Calculate the area and volume of revolution based on some data points
9 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Williams Dias
am 4 Okt. 2022
Bearbeitet: Torsten
am 4 Okt. 2022
I got the coordinates of an axisymmmetric salt cavern:
x = [0,38.161563,39.751307,40.878082,42.701189,44.524296,45.047229,45.990198,46.933167,46.785121,...
46.225562,41.318857,36.412152,35.7033,36.22748,37.398386,38.569291,39.069331,38.001499,29.200019,...
20.398538,20.034604,19.739322,16.900337,14.061351,13.944131,13.784996,6.892498,0];
y=[-3200.000000,-3185.689414,-3184.723755,-3183.243836,-3179.597622,-3175.951408,-3173.937343,-3152.720547,...
-3131.503752,-3130.054410,-3128.709270,-3120.531428,-3112.353586,-3109.972990,-3107.545039,-3105.203228,...
-3102.861417,-3100.098595,-3097.501874,-3086.500023,-3075.498173,-3074.978887,-3074.417715,-3068.076173,...
-3061.734631,-3061.537271,-3061.371840,-3055.685920,-3050.000000];
plot(x,y)
I would like to calulate the area and volume of revolution of it.
Thanks in advance,
William
0 Kommentare
Akzeptierte Antwort
Torsten
am 4 Okt. 2022
Bearbeitet: Torsten
am 4 Okt. 2022
y = [0,38.161563,39.751307,40.878082,42.701189,44.524296,45.047229,45.990198,46.933167,46.785121,...
46.225562,41.318857,36.412152,35.7033,36.22748,37.398386,38.569291,39.069331,38.001499,29.200019,...
20.398538,20.034604,19.739322,16.900337,14.061351,13.944131,13.784996,6.892498,0];
x=[-3200.000000,-3185.689414,-3184.723755,-3183.243836,-3179.597622,-3175.951408,-3173.937343,-3152.720547,...
-3131.503752,-3130.054410,-3128.709270,-3120.531428,-3112.353586,-3109.972990,-3107.545039,-3105.203228,...
-3102.861417,-3100.098595,-3097.501874,-3086.500023,-3075.498173,-3074.978887,-3074.417715,-3068.076173,...
-3061.734631,-3061.537271,-3061.371840,-3055.685920,-3050.000000];
plot(x,y)
fun = @(xq)interp1(x,y,xq,'spline');
V = pi*integral(@(x)fun(x).^2,x(1),x(end))
3 Kommentare
Torsten
am 4 Okt. 2022
Bearbeitet: Torsten
am 4 Okt. 2022
How do I get the area of revolution in this example (I'm a starter in matlab) ?
By integrating 2*pi* f(x)*sqrt(1+f'(x)^2) instead of pi*f(x)^2 in the limits x(1) to x(end):
y = [0,38.161563,39.751307,40.878082,42.701189,44.524296,45.047229,45.990198,46.933167,46.785121,...
46.225562,41.318857,36.412152,35.7033,36.22748,37.398386,38.569291,39.069331,38.001499,29.200019,...
20.398538,20.034604,19.739322,16.900337,14.061351,13.944131,13.784996,6.892498,0];
x=[-3200.000000,-3185.689414,-3184.723755,-3183.243836,-3179.597622,-3175.951408,-3173.937343,-3152.720547,...
-3131.503752,-3130.054410,-3128.709270,-3120.531428,-3112.353586,-3109.972990,-3107.545039,-3105.203228,...
-3102.861417,-3100.098595,-3097.501874,-3086.500023,-3075.498173,-3074.978887,-3074.417715,-3068.076173,...
-3061.734631,-3061.537271,-3061.371840,-3055.685920,-3050.000000];
dy = [(y(2)-y(1))/(x(2)-x(1)),(y(3:end)-y(1:end-2))./(x(3:end)-x(1:end-2)),(y(end)-y(end-1))/(x(end)-x(end-1))];
%fun =@(xq) interp1(x,y,xq,'spline').*sqrt(1+interp1(x,dy,xq).^2);
fun =@(xq) interp1(x,y,xq).*sqrt(1+interp1(x,dy,xq).^2);
A = 2*pi*integral(fun,x(1),x(end))
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Smoothing finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!