how to represent these data

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i need to make a graph on matlab of the material against the property.

here are my data:

eglass=1.38;
sglass= 1.85;
graphiteHiE = 1.1;
graphiteHiTS= 1.3;
Boron=1.1;
Silica = 2.65;
Tungsten= 0.22;
Beryllium = 0.71;
Kevlar49=1.87;

Traditional materials
steel=.043 - .27;
Al alloys= .052 - .23;
Glass=.028 - .84;
Tungs2=057 - .21;
Beryllium2=.38;


You can see that the traditional materials do not allow u to have the
same data type.

i am plotting the material against the property (the numbers)

what is the right plot command to do it (already tried plot and it
didn't work..each material has to stand out - will Excel be better?)
and how best to represent the traditional materials.

thank u.
0
Reply cibeji (30) 8/24/2010 6:06:43 AM

matlab_learner <cibeji@gmail.com> wrote in message <b1e6a269-1d3c-4f48-9d72-529df2eb2082@b4g2000pra.googlegroups.com>...
> i need to make a graph on matlab of the material against the property.
> 
> here are my data:
> 
> eglass=1.38;
> sglass= 1.85;
> graphiteHiE = 1.1;
> graphiteHiTS= 1.3;
> Boron=1.1;
> Silica = 2.65;
> Tungsten= 0.22;
> Beryllium = 0.71;
> Kevlar49=1.87;
> 
> Traditional materials
> steel=.043 - .27;
> Al alloys= .052 - .23;
> Glass=.028 - .84;
> Tungs2=057 - .21;
> Beryllium2=.38;
> 
> 
> You can see that the traditional materials do not allow u to have the
> same data type.
> 
> i am plotting the material against the property (the numbers)
> 
> what is the right plot command to do it (already tried plot and it
> didn't work..each material has to stand out - will Excel be better?)
> and how best to represent the traditional materials.
> 
> thank u.

Analyze this code, I hope that it will be useful:

M = [
1.38%eglass
1.85%sglass
1.1%graphiteHiE
1.3%graphiteHiTS
1.1%Boron
2.65%Silica
0.22%Tungsten
0.71%Beryllium
1.87%Kevlar49
];
names = {
'eglass'
'sglass'
'graphiteHiE'
'graphiteHiTS'
'Boron'
'Silica'
'Tungsten'
'Beryllium'
'Kevlar49'
};

pie(M)
legend(names)

%Traditional materials
steel=[.043  .27];
Al_alloys= [.052  .23];
Glass=[.028  .84];
Tungs2=[.057  .21];
Beryllium2=.38;

figure
bar(1,steel(2),'BaseValue',steel(1))
hold on
bar(2,Al_alloys(2),'r','BaseValue',Al_alloys(1))
bar(3,Glass(2),'g','BaseValue',Glass(1))
bar(4,Tungs2(2),'k','BaseValue',Tungs2(1))
legend('steel','Al alloys','Glass','Tungs2')

figure
m = [ .043  .27
      .052  .23
      .028  .84
      .057  .21];
  
errorbar(1:4,mean(m,2),m(:,2)-mean(m,2),'LineStyle','none')  
set(gca,'xtick',1:4,'xticklabel',{'steel','Al alloys','Glass','Tungs2'})
0
Reply Grzegorz 8/24/2010 7:06:04 AM


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