front sight, MOA?
jmi...@prml.com
If my front sight is 25 inches from my eyeball, then I can
approximately calculate the angle in MOA that it obstructs from view with
this math, right?
Distance to front sight is 25 inches.
That would be the radius of the circle that would be defined by 25 inches
x 2 x Pi. (Circumference = Pi 2d)
So the circle is about 157 inches in circumference.
One minute is one 60th of a degree, and a circle has 360 degrees so the
circle has 21,600 minutes of angle.
On a circle of 157 inches, a minute then spans about .00727 inches in
width of view on that circle.
If a front sight post is .075 inches, and it is 25 inches from the
eyeball, and if a minute of angle on that circle is .00727 inches, then
my front sight post covers up about 10 minutes of angle (MOA), right?
And at 100 meters, that front sight covers up about 10 minutes of angle,
or about 10 inches width on the target, right?
-----------------------------------------------------------
Learn about rec.guns at http://www.recguns.com
-----------------------------------------------------------
Peter McMullen
jmi...@prml.com wrote:
# Check my math here.
#
# If my front sight is 25 inches from my eyeball, then I can
# approximately calculate the angle in MOA that it obstructs from view with
# this math, right?
#
# Distance to front sight is 25 inches.
# That would be the radius of the circle that would be defined by 25 inches
# x 2 x Pi. (Circumference = Pi 2d)
#
I think you mean Pi*2r
# So the circle is about 157 inches in circumference.
#
# One minute is one 60th of a degree, and a circle has 360 degrees so the
# circle has 21,600 minutes of angle.
#
# On a circle of 157 inches, a minute then spans about .00727 inches in
# width of view on that circle.
#
# If a front sight post is .075 inches, and it is 25 inches from the
# eyeball, and if a minute of angle on that circle is .00727 inches, then
# my front sight post covers up about 10 minutes of angle (MOA), right?
#
# And at 100 meters, that front sight covers up about 10 minutes of angle,
# or about 10 inches width on the target, right?
Sounds close enough
--
"At this time, Google policy does not permit the
advertisement of websites that contain 'firearms and
ammunition'."
BOYCOTT GOOGLE
:-)
Jerry Houston
you've said so far, not the diameter.
<jmi...@prml.com> wrote in message
news:ccsiuk$7c1$1...@grapevine.wam.umd.edu...
# Check my math here.
#
# If my front sight is 25 inches from my eyeball, then I can
# approximately calculate the angle in MOA that it obstructs from view with
# this math, right?
#
# Distance to front sight is 25 inches.
# That would be the radius of the circle that would be defined by 25 inches
# x 2 x Pi. (Circumference = Pi 2d)
Dale Alexander
Lincoln Rifle Club (in California) yesterday sighting in my M1-A and the
front sight is wider than the complete black circle at 100 yards. I don't
have the targets with me right now, but a guesstamate might be 10" wide or
10 MOA.
Works for me...
Dale Alexander
<jmi...@prml.com> wrote in message
news:ccsiuk$7c1$1...@grapevine.wam.umd.edu...
> ...
jmi...@prml.com
jerry....@comcast.net says...
> ...
Yes, my calc is for 2 x radius, I should have written 2r instead of 2d.
Is my calc otherwise right, that a .075 wide front sight at 25 inches
from the eyeball spans 10 minutes of angle?
> ...
JoeA
keep talking about! Gotta get me a thinner front sight.
JoeA
Alf Sauve
Just picked up a BSA scope for my air rifle and the instructions highlighted
the fact that the thin reticule covered 30MOA and was useful for
rangefinding (30" @ 100yds).
I've been shooting and reloading for over 30 years and that's one of the, "I
never knew that" things.
Alf
<jmi...@prml.com> wrote in message
news:ccsiuk$7c1$1...@grapevine.wam.umd.edu...
> ...
