I live under the flightpath of a local airport. At night, the planes come flying in towards my house from over the Irish sea. Suppose that I am watching a plane approaching. Initially, the intensity of the headlights and the distance that the plane is away cause the individual lights to blur into a single light source. However, as the plane approaches, the lights separate into two light sources.
Assuming I have decent vision, that the headlights are 10 metres (accurate for an airliner?) apart and that any other variable that may affect the calculations is average for the north of England (for example, night time temperatures at ground level at this time of year are around 0 degrees Celsius), what are the steps that I need to carry out to calculate how far away a plane is at the moment that I can distinguish two light sources instead of one?
The width between the lights can vary widely between aircraft. Recognition and landing lights can be mounted on the nose gear, main gear, in the nose, wing roots (near the fuselage), wingtips, or just about anywhere else the manufacturer likes. Most airplanes have lights mounted in a combination of these locations.
I was flying a Piper Seminole the other day with recognition lights set in the wingtips, somewhere around 12m apart. I’m not sure of what large aircraft have recogs in the tips, but just to give you some idea, the Airbus A380 (which may or may not have them in the tips) has a wingspan of nearly 80m.
OK, let’s assume that the airport only allows A380’s to land, and that they have their headlights mounted on the wingtips. What are the steps needed to calculate the distance?
20/20 vision is the ability to discern points down to one arc-minute (1/60th of a degree). So if we assume the airplane is headed directly at the observer, we can set up an isosceles triangle with one angle at each wingtip and the last angle at the observer. We’ll use 80m for the short side of the triangle (side A), and 1/60th of a degree for the angle at the observer (angle A). With this information we can solve for the long sides of the triangle with simple math.
Angle A = 0.01667 degrees
Angles B&C=89.99167 degrees (the three angles have to add up to 180)
Side A= 80m
Sides B&C=275,000m
So, somebody with 20/20 vision should be able to discern the two lights as separate at a distance of 275 km. Of course, this doesn’t take into account atmospheric distortion. Temperature gradients, water vapor content, and airborne particulates will all have an effect. I have got to think that these factors will reduce the distance significantly.
Not to mention the fact that most airliners would have not even begun their descent at that distance, so recogs and landing lights would still be off. I have no idea how to calculate the atmospheric distortion.