What am i doing wrong? This expression is violating basic Mathematics.

Question

Sandip Ghatge 2020-5-12

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https://ww2.mathworks.cn/matlabcentral/answers/524985-what-am-i-doing-wrong-this-expression-is-violating-basic-mathematics

评论： Sandip Ghatge 2020-5-13

采纳的回答： Stephen23

在 MATLAB Online 中打开

If you run this code, you will get

SI(1,1) = -1 [SI is the sine term of the main formula mentioned below]

sqroot(1,1) = 0.0026

MI = 0.0026

Hence, according to %Main Formula

pos1(1,1) = 0.0026 + 0.0026*(-1) = 0

But the answer matlab is returning me is 0.002

How is this even possible?

Is there any error in the for loop?

%Defining values of parameters
Fg = 2.1188;
Fi = 2.7834;
Ft = Fg + Fi;
h = 0.002;
k = 1864;
e = 0.7;
m = 0.001;
% Initializing for x
x = zeros(11,1);
x(1,1) = 0;
l = 0.002;
%Formula 1 - determining new positions
for i = 1:10
    con1 = ((Ft*h)-(k*(h^2)*0.5))*(1-(e^2));
    x(i+1,1) = (Ft-(sqrt((Ft^2)-(2*k*((con1)+((e^2)*(Ft*x(i,1)-(k*0.5*(x(i,1)^2)))))))))/k;
end
%Initializing for constants N & S
N = zeros(10,1);
S = zeros(10,1);
W = (2*Ft)/m;
%Calculating constants N & S
for i = 1:10
    S(i) = (((W*l)-((k*(l^2))/m))*((e^2-1))-((e^2)*(x(i,1))*(W+((k*(x(i,1)))/m))));
end
sininv = zeros(10,1);
sqroot = zeros(10,1);
pos1 = zeros(5000,1);
pos2 = zeros(5000,1);
time = zeros(5000,1);
M = 2*Ft; 
time(1,1) = 0;
for i = 1:10
     N(i,1) = ((k*((x(i,1))^2)) - (2*Ft*(x(i,1))));
     sininv(i,1) = asin((x(i,1)-(M/2*k))/(sqrt(((M/2*k)^2)+(N(i,1)/k))));
     sqroot(i,1) = sqrt((((M/(2*k))^2)+(N(i,1)/k)));
    for j = 1:5000
        
        time(j+1,1) = time(j,1)+0.0001;
        
        %Formula 2 - pos1
        SI = (sin(((sqrt(k/m))*time(1,1))+sininv(1,1)));
        MI = M/(2*k);
        
        %MAIN FORMULA
        pos1(j,1) = MI+(sqroot(i,1))*(sin(((sqrt(k/m))*time(j,1))+sininv(i,1)));
    end
end

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Walter Roberson 2020-5-12

在 MATLAB Online 中打开

SI = (sin(((sqrt(k/m))*time(1,1))+sininv(1,1)));

Why are you calculating SI there? You do not use the value

time(j+1,1) = time(j,1)+0.0001;

It would be easier to go before the loops and define

time = (0:4999).' * 0.0001;

Sandip Ghatge 2020-5-12

在 MATLAB Online 中打开

I was just checking for SI Value, whether it gets -1 or not

and regarding time i tried multiple times as below, but still the issue remains unresolved,

%Defining values of parameters
Fg = 2.1188;
Fi = 2.7834;
Ft = Fg + Fi;
h = 0.002;
k = 1864;
e = 0.7;
m = 0.001;
% Initializing for x
x = zeros(11,1);
x(1,1) = 0;
l = 0.002;
%Formula 1 - determining new positions
for i = 1:10
    con1 = ((Ft*h)-(k*(h^2)*0.5))*(1-(e^2));
    x(i+1,1) = (Ft-(sqrt((Ft^2)-(2*k*((con1)+((e^2)*(Ft*x(i,1)-(k*0.5*(x(i,1)^2)))))))))/k;
end
%Initializing for constants N & S
N = zeros(10,1);
S = zeros(10,1);
W = (2*Ft)/m;
%Calculating constants N & S
for i = 1:10
    S(i) = (((W*l)-((k*(l^2))/m))*((e^2-1))-((e^2)*(x(i,1))*(W+((k*(x(i,1)))/m))));
end
sininv = zeros(10,1);
sqroot = zeros(10,1);
% pos1 = zeros(5000,1);
pos2 = zeros(5000,1);
M = 2*Ft; 
time = (0:4999).' * 0.0001;
for i = 1:10
     N(i,1) = ((k*((x(i,1))^2)) - (2*Ft*(x(i,1))));
     sininv(i,1) = asin((x(i,1)-(M/2*k))/(sqrt(((M/2*k)^2)+(N(i,1)/k))));
     sqroot(i,1) = sqrt((((M/(2*k))^2)+(N(i,1)/k)));
     
     MI = M/(2*k);
     
     pos1 = MI+(sqroot(i,1))*(sin(((sqrt(k/m)).*time)+sininv(i,1)));
end

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Answer 1

Stephen23 2020-5-12

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https://ww2.mathworks.cn/matlabcentral/answers/524985-what-am-i-doing-wrong-this-expression-is-violating-basic-mathematics#answer_432089

编辑：Stephen23 2020-5-12

在 MATLAB Online 中打开

Your code has two loops, loop j is nested inside loop i.

You allocate data to pos1 only using index j, which means that you replace the pos1 data ten times (once for each i iteration), and only keep the last value assigned. Here is a small peice of code to place just after the pos1 allocation and you will see the ten values that you allocate and then replace, once for each i iteration:

if j==1
    disp(pos1(1))
end

When I add that code, these are the ten values that you allocate to pos1(1):

0
0007348123729260462
001229147082940499
001551112338708252
001750919824749765
001867711745054821
001932038559298222
001965829383562728
001983030898594019
001991628840290395

The first nine are simply overwritten by the next value. Only the last value is finally stored. That value corresponds to j==1 and i==10, which you did not take into account when you did the calculation by hand. Using the correct i and j values gives exactly the same value as pos1:

>> tmp = (sqroot(1,1))*(sin(((sqrt(k/m))*time(1,1))+sininv(10,1))) % j==10
tmp = -0.002629935049104

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Stephen23 2020-5-12

在 MATLAB Online 中打开

Possibly you should use the i index as well:

pos1 = zeros(5000,10);
...
    pos1(j,i) = ...
...
plot(pos1)

But only you know if this is correct for your algorithm.

Sandip Ghatge 2020-5-13

Yes, i used the i index and was able to plot. Thank you

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Answer 2

Cris LaPierre 2020-5-12

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编辑：Cris LaPierre 2020-5-12

Nothing is wrong with your equation. However, think through your code. The nested for-loops are messing up your logic. The variable pos1 gets overwritten everytime the outerloop increments. You are checking the value of pos1(1,1) once your code has finished executing. That means you should be checking the math using sqroot(i,1) where i=10 (last value of outer loop counter).

SI(1,1) = -1

sqroot(10,1) = 0.0006

MI = 0.0026

pos1(1,1) = 0.0026 + 0.0006*(-1) = 0.002