-----------------------------------------------------------
chucks
#
--- snip ---
<GEEK_MODE_ON>
This fear of trig functions is too much to stand....
tan t = y/x where
t = angle (MOA/60 = degrees)
x = distance to target
y= the height of the vertical thing (sight, target, whatever)
solve for y:
y = x tan t
examples:
1 MOA at 100 Yds.
y= (100*3*12)" tan (1/60) = (3600) * (2.91 x 10-4) = 1.05 inches
30 MOA at 100 Yds.
y= 3600 tan 30/60 = 3600 tan 1/2 = 3600 * 0.0087=31.42 inches
Just make sure all the units for x and y are the same, most likely inches
</GEEK_MODE_ON>
Bob Holtzman
# approximately calculate the angle in MOA that it obstructs from view with
#
# That would be the radius of the circle that would be defined by 25 inches
# x 2 x Pi. (Circumference = Pi 2d)
#
# So the circle is about 157 inches in circumference.
#
# One minute is one 60th of a degree, and a circle has 360 degrees so the
# circle has 21,600 minutes of angle.
#
# On a circle of 157 inches, a minute then spans about .00727 inches in
# width of view on that circle.
#
# If a front sight post is .075 inches, and it is 25 inches from the
# eyeball, and if a minute of angle on that circle is .00727 inches, then
# my front sight post covers up about 10 minutes of angle (MOA), right?
#
# And at 100 meters, that front sight covers up about 10 minutes of angle,
# or about 10 inches width on the target, right?
Out of curiosity, why are you taking such a round about way of finding the
solution? Just constuct 2 proportional equilateral triangles, one of 25"
hight and with a base of .075, and another overlayed on the first except
of 100 yd hight and solve for the base length.
the base
--
Bob Holtzman
"If you think you're getting free lunch,
......check the price of the beer!"
Bart B.
subtends 1 inch for every 100 yards of range.
Although the trig value based on the sine of 1/60th of a degree which
is 1.047197536428328546947470696664 inch per hundred yards of range is
often quoted, it's not reality. Most american bullseye targets'
scoring rings are based on the traditional value. And scope sights'
adjustments are based on the traditional standard (go check their
specs).
And the standard micrometer aperture (peep) rear sight in the USA has
40 threads per inch in its adjustment screws. When used with the
original 30-inch sight radius standard and moving the aperture
.025-inch per turn of 12 clicks, 4 of those clicks moved the rear
sight 0.008333 inch. Dividing 30 inches by 3,600 results in .008333
inch. One click moves the sight with this setup moves the impact
exactly 1/4th inch for every 100 yards of range.
So, to simplify this issue, just divide the distance by 3,600 to find
out how much a MOA is worth on your shooting issues. Why 3,600?
That's how many inches there are in 100 yards.
Dividing 25 by 3,600 gives 0.0069444444444444444444444444444444 inch.
Gees, let's round it off to .007 inches. That's a good James Bond
number anyway.
chucks
"Bart B." <bar...@aol.com> wrote in message
news:cd1p81$g58$1...@grapevine.wam.umd.edu...
# The shooting minute of angle has traditionally been that which
# subtends 1 inch for every 100 yards of range.
#
--- snip ---
I guess I can live with the 0.05" error ;-)
The only reason I ever gave this any thought was in preparation of a Docter
(http://www.docteropticsusa.com/) sight that I put on a pistol. I was
curious on how large of a target area the 7 MOA dot covered. Being an
engineer (read anal) I made a spreadsheet of various distances and sizes.
The sight comes with a dial you can push on to a screwdriver to use to
adjust offsets in windage and elevation. I took the geek approach and fired
the gun off sand bags and then measured the offset of the holes from the
point of aim, dialed in the adjustments, and damn, it was pretty close. I
only needed to shoot 2 groups of 5 rounds to get the gun zeroed into what my
best (least wobble) bag shooting could yield. I used to work for a company
that made Head Up Displays so the theory is pretty much the same. We just
used miliradians as a unit of angle instead of MOA. Fun stuff
Good shooting,
Chuck
Peter McMullen
chucks wrote:
