problem to solve incorrrect

14 ビュー (過去 30 日間)
Crz
Crz 2025 年 2 月 1 日 0:05
コメント済み: Walter Roberson 2025 年 2 月 1 日 0:33
Ran in:
%% R-RRR
clear all; clc; close all
% Input data
AB=0.325; %(m)
BC=0.938; %(m)
CD=0.675; %(m)
CE=0.6; %(m)
xD=0.8; %(m)
yD=-0.432; %(m)
phi =pi; %(rad)
xA = 0; yA = 0;
rA = [xA yA 0];
rD = [xD yD 0];
xB = AB*cos(phi); yB = AB*sin(phi); rB = [xB yB 0];
% Position of joint C
%
eqnC1 = '(xCsol - xB)^2 + (yCsol - yB)^2 = BC^2 ';
% Distance formula: CD=constant
eqnC2 = '(xCsol - xD)^2 + (yCsol - yD)^2 = CD^2 ';
% Simultaneously solve above equations
solC = solve(eqnC1, eqnC2, 'xCsol, yCsol');
Incorrect number or types of inputs or outputs for function solve.
% Two solutions for xC - vector form
xCpositions = eval(solC.xCsol);
% Two solutions for yC - vector form
yCpositions = eval(solC.yCsol);
% Separate the solutions in scalar form
% first component of the vector xCpositions
xC1 = xCpositions(1);
% second component of the vector xCpositions
xC2 = xCpositions(2);
yC1 = yCpositions(1);
% second component of the vector yCpositions
yC2 = yCpositions(2);
% Select the correct position for C
% for the given input angle
if xC1 > xD
xC = xC1; yC=yC1;
else
xC = xC2; yC=yC2;
end
rC = [xC yC 0]; % Position vector of C
% Position of joint E
% Distance formula: CE=constant
eqnE1='(xEsol-xC)^2+(yEsol-yC)^2=CE^2';
% Slope formula:
% E, C, and D are on the same straight line
eqnE2='(yD-yC)/(xD-xC)=(yEsol-yC)/(xEsol-xC)';
solE = solve(eqnE1, eqnE2, 'xEsol, yEsol');
xEpositions = eval(solE.xEsol);
yEpositions = eval(solE.yEsol);
xE1 = xEpositions(1); xE2 = xEpositions(2);
yE1 = yEpositions(1); yE2 = yEpositions(2);
if xE1 < xC
xE = xE1; yE=yE1;
else
xE = xE2; yE=yE2;
end
rE = [xE yE 0]; % Position vector of E
% Angles of the links with the horizontal
phi2 = atan((yB-yC)/(xB-xC));
phi3 = atan((yD-yC)/(xD-xC));
fprintf('Results \n\n')
fprintf('rA = [ %g, %g, %g ] (m)\n', rA)
fprintf('rD = [ %g, %g, %g ] (m)\n', rD)
fprintf('rB = [ %g, %g, %g ] (m)\n', rB)
fprintf('rC = [ %g, %g, %g ] (m)\n', rC)
fprintf('rE = [ %g, %g, %g ] (m)\n', rE)
fprintf('phi2 = %g (degrees) \n', phi2*180/pi)
fprintf('phi3 = %g (degrees) \n', phi3*180/pi)
% Graphic of the mechanism
plot([xA,xB],[yA,yB],'k-o','LineWidth',1.5)
hold on % holds the current plot
plot([xB,xC],[yB,yC],'b-o','LineWidth',1.5)
hold on
plot([xD,xE],[yD,yE],'r-o','LineWidth',1.5)
% adds major grid lines to the current axes
grid on,...
xlabel('x (m)'), ylabel('y (m)'),...
title('positions for \phi = 45 (deg)'),...
text(xA,yA,'\leftarrow A = ground',...
'HorizontalAlignment','left'),...
text(xB,yB,' B'),...
text(xC,yC,'\leftarrow C = ground',...
'HorizontalAlignment','left'),...
text(xD,yD,'\leftarrow D = ground',...
'HorizontalAlignment','left'),...
text(xE,yE,' E'), axis([-2 3 -2 3])
  1 件のコメント
Walter Roberson
Walter Roberson 2025 年 2 月 1 日 0:33
Ran in:
With minimal changes
