fimplicit increasing number of points

5 ビュー (過去 30 日間)
Maxim Bogdan
Maxim Bogdan 2021 年 4 月 17 日
コメント済み: Walter Roberson 2021 年 4 月 17 日
I wonder how can I increase the number of points used by the fimplicit command to plot an implicit curve ?
From this question that I asked before I understand that it is possible to get the points used for the plot as a matrix with two lines:
But there seem to be about 500 points (I made more simulations with various curves and the number of points is approximately 500).
My question is: How can I increase that number to get a more precise representation?
I know that I can make changes in the fimplicit.m original matlab file but what is written there is super-professional and I cannot identify where in the code is set the precision.
P.S. My aim is to make a program that computes the perimeter of an implicit curve so for a better precision I need more points.

採用された回答

Walter Roberson
Walter Roberson 2021 年 4 月 17 日
fp = fimplicit(@(x,y) x.^2 + 2*y.^2 - 1); %Example
Points1 = [fp.XData;fp.YData];
axis equal
title('mesh default = 151')
fp = fimplicit(@(x,y) x.^2 + 2*y.^2 - 1, 'meshdensity', 255); %Example
Points2 = [fp.XData;fp.YData];
axis equal
title('mesh 255')
size(Points1)
ans = 1×2
2 587
size(Points2)
ans = 1×2
2 1011
  2 件のコメント
Maxim Bogdan
Maxim Bogdan 2021 年 4 月 17 日
That's awesome! From where do you know these stuff? I didn't find this in the documentation of fimplicit...
Walter Roberson
Walter Roberson 2021 年 4 月 17 日

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeParticle & Nuclear Physics についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by