tag:blogger.com,1999:blog-2807403883562053852.post6133458019196438708..comments2020-08-03T21:06:54.111-06:00Comments on Clayton Cramer.: Somehow I Received an A in Both 2nd Semester and 3rd Semester Algebra Clayton Cramerhttp://www.blogger.com/profile/03258083387204776812noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-2807403883562053852.post-22625768664933745272019-06-26T10:52:09.288-06:002019-06-26T10:52:09.288-06:00z = sqrt(r^2 -(x-h)^2) + kz = sqrt(r^2 -(x-h)^2) + kNilshttps://www.blogger.com/profile/09679998177431944022noreply@blogger.comtag:blogger.com,1999:blog-2807403883562053852.post-7115990866494240892019-06-23T23:19:31.520-06:002019-06-23T23:19:31.520-06:00OK, your formula for a circle is
(x-h)² + (y-k)² =...OK, your formula for a circle is<br />(x-h)² + (y-k)² = r²<br />This constrains your (x, y) pairs to be along a circle of radius r, since the further x gets from h, the closer y has to be to k.<br />In an x, y, z coordinate system, movement along the z axis is independent of what's happening with x and y. So you would excavate your circle (or your half circle, constraining either x or y to be non-negative), then lower your tool an increment of z before excavating again. And repeat until it's deep enough.<br /><br />It might be easier if you can rewrite your code in polar coordinates.<br />Karlhttps://www.blogger.com/profile/09767954910569442805noreply@blogger.com