Hammbone Posted February 6, 2013 Share Posted February 6, 2013 Perhaps I'm over analyzing this: I just shot up 6 targets doing accuracy tests with some experimental reloads. Once I was done, I drew a verticle and horizontal centerline through the bullseye. The I measured the distace from "x" and "y" axis for each hole and created a table of coordinate points in Excel...assigning quadrants by dictating (+,+) for quad 1, (-,+) for quad 2, (-,-) for quad 3, and (+,-) for quad 4...as we're taught with the unit circle in calculus. Here's what I can't figure out; How do I mathematically calculate the diameter of the smallest fit circle for a given set of 2D coordinate points? Link to comment Share on other sites More sharing options...
Sarge Posted February 6, 2013 Share Posted February 6, 2013 Link to comment Share on other sites More sharing options...
Corey Posted February 6, 2013 Share Posted February 6, 2013 I just look for the group where the holes look closer to each other than others, haha. Link to comment Share on other sites More sharing options...
JerryShoots Posted February 6, 2013 Share Posted February 6, 2013 If the bullseye is the center then wouldn't the distance between 0,0 and center of the farthest hole from (0,0) be the radius of the smallest circle? Link to comment Share on other sites More sharing options...
JerryShoots Posted February 6, 2013 Share Posted February 6, 2013 Unless your talking spread of group from hole to hole. Then wouldn't it be center of hole to farthest hole distant is the diameter of smallest circle? Link to comment Share on other sites More sharing options...
Steve RA Posted February 6, 2013 Share Posted February 6, 2013 Get a program called "OnTarget precision calculator" - works very nicely. Link to comment Share on other sites More sharing options...
56hawk Posted February 6, 2013 Share Posted February 6, 2013 Not sure I follow what you are trying to accomplish. The smallest circle is going to be the one that intersects the two holes that are furthest from each other. You could use the distance formula to find these two if it's not obvious. Link to comment Share on other sites More sharing options...
Hammbone Posted February 6, 2013 Author Share Posted February 6, 2013 JereyShoots - You'd think that, but its not true. First of all, I'm looking at group sizes for now. Ignoring relation to center of bullseye. However, I did use the bullseye as coordinate (0, 0). Then I shifted the center to the average center of the group (turns out mathematicians have multiple methods for calculating this). I quickly saw the flaw with my method, but I'm sticking with it for now. The biggest problem with measuring the two outermost holes is that you need 3 points to calculate a circle. You can do it with a center point and two circumference points, or 3 circumference points. The problem with using the center as one point is that there is error induced by assuming the center is the average of all points (especially if there is a "flier" or two). The problem with using 3 circumference points is that I don't know how to mathematically select the optimum 3 points. SteveRA - Is this software free? I have experience writing programming code in Fortran, m-code in MatLab, Engineering Equation Solver, and of course Excel...so I'm not affraid to get my hands dirty per say. Just trying to figure out the math behind it. Link to comment Share on other sites More sharing options...
Steve RA Posted February 6, 2013 Share Posted February 6, 2013 Far as I know it's a free program, my shooting partner uses it, although that is a picture of one of my targets. Based on your qualifications, it would be a piece of cake for you. Link to comment Share on other sites More sharing options...
Steve RA Posted February 6, 2013 Share Posted February 6, 2013 Here is the web site, looks like it is $11.99, but, you can get a free two week trial it seems. http://ontargetshooting.com/ It seems as though it would answer the questions you were attempting to solve in your first post. Link to comment Share on other sites More sharing options...
Hammbone Posted February 7, 2013 Author Share Posted February 7, 2013 I found a few algorithms, but they gave me a headache. I'm just going to this: -Delete the two farthest from the group (of 10 shots) -Then shift the center point from the bullseye to the average of the remaining group -Then calculate the standard deviate from the new centroid -Then multiply stdev x 3 ...that will give me a radius for 99.7% of the data points. I am just doing this for my own personal record keeping for developing new loads. Link to comment Share on other sites More sharing options...
Steve RA Posted February 7, 2013 Share Posted February 7, 2013 Why delete the two farthest ?? Seems as though you would be going for an 8 round group then. Don't think that would be very accurate if you throw out two fliers. Link to comment Share on other sites More sharing options...
Hammbone Posted February 7, 2013 Author Share Posted February 7, 2013 I thought about that after I did it. I ended up putting those two back in. ...gotta keep myself honest! Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now