- #36
gary350
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gary350 said:
How accurate can the shadow of a large sundial be?
- Thread starter Jonathan212
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In summary, a larger sundial will be more accurate but there is a trade-off between shadows getting less sharp at a distance and the larger sundial becoming more inaccurate over time.
- #37
Baluncore
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Take a look at this page on wikpedia. It shows the time corrections and height of the Sun.gary350 said:
https://en.wikipedia.org/wiki/Equation_of_time#Graphical_representation
- #38
Jonathan212
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Answering my own question, the -6 dB is at v = 0 below. Ie even in the straight line from a point source to the knife edge and to an aligned observer there is already attenuation by a factor of 4.
where v is defined as
where in the case of the sundial we plug in d1 = infinity. Source: http://www.waves.utoronto.ca/prof/svhum/ece422/notes/19-diffraction.pdf
But what happens at v = 1 where diffused light is 14 dB or 25 times weaker? The v = 1 isosurface of d2-versus-h is actually parabolic, not flat (d2 = 2/wavelength * h^2). A bit of a surprise:
So putting the observer a long way from the edge does not spread light very much, h does not increase in proportion. At d2 = 10 metres, h is still just 2 millimeters for red light. The sun's transition with the sun's diameter of 0.5 degrees would be 87 mm wide at the same distance.
where v is defined as
where in the case of the sundial we plug in d1 = infinity. Source: http://www.waves.utoronto.ca/prof/svhum/ece422/notes/19-diffraction.pdf
But what happens at v = 1 where diffused light is 14 dB or 25 times weaker? The v = 1 isosurface of d2-versus-h is actually parabolic, not flat (d2 = 2/wavelength * h^2). A bit of a surprise:
So putting the observer a long way from the edge does not spread light very much, h does not increase in proportion. At d2 = 10 metres, h is still just 2 millimeters for red light. The sun's transition with the sun's diameter of 0.5 degrees would be 87 mm wide at the same distance.
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