我想在同一图表上绘制电动机的功率和扭矩。
我使用一个循环绘制三角星配置,还有一个双齿轮系统,所以我想绘制每个配置的2系列数据(功率和扭矩)。
我尝试了这个,但我注意到在第二个循环中它绘制在第二个轴,而不是在第一个轴。我使用matlab 2015,所以我没有yyaxis function
clc; clear all;close all
tau=[8.823;1.604]; %in ordine decrescente
n_gamme=length(tau);
data=[22 22;575 1200;1500 6000;365 175];%P,Ginocchio,rpm max,C
names={'Star';'Delta'};
nN=zeros(2);nmax=zeros(2);M_N=zeros(2); %inizializzo
for i=1:n_gamme %contatore gamme
% j=1; se non ho stella-triangolo
for j=1:2 %contatore stella-triangolo
P=data(1); %potenza max
nN(i,j)=data(2,j)./tau(i); %rpm, ginocchio
nmax(i,j)=data(3,j)/tau(i); %giri max
M_N(i,j)=data(4,j)*tau(i); %coppia max
end
end
PL=['r','r-';'g','g-'];
% j=1; %se non ho stella-triangolo
for j=1:2 %contatore stella-triangolo
% subplot(length(tau),1,j);
% figure
x1(j,:) = linspace(0,nN(1,j)); %numero di giri da 0 al ginocchio 1
x2(j,:) = linspace(nN(1,j),nmax(1,j)); %da ginocchio 1 a max 1
x3(j,:) = linspace(0,nN(2,j)); %da 0 a ginocchio 2
x4(j,:) = linspace(nN(2,j),nmax(2,j)); %da ginocchio 2 a max 2
C_1(1,j)=M_N(1,j); %coppia costante 1 marcia
C_2(j,:)=30*P*1000./(pi*x2(j,:)); %coppia 1 marcia
C_3(2,j)=M_N(2,j); %coppia costante 2 marcia
C_4(j,:)=30*P*1000./(pi*x4(j,:)); %coppia 2 marcia
ax1 = gca;
plot(ax1,[0,nN(1,j)],[C_1(1,j) C_1(1,j)],PL(1,j),'LineWidth',2/j^2) %I coppia costante
hold on
plot(ax1,x2(j,:),C_2(j,:),PL(2,j),'LineWidth',2/j^2) %coppia I
plot(ax1,[max(x2(j,:)),nN(2,j)],[C_3(2,j) C_3(2,j)],PL(1,j),'LineWidth',2/j^2) %II coppia costante
plot(ax1,x4(j,:),C_4(j,:),PL(2,j),'LineWidth',2/j^2) %coppia II
ylim([0 3500])
ylabel(ax1,'Torque [Nm]'); %# Add a label to the left y axis
set(ax1,'Box','off');%# Turn off the box surrounding the whole axes
axesPosition = get(gca,'Position');
ax2 = axes('Position',axesPosition,...
'YLim',[0 30],... %# and a different scale
'XAxisLocation','top',...
'YAxisLocation','right',...
'Color','none',...
'XColor','b','YColor','b',...
'Box','off'); %# ... and no surrounding box
hold on
m=P./nN; %pendenza potenza
P_1(j,:)=m(1,j).*x1(j,:); %potenza 1 marcia
P_2(1,j)=P; %Potenza costante 1 marcia
P_3(j,:)=m(2,j).*x3(j,:); %Potenza 2 marcia
P_4(2,j)=P; %Potenza costante 2 marcia
plot(ax2,x1(j,:),P_1(j,:),'b--')
plot(ax2,[max(x1(j,:)),nmax(1,j)],[P P],'b--')
plot(ax2,x3(j,:),P_3(j,:),'b--')
plot(ax2,[max(x3(j,:)),nmax(2,j)],[P P],'b--')
ylabel(ax2,'Power [kW]');
linkaxes([ax1 ax2],'x'); %lega tra di loro gli assi,
grid on
grid minor
end
我如何以相同的比例绘图
答案 0 :(得分:1)
在R2016a之前,你应该使用plotyy
来表示你想要的那种情节。
答案 1 :(得分:1)
原因是您使用ax1 = gca
。 gca
是当前轴,在完成第一次迭代后,当前轴是第二个,因为它是您绘制的最后一个轴。
要解决此问题,请在循环之前添加轴创建ax1 = axes
,并将ax1 = gca
替换为axes(ax1)
,以便ax1
成为当前轴。
注意:另请注意,您要创建ax2
两次。这意味着实际上有两个相同的轴具有完全相同的位置和属性。更好的做法是在for
循环之前移动所有轴,并且循环仅用于绘制它们。我会在循环之前写下以下内容:
ax1 = axes;
hold on;
ylim([0 3500])
ylabel(ax1,'Torque [Nm]'); %# Add a label to the left y axis
set(ax1,'Box','off');%# Turn off the box surrounding the whole axes
axesPosition = get(ax1,'Position');
ax2 = axes('Position',axesPosition,...
'YLim',[0 30],... %# and a different scale
'XAxisLocation','top',...
'YAxisLocation','right',...
'Color','none',...
'XColor','b','YColor','b',...
'Box','off'); %# ... and no surrounding box
hold on;
ylabel(ax2,'Power [kW]');
linkaxes([ax1 ax2],'x'); %lega tra di loro gli assi,
xlim([0,3800]);
grid on
grid minor
现在循环就是这样:
for j=1:2 %contatore stella-triangolo
% subplot(length(tau),1,j);
% figure
x1(j,:) = linspace(0,nN(1,j)); %numero di giri da 0 al ginocchio 1
x2(j,:) = linspace(nN(1,j),nmax(1,j)); %da ginocchio 1 a max 1
x3(j,:) = linspace(0,nN(2,j)); %da 0 a ginocchio 2
x4(j,:) = linspace(nN(2,j),nmax(2,j)); %da ginocchio 2 a max 2
C_1(1,j)=M_N(1,j); %coppia costante 1 marcia
C_2(j,:)=30*P*1000./(pi*x2(j,:)); %coppia 1 marcia
C_3(2,j)=M_N(2,j); %coppia costante 2 marcia
C_4(j,:)=30*P*1000./(pi*x4(j,:)); %coppia 2 marcia
plot(ax1,[0,nN(1,j)],[C_1(1,j) C_1(1,j)],PL(1,j),'LineWidth',2/j^2) %I coppia costante
plot(ax1,x2(j,:),C_2(j,:),PL(2,j),'LineWidth',2/j^2) %coppia I
plot(ax1,[max(x2(j,:)),nN(2,j)],[C_3(2,j) C_3(2,j)],PL(1,j),'LineWidth',2/j^2) %II coppia costante
plot(ax1,x4(j,:),C_4(j,:),PL(2,j),'LineWidth',2/j^2) %coppia II
m=P./nN; %pendenza potenza
P_1(j,:)=m(1,j).*x1(j,:); %potenza 1 marcia
P_2(1,j)=P; %Potenza costante 1 marcia
P_3(j,:)=m(2,j).*x3(j,:); %Potenza 2 marcia
P_4(2,j)=P; %Potenza costante 2 marcia
plot(ax2,x1(j,:),P_1(j,:),'b--')
plot(ax2,[max(x1(j,:)),nmax(1,j)],[P P],'b--')
plot(ax2,x3(j,:),P_3(j,:),'b--')
plot(ax2,[max(x3(j,:)),nmax(2,j)],[P P],'b--')
end