How to fix Error using atan2 Inputs must be real?

Function

function ConcLin = Line(A,B)
%% Linear Moving along desired path
%  **** Linear First Movement ***

% Calling Inverse Kinematics to


%% number of increments 
ResLin = 100;

  %% Equations
DeltaX = (B(1,1) - A(1,1)) / ResLin;
DeltaY = (B(1,2) - A(1,2)) / ResLin;
DeltaZ = (B(1,3) - A(1,3)) / ResLin;


%% Looping through each point of the line
tic;
AnglesLin = zeros(6); %preallocating memory

for f = 1:ResLin
   
    %% calculating actual point  
   A(1,1) = A(1,1) + DeltaX * f; 
   A(1,2) = A(1,2) + DeltaY * f;
   A(1,3) = A(1,3) + DeltaZ * f;
   AnglesLin(f,:) = IKine(A); 
    
end
toc

%% Increments of Time
tic;
TimeLin = zeros(1); %preallocating memory
for clockLin = 1:ResLin
    TickTockLin = clockLin*0.1+10;
    TimeLin(clockLin,:) = TickTockLin;
end
toc

%% Concatenating Time & Angles. Time will be the first column
ConcLin = [TimeLin,AnglesLin];

Main

%% Coordinates Input
% LINE desired Paths. If Input method is used, comment out this block of code.
% *****We will define these as one matrix 1x12*****
Lin1 = [750, -75,  670, 0, 0, 1, 0, -1, 0, 1, 0,  0];%A
% Xa = Lin1(1,1);  Ya = Lin1(1,2); Za = Lin1(1,3); 
Lin2 = [750, -75,  550, 0, 0, 1, 0, -1, 0, 1, 0,  0];%B
% Xb = Lin2(1,1);  Yb = Lin2(1,2); Zb = Lin2(1,3);


Lin3 = [750,  75,  550, 0, 0, 1, 0, -1, 0, 1, 0,  0];
Lin4 = [750,  75,  670, 0, 0, 1, 0, -1, 0, 1, 0,  0]; 
Lin5 = [750,   0,  700, 0, 0, 1, 0, -1, 0, 1, 0,  0];
Lin6 = [750,   0,  550, 0, 0, 1  0, -1, 0, 1, 0,  0]; 

%% Calling Linear Motion
ConcLin = Line(Lin1, Lin2);

[Torsten]

Make sure that the two parameters you pass to atan2 are real, not complex.

E.g. the sqrt from the screenshot can become complex if the argument is < 0.

[Konard Adams]

@Torsten Can you please give an example?

Is there a method or built-in-function that can be use to keep or hold it as a real number?

[Torsten]

1-(K/p3)^2 becomes negative, so the sqrt of this expression gives a complex result.

Since I don't know what this expression means, I can give no further advice.

[Steven Lord]

You're assuming d3/p1 is strictly less than or equal to 1. What guarantee do you have that this is the case?

If it was ever so slightly greater than 1:

d3 = 2;
p1 = 2 - eps(2); % Just barely less than 2
s = (d3/p1)

s = 1.0000

s > 1 % true

ans = logical
   1

sqrt(1-s^2) % complex

ans = 0.0000e+00 + 2.1073e-08i

You can use min and max to ensure s is strictly in a desired range.

[Konard Adams]

Thank you. Here is the solution.:

When we move from previous point to next point, the coordinate equals to the previous coordinate plus the Delta, so we don't need to multiply f.

Before:

for f = 1:ResLin
   
    %% calculating actual point  
   A(1,1) = A(1,1) + DeltaX * f; 
   A(1,2) = A(1,2) + DeltaY * f;
   A(1,3) = A(1,3) + DeltaZ * f;
   AnglesLin(f,:) = IKine(A);

Correct:

   for f = 1:ResLin
   
    %% calculating actual point  
   A(1,1) = A(1,1) + DeltaX; 
   A(1,2) = A(1,2) + DeltaY;
   A(1,3) = A(1,3) + DeltaZ;
   AnglesLin(f,:) = IKine(A);

Answer 1

[Konard Adams]

When we move from previous point to next point, the coordinate equals to the previous coordinate plus the Delta, so we don't need to multiply f.

Before:

for f = 1:ResLin
   
    %% calculating actual point  
   A(1,1) = A(1,1) + DeltaX * f; 
   A(1,2) = A(1,2) + DeltaY * f;
   A(1,3) = A(1,3) + DeltaZ * f;
   AnglesLin(f,:) = IKine(A);

Correct:

     for f = 1:ResLin
   
    %% calculating actual point  
   A(1,1) = A(1,1) + DeltaX; 
   A(1,2) = A(1,2) + DeltaY;
   A(1,3) = A(1,3) + DeltaZ;
   AnglesLin(f,:) = IKine(A);
   

Now it works. @Torsten @Steven Lord Thanks for helping. I learn a lot from these discussions. Cheers!