# <jmi...@prml.com> wrote in message
# news:ccsiuk$7c1$1...@grapevine.wam.umd.edu...
# # Check my math here.
# #
# # If my front sight is 25 inches from my eyeball, then I can
#
# --- snip ---
#
# <GEEK_MODE_ON>
#
# This fear of trig functions is too much to stand....
#
# tan t = y/x where
# t = angle (MOA/60 = degrees)
# x = distance to target
# y= the height of the vertical thing (sight, target, whatever)
# solve for y:
# y = x tan t
#
# examples:
#
# 1 MOA at 100 Yds.
#
# y= (100*3*12)" tan (1/60) = (3600) * (2.91 x 10-4) = 1.05 inches
#
# 30 MOA at 100 Yds.
#
# y= 3600 tan 30/60 = 3600 tan 1/2 = 3600 * 0.0087=31.42 inches
#
# Just make sure all the units for x and y are the same, most likely inches
#
# </GEEK_MODE_ON>
#
Most folks here know that 1 MOA at 100 yards equals about 1
inch.
So how does your equation solve for how much of the target
width, in inches, is covered by the front sight width, if
the eye/site radius is 25 inches, the front sight width is
.075 inches, and the target distance is 100 yards???
I think you need to put on your thinking cap again...
--
"At this time, Google policy does not permit the
advertisement of websites that contain 'firearms and
ammunition'."
BOYCOTT GOOGLE
:-)
Peter McMullen
Bart B. wrote:
# The shooting minute of angle has traditionally been that which
# subtends 1 inch for every 100 yards of range.
#
# Although the trig value based on the sine of 1/60th of a degree which
# is 1.047197536428328546947470696664 inch per hundred yards of range is
# often quoted, it's not reality. Most american bullseye targets'
# scoring rings are based on the traditional value. And scope sights'
# adjustments are based on the traditional standard (go check their
# specs).
#
# And the standard micrometer aperture (peep) rear sight in the USA has
# 40 threads per inch in its adjustment screws. When used with the
# original 30-inch sight radius standard and moving the aperture
# .025-inch per turn of 12 clicks, 4 of those clicks moved the rear
# sight 0.008333 inch. Dividing 30 inches by 3,600 results in .008333
# inch. One click moves the sight with this setup moves the impact
# exactly 1/4th inch for every 100 yards of range.
#
# So, to simplify this issue, just divide the distance by 3,600 to find
# out how much a MOA is worth on your shooting issues. Why 3,600?
# That's how many inches there are in 100 yards.
#
# Dividing 25 by 3,600 gives 0.0069444444444444444444444444444444 inch.
# Gees, let's round it off to .007 inches. That's a good James Bond
# number anyway.
#
I think you are overlooking the goal. If I am not mistaken,
jmills wants a formula to calculate how much of the target
width, in MOA *or* inches, is covered by the width of his
front site blade, for a given eye/sight radius, sight width,
and target distance. Trig would work, but its been 30+ years
for me...
---
"At this time, Google policy does not permit the
advertisement of websites that contain 'firearms and
ammunition'."
BOYCOTT GOOGLE
:-)
chucks
"Peter McMullen" <pmcm...@san.rr.com> wrote in message
news:cd319h$2sk$1...@grapevine.wam.umd.edu...
#
#
# chucks wrote:
--- snip ---
#
# width, in inches, is covered by the front sight width, if
# the eye/site radius is 25 inches, the front sight width is
# .075 inches, and the target distance is 100 yards???
#
# I think you need to put on your thinking cap again...
#
--- snip ---
Distance from the eye to the sight is 25". That will be the adjacent side of
a right triangle (x). The opposite side of the right triangle is 1/2 sight
width or 0.075/2 = 0.0375" (y). Solving for the angle, t: tan t = y/x, t
= atan y/x =0.0859 degrees. Let's take it out to 100 yds or 3600" and solve
for y.:
y = x tan t = 3600 * 0.0015 = 5.4". If your sight was a circle, that number
would be the radius of the circle on the 100 yd target. The width of the
area on the target would be twice that. Can't really solve for the area
since the height of the sight was not given.
no...@bouvt.com
says...