%% R-RRR
clear all; clc; close all
% Input data
AB=0.325; %(m)
BC=0.938; %(m)
CD=0.675; %(m)
CE=0.6; %(m)
xD=0.8; %(m)
yD=-0.432; %(m)
phi =pi; %(rad)
xA = 0; yA = 0;
rA = [xA yA 0];
rD = [xD yD 0];
xB = AB*cos(phi); yB = AB*sin(phi); rB = [xB yB 0];
% Position of joint C
%
eqnC1 = '(xCsol - xB)^2 + (yCsol - yB)^2 = BC^2 ';
% Distance formula: CD=constant
eqnC2 = '(xCsol - xD)^2 + (yCsol - yD)^2 = CD^2 ';
% Simultaneously solve above equations
solC = solve(str2sym(eqnC1), str2sym(eqnC2), sym('xCsol'), sym('yCsol'));
% Two solutions for xC - vector form
xCpositions = eval(solC.xCsol);
% Two solutions for yC - vector form
yCpositions = eval(solC.yCsol);
% Separate the solutions in scalar form
% first component of the vector xCpositions
xC1 = xCpositions(1);
% second component of the vector xCpositions
xC2 = xCpositions(2);
yC1 = yCpositions(1);
% second component of the vector yCpositions
yC2 = yCpositions(2);
% Select the correct position for C
% for the given input angle
if xC1 > xD
xC = xC1; yC=yC1;
else
xC = xC2; yC=yC2;
end
rC = [xC yC 0]; % Position vector of C
% Position of joint E
% Distance formula: CE=constant
eqnE1='(xEsol-xC)^2+(yEsol-yC)^2=CE^2';
% Slope formula:
% E, C, and D are on the same straight line
eqnE2='(yD-yC)/(xD-xC)=(yEsol-yC)/(xEsol-xC)';
solE = solve(str2sym(eqnE1), str2sym(eqnE2), sym('xEsol'), sym('yEsol'));
xEpositions = eval(solE.xEsol);
yEpositions = eval(solE.yEsol);
xE1 = xEpositions(1); xE2 = xEpositions(2);
yE1 = yEpositions(1); yE2 = yEpositions(2);
if xE1 < xC
xE = xE1; yE=yE1;
else
xE = xE2; yE=yE2;
end
rE = [xE yE 0]; % Position vector of E
% Angles of the links with the horizontal
phi2 = atan((yB-yC)/(xB-xC));
phi3 = atan((yD-yC)/(xD-xC));
fprintf('Results \n\n')
Results
fprintf('rA = [ %g, %g, %g ] (m)\n', rA)
rA = [ 0, 0, 0 ] (m)
fprintf('rD = [ %g, %g, %g ] (m)\n', rD)
rD = [ 0.8, -0.432, 0 ] (m)
fprintf('rB = [ %g, %g, %g ] (m)\n', rB)
rB = [ -0.325, 3.9801e-17, 0 ] (m)
fprintf('rC = [ %g, %g, %g ] (m)\n', rC)
rC = [ 0.589353, 0.20929, 0 ] (m)
fprintf('rE = [ %g, %g, %g ] (m)\n', rE)
rE = [ 0.402112, 0.779326, 0 ] (m)
fprintf('phi2 = %g (degrees) \n', phi2*180/pi)
phi2 = 12.8926 (degrees)
fprintf('phi3 = %g (degrees) \n', phi3*180/pi)
phi3 = -71.816 (degrees)
% Graphic of the mechanism
plot([xA,xB],[yA,yB],'k-o','LineWidth',1.5)
hold on % holds the current plot
plot([xB,xC],[yB,yC],'b-o','LineWidth',1.5)
hold on
plot([xD,xE],[yD,yE],'r-o','LineWidth',1.5)
% adds major grid lines to the current axes
grid on,...
xlabel('x (m)'), ylabel('y (m)'),...
title('positions for \phi = 45 (deg)'),...
text(xA,yA,'\leftarrow A = ground',...
'HorizontalAlignment','left'),...
text(xB,yB,' B'),...
text(xC,yC,'\leftarrow C = ground',...
'HorizontalAlignment','left'),...
text(xD,yD,'\leftarrow D = ground',...
'HorizontalAlignment','left'),...
text(xE,yE,' E'), axis([-2 3 -2 3])

