How do I fix my display of the x and y values? They should be x = 1.15 and y = 3.85.

%code to calculate volume of water in a cylinder tank
voltage = 1.15; %input voltage from floater when tank is empty

%maximum voltage is 5V

%tank dimensions in metres
A = 2.63;
B = 3.945;
C = 5.9175; 

x = minimum_voltage == acosd(B/2*A)*5/180;
y = maximum_voltage == ((180 - angle)/180)*5;

disp(x)
disp(y)

%Convert voltage to angle
angle = (voltage*180)/5;
disp(angle)

%Check input voltage is valid 
if voltage < minimum_voltage 
    disp('ERROR') 
elseif voltage > maximum_voltage 
    disp('ERROR')
    volume = 'ERROR';
else
    disp('Voltage seems sensible') 
end 

>> VolumeCalculations
   1

   0

    41.4000

Voltage seems sensible

Answer 1

[Image Analyst]

Since you have not yet defined minimum_voltage and maximum_voltage, these lines will throw an error saying they are not defined. And if you had defined them, x and y would be either true or false since x and y are both set equal to a comparison condition.

x = minimum_voltage == acosd(B/2*A)*5/180;
y = maximum_voltage == ((180 - angle)/180)*5;

Simply delete all references to x and y and have this:

minimum_voltage = acosd(B/2*A)*5/180;
maximum_voltage = ((180 - angle)/180)*5;
fprintf('Min voltage = %f.  Max voltage = %f.\n', minimum_voltage, maximum_voltage);

Answer 2

[Chad Greene]

The double equals is defining x and y as a boolean. It's saying x = 1 if minimum_voltage equals acosd(B/2*A)*5/180. Let's find out what acosd(B/2*A)*5/180 actually equals:

voltage = 1.15; 
A = 2.63;
B = 3.945;
C = 5.9175; 
acosd(B/2*A)*5/180

ans = 0.0000 + 3.7083i

The way you have written your code, it says x = 1 if minimum_voltage equals 0.0000 + 3.7083i. Otherwise, x = 0. If you want x to be 1.15, write

x = 1.15;