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Is there a device available that can measure shadow lengths for an object x? If not, how difficult would it be to manufacture a device that measures shadow lengths?

Posted on 2009-02-22
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If there was an object that was for example 2cm, is it possible to measure the shadow cast by it during different parts of the day?
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Question by:fiazf
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Thibault St john Cholmondeley-ffeatherstonehaugh the 2nd earned 240 total points
ID: 23704510
you could draw a set of circles around your object at measured intervals and use a time-lapsed camera to photograph the shadow through the day.
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by:MrWid
MrWid earned 120 total points
ID: 23704511
Being a keen Gizmologist, I have never seen an automatic shadow meter. The biggest problem is how accurate do you want the results. If an accuracy of 5mm is OK then an array of LDR's "Light Dependent Resistors" could be laid for the shadow to fall on and this information fed through an electronic decoder to give an answer. However If I was to make a more accurate one (say >1mm), I would use a motorised mechanical screw with the LDR attached so that the  LDR could be moved back and forth until the shadow end is found. This would also need a form of indexing to work out the distance. Also if the shadow moves sideways during the day, then a way to track the shadow will be needed (LDR's and motors again). This will all need to be controlled preferably by a microprocessor or computer. Quite a complex project.
I would also look in to computer video processing. If the camera was above the object and the object is on a plain background, then it would not be too hard to make a program that can work out the length of shadow. Again, not a quick solution.

Hope these ideas help.
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by:aburr
ID: 23706405
"Is there a device available that can measure shadow lengths for an object x?"
yes. It is called a ruler.
RobinD has the right idea. Another possibility is to photograph some graph paper with the object place at the origin. The end of the shadow can then be read out in x, y coordinates and converted into distance. Use a multiple exposure to get all the shadow ends on the same photo.
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Author Comment

by:fiazf
ID: 23706662
If it was just a case of measuring with a ruler, then I don't think I would have asked the question.
The shadows start long, become short and then elongate again over a period of many hours. Although the times that I am interested in are for about 3-6 hours, depending upon the time of the year, which is labour intensive. So I could stand there watching the shadows and measuring them with a ruler or if there was a device that did all the "hard" work for me, then I would be done.
The thing with the camera is useful, but don't know how much one of those cameras costs AND it is still quite labour intensive. Mr Wild's idea sounds more automated, but also probably more difficult to put together.
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LVL 17
ID: 23707043
What accuracy are you trying to measure to?
Your question initially sounded like a school project except for the mention of the 2cm height. Is this to be measured on planet earth? if so I would think that calculating the length from the sun's position would be more accurate than looking for the end of a fuzzy shadow. If off planet then you will need a more robust, complex and expensive solution.
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by:aburr
ID: 23707457
"The thing with the camera is useful, but don't know how much one of those cameras costs"
A camera for photographing stick and graph paper without multiple exposures would cost about $9.99. You would need a tripod and would have to take a picture however often you want.
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by:d-glitch
d-glitch earned 390 total points
ID: 23711597
The position of the sun [and therefore the length of a shadow], depends on latitude,
longitude, time of day, and time of year.

These things can all be calculated precisely.  You don't have to measure them.
You might want to check out the US Naval Observatory website:

     http://www.usno.navy.mil/
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by:d-glitch
ID: 23711687
Here is a sun position calculator:

     http://sunposition.info/sunposition/index.php
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by:d-glitch
d-glitch earned 390 total points
ID: 23711871
Here is another calculator:

     http://www2.arnes.si/~gljsentvid10/nebes_pod/legasonca.html

For a vertical stake of height h, the length of the shadow depends on the elevation angle of the sun which you can calculate using the link above.
 
                   h
       len = -----------
              tan(elev)
 
Be careful with degrees and radians.

Open in new window

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LVL 17
ID: 23714183
RazF might be trying to measure this precisely, perhaps the latitude isn't known?
On Earth though the refraction distortion at the edge of a shadow will make it hard to decide the exact length of a shadow cast by a 2cm object.
Can we have more information please RazF?
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by:d-glitch
d-glitch earned 390 total points
ID: 23714678
RobinD is correct.  More information would be helpful.

And I also note that she wast the first one to suggest tracking the sun
rather than measuring a shadow directly.

