Over the past few weeks I have graduated in distance from 10m at the start of the beginners course up to 20m indoors and 60yds on the outdoor field. Before last week, the furthest distance I had attempted was 50m in a Short Metric round, but at the weekend I attempted 60 yards for the first time in a Warwick round. 60yds is just under 55m so I was confident of reaching the target, but I was initially surprised that scoring seemed a little easier despite the increase in distance.
The reason, of course, is perfectly clear : the Short Metric round is shot on an 80cm face whereas the Warwick round is shot on a 122cm face. This got me thinking about how much easier it is to shoot at a larger face and how far away does a 122cm target need to be to look like an 80cm target at 50m? To answer these questions, I had to dig into my memory for some high school trigonometry and algebra, as well as some basic physics. Here is my take on the problem.
The eye and brain interpret the size of an object from the angle it makes at the objective (the eyeball), so the wider the angle, the larger it looks and vice versa. Imagine drawing lines from the outside edges of an object back to the eyeball - the angle between the lines defines how large the object looks. A larger object far away can appear to be the same size as a smaller object which is closer. Father Ted explains it well, but Fr. Dougal doesn't quite get it...
I calculated the theoretical angle at the eye for an 80cm target at 50m, and a 122cm target at 60yds (54.86m) with the following results:
80cm face at 50m : 0.92°
122cm face at 60yds : 1.27°
This means that the larger face at 60yds appears almost 40% larger than the smaller face at 50m. No wonder it seemed easier! What has really surprised me is how small these angles are.
The other question was how far away would a 122cm face need to be to look the same size as an 80cm face at 50m. Doing the maths (math for my North American readers!) gives a result of 76m/83yds, but I am nowhere near being able to shoot that far yet, so won't be testing the theory in practice for a while yet.
Interesting. When we shoot a 12 dozen competition Gents FITA round we do 90/70 metres on a 122cm face, then 50/30 metres on an 80cm face.
ReplyDeleteThe change in face sizes invariably concides with a lunch break so there is often up to an hour between shooting.
My gut feeling is always that 50 metres on an 80cm face is about the same or a bit harder than 70 metres on a 122cm face. Certainly my 50cm scores are often about the same or worse than my 70 metre scores. This seems to confirm your maths though I suspect the lunch break may also have an impact!
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ReplyDeleteI think the ratio is more than that, because you're dealing with solid angles; it's the apparent area not just the angle subtended. I will make the assumption that I can ignore the difference between a flat target at 50m and one ever-so-slightly curved around a theoretical 50m sphere :). I can then use the solid angle formula: the proportion of the surface area of that large sphere covered by the target. Or in archery terms, the areay you are aiming at in proportion to the full area you COULD theoretically hit, if you could turn 360 degrees and shoot into the ground :)
ReplyDeleteFor the short metric:
Area of target = 0.16 Pi square metres
Surface area of sphere with radius 50m = 10000 Pi square metres
Solid angle 0.000016
For the Warwick (I am using metric throughout):
Area of target = 0.372 Pi square metres
Surface area of sphere with radius 54.86m = 12038 Pi square metres
Solid angle 0.000031
So it's almost twice as easy to hit! Interestingly, as a sanity check I squared those angles you calculated and you get the same ratio. I think this is because the sine function is almost linear at small angles.
Enjoying your blog - hopefully see you later today to shoot.
Martin R is absolutely right - the area of the target is important, not the dimensions, when you are shooting arrows at it. I really wanted to keep things simple and the linear dimensions seemed to be a good way of showing change in perceived size with distance. I did consider applying solid angles, but the thought of looking up what a steradian is after 30-odd years out of mathematics was too much to contemplate. I claim the Fr Dougal defence.....
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