How accurate can the shadow of a large sundial be?

[Jonathan212]

Saw a sundial the other day and it occurred to me that a larger one would be more accurate. But then shadows get less sharp at a distance so there is a trade-off.

What would be the maximum accuracy that can be achieved by making a sundial larger?

 

[Ranger Mike]

http://www.planetary.org/explore/projects/earth-dial/how-to-read-a-sundial.html
very good read on sundial vs clock time.

 

[Gigaz]

Here's what I would expect is the biggest source of uncertainty:

Earth rotates around its own axis every 23h 56m 4.1s
The day is longer because the Earths move a bit further in its orbit around the Sun, so the Sun is at a slightly different angle on the next day. This adds 235.9 seconds to the day.
The orbit of the Earth around the Sun is has an eccentricity of 0.0167. This means that the distance to the Sun changes by 3.3% over the year between aphelion and perihelion. According to Keplers second law, this means that the Earth is 3.3% faster on its orbit on perihelion, compared to aphelion. This effect makes an ideal sun dial error oscillate with an amplitude of about +-4 seconds over the year.

After further thinking, I figured I messed up. The effect is cumulative over months, so the error builds to at least a few minutes.

 

[Baluncore]

What would be the maximum accuracy that can be achieved by making a sundial larger?

The Earth rotates relative to the Sun at 15° per hour. It takes 4 minutes to move by 1°. The Sun has a diameter of 0.5° so it appears to move one diameter in 2 minutes. That limits the ability to read the position of the shadow on a sundial.
Accurate solar times can be read using a simple optical system like a sextant, where the angle to the edges of the solar disk are read.

You need to know the date so you can correct the ±15 minutes for the equation of time.
https://en.wikipedia.org/wiki/Equation_of_time#Graphical_representation
https://en.wikipedia.org/wiki/Analemma

 

[Jonathan212]

There must be another factor too when all others have been taken into account, and that is light diffusion. A skyscraper's shadow must not be as sharp as the size of the sun dictates but less.

 

[Baluncore]

Heat haze along the atmospheric path of the light from the Sun may cause some shimmer, but that should be small compared with the angular diameter of the Sun, it is not a problem when using a sextant for navigation.

A bigger effect will be the elevation of the sun near dawn and dusk due to difrefraction by the atmospheric pressure gradient. That difrefraction should not be a problem near midday.

https://en.wikipedia.org/wiki/Atmospheric_refraction#Values

 

[Jonathan212]

I have this picture in my mind of seawaves entering a harbour and becoming less straight at the corners. Aren't light waves that have just cleared an object less flat than unopposed light waves entering a sextant? We have the numbers for sun diameter etc, what are the numbers for the effect due to the wave nature and wavelength of light?

 

[Baluncore]

Optical telescopes, camera apertures and lenses work OK, so why should a sextant or sundial be different?

Knife-edge diffraction bleeds a small amount of energy passing a knife edge into the shadow zone. The energy affected is only that which passes a few wavelengths from the knife edge. The wavelength of light is extremely small compared with the size of a gnomon, or a building, so the proportion of light diffracted is minuscule.

The Sun produces a very wide spectrum of light. Knife edge diffraction will really only be visible for monochromatic light passing within a few wavelengths of a sharp edge.

 

[Baluncore]

While the Sun is above the horizon, light from the Sun, (mostly blue), is scattered by the atmosphere which makes the sky look blue. That diffuse blue light illuminates exposed shadows that the direct sunlight cannot reach. Rose coloured glasses should increase the contrast, and your belief in your ability to accurately read a sundial. Why do we not paint sundials red to get a better contrast?

 

[Keith_McClary]

The Binoy Dial uses a lens on the gnomon to focus sunlight. Seems like that could be more accurate.

 

[Jonathan212]

Light turning a few wavelengths to the left means several meters for detectors far enough from the edge.

We have the numbers for sun diameter etc which translate into an angle and the angle translates into 2 minutes of time precision. What are the numbers for the effect due to the wave nature and wavelength of light?

 

[Jonathan212]

That's like asking, if the sun were a point source, what would be the maximum accuracy of the sundial once all other factors like analemma etc were corrected for?