# I think you are overlooking the goal. If I am not mistaken,
# jmills wants a formula to calculate how much of the target
# width, in MOA *or* inches, is covered by the width of his
# front site blade, for a given eye/sight radius, sight width,
# and target distance. Trig would work, but its been 30+ years
# for me...
#
Here is the answer.
Cos of the angle we are solving for =
(25 inches squared + 25 inches squared)- (.075 inches squared)
divided by
2 X (25 inches x 25 inches)
So Cos of the angle we are solving for =
(1,249.994375
divided by
1,250)
That gives us a Cos of the angle we are solving for of .9999955
That equals 0.003000001124989 radians.
That equals 0.171887402996349 degrees.
That equals 10.313244179781000 minutes of angle.
So if the sight distance from eyeball to front sight is 25 inches, and
the width of the front sight is .075 inches, then the sight covers
10.3132 MOA, which is a little more than 10 inches in width at 100 yards
(caculation for the last point is not shown).
Since the front sight is .075 inches wide, each 1 MOA is represented on
that front sight by about 73 ten-thousandths of an inch.
Can anyone find an error with this trig?
Strider
<chu...@verizon.com> wrote:
#
#"Bart B." <bar...@aol.com> wrote in message
#news:cd1p81$g58$1...@grapevine.wam.umd.edu...
## The shooting minute of angle has traditionally been that which
## subtends 1 inch for every 100 yards of range.
##
#--- snip ---
#
#I guess I can live with the 0.05" error ;-)
#
#The only reason I ever gave this any thought was in preparation of a Docter
#(http://www.docteropticsusa.com/) sight that I put on a pistol. I was
#curious on how large of a target area the 7 MOA dot covered. Being an
#engineer (read anal) I made a spreadsheet of various distances and sizes.
#The sight comes with a dial you can push on to a screwdriver to use to
#adjust offsets in windage and elevation. I took the geek approach and fired
#the gun off sand bags and then measured the offset of the holes from the
#point of aim, dialed in the adjustments, and damn, it was pretty close. I
#only needed to shoot 2 groups of 5 rounds to get the gun zeroed into what my
#best (least wobble) bag shooting could yield. I used to work for a company
#that made Head Up Displays so the theory is pretty much the same. We just
#used miliradians as a unit of angle instead of MOA. Fun stuff
#
#
#Good shooting,
#Chuck
#
Why not just take a big sheet of paper and cover it in lines spaced
one inch apart.. Use really dark lines.Then look through the sight
and see how many of the blocks are covered at 100 yards. That way,
moa conversion would be simple and not a problem fit for an IBM
mainframe.
Strider
Peter McMullen
jmi...@prml.com wrote:
# Check my math here.
#
# If my front sight is 25 inches from my eyeball, then I can
# approximately calculate the angle in MOA that it obstructs from view with
# this math, right?
#
ok...
I think I remember my trig well enough to figure this out.
Essentially, angle from the eyeball to each side of the
sight blade (and each side of the target) is an isosceles
triangle something like:
c1
c
a b b1
d
d1
sight target
hopefully this looks right in the post
So since trig deals with right triangles, we are really
talking about two right angles back to back.
So we can exploit the tangent function and simply
divide the sight width "cd" by the sight radius (from the
eye) "ab", and multiply by the distance to target "ab1" to
get the target area covered c1d1
or .075/25*3600 (for 100 yards)
equals 10.8 inches
so you could make up a simple formula of sight blade width,
divided by the sight radius (from your eye) times the
distance to target or
S / R * D = Ac (area covered)
--
"At this time, Google policy does not permit the
advertisement of websites that contain 'firearms and
ammunition'."
BOYCOTT GOOGLE
:-)
Michael
presumably awesomely outrageous dude, wrote the following in rec.guns:
# I guess I can live with the 0.05" error ;-)
It's that kind of sloppy, "close enough for government work" attitude that
permits Replicators to overrun the Asgard homeworld, to say nothing of
leading Teal'c to believe that the stubble's not so bad and he can go
another couple of days without shaving his head.
--
Michael
Bart B.
divide .075 inch by 0.00694 and get about 10.81 moa. That front sight
will cover almost 11 inches at 100 yards.
