plotting a spherical segment

33 ビュー (過去 30 日間)
Gilles
Gilles 2016 年 11 月 24 日
編集済み: isku 2020 年 7 月 24 日
Hi, I need to plot a spherical segment (blue region of the sphere in the attached picture).
Do you have any suggestions?
Gilles

採用された回答

KSSV
KSSV 2016 年 11 月 24 日
clc; clear all ;
[X,Y,Z] = sphere(200) ;
surf(X,Y,Z) ;
r = 1; b = 0.9 ; h = 1 ;
%%Get radius
R = sqrt(X.^2+Y.^2) ;
X1 = X ; Y1 = Y ; Z1 = Z ;
X1(Z<r/2) = NaN ; Y1(Z<r/2) = NaN ; Z1(Z<r/2) = NaN ;
X1(Z>b) = NaN ; Y1(Z>b) = NaN ; Z1(Z>b) = NaN ;
surf(X1,Y1,Z1) ;
axis equal
% shading interp ;

その他の回答 (1 件)

isku
isku 2020 年 7 月 24 日
編集済み: isku 2020 年 7 月 24 日
My code:
>> clear; syms t x y z real;
r = 1
a = .4*r % 40 pct. of r
b = .6*r % 60 pct. of r
eq = x^2+y^2+z^2 == r^2 % equation of a sphere with center at the origin
% define high by projecting on xz-plane by setting y=0. And then x=0 e,g. the high on z-axe
za = solve( subs( eq, [x^2, y^2,r^2], [0,0, (r^2-a^2)] ), z ) % by Pythagoras, see figure under
za = za(2) % only positive value, upper half of the sphere.
% Similarly
zb = solve( subs( eq, [x^2, y^2,r^2], [0,0, (r^2-b^2)] ), z ) % by Pythagoras
zb = zb(2) % only positive value
h = za - zb
% plotting
>> clear x;
x(t) = sin(t) *r ; % -1<x<1
[X,Y,Z] = sphere;
s = mesh(X,Y,Z, 'edgecolor', 'k', 'FaceAlpha',0.3) ;
hold on
yApos = sqrt( r^2 -x^2-za^2)
yAneg = -yApos
zA(t) = 0*t + za;
fplot3( x, yApos, zA, [-pi, pi], 'red', 'LineWidth', 2 ),
fplot3( x, yAneg, zA, [-pi, pi], 'red', 'LineWidth', 2 )
yBpos = sqrt( r^2 -x^2-zb^2)
yBneg = -yBpos
zB(t) = 0*t+ zb
fplot3( x, yBpos, zB, [-pi, pi], 'red', 'LineWidth', 2 )
fplot3( x, yBneg, zB, [-pi, pi], 'red', 'LineWidth', 2 )
xlabel('x'), ylabel('y'), zlabel('z'), axis equal , grid on
axis( [-1,1, -1,1, -1,1] *r )
campos( [5,5,5])
Using Pythagoras to define the hight at given a and b.
The high on z-axe at b:
zb^2 = r^2 - b^2 % b= 0.6*r

カテゴリ

Help Center および File ExchangeSurface and Mesh Plots についてさらに検索

タグ

Community Treasure Hunt

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

Start Hunting!

Translated by