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

Bob Camp lists at rtty.us
Wed Nov 3 11:01:15 UTC 2010


Hi

I have indeed seen people try to do this with 18 ~ 24" aperture optics. They don't seem to do a lot better than the smaller stuff spotting holes at distance. They do get a nice bright image though. 

Bob

On Nov 3, 2010, at 12:15 AM, jimlux wrote:

> Robert Darlington wrote:
>> Hi Jim,
>> This doesnt' look right to me.
> 
> you're right.. I forgot inches/feet.. divide my 0.08 by 12..
> .007 mrad..
> 
> Now we're talking big, big aperture.. instead of 2-3 inches, 2-3 feet.
> 
>  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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> 
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