| Ring | Metric score | Imperial score | Outside dia(cm) | Inside dia(cm) | Area (sq.cm) | %area (metric) | %area (imperial) |
| Outer white | 1 | 1 | 122 | 109.8 | 2221.07 | 19% | 36% |
| Inner White | 2 | 1 | 109.8 | 97.6 | 1987.28 | 17% | |
| Outer black | 3 | 3 | 97.6 | 85.4 | 1753.48 | 15% | 28% |
| Inner black | 4 | 3 | 85.4 | 73.2 | 1519.68 | 13% | |
| Outer blue | 5 | 5 | 73.2 | 61 | 1285.89 | 11% | 20% |
| Inner blue | 6 | 5 | 61 | 48.8 | 1052.09 | 9% | |
| Outer red | 7 | 7 | 48.8 | 36.6 | 818.29 | 7% | 12% |
| Inner red | 8 | 7 | 36.6 | 24.4 | 584.49 | 5% | |
| Outer gold | 9 | 9 | 24.4 | 12.2 | 350.7 | 3% | 4% |
| Inner gold | 10 | 9 | 12.2 | 0 | 116.9 | 1% | |
| 11689.87 |
Before looking at the results, it is worth looking at how the calculations are done. I decided to work with a 122cm diameter face, the largest in common usage in target archery. There are, of course, ten scoring rings on a standard face, two each of white, black, blue, red and gold, working from the outside inwards. Each of these rings is the same width so it is easy to find that for a 122 cm face, each ring is 6.1cm wide (122/20). The inner gold is the only complete circle, having a diameter of 12.2cm.
To work out the area covered by each of the rings, simply calculate the area of the circle from the outside diameter of the ring then subtract the area of the circle from the inside diameter of the ring. If you cast your memory back to your schooldays you will recall that the area of a circle is Pi times the radius squared. Using a spreadsheet means that you can do this fairly easily, but it might help to know that in Excel, the function PI() can be used in a function instead of typing 3.1.4159265.....
The actual areas of each of the rings are shown in the table above. I then summed them before calculating the percentage of the total target area for each ring.
As I said earlier, the results surprised me. What immediately stood out was the difference in percentage area of the outer white (19%) and inner gold (1%). The 19% figure wasn't that much of a shock, but the 1% definitely was. To score a ten, you need to almost twenty times as accurate as scoring a one. That probably explains why I get significantly more ones than tens, and has nothing at all to do with my ability!
The other oddity I noticed was that the percentage areas form a linear progression: 1%, 3%, 5%, 7%, 9%, 11%, 13%, 15%, 17% and 19%. There is probably a sound mathematical reason for this but I cannot work it out. let me know if you find out!
The above calculations are based on ten-zone, metric scoring, but you can easily find out the area percentages by summing the two figures for each of the coloured scoring zones. I have included these figures in the table above.
So, what about the smaller targets? Clearly, the areas will be smaller, but the proportions of each of the scoring rings are identical to the 122cm face, meaning that the percentages will be the same regardless of the overall diameter.
The actual areas of each of the rings are shown in the table above. I then summed them before calculating the percentage of the total target area for each ring.
As I said earlier, the results surprised me. What immediately stood out was the difference in percentage area of the outer white (19%) and inner gold (1%). The 19% figure wasn't that much of a shock, but the 1% definitely was. To score a ten, you need to almost twenty times as accurate as scoring a one. That probably explains why I get significantly more ones than tens, and has nothing at all to do with my ability!
The other oddity I noticed was that the percentage areas form a linear progression: 1%, 3%, 5%, 7%, 9%, 11%, 13%, 15%, 17% and 19%. There is probably a sound mathematical reason for this but I cannot work it out. let me know if you find out!
The above calculations are based on ten-zone, metric scoring, but you can easily find out the area percentages by summing the two figures for each of the coloured scoring zones. I have included these figures in the table above.
So, what about the smaller targets? Clearly, the areas will be smaller, but the proportions of each of the scoring rings are identical to the 122cm face, meaning that the percentages will be the same regardless of the overall diameter.
Thanks, Mark. I'd wondered why such a high percentage of my scoring shots were 1s, and this table has helped me understand it!
ReplyDeleteBTW, you forgot to mention that you got a very creditable 3rd class in very windy conditions at 80/60.
The area of a circle is πr². If you count the 10-ring as having a radius (or r) of 1, the radius of each subsequent ring going out are therefore multiples of r.
ReplyDeleteThe area of the 10-ring is πr². To find the area of the 9-ring, you find the area of the 9-circle and take away the area of the the 10-circle. The equation for this is π(2r)²- πr². The 2r accounts for the 9-ring having a radius twice that of the 10-ring. We can take out both π symbols as they cancel each other out. This leaves (2r)²- r², which is expanded to 4r²- r² which then equals 3r2. So the 9-ring is 3 times bigger than the 10 ring.
Let’s expand this to the 8-ring, which has a diameter of 3r. The equation for the area is the 8-circle minus the 9-circle, or (3r)²- (2r)² which equals 9r²-4r² which equals 5r². So, this shows the 8-ring is 5 times bigger than the 10-ring. The changes in proportional size basically come from the increasing (gaps) between square numbers, 1 (3) 4 (5) 9 (7) 16 (9) 25…. and so on. Confused yet?