Why fimplicit3 draw the same function differently
显示 更早的评论
f=@(x,y,z) (x+y+z).^2
fimplicit3(f)
We actually know that this equals to draw the graph of 
，but the code turns out to be this:
，but the code turns out to be this:What happened to the function？
1 个评论
Seems like a bug -
f=@(x,y,z) (x+y+z).^2;
fun=fimplicit3(f)
No output if Mesh Density is changed, and the axes limits go from the default [-5 5] to [0 1];
figure
fimplicit3(f,'MeshDensity',50)
Specifying interval gives the same result as above
figure
fimplicit3(f,[-5 5],'MeshDensity',50)
I tested the same in my R2021a and it has the same behaviour as above.
采纳的回答
This is NOT a bug.
fimplicit3 plot the surface f(x,y,z) = 0, do you must help it by having a function that explicitly cross the level f=0, because it approximate by a tensorial tri-linear and find where it corss the level. But your function is defined as (x+y+z).^2 is always >= 0. Thus fimplicit have a hard time to estimate the surface.
If you add a tiny positive error in your square function it does not cross at all the level f=0.
If you want to determine the surface x+y+z = 0, then do it straight manner
h = @(x,y,z) x+y+z;
fimplicit3(h)
Do not square the function, this is stupid way to chanllenge fimplicit3 numerically.
2 个评论
"Do not square the function, this is stupid numerically."
How about symbolically?
syms f(x,y,z)
h = (x+y+z)^2;
fimplicit3(h)
"because it approximate by a tensorial tri-linear and find where it corss the level."
Do you have a source for this?
Bruno Luong
2023-9-7
Under the hood is numerical surface level, regradless where as your function is symbolic or not.
No I don't have a source but I program myself such level surface. There is nothing sophisticate behind, one just need to decide the cube in voxel and look where the edges cross the level, just like contour in 2D.
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