Peter McMullen
chucks wrote:
# "Peter McMullen" <pmcm...@san.rr.com> wrote in message
# news:cd319h$2sk$1...@grapevine.wam.umd.edu...
# #
# #
# # chucks wrote:
# --- snip ---
# #
# # So how does your equation solve for how much of the target
# # width, in inches, is covered by the front sight width, if
# # the eye/site radius is 25 inches, the front sight width is
# # .075 inches, and the target distance is 100 yards???
# #
# # I think you need to put on your thinking cap again...
# #
# --- snip ---
#
# Distance from the eye to the sight is 25". That will be the adjacent side of
# a right triangle (x). The opposite side of the right triangle is 1/2 sight
# width or 0.075/2 = 0.0375" (y). Solving for the angle, t: tan t = y/x, t
# = atan y/x =0.0859 degrees. Let's take it out to 100 yds or 3600" and solve
# for y.:
# y = x tan t = 3600 * 0.0015 = 5.4". If your sight was a circle, that number
# would be the radius of the circle on the 100 yd target. The width of the
# area on the target would be twice that. Can't really solve for the area
# since the height of the sight was not given.
#
Thanks Chuck. I dusted off my gray cells and figured it out.
Regarding the "area", I actually was not clear in my terms.
What I should have specified was the width of the target
surface covered by the sight blade silhouette.
--
"At this time, Google policy does not permit the
advertisement of websites that contain 'firearms and
ammunition'."
BOYCOTT GOOGLE
:-)
John Chase
# "Bart B." <bar...@aol.com> wrote in message
# news:cd1p81$g58$1...@grapevine.wam.umd.edu...
# # The shooting minute of angle has traditionally been that which
# # subtends 1 inch for every 100 yards of range.
# #
# --- snip ---
#
# I guess I can live with the 0.05" error ;-)
Yeah, but at a thousand yards that's a half-inch. Could be the
difference between an "X" and "just a 10". :-)
-jc-
Bart B.
# Yeah, but at a thousand yards that's a half-inch. Could be the
# difference between an "X" and "just a 10".
Of all the metallic (iron) sighted rifles used in high power
competition, there's probably less than .05% of 'em whose
"quarter-minute" or "half-minute" rear sight actually moves bullet
impact 2.5 or 5 inches at 1000 yards. Most of them have sight
radiuses of more than the 30 inches these sights are set up for.
That's because their barrels are longer than the 24-inch ones these
sights were first mounted on about 90 years ago. For example, a
Palma rifle with its 30-inch tube will have about 36 inches of sight
radius which gives 5 clicks per actual (shooter's) minute of angle
with the traditional quarter-minute sight.
For those interested, they might measure the sight radius on their M1
or M1A US service rifle with the rear sight raised 20 clicks up from
bottom. That's from the rear of front sight blade's top to rear of
the actual aperture in the elevation slide or NM hood. Divide that
sight radius by 3,600. Then note that the service (battle) sight's 32
tpi windage screw moves the rear sight base 0.0078125 per click. The
NM rear sights have 64 tpi for windage, so it takes two clicks on the
windage knob to move it 1 moa. The movement on both ain't exactly
1.0000000 minute of angle 'cause the sight radius ain't 28.125 inches.
Shucks, I'm gonna really stir the pot now. If you look at the M1 and
M14 rear sight, you'll note the elevation slide is curved. It's
curved in such a way that when near the bottom, the aperture moves up
less per elevation click than when near the top. One gets more
movement per click when zeroed for 1000 yards (about 45 to 50 clicks
up) than when zeroed for 100 yards.
What's this I just stirred up in the service sight pot? Why, it's a
pair of National Match apertures!! Careful examination shows their
.0595- and .520-inch apertures are drilled .0020-inch off center in
the direction of the notch in the hood's rear edge. That's so they
can be rotated in the elevation slide from notch up to notch down and
make a "half-moa" elevation correction. But it just ain't exactly
that for several reasons you probably have already figured out. It
is, however, pretty darned close; close enough for government work.
