[SOLVED] Math help! Accurately reducing edge length
polycounter lvl 7
Edit: solved it, see: below
I have a cone and I need the faces to reduce by x down the length of it. The have to reduce so that each face is as close to a square as possible. I don't know the rate of the reduction or the number of faces, so I'm stuck with algebraic formulas and I think I need something calculus. It's got to generate a vaguely L shaped curve.
If it helps the starting edge is .313 cm long and the ending edge is .031 cm long. The entire length of the cone is 40 cm.
Replies
Wouldn't you be able to add a bunch of loops and then run a relax on them, which averages their distances, thus creating equally spaced squares?
That's a really good idea! The only problem though is that I need the diameter of the cone to remain the same (because I'm a tight ass) and that would flatten out the squares.
Here's what I did instead: I used a measure tool plugin to get the distance of one edge at the base of the cone (.313 cm). Then I made a plane with the width as the same dimension positioned the pivot point of the plane to one edge and then snapped it to the leading vertex on the base of the cone. Then I used the edge loop tool and an orthographic veiw to get as close as possible to an edge loop at that distance. Then I took the distance between two vertices on the edge loop (.311 cm). 311/313=.993
Then I made a plane and used the duplicate with transform option. I set the z transform to .313 and the z scale to .993 and guessed at duplicating it 80 times. The planes matched the edgeloops that I had already make, so I'm figureing it worked.
Combined the 80 planes into a mesh and then booleaned it with the cone, and I've got all my edge loops figured out. Hopefully
Ups.. i did this a few days ago but didn't hit the post reply button..😅..
If you have a cone then the length of the edges depends on the number of divisions (or vertices) of the round border (radius) because: Initially thinging of a main radius of 0.5 will give a circumference of 2*pi*0.5 = pi = 3.1415. Using 4 vertices this gives a pyramide with an edge of 0.707 = sqrt(2)/2.. Using 6 vertices gives 0.5.. 8: 0.383, 9: 0.342 and 10: 0.309 (leading to an approximation of pi with 4*0.707= 2.828, 6x0.5=3, 8x0.383=3.064, 9x0.342=3.078, 10x0.309=3.09 ..infinitely = 3.1415.. = pi). So the needed length to the top (not height) varies.. means: depends on the radius subdivisons ... so i'm wondering where your 0.313 comes from and why this would give an L-shape ?? I assume your radius is not 0.5 and you didn't mention the subdivison.
Anyway: Using almost square like faces leads to an slighty smaller upper edge and also it doesn't fit exactly at the top so it may has to be equalised over it's length ? Here using 16 vertices and a radius of 0.5, height of 0.4 and a top radius of 0.05 (according to your starting and ending edge):
Of course for different heights this also differs. So you may just slide some edges manually?