 

[Jonathan212]

Wow! That angle is big.

https://en.wikipedia.org/wiki/Knife-edge_effect

 

[Baluncore]

We have the numbers for sun diameter etc which translate into an angle and the angle translates into 2 minutes of time precision. What are the numbers for the effect due to the wave nature and wavelength of light?

The wavelength of light is so short, and the spectrum so wide, that it has no effect on the accuracy of a sundial. You can improve the accuracy by replacing the plate gnomon with a cylindrical style, and replacing the flat plate with part of a graduated cylinder about the style axis. If the diameter of the style is selected to be 0.5° when viewed from the scale, the shadow of the style will not be asymmetric but will have the mid-line darkest, with light either side. That will improve the resolution by a factor of 4 or more. It would cancel edge diffraction effects, if they were visible, but it cannot reduce the horizon refraction in the few minutes after dawn or before dusk.

Wow! That angle is big.

That angle will be 90°, but it effects only a thin skin about one micron thick, and then only if the edge is optically straight, and a perfect conductor. Illumination falling in the shadow will be less than 0.1% of that of the sunlit area, so you will not notice it.

 

[Jonathan212]

No it won't be 0.1%, it'll be 0.01% at 120 degrees. Without sources to back them up, such numbers are not appreciated.

 

[Jonathan212]

Image processing software could easily find the location of the mid point between fully lit and fully shadowed to high precision but not the human eye, unless specially trained and if the sundial is the right size.

 

[Baluncore]

No it won't be 0.1%, it'll be 0.01% at 120 degrees. Without sources to back them up, such numbers are not appreciated.

You are saying it would be harder to see than I am.
But how can the angle be greater than 90 degrees? If it was greater it would hit the back of the knife edge that diffracted it.
I got the 0.1% from a conservative -60dB on a diffraction chart.
Where did you get your 0.01% ?

 

[Jonathan212]

The 0.01% is arbitrary of course, and so is the 120 angle measured from the plane of the knife. Where is the diffraction chart?

 

[Baluncore]

Image processing software could easily find the location of the mid point between fully lit and fully shadowed to high precision but not the human eye, unless specially trained and if the sundial is the right size.

There is no such thing as fully shadowed. A cloud in the sky to one side and clear of the sun, a nearby building, or an observer wearing a white shirt, will change the level of illumination of the shadow. That will change the critical threshold.

Don't put your faith in DSP until you have discovered the limitations of rose coloured glasses.

 

[Baluncore]

The 0.01% is arbitrary of course, and so is the 120 angle measured from the plane of the knife.

The light is diffracted as it crosses the knife edge. The light is initially at 90 degrees to the plane of the knife. It then turns to fill the shadow behind the knife. If you measure the angle relative to the plane of the knife the direction becomes 0° or 180°, not 120°. The light still only changes direction by 90° on passing the edge.

Where is the diffraction chart?

“Electronic Transmission Technology; lines, waves and antennas”. By William Sinnema. 1979. Chapter 8.

 

[marcusl]

The world's largest sundial is at Jantar Mantar in India.
https://en.wikipedia.org/wiki/Jantar_Mantar,_Jaipur
http://www.jantarmantar.org/learn/observatories/instruments/samrat/index.html
It was built in 1734 and is said to tell local time accurately to within 2 seconds.

 

[Jonathan212]

Where is the -3 dB or -6 dB point in that diffraction chart in the Sinnema book?

Stating the obvious once again, those 0.01% and 120 figures are just fake figures made up to help you feel how people might feel when unsubstantiated figures are thrown at them by a stranger online - trying to prove them wrong is missing the point, the point is your figures need backing. Ideally online but that book is ok too, sort of, as long as you're honest. Fully shadowed in the context of the conversation means no straight light from the sun but even with all diffractions taken into account the image processing would be trivial: just fit an expected parametric curve to pixel values along a line in a photograph of the sundial. But we must stay low tech like the eye or it's pointless. Maybe lenses too.

Marcusl, I can΄t find how that big Indian sundial achieves 2 second accuracy. This contradicts calculations by people above, any idea how it might be done?

