how to convert vector to scalar to get plot as line
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paripavithran sampathu
2016 年 12 月 7 日
コメント済み: Rena Berman
2017 年 9 月 14 日
clear all;
clc;
syms q1 q2 q3 q4 qt1 qt2 qt3 qt4 m1 m2 m3 m4 l1 l2 l3 l4 h1 h2 h3 h4 g x
syms I1 I2 I3 I4 Q1 Q2 Q3 Q4 qtt1 qtt2 qtt3 qtt4 t real
l1=27;
l2=23;
l3=19.8;
l4=5;
g=9.80665;
m1=23.687;
m2=17.15;
m3=2.0125;
m4=0.4179;
h1=765;
h2=568;
h3=418;
h4=220;
I1=0.099379;
I2=0.050526;
I4=0.000405;
I3=0.000702;
for x=0:0.1:10;
% z=0:0.1:10;
% qt1=x;
q1=3-(3*cos(pi*x/10));
qt1=3*sin(pi*x/10)*pi/10;
qtt1=3*cos(pi*x/10)*(pi^2/100);
q2=3-(3*sin(pi*x/10));
qt2= -3*cos(pi*x/10)*(pi/10);
qtt2=3*sin(pi*x/10)*(pi^2/100);
qt3=0.198*x;
qtt3=0.198*x;
qtt4=sin(x)+cos(x);
Q1=qtt2*((l1*l2*m2*cos(q2))/2 + l1*l2*m3*cos(q2)) - qt2*((l1*l2*m2*qt2*sin(q2))/2 + l1*l2*m3*qt2*sin(q2)) + qtt1*(2*I1 + (2*l1^2*m1)/3 + l1^2*m2 + l1^2*m3);
Q2=qtt2*(2*I2 + (l2^2*m2)/4 + l2^2*m3) + qtt1*((l1*l2*m2*cos(q2))/2 + l1*l2*m3*cos(q2)) - qt2*((l1*l2*m2*qt1*sin(q2))/2 + l1*l2*m3*qt1*sin(q2)) + (l1*l2*m2*qt1*qt2*sin(q2))/2 + l1*l2*m3*qt1*qt2*sin(q2);
Q3=g*m3*qtt3 - g*m4 - (g*m3)/2;
Q4=l4^2*m4*qtt4;
xa= l1* cos(Q1);
ya=l1* sin(Q);
plot(xa,ya)
xb=(l1+l2)* cos(Q2);
yb=(l1+l2) * sin(Q2);
plot(xb,yb)
xc=(l1+l2)* cos(Q3);
yc= qtt3*g* sin(Q3);
plot(xc,yc)
xd=(l1+l2+l4) * cos(Q4);
yd=(l1+l2+l4) * sin(Q4);
plot(xd,yd
hold on
end
2 件のコメント
採用された回答
Walter Roberson
2016 年 12 月 7 日
You want to convert scalar to vector (like your tag), not vector to scalar (like your title)
See attached.
2 件のコメント
Walter Roberson
2016 年 12 月 13 日
You did not have a vector in your code, so there was no vector to convert into scalar.
その他の回答 (1 件)
Walter Roberson
2017 年 9 月 10 日
Please stop editing away your question.
The people who answer questions here are volunteers. We answer questions so that everyone can learn from what we write. When you edit away your question, you leave us feeling as if you have taken advantage of us for some free private consulting.
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