# Calculating left-right Horizon Drop

The following sketch shows how the earth's horizon (red arc) appears to curve down left and rigth in the frame of view with a horizontal viewing angle α. The downward curve has to do with the perspective transformation of the horizon onto a flat display. If we look down from a certain height to the horizon, the horizon at the edges of the frame appears lower than at the center of the frame.

We can calculate the anglular drop δ between the center of the horizon and the edges and using the distance to the center of the horizon v we can calculate from this angle the apparent drop left to right p as it were measured by a scale at the horizon. To look at the horizon on a spherical earth we have to look down from eye level a certain angle. This angle is called the drop angle φ. This angle is the same as the angle between the lines AZ and CZ, where Z is the center of the earth. Because the triangle ACZ is a right angled triangle with the right angle at C and the angle φ at Z, we can use trigonometry to calculate the angle:

(1)
where'
 $\varphi$ ' =' 'drop angle in radian $R$ ' =' '6371 km = radius of the earth $h$ ' =' 'observer height

The distance v from the observer's eye to the horizon can be calculated using Pythagoras, because we have a right angle triangle ACZ, where Z is the center of the earth and the right angle is at C. So AZ = R + h and CZ = R and v = AC:

(2)
where'
 $v$ ' =' 'distance from observer's eye to the horizon $R$ ' =' 'radius of the earth $h$ ' =' 'observer height

To calculate the line x = BE = BC we can use similar triangles. The triangles ACZ and ABC are both right angle triangles and share the same angle φ at Z and at C respectively. We can equate the ratio adjacent : hypotenuse = CZ : AZ = BC : AC and solve for BC = x:

(3)
where'
 $x$ ' =' 'line BC = line BE $v$ ' =' 'line AC = distance to horizon $R$ ' =' 'line CZ = radius of the earth $R+h$ ' =' 'line AZ = distance of observer from earth's center

To calculate y we look at the right angle triangle AED, where y = ED and the angle at A is half the horizontal field of view angle α. Using trigonometry we get:

(4)
where'
 $y$ ' =' 'line ED $v$ ' =' 'line AE = distance to horizon $\alpha$ ' =' 'horizontal field of view angle

The field of view angle θ is commonly refered to the diagonal of the viewing frame with aspect ratio r. So we have to calculate the horizontal field of view angle as follows:

(5)
where'
 $\alpha$ ' =' 'horizontal field of view angle $\theta$ ' =' 'diagonal field of view angle, depending on the focal length $r$ ' =' 'w / h = aspect ratio of viewing frame

The rest of the calculations can only be executed if y < x, i.e. the earth does not fit as a whole in the frame of view, so there is an intersection between the left-right border of the frame and the horizon at all.

Now we need the length of the line AB = m + h. We can calculate using Pythagoras the line AB from AE and BE:

 (6)

And we need the line BD = t from the right angled triangle BED with the right angle at D:

 (7)

Now we can see what happens if the earth fits as a whole in the frame. In this case y > x so the argument of the square root gets negative and we have no real solution for t.

Now we can calculate the angle β of the right angled triangle ABD with the right angle at B and the angle β at A:

 (8)

Finally we can calculate the horizon left-right drop angle δ:

 (9)

and the horizon left-right drop as measured on a scale at the horizon:

 (10)

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