Field Archery Target Sizes

dvd8n

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I'm trying to figure out the target ring sizes for field - the 20, 40, 60, 80cm black ones with 4 black rings and a gold 5 and 6 in the middle.

The outer rings are obviously all the same width but I can't find it documented how the golds are sized - for example do the 5 and 6 both have full width or is the 6 just an inner zone on the 5? I can't find it documented and it's not obvious from pictures.
 

bimble

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the six is just an inner zone (used to be an 'X' worth five), so centre to outer edge of 5, equals width of one of the black scoring rings. 8/6/4/2cm for respective faces.
 

Valkamai

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The six is the inner section of the five, so the five and six are half the width of the other scoring rings.

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dvd8n

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The six is the inner section of the five, so the five and six are half the width of the other scoring rings.
the six is just an inner zone (used to be an 'X' worth five), so centre to outer edge of 5, equals width of one of the black scoring rings. 8/6/4/2cm for respective faces.
So the width of a ring is the target diameter divided by 10? I think that even I can cope with that. If only I didn't need to convert to imperial too :D
 

bimble

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So the width of a ring is the target diameter divided by 10? I think that even I can cope with that. If only I didn't need to convert to imperial too :D
There is a reason why it's easier to work in the units they do to set the targets... ;) saying that, rounds like the York, Hereford, St George, etc are all official rounds in New Zealand... but they are a metric country, so when you look at the rules you'll see the York round is 72 arrows at 91m, 48 at 73m and 24 at 55m...
 

dvd8n

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Yeah, I understand. However my local field club was all imperial when I joined. It's all umarked 2D now, which isn't any units at all, but it's too late. I can't help but think in imperial for archery.
 
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