循環播放不同尺寸的矢量

Question

該代碼旨在計算具有疊加原理的光束的偏轉，其中對於光束的每個給定位置，計算所有單獨的力的影響，然後相加在一起。

function deflection = beamSuperposition(positions, beamLength,loadPositions, loadForces, beamSupport) 
%If several loads are applied, calculate each one individually and end with sum 

x = positions; 
l = beamLength; 
a = loadPositions; 
W = loadForces; 
E = 200*10^9; 
I = 0.001; 

%For every position(x) element calculate the expression for all 
%loadPositions(a) and their respective loadForce(W) 

%Make sure a and W are of same size 
%Make sure neither x or a ever surpass l 

%Go trhough each element of x 
for i = 1:length(x) 
    %For beamSupport = 1 only 
    while beamSupport == 1 
     %Go through each element of 'a' and calculate the deflection 
     for n = 1:length(a) 
      %There are two different equations to calculating the deflection depending on the position of the loadForce compared to the position of interest 
     while true 
      if x(i) < a(n) 
       y = (W(n)*(l-a(n))*x(i))*(l^2-x(i)^2-(l-a(n))^2)/(6*E*I*l); 
      end 
      if x(i) >= a(i) 
       y = (W(n)*a(n)*(l-x(i)))*(l^2-(l-x(i))^2-a(n)^2)/(6*E*I*l); 
      end 
      break 
     end 
     end 
     break 
    end 
%Sum the values for each y element?!? 
    deflection = y 
    end 

我該如何按預期做這項工作？輸出應該是一個矢量y，其大小與x相同，每個值的總和偏差爲x。

使用中;

beamSuperposition([1,2,5],10,[6,5,3],[10,15,20],1) 

就回去了，如果deflection = y不supressed，

deflection = 

    5.8333e-07 


deflection = 

    1.0967e-06 


deflection = 

    1.6500e-06 


ans = 

    1.6500e-06 

但首先應該只

Answer 1

我做了一些小的變化。首先，我已將中央兩個if聲明更改爲單個if else聲明。此刪除你有

if x(i) >= a(i) 

裏面應該有可能讀

if x(i) >= a(n) 

雖然有我還刪除了一些我沒有很明白的目的，你while break控制流的一個錯誤。你可能想要比較。

其次，我將你的輸出保存到y的i-條目。我也預分配它，所以它不會改變循環中的大小。

最後，如@StackPlayer所示，我用循環變量ii代替i。關於Using i and j as variables in Matlab的問題有一些討論。

function deflection = beamSuperposition(positions, beamLength,loadPositions, loadForces, beamSupport) 
%If several loads are applied, calculate each one individually and end with sum 

x = positions; 
l = beamLength; 
a = loadPositions; 
W = loadForces; 
E = 200*10^9; 
I = 0.001; 

%For every position(x) element calculate the expression for all 
%loadPositions(a) and their respective loadForce(W) 

%Make sure a and W are of same size 
%Make sure neither x or a ever surpass l 

y = zeros(size(x)); 
%Go through each element of x 
for ii = 1:length(x) 
    %For beamSupport = 1 only 
    if (beamSupport == 1) 
     %Go through each element of 'a' and calculate the deflection 
     for n = 1:length(a) 
      %There are two different equations to calculating the deflection depending on the position of the loadForce compared to the position of interest 
      if x(ii) < a(n) 
       y(ii) = (W(n)*(l-a(n))*x(ii))*(l^2-x(ii)^2-(l-a(n))^2)/(6*E*I*l); 
      else 
       y(ii) = (W(n)*a(n)*(l-x(ii)))*(l^2-(l-x(ii))^2-a(n)^2)/(6*E*I*l); 
      end 
     end 
    end 
end 
%Sum the values for each y element?!? 
deflection = y; 
end 

實施例使用：

>> beamSuperposition([1,2,5],10,[6,5,3],[10,15,20],1) 

ans = 

    1.0e-05 * 

    0.0583 0.1097 0.1650 

Answer 2

第一件事返回值作爲載體的，而不是最後的值，避免使用I和J MATLAB（它們用於虛數）。

在你的第二個while循環，你如果不是 「X（I）」 的 「X」 條件