 

[Jonathan212]

As low tech as it gets...

It says where the twig's shadow disappears, that's what the time is (marks are every 2 seconds). And you then correct for analemma with a calculator? But the big shadow is much sharper than the 2 minutes of width expected based on the diameter of the sun.

 

[Jonathan212]

A partially occluded sun would have a visible area like this:

Source: http://mathworld.wolfram.com/CircularSegment.html

Therefore the plot of light from fully shadowed to fully lit should look like this:

Let's add the diffraction chart on top. Where does -3 dB or -6 dB fall?

 

[gary350]

I built a sun dial it is very accurate. Your right a tiny one to fit on a wrist watch is hard to see. A 12" sun dial is easy to read. A 4 ft sun dial shadow is fuzzy but still easy to read. I have not found information for angle of the pointer but the sun dials I have seen were 40 and 45 degree angle. June 21/22 sun is 76 degrees at my house. Dec 21/22 sun is 34 degrees at my house. Angle of sun dial pointer might be best set at 76+34/2=55 degrees for my house but not sure.

 

[Baluncore]

I have not found information for angle of the pointer but the sun dials I have seen were 40 and 45 degree angle.

The edge (style) of the pointer (gnomon) should be parallel with the axis of Earth's rotation. The slope angle of the gnomon for your location should be your latitude on Earth.

Your June solstice is 76°, while Dec is 34°, so you must be in the northern hemisphere.
The sum and difference of those two solstice elevation angles is interesting.

Split the difference to get the equinox angle (76°+34°)/2 = 55° and we have your latitude on the face of the Earth as 55°North.

The seasons are caused by the Earth's rotation axis being tilted by 23.4° relative to the axis of the Earth's orbital plane about the Sun. (76°–34°)/2 = 21° so there is a 2° error in your solstice measurements. Accurate values should be your latitude ±23.4°.

 

[gary350]

The edge (style) of the pointer (gnomon) should be parallel with the axis of Earth's rotation. The slope angle of the gnomon for your location should be your latitude on Earth.

Your June solstice is 76°, while Dec is 34°, so you must be in the northern hemisphere.
The sum and difference of those two solstice elevation angles is interesting.

Split the difference to get the equinox angle (76°+34°)/2 = 55° and we have your latitude on the face of the Earth as 55°North.

The seasons are caused by the Earth's rotation axis being tilted by 23.4° relative to the axis of the Earth's orbital plane about the Sun. (76°–34°)/2 = 21° so there is a 2° error in your solstice measurements. Accurate values should be your latitude ±23.4°.

I found answers online but they don't add up? I am at zip code 37129. Online says angle of the sun will be same as my latitude on June 21/22.

Online says, my latitude is 86.39 degrees so sun should be 68.39 degree at solar 12 noon.

Online say, solar 12 noon is 12:50 pm by the clock.

I hammered a stake into the ground at an angle so it makes no shadow at 12:50 pm. Taking a measure off of a level it is not 86 degrees not even close?

 

[Kenwstr]

A sundial gives solar time. Think of it in those terms. Basically, your local time is calibrated by the sun passing over your local meridian being midday. This is true wherever you are. You may correct this for the difference between local and time zone meridians to obtain the local time zone solar time etc. However, the position of the shadow is subject to atmospheric refraction and that is not a constant but largely influenced by temperature & pressure. When I did geodesy, we did these correction for both astro obs (Laplace) and zenith to distant geodetic stations. However large the sun dial, it will be subject to this variation. I expect this will be a major factor in it's direct reading accuracy.

 

[Baluncore]

I found answers online but they don't add up? I am at zip code 37129.

I google 'latitude of zip 37129' and get 35.9173° N, 86.4491° W

Your longitude is 86.45° West of Greenwich, London, England. That sets your time zone.
Your latitude is 35.9° N, so you are 35.9° North of the equator.

The gnomon of your sundial should have a slope angle of your latitude, 35.9° downwards towards the south, relative to the horizontal.

The solstices are on the 21 or 22, of June or December.
The equinoxes are on the 22 or 23, of March or September.

