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

[Robert Darlington]

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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