Bob Holtzman
# Why not just take a big sheet of paper and cover it in lines spaced
# one inch apart.. Use really dark lines.Then look through the sight
# and see how many of the blocks are covered at 100 yards. That way,
# moa conversion would be simple and not a problem fit for an IBM
# mainframe.
At 100 yds w/ no magnification (do I misunderstand?)? You had better
use a paint brush!
--
Bob Holtzman
"If you think you're getting free lunch,
......check the price of the beer!"
Bob Holtzman
# chucks wrote:
#
# # "Bart B." <bar...@aol.com> wrote in message
# # news:cd1p81$g58$1...@grapevine.wam.umd.edu...
# # # The shooting minute of angle has traditionally been that which
# # # subtends 1 inch for every 100 yards of range.
# # #
# # --- snip ---
# #
# # I guess I can live with the 0.05" error ;-)
# Yeah, but at a thousand yards that's a half-inch. Could be the
# difference between an "X" and "just a 10". :-)
Go ahead and laugh but matches are won or lost by that much.
--
Bob Holtzman
"If you think you're getting free lunch,
......check the price of the beer!"
Bob Holtzman
#
# width, in MOA *or* inches, is covered by the width of his
# front site blade, for a given eye/sight radius, sight width,
# and target distance. Trig would work, but its been 30+ years
# for me...
Proportional triangles and simple plane geometry are your friends.
--
Bob Holtzman
"If you think you're getting free lunch,
......check the price of the beer!"
Tom Gibson
[snip]
# Distance from the eye to the sight is 25". That will be the adjacent side of
# a right triangle (x). The opposite side of the right triangle is 1/2 sight
# width or 0.075/2 = 0.0375" (y). Solving for the angle, t: tan t = y/x, t
# = atan y/x =0.0859 degrees. Let's take it out to 100 yds or 3600" and solve
# for y.:
# y = x tan t = 3600 * 0.0015 = 5.4". If your sight was a circle, that number
# would be the radius of the circle on the 100 yd target. The width of the
# area on the target would be twice that. Can't really solve for the area
# since the height of the sight was not given.
So the 10" estimate of the original poster was .8" on the small side.
Trig refresher aside, I'd say that a simple "yes" would have been
sufficient.
Using the 1MOA = 1" @ 100yds 'rule' that most scope manufacturers and
shooter live by, the guy did a few simple calculations and arrived at
a very good estimation of the 'correct' answer, IMO.
For those of us without a scientific caluclator in our range bags...
For any given Sight Radius (distance from eyeball to front sight):
Front Bead coverage @ 100yds = Diameter of Bead÷.00727
examples:
1/16" bead (.0625"), the sight covers approx. 8.6" @ 100yds
3/32" bead (.09375) = nearly 13" @ 100yds
For all practical purposes, you can probably remember .007 (think
James Bond, OK?). The resulting 'error' for rounding off the .00027
is less than 4% or less than 1/2" @ 100yds for the 3/32" sight.
All of a sudden, those 4" groups @ 100yds are looking mighty good.
He-double hockey sticks, I'm consistently shooting sub .5 MOA groups
with my 45-70! I had no idea I was such a good shot.<BSEG>
The things you learn on Usenet.
Tom G
Dave Hinz
# In article <cd5q61$bia$1...@grapevine.wam.umd.edu>, John Chase wrote:
# #
#
# Go ahead and laugh but matches are won or lost by that much.
Right, but matches are scored on "does it touch the line", not calculating
the MOA of deviation from the center of the bull. 1/2" in calculation error
of "MOA of accuracy" isn't relevant, it's where the hole really goes.
Strider
<hol...@sonic.net> wrote:
#In article <cd4gqu$nh9$1...@grapevine.wam.umd.edu>, Strider wrote:
## Why not just take a big sheet of paper and cover it in lines spaced
#
## one inch apart.. Use really dark lines.Then look through the sight
## and see how many of the blocks are covered at 100 yards. That way,
## moa conversion would be simple and not a problem fit for an IBM
## mainframe.
#
#At 100 yds w/ no magnification (do I misunderstand?)? You had better
#use a paint brush!