A shadow can be fuzzy, from atmospheric effects at dawn and dusk, and from
umbra/penumbra effects (since the sun is not a point source).

Are you trying to make a sundial?
Or determine a layout for your vegetable garden?
Or adjust a solar cell array for max output?

What kind of resolution do you need?

There are linear photodiode arrays with resolutions of 0.14 mm.  
They cost on the order of $500-1000 depending on what size you need.

     http://optoelectronics.perkinelmer.com/catalog/Product.aspx?ProductID=RL1024PAQ-021

Do you want data or some sort of signal to drive another box?
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by:d-glitch
ID: 23714781
>>  There are linear photodiode arrays with resolutions of 0.014 mm.
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LVL 17
ID: 23715387
d-glitch, I'm sure you meant to type 'he', but just to save any confusion I am male. At least I was when I last looked!

Just wondering if it's possible to use the arrays in a dismantled LCD monitor to measure light- like the effect you get by ripping the top off an old transistor. Those old junctions varied their resistance when exposed to light. If you could read the light from an array like a monitor then recording a shadow from a 2cm object might be possible.
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by:d-glitch
ID: 23715782
Sorry  Robin.  
Very bad assumption on my part.
Maybe due to some transatlantic bias.





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LVL 17
ID: 23715949
> transatlantic bias
lol, blame my accent.
 
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Author Comment

by:fiazf
ID: 23716989
Interesting bit of discussion so far. Unfortunately or fortunately (depending on which way you look at it), I'm sitting in the UK so there is a bit of time lag before I can read your comments. The edge of a shadow is indeed very fuzzy at times, but this also in my experience depends on the type of object being used. Accuracy is important, but the shadow lengths that I am mainly interested in are after noon and what I found is that a short while after noon the shadow length increases quite rapidly. To give you some crude readings for a 2cm object (it doesn't have to be 2cm, it could have been smaller or longer) - I just happened to be using a 2 cm long screw! At 11:45 the observed shadow length was 6.0cm then an hour later at 12:45pm the shadow length had only increased to 6.4cm. However, later at 1:40pm it was 7.6cm and at 1:50pm it was 8.2cm. In other words the rate of change around noon is slow, but later on it gets more.
The accuracy I am needing is roughly to within 0.1cm (1mm). I am interested in determining the zenith shadow, which can easily be done using the method of the Indian circle. Then I need to measure the shadow of an object after noon until the length of the shadow of object x becomes as long as the length of object x, plus the zenith shadow. In other words until the shadow length = object + zenith shadow.
I need a relatively automated solution, although this sounds like its going to be quite expensive.
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LVL 17
ID: 23719635
fiazf, apologies for getting your name wrong - I'll blame it on my failing eyesight and the tiny font on my pda screen.
There is a fuzziness on shadow edges blamed on variously named phenomena, the most common being 'knife-edge diffraction'. There's a sort of explanation here:
http://wapedia.mobi/en/Knife-edge_effect
With the size of object you are measuring I think you could have difficulty finding the edge of the shadow within 1mm. I am in the UK as well and so I can't check this for certain as the sky is most definitely overcast. I thnk there are added atmospheric effects that increase the distortion as well - I do remember on the moon landings that they said how clear the edges of the shadows were.
The change is sinusoidal-ish, a complicated calculation but would be possible with the sun position calculators that d-glitch posted.
Another thing for you to consider are keeping your measuring surface level. Sounds easy but what are you going to keep it level with? The earth's centre of gravity is not in the middle of the ball and in fact moves about. The shape of the earth has best been described as 'like a potato', ie it is not that nice oblate spheroid that all the maps are drawn on. There is an awful lot of assumption made when overlaying maps onto gps coordinates, it works great for finding your way around but it is only an approximation. Nobody really knows what sea-level is, remember that coastlines dip and rise an inch or two in a decade so a sea-level marked on a cliff ten years ago may not be at the same height now.
I'm not trying to put you off your project, just trying to make sure you are aware of some of the considerations before you spend time and effort on an expensive solution.
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Assisted Solution

by:d-glitch
d-glitch earned 390 total points
ID: 23722011
Hello fiazf,

You seem to know what you want, but I still can't tell why.

