Friend Waterman,

I have an old beat up 3rd Model Winder which when I bought it years ago had the original 22 Short barrel which is not in the best of condition and was, at some time in the past, re-chambered to "22 Long R" probably by R. F. Sedgley, and to add insult to injury it has a ring in the middle of it. Now, regardless of the barrel's less than perfect condition, using CCI 22 Short Target ammo, I shot a 25 yard 10 shot group that I could cover with the tip of my little finger, as I recall I was using a Lyman 438 scope. At that point I stopped looking for 22 Short ammo and stocked up on the CCI Short Target.

Bob

PS I have had very bad luck with the Aguila Golden Eagle Target ammo.

Richard,

I'm about to leave to go shooting at "Windy Acres Gun Club," unfortunatly, we don't get enough rain out here to conduct a proper deflection test, perhaps Joe "Westerner" or some of his buds from the Pacific Northwest could figure it out, I did want to say that Midway has a sale on 22 Ammunition and has the best prices I've seen recently. Such is life in the far west.

Bob

If it was caused by deflection, then why would it be 3" low? Some of those drops had to be hitting the underside of the bullet nose, which would have driven the bullet upwards. Something going on with all shots being 3" low.

For the raindrop to impact the bottom of the bullet, the bullet would have to hit the raindrop.  Raindrops are "raindrop" shaped, so if the bottom of the bullet hit the raindrop, it would only hit the tapered upper end of a downward moving mass. 

Assume a velocity of 750 fps for the 22 Short bullet.  Ignore deceleration (since I can't deal with it) and we have a time of flight over 25 yards of 0.1 seconds.

The bullet can hit any raindrop (or vice versa) within a path that is one bullet diameter plus one-half a uniform raindrop diameter on either side of the bullet.  With big uniform raindrops, that diameter is .25".  So the path width is 0.47"

With a steady one-hour rain dropping 2" of rain, a path 75 feet (900 inches) long x 0.47" wide will be hit by 110,122 raindrops in one hour.  Neglecting gravitational acceleration, bullet deceleration, bullet spin (450 revs per second) and the bullet trajectory (all beyond my elderly math skills), the bullet will be hit by (or will hit) 3 raindrops on its path to the target.

I think that is enough to deflect the bullet, but would the deflection be 3 inches?  Without getting into vectors, and again, I have fuzzy math skills, if a single raindrop with a downward velocity of 24 fps hits a bullet with a velocity of 750 fps, the result can be thought of a change in direction and can be thought of as a tangent function.  The arc tangent of 24/750 is one degree and 50 minutes.  If the raindrop was a solid of the same density as the bullet and if the bullet was not spinning, the bullet would be deflected by almost 2 degrees, or 27" at 25 yards.  Since the raindrop is much less dense and since the bullet is spinning, the deflection will be much less.  But I think I can believe a 3" deflection, especially if we include the downdraft factor.

Downdraft must also be a factor.  As must the angle between the rifle barrel and the target.  I made no effort to correct for that angle, which could well have been negative.  I often staple 3 targets in a vertical column.

Richard