[time-nuts] A real world project need for timing accuracy...

[Jean-Louis Oneto]

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" <lists at rtty.us>
To: "Discussion of precise time and frequency measurement" 
<time-nuts at febo.com>
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 <jimlux at earthlink.net> 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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