf {x} _{1},\mathbf {x} _{2}}$ as common foci.

Iteration as fixpoint determination for the example: starting point ${\displaystyle \mathbf {v} _{0}=(25,15)}$ (green), starting point ${\displaystyle \mathbf {v} _{m}}$ (blue) is the center of mass

Replacing in formula (2) vector ${\displaystyle \mathbf {v} }$ in the nominator by ${\displaystyle \mathbf {v} _{k+1}}$ and in the denominator by ${\displaystyle \mathbf {v} _{k}}$ and solving the equation for ${\displaystyle \mathbf {v} _{k+1}}$ one gets:[3]

(4):${\displaystyle \quad \mathbf {v} _{k+1}=\sum _{i=1}^{n}{\frac {m_{i}\mathbf {x} _{i}}{\|\mathbf {x} _{i}-\mathbf {v} _{k}\|}}{\big /}\sum _{i=1}^{n}{\frac {m_{i}}{\|\mathbf {x} _{i}-\mathbf {v} _{k}\|}}}$

which describes an iteration. A suitable starting point is the center of mass with mass ${\displaystyle m_{i}}$ in point ${\displaystyle \mathbf {x} _{i}}$:

${\displaystyle \mathbf {v} _{0}={\frac {\sum _{i=1}^{n}m_{i}\mathbf {x} _{i}}{\sum _{i=1}^{n}m_{i}}}}$.

This algorithm is called Weiszfeld-algorithm.[4]

Formula (4) can be seen as the iteration formula for determining the fixed point of function

(5)${\displaystyle \quad \mathbf {G} (\mathbf {x} )=\sum _{i=1}^{n}{\frac {m_{i}\mathbf {x} _{i}}{\|\mathbf {x} _{i}-\mathbf {x} \|}}{\big /}\sum _{i=1}^{n}{\frac {m_{i}}{\|\mathbf {x} _{i}-\mathbf {x} \|}}}$

with fixpoint equation

${\displaystyle \quad \mathbf {x} =G(\mathbf {x} )}$

(see fixed point)

Remark on numerical problems:
The iteration algorithm described here may have numerical problems if point ${\displaystyle \mathbf {v} _{k}}$ is close to one of the points ${\displaystyle \mathbf {x} _{1},...\mathbf {x} _{n}}$.

• Fermat point (case ${\displaystyle n=3,m_{1}=m_{2}=m_{3}=1}$)
• Varignon's theorem
• geometric median

• MathePrisma Uni Wuppertal: Lösung eines Standort-Problems mit GeoGebra
• The Frame of Varignon: Mathematical Treatment

• Uwe Götze: Risikomanagement, Physica-Verlag HD, 2013, ISBN 978-3-642-57587-7, S. 268
• Andrew Wood, Susan Roberts : Economic Geography, Taylor & Francis, 2012, ISBN 9781136899478, p. 22
• H. A. Eiselt, Carl-Louis Sandblom :Operations Research, Springer Berlin Heidelberg, 2010, ISBN 9783642103261, p. 239
• Robert E. Kuenne: General Equilibrium Economics, Palgrave Macmillan UK, 1992, ISBN 9781349127528, p. 226