In the atmosphere, if your aperture is smaller than the R_0 (Fried
parameter, typically about 1~2" in daylight, can be several feet by a still
night), your resolution is similar to that corresponding at your aperture,
and the whole image will seems to move. If your aperture is greater than
R_0, the image will explode in several "speckles", each with a size
corresponding to the diffraction limit of your aperture, grouped in a
cluster corresponding to the diffranction limit of R_0, and the pattern of
the speckle will change rapidly (several 10's of Hz). You can exploit the
theoritical diffraction limit by using a fast camera, processing each frame
(autocorrelation or FFT) and then averaging the results. (search "Speckle
Interferometry" for details, that will hardly be low-cost or easy...)
HTH,
Regards,
Jean-Louis
----- Original Message -----
From: "Bob Camp" <[email protected]>
To: "Discussion of precise time and frequency measurement"
<[email protected]>
Sent: Wednesday, November 03, 2010 12:04 AM
Subject: Re: [time-nuts] A real world project need for timing accuracy...
Hi
The Wikipedia numbers would all work out just fine in a vacuum or in
*very* still air. I have yet to find a real world situation (daylight)
where you are anywhere near those conditions.
Bob
On Nov 2, 2010, at 1:16 PM, Robert Darlington wrote:
Hi Jim,
This doesnt' look right to me. I'm getting roughly 2.3 inches at 2400
feet
is 0.08 miliradians. 0.01 miliradians (1*10^-5 radians) at 2400 feet
is
0.288 inches (roughly 30 caliber). Wikipedia says that to resolve 0.01
miliradians you need:
R (in radians) = lambda / diameter (of scope) (aka, Dawes Limit if you
use
562nm light)
1 * 10^-5 radians = 562nm (green) / X
X= 5.62cm aperture or 2.2". This is what it comes to on paper, in
practice you'd probably need something bigger because of atmospheric
effects, lens quality, and the like.
That being said, I can't see my holes at 300 yards with my Leupold scope
with an opening greater than an inch. I can just barely make them out at
200 yards. See http://en.wikipedia.org/wiki/Angular_resolution - Also,
somebody please double check my math.
-Bob
On Tue, Nov 2, 2010 at 7:28 AM, jimlux <[email protected]> wrote:
Bob Camp wrote:
Hi
Ok, I mis-understood the question.
In my experience, you can have big buck (as in many thousands of
dollars)
optics and not see .2" holes at 800 yards. The bull's eye is a *lot*
bigger
than the hole the bullet made.
0.2" at 2400 ft is about 0.08 milliradian.. or 0.3 minutes of arc.
Your
eye can resolve about 1 minute of arc... I'm not questioning your
experience, but it seem that even a moderate power scope should allow
you to
see the holes. As I recall, the Rayleigh limit for resolution is
something
like 0.7 milliradian/mm of aperture, so 10-15 mm aperture would be in
the
right ballpark..
I can imagine needing more aperture than 3", though.. you're not
interested
in resolving a star, but something more akin to separating dots.
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