Calculation of Visual Angle

Image visangle
Figure 1: Visual angles for an object on the horizon.

Draw the principle rays from the ends of the inverted arrow bc through the nodal point n and extend these principle rays to the retina. Two similar triangles are formed. The first Dbcn has exactly the same angles as Db'c'n. The visual angle of the image on the retina can be calculated by measuring the angle subtended by the object outside of the eye. This is true because q = q'.

Angle q is contained in Dbcn . However, in trigonometry it is convenient to work with right angles. So let's bisect line bc, placing a at its midpoint. q1 is contained in Dban and q2 is contained in Dacn. So, q1 = q2, q = q1 + q2, hence  q = 2 x q1    .

q  is calculated by dividing line bc in half and drawing the red line to n, thus forming a right triangle (Dban). In other words, distance |ba| = |ac|. The length of the red line is d. Then q1 is calculated by dividing distance |ba| by distance d to get tan(q1), and then taking the arctangent to get angle q1and finally  q = 2 x q1. These calculations can be easily performed with a calculator that has trigonometric functions.

tan(q1) = (|ba| ÷ d) (1)    
q1 = arctan(|ba| ÷ d) (2)    
 q = 2 x arctan(|ba| ÷ d) (3)    

If you want to know the distance between bc given that q  is a given angle, take the tangent of q1and multiply it by the distance d. This calculation gives you the distance from a to b  To determine the distance between b and c, multiply the calculated distance from a to b by 2 and you know the distance bc.

|ba| = tan( q1) x d (4)    
|ba| = tan(q ÷2) x d (5)    
|bc| = 2 x |ba| (6)    
|bc| = 2 x (tan(q ÷ 2) x d) (7)    

Thanks to Dr. Darren Weber for tidying the trigonometric notation