Hello Matlab Community, Can someone write sum loop for me as menstion in the image and how to use subscript 1 with Q..

Answer 1

[Walter Roberson]

A = rand(1,20); B = rand(1,20);
X = A./B;

%looped sum
total_looped = 0;
for Index = 1 : numel(X)
    total_looped = total_looped + X(Index);
end

%Unlooped sum
total_unlooped = sum(X);

format long g
total_looped

total_looped = 
          35.8484976612101

total_unlooped

total_unlooped = 
          35.8484976612101

difference = total_looped - total_unlooped

difference = 
     0

[Umar Farooq]

thank you dear for your kind and reply. actually i have write in this way is it right?

w=[1.000 0.993 0.984 0.968 0.938 0.870 0.787 0.766 0.660 0.553 0.383 0.298 0.234 0.191 0.085]

d=[2.000 1.000 0.500 0.250 0.150 0.053 0.054 0.039 0.023 0.013 0.008 0.006 0.005 0.004 0.001]

di=d.*100

wi=(w.*1000)

%----------------------------------------------------------------------------------------------

ps=2.65; %partical desnity

eb=1.25; %bulk density

e=(ps-eb)/eb

n=15; %number of partical

Vvi=(wi/ps)*e; % pore volume

summation =0.0;

Vvi=[422.642 419.854 415.870 408.982 396.447 367.671 332.718 323.725 278.764 233.802 161.863 125.893 98.916 80.931 35.969]

Vb=[800 794.723 787.1829 774.144 750.417 695.948 629.787 612.7659 527.659 442.553 306.382 238.297 187.234 153.1914 68.085]

for i=1:n

Vb=summation.*(wi*(i))./eb;

summation = summation+Vvi*(i)/eb

teta_vi=summation.*(Vvi*(i)/Vb); %volumetric water content

end

theta=(teta_vi+teta_vi+1)/2; % Y- Grapgh volumtric water content

Answer 2

[Umar Farooq]

Erro

Index in position 1 exceeds array bounds. Index must not exceed 1.

Error in New_Arya_Paris (line 27)

RI(i,j)=(di(i,j)+ di(i+1,j))/4000;

-------------------------------------------------------

Problm

w=[1.000 0.993 0.984 0.968 0.938 0.870 0.787 0.766 0.660 0.553 0.383 0.298 0.234 0.191 0.085]

d=[2.000 1.000 0.500 0.250 0.150 0.053 0.054 0.039 0.023 0.013 0.008 0.006 0.005 0.004 0.001]

di=d.*100

wi=(w.*1000)

ni=15

for j=1:1;

for i=1:(ni-1)

RI(i,j)=(di(i,j)+ di(i+1,j))/4000;

end

end