Calculation of a symmetric Gnomon that approximately corrects the Equation of Time. Werner Riegler,
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1 Calculation of a symmetric Gnomon that approximately corrects the Equation of Time,
2 The Gnomon of a sundial can be shaped such that the Equation of Time is automatically corrected and the Sundial shows the Civil Time. Because the sun has the same declination twice a year, but the Equation of Time is different at these dates, the gnomon has to be changed twice a year, like it is e.g. the case for the dial of Martin Bernhardt: In order to avoid changing the gnomon twice a year I developed a semi transparent double gnomon that incorporates both: The absolute value of the equation of time at a given declination is not very different, so by approximating the analemma with a symmetric curve one can arrive at a single Gnomon that doesn t have to be changed. The error introduced by this approximation is at maximum 1.7 minutes. The calculation of this Gnomon is described in the next pages. 2
3 Equation of Time (Degrees) Equation of Time (Minutes) Solar Declination (Degrees) Solar Declination Days after December 21 st Equation of Time (EoT): Civil Time = Sundial Time + EoT Days after December 21 st Equation of Time in Degrees =EOT (Minutes)/4 Days after December 21 st 3
4 Equation of Time Error (Minutes) Equation of Time (Degrees) Equation of Time (Degrees) EoT vs. Declination Dec. 21 to Jun. 21 Jun. 21 to Dec. 21 Solar Declination (Degrees) EoT vs. Declination Averaged to make it symmetric Solar Declination (Degrees) Error from Symmetrization Maximum 1.7 Min. around June 21 Solar Declination (Degrees) 4
5 Symmetrized Equation of Time τ (Degrees) Using the symmetric Approximation the Equation of Time τ is now a unique function of the Solar Declination δ. We call the function absolute value of the approximated equation of Time τ (δ). Solar Declination δ (Degrees) 5
6 Calculation of the Gnomon: At 12:00 + EoT the Sun is exactly in southern direction A Ray connected to Point P and pointing South with a declination of δ must touch the Gnomon. This defines a ray for every declination δ y South R x P x=-r*sin(τ (δ)) y=-r*cos(τ (δ)) 12:00 EoT (degrees) 6
7 Parametric Representation of the sunrays: South x=- R*Sin(τ (δ)) + k*0 y=- R*Cos(τ (δ)) + k*1 z= 0 + k*tan(δ) δ = to k= 0 to 1500 mm Magnifications: 7
8 Sunrays: x (k, δ)=- R*Sin(τ (δ)) + k*0 y (k, δ) =- R*Cos(τ (δ)) + k*1 z (k, δ) = 0 +k*tan(δ) δ = to k= to 1500 mm Gnomon: The Gnomon must be the rotationally symmetric body that touches the surface of sun rays. How can we calculate this body? To find the radius of the Gnomon at height z0 we cut the sunrays with a plane at z=z0 which gives a 1-dimensional curve. Example z0=150mm Then we find the circle centered at zero that touches this curve. The radius r0 of this circle is the radius of the Gnomon at height z0. The parametrization of this curve at z=z0 is: z0= k*tan(δ) k=z0/ Tan(δ) X(δ)=- R*Sin(τ (δ)) y (δ) =- R*Cos(τ (δ)) + z0/ Tan(δ) δ = to To find the circle, centered at zero, that touches this curve, we simply have to find the minimum distance between the point (0,0) and the curve, i.e. r(δ) 2 = x(δ) 2 +y(δ) 2 = = R 2 *Sin(τ (δ)) 2 +[R*Cos(τ (δ)) -z0/tan(δ)] 2 Minimum r0 8
9 On the following pages the touching circles at different heights z0 above the equatorial plane are shown. The radius R of the dial is assumed to be 500mm. 9
10 Z0=1mm r(δ) δ 10
11 Z0=50mm r(δ) δ 11
12 Z0=100mm r(δ) δ 12
13 Z0=150mm r(δ) δ 13
14 Z0=200mm r(δ) δ 14
15 Z0=216mm r(δ) δ 15
16 Z0=-50mm r(δ) δ 16
17 Z0=-100mm r(δ) δ 17
18 Z0=-150mm r(δ) δ 18
19 Z0=-200mm r(δ) δ 19
20 Z0=-213mm r(δ) δ 20
21 Above and below a certain limiting declination there is no solution to the minimization problem. The reason is that the correction for one day is shadowing the correction of the next day, so the correcting Gnomon can only exist in a certain range of declinations. The solar declination ranges from to degrees. The rage where a solution for the correcting Gnomon can be found is to degrees. This means that only ±10 days around Decmeber 21 st and ±2 days around June 21 st the indicator isn t precise. The optimum shape to minimize the error during these days is to simply cut the gnomon at this height. 21
22 Limit at positive declination: degrees, z0=216mm r(δ) Z0=-213mm δ Limit at negative declination: 23.3 degrees, z0=-213mm r(δ) δ The total height of the gnomon is therefore = 429mm 22
23 Final Gnomon 23
24 Final Gnomon 24
25 Final Gnomon z 0 (mm) r 0 (mm) z 0 (mm) r 0 (mm) z 0 (mm) r 0 (mm) z 0 (mm) r 0 (mm)
26 Final Gnomon z 0 (mm) r 0 (mm) z 0 (mm) r 0 (mm) z 0 (mm) r 0 (mm) z 0 (mm) r 0 (mm)
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