Solid lines of appropriate width might work. i.e. a painted line 4"
wide on one sheet, 5" on the next, etc. and see which one the sight
covers.
Strider
Peter McMullen
Bob Holtzman wrote:
# In article <cd319i$2sl$1...@grapevine.wam.umd.edu>, Peter McMullen wrote:
# #
# # I think you are overlooking the goal. If I am not mistaken,
# # jmills wants a formula to calculate how much of the target
# # width, in MOA *or* inches, is covered by the width of his
# # front site blade, for a given eye/sight radius, sight width,
# # and target distance. Trig would work, but its been 30+ years
# # for me...
#
# Proportional triangles and simple plane geometry are your friends.
#
Which would make it about 35 years...
--
"At this time, Google policy does not permit the
advertisement of websites that contain 'firearms and
ammunition'."
BOYCOTT GOOGLE
:-)
Bob Holtzman
# On Fri, 16 Jul 2004 11:17:47 +0000 (UTC), Bob Holtzman <hol...@sonic.net> wrote:
# # In article <cd5q61$bia$1...@grapevine.wam.umd.edu>, John Chase wrote:
# # #
# # # Yeah, but at a thousand yards that's a half-inch. Could be the
# #
# # Go ahead and laugh but matches are won or lost by that much.
# Right, but matches are scored on "does it touch the line", not calculating
# the MOA of deviation from the center of the bull. 1/2" in calculation error
# of "MOA of accuracy" isn't relevant, it's where the hole really goes.
I know. I was referring to the phrase about .05" error.
--
Bob Holtzman
"If you think you're getting free lunch,
......check the price of the beer!"
Bart B.
bullet holes regardless of caliber used. Which means a hole from a
.224-in bullet that's .040 inch from touching the line of the higher
scoring ring will have the 30 caliber plug's .308-in. diameter flange
now go .002-inch into the higher scoring ring and the shot be given
the higher value.
This rule bothered lots of folks who think the bullet must touch the
scoring ring in order to be given the higher value. It's like
shooting balloons; if you don't touch the balloon, it don't pop. But
the NRA's position is that everybody shooting sub 30 caliber bullets
will be scored according to where the bullet center is so all holes
will be plugged with a 30 caliber plug.
Interesting reasoning, but that's the way it is.
Douglas Skalitzky
# In highpower competition, a 30 caliber plug is put in all those close
# bullet holes regardless of caliber used. Which means a hole from a
# .224-in bullet that's .040 inch from touching the line of the higher
# scoring ring will have the 30 caliber plug's .308-in. diameter flange
# now go .002-inch into the higher scoring ring and the shot be given
# the higher value.
#
# This rule bothered lots of folks who think the bullet must touch the
# scoring ring in order to be given the higher value. It's like
# shooting balloons; if you don't touch the balloon, it don't pop. But
# the NRA's position is that everybody shooting sub 30 caliber bullets
# will be scored according to where the bullet center is so all holes
# will be plugged with a 30 caliber plug.
#
# Interesting reasoning, but that's the way it is.
I thought the .30 caliber rule was dropped a few years ago. Mouse guns
don't seem to need any help to win matches these days.
If it was up to me, all the dinosaurs hauling .30 cals to the line would get
20 bonus points for conspicuous intrepidity.
Doug S.
Milwaukee, WI USA
Bart B.
# Mouse guns don't seem to need any help to win matches these days.
They indeed do very well through 600 yards. Back in '71 when I was
the first person in the USA to shoot an M16 in a highpower match
(happened at the Nationals on rifle leg day), we knew that rifle would
eventually be tricked out to shoot very well indeed. But at long
range, the 22 caliber rifles just haven't quite fared as well as 26
through 30 caliber.
Interesting you call them 'mouse' guns. When the US Navy Rifle Team
took 12 M16's fitted with Clerke barrels and Redfield International
rear sights to the 1971 Nationals, little did we know that just
because we wore Mickey Mouse sweatshirts that the animal's name would
be forever linked to the M16.