constant vs linear triangle elements for BEM

[mark smith]

I appreciate this is not specifically an Elmer question but maybe someone can help?
I have a BEM electrostatic solver where I have used constant elements and would like to know what accuracy improvement I might get if i implemented linear elements with 3 collocation points if i maintain the number of degrees of freedom?
I'm assuming N/3 linear triangles will give me better accuracy than N constant triangles? & if so how many constant triangles would I need to give me the same accuracy as N/3 linear ones? I ask because I want to know if the effort is worth it rather than just using lots of constant triangles.
Thanks in advance
regards
mark

[hazelsct]

Mark, once the variations are reasonably well resolved, the error in the discretization should go as grid spacing h (i.e. 1/N) for constant and h^2 (i.e. 1/N^2) for linear elements. So as you increase N, the N linear elements become much more accurate than the N/3 constant elements. There's no general proportionality for the accuracy, it depends strongly on the problem and mesh node distribution (uniform/non/etc.) and even on N.