hi everybody,
how can i get X,Y values from the following matlab-script?
X = @(t) 0.5 + 0.5*(((abs(cos(t))).^B)/cos(t))
Y = @(t) 0.5*T*(((abs(sin(t))).^B)/sin(t))*(1 - (0.5 + 0.5*(((abs(cos(t))).^B)/cos(t))).^P) + C*sin(((0.5 + 0.5*(((abs(cos(t))).^B)/cos(t))).^E * pi)) + R*sin((0.5 + 0.5*(((abs(cos(t))).^B)/cos(t)))*2*pi)
t = [0.01:0.01:2*pi]
B = 1.75
T = 0.18
P = 1.1
C = 0.11
E = 0.8
R = 0.0
fplot(X,Y)
As you can see in the attaches photo, I couldn't have then. I just could plot them

 채택된 답변

Voss
Voss 2022년 2월 13일

0 개 추천

t = [0.01:0.01:2*pi]
t = 1×628
0.0100 0.0200 0.0300 0.0400 0.0500 0.0600 0.0700 0.0800 0.0900 0.1000 0.1100 0.1200 0.1300 0.1400 0.1500 0.1600 0.1700 0.1800 0.1900 0.2000 0.2100 0.2200 0.2300 0.2400 0.2500 0.2600 0.2700 0.2800 0.2900 0.3000
B = 1.75
B = 1.7500
T = 0.18
T = 0.1800
P = 1.1
P = 1.1000
C = 0.11
C = 0.1100
E = 0.8
E = 0.8000
R = 0.0
R = 0
X = @(t) 0.5 + 0.5*(((abs(cos(t))).^B)./cos(t))
X = function_handle with value:
@(t)0.5+0.5*(((abs(cos(t))).^B)./cos(t))
Y = @(t) 0.5*T*(((abs(sin(t))).^B)./sin(t)).*(1 - (0.5 + 0.5*(((abs(cos(t))).^B)./cos(t))).^P) + C*sin(((0.5 + 0.5*(((abs(cos(t))).^B)./cos(t))).^E * pi)) + R*sin((0.5 + 0.5*(((abs(cos(t))).^B)./cos(t)))*2*pi)
Y = function_handle with value:
@(t)0.5*T*(((abs(sin(t))).^B)./sin(t)).*(1-(0.5+0.5*(((abs(cos(t))).^B)./cos(t))).^P)+C*sin(((0.5+0.5*(((abs(cos(t))).^B)./cos(t))).^E*pi))+R*sin((0.5+0.5*(((abs(cos(t))).^B)./cos(t)))*2*pi)
fplot(X,Y)
X = X(t)
X = 1×628
1.0000 0.9999 0.9998 0.9997 0.9995 0.9993 0.9991 0.9988 0.9985 0.9981 0.9977 0.9973 0.9968 0.9963 0.9958 0.9952 0.9946 0.9939 0.9932 0.9925 0.9917 0.9909 0.9901 0.9892 0.9883 0.9873 0.9864 0.9853 0.9843 0.9832
Y = Y(t)
Y = 1×628
0.0000 0.0000 0.0000 0.0001 0.0001 0.0002 0.0003 0.0003 0.0004 0.0006 0.0007 0.0008 0.0009 0.0011 0.0013 0.0014 0.0016 0.0018 0.0021 0.0023 0.0025 0.0028 0.0031 0.0033 0.0036 0.0040 0.0043 0.0046 0.0050 0.0053

댓글 수: 2

Hasan Ballouk
Hasan Ballouk 2022년 2월 14일
perfect, thank you very much!
Voss
Voss 2022년 2월 15일
You're welcome!

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

도움말 센터File Exchange에서 MATLAB에 대해 자세히 알아보기

질문:

2022년 2월 13일

댓글:

2022년 2월 15일

Community Treasure Hunt

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

Start Hunting!

Translated by