What am i doing wrong? This expression is violating basic Mathematics.
显示 更早的评论
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
2 个评论
Walter Roberson
2020-5-12
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
采纳的回答
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
0.0007348123729260462
0.001229147082940499
0.001551112338708252
0.001750919824749765
0.001867711745054821
0.001932038559298222
0.001965829383562728
0.001983030898594019
0.001991628840290395
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
3 个评论
Sandip Ghatge
2020-5-12
Stephen23
2020-5-12
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
更多回答（1 个）
Cris LaPierre
2020-5-12
编辑：Cris LaPierre
2020-5-12
0 个投票
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
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