Problem with script in if -else conditions

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Sampath Dissanayake
评论： Star Strider 2015-2-20
I am PhD student. I have to calculate a parameter (settling velocity-ws ) using two different sets of equations selected according a criteria (sediment diameter - D50). Do not worry about the subject. But the idea is that ws depends on D50. For sand (D50>0.00005), I want to use one set of equations to calculate ws. For other sediment particles (D50<0.00005), I want to use another set of equations to calculate ws. I load all sediment diameter (D50) values using a text file.
I used if and else conditions (please see the script written by me; I am new to matlab, this may not be efficient).
However, the results from my script gives an answer but using only one set of equations to calculate settling velocity. There should be a problem of writing if condition in my script. Here is the sample data set.
D50=
0.0002626
0.0002626
0.000003504
0.0000108
0.0000985
This is the answer;
ws=
0.069
0.069
0.0000123
0.000117
0.009717
The correct answer should be;
ws=
0.03681
0.03681
0.0000123
0.000117
0.009717
This is my code for actual set of equations; I want the answers written into a text file.
%%%Settling velocity
g=9.81;
psed=2650;
pw=1027;
neu=0.000001004;
p=0.4;
load D50.txt
if D50 > 0.0000625
Dstar=D50*(g*((psed/pw)-1)/neu^2)^(1/3);
Dstarpower=Dstar.^3;
idleA=ones(size(D50));
idleB=107.3296*idleA+1.049*Dstarpower;
idleC=idleB.^0.5;
idleD=idleC-10.36*idleA;
idleE=neu*idleD;
ws=idleE./D50
else
ws=1000000*(D50.^2)
end
CoefA=ws/((1-p)*psed);
fileID1=fopen('ws.txt','w');
fprintf(fileID1,'%6.6f\r\n',ws);
fclose(fileID1);
fileID2=fopen('CoefA.txt','w');
fprintf(fileID2,'%6.6f\r\n',CoefA);
fclose(fileID2);
I am very grateful if you can sort out this problem. Do not bother about equations, if equations are complex please give me a sample code for a case
if D50>0.00005
D_STAR=100*D50;
ws=(D_STAR*100+10)/.D50
else
ws=1000000*(D50.^2)
end
Thank you. Ruwan Sampath

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采纳的回答

Andrew Newell
Andrew Newell 2015-2-20
The problem is that you are testing the entire array at once with your code. If you type
D50 > 0.0000625
you get
ans =
1
1
0
0
1
An if test is only true if all the elements are true. You need to create a loop to test each element separately:
ws = zeros(size(D50));
for ii = 1:numel(D50)
if false %D50(ii) > 0.0000625
Dstar=D50(ii)*(g*((psed/pw)-1)/neu^2)^(1/3);
Dstarpower=Dstar.^3;
idleA=ones(size(D50(ii)));
idleB=107.3296*idleA+1.049*Dstarpower;
idleC=idleB.^0.5;
idleD=idleC-10.36*idleA;
idleE=neu*idleD;
ws(ii)=idleE./D50(ii);
else
ws(ii)=1000000*(D50(ii).^2);
end
end
ws
The result is
ws =
0.036807126831593
0.036807126831593
0.000012278016000
0.000116640000000
0.007330330176007
The last entry disagrees with your expectation for the correct answer, but it is consistent with the test condition.

更多回答（1 个）

Star Strider
Star Strider 2015-2-20
I believe some of your ‘correct’ answers aren’t actually correct. The else block applies to some of the ‘D50’ elements.
The easiest solution (without disrupting your code because I don’t understand what you’re doing) is to use a for loop to iterate through your ‘D50’ vector:
for k1 = 1:length(D50)
D50v = D50(k1);
if D50v > 0.0000625
Dstar=D50v*(g*((psed/pw)-1)/neu^2)^(1/3);
Dstarpower=Dstar.^3;
idleA=ones(size(D50v));
idleB=107.3296*idleA+1.049*Dstarpower;
idleC=idleB.^0.5;
idleD=idleC-10.36*idleA;
idleE=neu*idleD;
ws(k1)=idleE./D50v;
else
ws(k1)=1000000*(D50v.^2);
end
end
This produces for ‘ws’:
ws =
36.8071e-003 36.8071e-003 12.2780e-006 116.6400e-006 7.3303e-003

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