what i have so far
r = linspace(0,2*pi);
th = 0:pi/6:2*pi ;
[R,T] = meshgrid(r,th) ;
X = R.*cos(T) ;
Y = R.*sin(T) ;
Z = R ;
surf(X,Y,Z)
What i'm wondering is, how do i make the length of each side of the base a user input,
and how do i make the height of the decagonal pyramid a user input as well.

1 Kommentar

Dyuman Joshi
Dyuman Joshi am 22 Okt. 2023
Use input
Though, with the given method of generating the pyramid, the lenght of each side of the base is the same.

Melden Sie sich an, um zu kommentieren.

 Akzeptierte Antwort

Matt J
Matt J am 22 Okt. 2023
Bearbeitet: Matt J am 22 Okt. 2023

0 Stimmen

sidelength=10; height=8; %User inputs
V=nsidedpoly(10,'Side',sidelength).Vertices;
V(end+1,3)=-height;
trisurf( delaunay(V(:,1:2)), V(:,1), V(:,2), V(:,3),'FaceColor','c')

3 Kommentare

Matthew
Matthew am 22 Okt. 2023
Bearbeitet: Matthew am 22 Okt. 2023
This is perfect, thank you.
I'm trying to create a user input program that creates a new shape that fits inside the decagon at the intersection of two sides. i added some code to create that angle in degrees based off the input.
sidelength = input('Enter Sidelength');
height = input('Enter Height');
V=nsidedpoly(10,'Side',sidelength).Vertices;
V(end+1,3)=-height;
trisurf( delaunay(V(:,1:2)), V(:,1), V(:,2), V(:,3),'FaceColor','c')
%% calculating angle for new shape based off sidelength and height input
x6 = (sidelength*3.078)/2
angle1 = atand(height/x6)
angle2 = 90-angle1
My question now is how do i plot a new (almost like an obtuse V) shape that would fit inside the decagon at the intersection of both these angles? the new shape doesn't need to run the length of both sides, and doesn't need to be very wide either.
I know this is alot, and i'm sorry if my question is enigmatic
Matt J
Matt J am 22 Okt. 2023
There's nothing wrong with asking more questions, but please Accept-click the answer to this question and post your new one in a new thread.
Matthew
Matthew am 22 Okt. 2023
ok will do, thank you.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Line Plots finden Sie in Hilfe-Center und File Exchange

Gefragt:

am 22 Okt. 2023

Kommentiert:

am 22 Okt. 2023

Community Treasure Hunt

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

Start Hunting!

Translated by