At the equinox, (or on average), the sun is above the Earth's equator, so it will be 35.9° from the vertical where you are. That will be 90° – 35.9° = 54.1° above your southern horizon.

At the equinoxes, or on average, the sun will reach 54.1° above your southern horizon.
At the summer solstice it will reach 54.1° + 23.4° = 77.5° above your southern horizon.
At the winter solstice it will reach 54.1° – 23.4° = 30.7° above your southern horizon.

 

[gary350]

I google 'latitude of zip 37129' and get 35.9173° N, 86.4491° W

Your longitude is 86.45° West of Greenwich, London, England. That sets your time zone.
Your latitude is 35.9° N, so you are 35.9° North of the equator.

The gnomon of your sundial should have a slope angle of your latitude, 35.9° downwards towards the south, relative to the horizontal.

The solstices are on the 21 or 22, of June or December.
The equinoxes are on the 22 or 23, of March or September.

At the equinox, (or on average), the sun is above the Earth's equator, so it will be 35.9° from the vertical where you are. That will be 90° – 35.9° = 54.1° above your southern horizon.

At the equinoxes, or on average, the sun will reach 54.1° above your southern horizon.
At the summer solstice it will reach 54.1° + 23.4° = 77.5° above your southern horizon.
At the winter solstice it will reach 54.1° – 23.4° = 30.7° above your southern horizon.

How to aim sun dial pointer at solar 12 noon? I think magnet north is wrong. True north is about 4 degree from magnet but which direction?

 

[tech99]

How to aim sun dial pointer at solar 12 noon? I think magnet north is wrong. True north is about 4 degree from magnet but which direction?

Magnetic Variation (UK) or Deviation (USA) is given on maps and sea charts. It varies a little with the exact location and also each year.

 

[Baluncore]

At any time at night, you could line the gnomon up with the pole star. If you extend the style of the gnomon as a straight line it should point directly at the pole star.

A magnetic compass will not always give good results. The easiest way to first set up an experimental sundial is simply to set the shadow to 12 noon at midday.
To do it more accurately you will need to know the time of your local solar noon. You would need to adjust for the equation of time (±15minutes) using the date and analemma diagram, and you need to adjust for your position (±30 minutes) relative to the reference longitude of your time zone.

 

[gary350]

At any time at night, you could line the gnomon up with the pole star. .

To do it more accurately you will need to know the time of your local solar noon. .

That is what I was asking, how to find solar noon at my location? Few days ago I looked up, sun rise & sun set at my zip code, chart shows time all day and at 12:50pm solar angle peaks. 12:50pm and 12:51pm are the same angle. I was looking for this chart again but can not find it I need to double check solar time.

I should have known to use north star I read in the Bible north star is only star in the sky that does not move. People at sea use north star for navigation. That will be hard in the dark and I live in TN we lots of tall trees. My back yard is donut circle of in center of trees.

 

[Baluncore]

Few days ago I looked up, sun rise & sun set at my zip code, chart shows time all day and at 12:50pm solar angle peaks.

That tells me that you now have "daylight saving", as the clock is 1 hour fast and reads about 1 PM when it is solar noon.
If you cannot align the gnomon with the pole star, then set the sundial to cast it's thin shadow on 12 noon when the sun is at it's highest point during the day.
Remember that a sundial reads local solar time which wanders through the year according to the analemma.

 

[gary350]

Can't find the original information but I found this, almost as good plus it lists solar noon as 12:50:54pm

https://sunrise-sunset.org/us/murfreesboro-tn

 

[gary350]

Can't find the original information but I found this, more accurate plus it lists solar noon as 12:50:54pm

It is interesting solar noon changes from 12:50 to 12:51 then 12:52 then 12:51 and so on. Some where I read this has something to do with Earth wobble & orgit not being perfect circle.

https://sunrise-sunset.org/us/murfreesboro-tn

 

[Baluncore]

Some where I read this has something to do with Earth wobble & orgit not being perfect circle.

Take a look at this page on wikpedia. It shows the time corrections and height of the Sun.
https://en.wikipedia.org/wiki/Equation_of_time#Graphical_representation

 

[Jonathan212]

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.