サインインしてコメントする。

サインインしてこの質問に回答する。

回答 (1 件)

Torsten
Torsten 2025 年 2 月 1 日 0:23
Ran in:
%% R-RRR
clear all; clc; close all
% Input data
AB=0.325; %(m)
BC=0.938; %(m)
CD=0.675; %(m)
CE=0.6; %(m)
xD=0.8; %(m)
yD=-0.432; %(m)
phi =pi; %(rad)
xA = 0; yA = 0;
rA = [xA yA 0];
rD = [xD yD 0];
xB = AB*cos(phi); yB = AB*sin(phi); rB = [xB yB 0];
% Position of joint C
%
syms xCsol yCsol
eqnC1 = (xCsol - xB)^2 + (yCsol - yB)^2 == BC^2 ;
% Distance formula: CD=constant
eqnC2 = (xCsol - xD)^2 + (yCsol - yD)^2 == CD^2 ;
% Simultaneously solve above equations
solC = solve([eqnC1, eqnC2], [xCsol, yCsol]);
% Two solutions for xC - vector form
xCpositions = eval(solC.xCsol)
xCpositions = 2×1
0.2143 0.5894
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
% Two solutions for yC - vector form
yCpositions = eval(solC.yCsol)
yCpositions = 2×1
-0.7675 0.2093
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
% Separate the solutions in scalar form
% first component of the vector xCpositions
xC1 = xCpositions(1);
% second component of the vector xCpositions
xC2 = xCpositions(2);
yC1 = yCpositions(1);
% second component of the vector yCpositions
yC2 = yCpositions(2);
% Select the correct position for C
% for the given input angle
if xC1 > xD
xC = xC1; yC=yC1;
else
xC = xC2; yC=yC2;
end
rC = [xC yC 0]; % Position vector of C
% Position of joint E
% Distance formula: CE=constant
syms xEsol yEsol
eqnE1 = (xEsol-xC)^2+(yEsol-yC)^2==CE^2;
% Slope formula:
% E, C, and D are on the same straight line
eqnE2=(yD-yC)/(xD-xC)==(yEsol-yC)/(xEsol-xC);
solE = solve([eqnE1, eqnE2], [xEsol, yEsol]);
xEpositions = eval(solE.xEsol)
xEpositions = 2×1
0.7766 0.4021
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
yEpositions = eval(solE.yEsol)
yEpositions = 2×1
-0.3607 0.7793
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
xE1 = xEpositions(1); xE2 = xEpositions(2);
yE1 = yEpositions(1); yE2 = yEpositions(2);
if xE1 < xC
xE = xE1; yE=yE1;
else
xE = xE2; yE=yE2;
end
rE = [xE yE 0]; % Position vector of E
% Angles of the links with the horizontal
phi2 = atan((yB-yC)/(xB-xC));
phi3 = atan((yD-yC)/(xD-xC));
fprintf('Results \n\n')
Results
fprintf('rA = [ %g, %g, %g ] (m)\n', rA)
rA = [ 0, 0, 0 ] (m)
fprintf('rD = [ %g, %g, %g ] (m)\n', rD)
rD = [ 0.8, -0.432, 0 ] (m)
fprintf('rB = [ %g, %g, %g ] (m)\n', rB)
rB = [ -0.325, 3.9801e-17, 0 ] (m)
fprintf('rC = [ %g, %g, %g ] (m)\n', rC)
rC = [ 0.589353, 0.20929, 0 ] (m)
fprintf('rE = [ %g, %g, %g ] (m)\n', rE)
rE = [ 0.402112, 0.779326, 0 ] (m)
fprintf('phi2 = %g (degrees) \n', phi2*180/pi)
phi2 = 12.8926 (degrees)
fprintf('phi3 = %g (degrees) \n', phi3*180/pi)
phi3 = -71.816 (degrees)
% Graphic of the mechanism
plot([xA,xB],[yA,yB],'k-o','LineWidth',1.5)
hold on % holds the current plot
plot([xB,xC],[yB,yC],'b-o','LineWidth',1.5)
hold on
plot([xD,xE],[yD,yE],'r-o','LineWidth',1.5)
% adds major grid lines to the current axes
grid on,...
xlabel('x (m)'), ylabel('y (m)'),...
title('positions for \phi = 45 (deg)'),...
text(xA,yA,'\leftarrow A = ground',...
'HorizontalAlignment','left'),...
text(xB,yB,' B'),...
text(xC,yC,'\leftarrow C = ground',...
'HorizontalAlignment','left'),...
text(xD,yD,'\leftarrow D = ground',...
'HorizontalAlignment','left'),...
text(xE,yE,' E'), axis([-2 3 -2 3])

カテゴリ

Help Center および File Exchange で Circuit Design and Analysis についてさらに検索

タグ

製品


リリース

R2021a

Community Treasure Hunt

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

Start Hunting!

Translated by