I will jump into the middle of this discussion to say:
Here is a web page with essentially the same result but with a coefficient
of 0.03679. Perhaps it uses a different wavelength or criterion.
http://members.rogers.com/penate/pinsize.htm
A Web search on "optimal pinhole" should find other pages. Pinhole camera
enthusiasts are concerned with the same problem.
I have not been following the discussion, but independent of the eye's
ability to determine the center of a disk I would think that a smaller
image disk would be better.
For determining the position of the sun I would call attention to Tycho
Brahe's method, which combines pinhole projection together with seeing
equal brightness in two slits on opposite sides of the solar image.
http://www.kb.dk/elib/lit/dan/brahe/engelsktekst/frame30-en.htm
Without looking it up (error analyses have been published) I am guessing
that he achieved precision on the order of one arc minute.
Gordon
At 21:21 3/22/03 -0500, you wrote:
> ................., I would be interested in what they
> consider the optimal dimensions for the pin hole vs. the
> size etc. I will install one tomorrow at my home,
> commemorating the Spring Equinox.
Here is perhaps a different way of thinking about the
problem of the hole diameter vs its distance from the
plane on which the spot of light lands.
As the hole becomes smaller, physical optics makes the
spot of light larger. However, as the hole becomes larger,
geometric optics makes the spot of light larger. So I think
the thing to do is set the equations for these two diameters
equal to each other to find the minimum*. When I do that, I
get:
D = .03 SQRT L
where
D = diameter of hole, in mm
L = distance from hole to image plane, in mm
Since one equation is going up and the other down, the belly
of the curve which is their sum is very broad. Therefore, I
would increase the .03 figure to .04 to get much more light
through with no visible degradation of the fineness of the
spot.
Reworking the equation, we can see that the angle the spot
subtends becomes smaller as L becomes larger. This argues
for a large L.
On the other hand, the human eye is very good at finding
the center of a spot of light, so perhaps I am attacking
entirely the wrong problem!!
Best,
John Bercovitz
* I used the diameter of the first dark ring in the diffraction
pattern. I used the sodium D line for the wavelength. I used
a point object (not the sun, which subtends half a degree);
however, what I'm actually trying to do is sharpen the image
of the sun, so believe my approach is correct.
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Gordon Uber [EMAIL PROTECTED] San Diego, California USA
Webmaster: Clocks and Time: http://www.ubr.com/clocks
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