[Archers] New Archery Rule Change (drawings?)

[Siegfried]

My most sincere apologies to whoever this other marshal is.  However,
those numbers are complete bunk/incorrect

Realize that the 'width' of the diamond forend of the safety zone has
not changed, at all, from the previous ruling, nor from current Society
recommendations.

The math goes as such:

You need to make a 30 degree angle out from the line between you, the
shooter, and the target.  Luckily for us this creates a special right
triangle for us, a 30/60/90 triangle, which has the properties of having
the sides match the ratio of 1:2:sqrt(3) (the latter being appro 1.73)

Therefore, for every 1.73 yards away a target is, you must have 1 yard
of clearance on either side of it.

So for a standard RR range, where you have a target 40 yards away ...
You need 40/1.73 = 23.12 yards clear on either side of it.  With a
little rounding, that means:

A 40yd target, needs a 46yd wide safety zone width from the point of the
target, and beyond.  Increase the width of this by the width of your
shooting line.   So if you had a 4yd wide shooting line (pretty normal).
 You end up with a 50yd wide cone.

Not the 140yds that was stated below.

The WIDEST cone you possibly need, for targets 50yds away, or farther,
is 58yds wide, plus the width of your shooting line.

I've quickly sketched this out, taken a picture and uploaded it as a
reference.  (Sorry for quality, was going it quick)

http://yfrog.com/86k7kj

Siegfried


On 3/4/10 11:02 AM, John Atkins wrote:
> My trigonometry is really rusty but these numbers were worked out by
> another marshal obviously more adept at trig than I.  I pass them on for
> consideration.
>  
> The rule:  At a minimum, each shootintg station will have a safety zone
> that extends 30 degrees from each end of the shooting line to a line
> even with the furthest target or 50 yards, whichever is closer...
>  
> Thus, taking a 30 degree tangent, 1 / 1.732 = 0.577, we can figure out
> the size field for a single shooter.  Assuming the shooting station is 2
> yards wide then at a 40 yard target, the clear width would be (2 * 40 /
> 0.577 ) + 2 = 140 yards wide.  Each additional shooter on the line would
> add 2 yards to this width.  A 50 + yard clout shoot would require a
> field 177 yards wide.
> 
> 
> 
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