Bifilar sundial

The first wire ${\displaystyle f_{1}\,}$ is orientated north-south at a constant distance ${\displaystyle h_{1}\,}$ from the dial plate ${\displaystyle \Pi \,}$
The second wire ${\displaystyle f_{2}\,}$ is orientated east-west at a constant distance ${\displaystyle h_{2}\,}$ from the dial plate ${\displaystyle \Pi \,}$ (thus ${\displaystyle f_{2}\,}$ is orthogonal to ${\displaystyle f_{1}\,}$ which lies on the plane of the meridian ).

In this proof ${\displaystyle \varphi }$ (pronounced phi) is the latitude of the dial plate.

Respectively, ${\displaystyle ({\mathcal {D}}_{1})}$ and ${\displaystyle ({\mathcal {D}}_{2})}$ are the vertical projections of wires ${\displaystyle f_{1}\,}$ and ${\displaystyle f_{2}\,}$ on the dial plate ${\displaystyle \Pi \,}$.

Point ${\displaystyle O\,}$ is the point on the dial plate directly under the two wires' intersection.
That point is the origin of the X,Y co-ordinate system referred to below.

The X-axis is the east–west line passing through the origin. The Y-axis is the north–south line passing through the origin. The positive Y direction is northward.

One can show that if the position of the sun is known and determined by the spherical coordinates ${\displaystyle t_{\odot }}$ et ${\displaystyle \delta \,}$ (pronounced t-dot and delta, respectively the known as the hour angle et declination), the co-ordinates ${\displaystyle x_{I}\,}$ and ${\displaystyle y_{I}\,}$ of point ${\displaystyle I\,}$, the intersection on the two shadows on the dial-plate ${\displaystyle \Pi \,}$ have values of :

${\displaystyle {\begin{matrix}x_{I}&=&h_{1}{\frac {\sin t_{\odot }}{\sin \varphi \ \operatorname {tan} \delta \ +\ \cos \varphi \cos t_{\odot }}}\\\ &\ &\ \\y_{I}&=&h_{2}{\frac {-\cos \varphi \ \operatorname {tan} \delta \ +\ \sin \varphi \cos t_{\odot }}{\sin \varphi \ \operatorname {tan} \delta \ +\ \cos \varphi \cos t_{\odot }}}\end{matrix}}}$

Eliminating the variable ${\displaystyle \delta \,}$ in the two preceding equations, one obtains a new equation defined for ${\displaystyle x_{I}\,}$ and ${\displaystyle y_{I}\,}$ which gives, as a function of the latitude ${\displaystyle \varphi }$ and the solar hour angle solaire ${\displaystyle t_{\odot }}$, the equation of the trace of the sun associated with the local apparent time. In its simplest form this equation is written:

${\displaystyle {\frac {x_{I}}{y_{I}+h_{2}/\operatorname {tan} \varphi }}={\frac {h_{1}\sin \varphi }{h_{2}}}\ \operatorname {tan} t_{\odot }}$

This relation shows that the hour traces are indeed line segments and the meeting-point of these line segments is the point ${\displaystyle C\,}$:

${\displaystyle x_{C}=0\,}$

${\displaystyle y_{C}=-h_{2}/\operatorname {tan} \varphi }$

In other words, point C is south of point O (where the wires intersect), by a distance of ${\displaystyle h_{2}/\operatorname {tan} \varphi }$, where ${\displaystyle \varphi }$ is the latitude.[1]

Special case

If one arranges the two wire heights ${\displaystyle h_{2}\,}$ and ${\displaystyle h_{1}\,}$ such :

${\displaystyle h_{2}=h_{1}\sin \varphi \quad }$

then the equation for the hour lines can be simply written as:

${\displaystyle {\frac {x_{I}-x_{C}}{y_{I}-y_{C}}}=\operatorname {tan} t_{\odot }}$

at all times, the intersection ${\displaystyle I\,}$ of the shadows on the dial plate ${\displaystyle \Pi \,}$ is such that the angle ${\displaystyle {\widehat {OCI}}}$ is equal to the hour angle ${\displaystyle t_{\odot }}$ of the sun so thus represents solar time.

So provided the sundial respects the la condition ${\displaystyle h_{2}=h_{1}\sin \varphi \quad }$ the trace of the sun corresponds to the hour-angle shown by lines (rays) centred on the point ${\displaystyle C\,}$ and the 13 rays that correspond to the hours 6:00, 7:00, 8:00, 9:00... 15:00, 16:00, 17:00, 18:00 are regularly spaced at a constant angle of 15°, about point C. [lower-alpha 1][1]

A London dial is the name given to dials set for 51° 30' N. A simple London bifilar dial has a dial plate with 13 line segments drawn outward from a centre-point C, with each hour's line drawn 15° clockwise from the previous hour's line. The midday line is aligned towards the north.

The north–south wire is 10 cm (${\displaystyle h_{1}}$) above the midday line. That east-west wire is placed at a height of 7.826 (${\displaystyle h_{2}}$) centimeters- equivalent to 10 cm x sin(51° 30'). This passes through C. The east–west wire crosses the north–south wire 6.275 cm north of the centre-point C- that being the equivalent of - 7.826 (${\displaystyle h_{2}}$) divided by tan (51° 30').