Talk:Ragged holes
You can't really use CEP(p) since you have no idea of what percentage of holes you'll be able to find if you shoot another target.
Consider this situation. You make 100 shots. There are enough shots for which a position can be determined so that you can estimate the group center. Now you count how many shots you can find (say 30). Take half of that number-15. Now use CEP(85). Sounds good right? Dah... How many different circles can you draw that eliminate 15 of the holes? A lot. It would conceivable to find all such circles and take average position and size. Not possible work for a human, but a computer doesn't care.
If done right ragged hole measurements aren't crippling.
Herb (talk) 18:12, 1 June 2015 (EDT)
- Herb: Since you took out the following you need to provide some sort of substantial criticism of it. I can't tell if what you say above is an attempt to explain it or disprove it? More than one person has agreed that this is the best practical solution to the problem it addresses:
- In practice, given a target with a ragged hole and a small number of "censored" shots, it is probably adequate to place them evenly inside the hole. If the number of censored shots is large a better solution is to:
- Set p = proportion of shots that were censored.
- Find the smallest sigma such that CEP(p) covers the ragged hole.