I ran the numbers for London today through the solar calculator and plotted the
results.
     
     http://www2.arnes.si/~gljsentvid10/nebes_pod/legasonca.html

Plugging numbers by hand is hardly ideal.  You can probably find the general equation for apparent solar motion online somewhere.  

If you really need to do this in real time, the best approach would probably be to
set up a camera over your sundial and use image processing software to extract
the shadow features.  

Matlab is the industry standard for this type of software task, but there are free,
open source substitutes as well.

Probably looking at $500 to $1000 for hardware nd 50 to 200 hours of programming.
Time	Elev	Shadow
		
10:00	29.17	4.10
10:30	32.57	3.72
11:00	35.34	3.46
11:30	37.35	3.30
12:00	38.52	3.21
12:23	38.80	3.19 <== Local Solar Noon
12:30	38.75	3.20
13:00	38.12	3.24
13:30	36.58	3.36
14:00	34.23	3.56
14:30	31.18	3.86
15:00	27.52	4.33

Open in new window

Shadow.bmp
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Expert Comment

by:MrWid
ID: 23724129
RobinD:
"The earth's centre of gravity is not in the middle of the ball and in fact moves about."

Wow can you point me to where I can read more about this? Ta!

The idea of using a ragged TFT screen in reverse is novel but I can see that interfacing to the screen being complex.
I would guess that a good webcam positioned directly above the screw and a good security time lapse program like:

http://www.luxriot.com

Coupled with graph paper or concentric circles to grade the shadow would give results to 1mm. I am sure that someone with a bit of programming knowledge could analyse the images and automate the whole thing.
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LVL 17
ID: 23725656
>Wow can you point me to where I can read more about this? Ta!

Tidal forces affect the apparent centre of gravity, the seas move around whereas the continents don't move so fast on a daily basis - they do move, floating about on that big core of molten metal that we are resting on. As the molten core and the tectonic plates move around so does the centre of gravity. Remember that the earth is just part of a system and reletively close to the moon, this affects the centre of mass as it swings round us.
http://www.sciencedaily.com/releases/2006/11/061126121122.htm
http://www.space.com/scienceastronomy/planetearth/core_cyclones_991110.html
The concept of the earth as a smooth ball of even texture right through it's layers is not at all correct.
 
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LVL 17
ID: 23740218
I plotted shadow lengths once for my daughter's homework project. I plotted them on a 24 hour polar graph and the result looked very much like a straight line. I've never been able to determine whether this was an accident of date and longitude or whether you always get a straight line graph. If you do always get a straight line then calculating the lengths should be quite easy.
 
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Expert Comment

by:d-glitch
ID: 23745026
The shadow length depends on the position of the sun in the sky.
Can you invert the problem and track the position of the sun directly?

This is a much simpler problem, and there are commercial solutions available.
This is the sort of thing you would use to align a solar array for max efficiency.

     http://www.ecobusinesslinks.com/solar_tracking_sun_trackers.htm

The simplest electromechanical implementation would use two servo motors to
control heading and elevation.  The sensor would use an X-shaped extrusion with
four solar cells, one in each quadrant.

When the X is pointed at the sun, each quadrant is fully iluminated.  The servos
adjust the heading and elevation to maintain tracking.  The data can be read out
directly from the servo position potentiometers.

The cost here is probably on the order of $100 and the level of difficulty is simple
high school science fair.



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Author Comment

by:fiazf
ID: 23751045
Not sure I understand the last suggestion from d-glitch.
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Expert Comment

by:d-glitch
ID: 23751176
I am not sure what you are trying to do.
What is your actual goal in all this?

Are you really trying to measure a shadow??

Or are you measuring a shadow in order to infer something about the position of the sun?
If this is the case, then my last post suggests a method for measuring the sun's position directly.


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Author Comment

by:fiazf
ID: 23758023
I am not doing this to determine the sun's position, but just seeking to measure the shadows directly.
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Expert Comment

by:MrWid
ID: 23765083
Thanks for the links RobinD :)
It is somthing I never realy thought about. It could explain why I have days where thing fall out of cupboards or fall over. :)
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Author Closing Comment

by:fiazf
ID: 31549761
Some things were probably simple for the experts but too technical for me.
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