Dear Captain Lim,
Before I ask my question, I have done some homework to come out with some data first to determine visibility of a place on ground from a height, and also the distances between two locations along the curvature of the Earth. My explanation first, and my question comes last below.
The Explanation:
The distance d in miles to the true horizon on earth seen from a plane or from any height is approximately:
First Equation: d = ?? (1.5 h), where h = height in feet of the eye (?? = square root)
This is a very simple formula which is applicable from most heights. Thus from an aircraft flying at 33,000 feet, then the distance to the true horizon seen by a pilot or a passenger is 222.485 miles (358 km) away.
Standing on a hill or tower of 100 feet high or even from an aircraft, the height (h) is much smaller than the equatorial radius (R) of the Earth of 3963.189 statute miles (6,378.135 km).
The exact formula for distance from the viewpoint to the horizon, applicable even for satellites, is:
Second Equation: d = ?? (2Rh + h2), where, (?? = square root)
d = distance is also the true distance to the horizon from a height
R = Equatorial radius of the Earth (6,378,135 metres)
Applying the second equation, the horizon seen from an aircraft flying at 10 km high would also be 357.299 km (222.0154 miles) away.
The calculations using the above two formulae give the straightline distance from the plane cockpit, passenger pothole, or a hill to the horizon, and NOT the distance to the horizon along the ground which would be longer for low heights such as seen from a hill.
In the second equation, both the radius (R) of the Earth, and the height (h) of the observer must be given in the same units (e.g. kilometers), but any consistent units will work.
The above two formula for d is only for the straight line of sight distance to the object of view, say the horizon.
A different relationship involves the arc length distance s along the curved surface of the Earth to the bottom of object. In this case, we apply:
The Third Equation:
l = cos1 [r ?? (r+h)] x [2?? r ?? 360]
(?= square root)
where,
l = the distance from the observer to the horizon along the curve (Great Circle) of the planet (Earth), along the ground
r (equatorial radius of Earth) = 6,378.135 km
h (height of plane) = 10,000 metres (10 km)
(r+h) = 6388.135 km
r ?? (r+h) = 0.998 434 597
Circumference of Earth (2?? x 6378.135 km) = 40075 km
2?? r ?? 360 = 111.3194559 km
Cos1 [r ?? (r+h)] is the angle in degree between the observer and the horizon, measured from the centre of Earth = 3.206324 degrees
Thus at an altitude of 10 km (33.000 ft) or 6.25 miles, the typical ceiling altitude of an jet airliner, the actual arc length (l) along the curved surface of the Earth to the bottom of an object (say a town) is 356.926 km (221.78 miles) away.
The distances along the curvature of Earth for low heights less than that of a jet plane at 10,000 meters, it would be greater than that of a direct lineofvision. The ceiling altitude of 10 km of a jet liner is about the limit where the straight lineofvision of a pilot to the horizon slightly exceeds the actual arc distance measured on the ground. However, several calculations using varying measurements for the radius of the Earth showed that the differences were less than a kilometer provided the observer did not fly above that of a commercial jet plane.
The Limit of Comparative Distances:
At much further distances, say 150 ?C 300 km away as seen from a satellite, or even further out as seen by a space traveler from outer space, say from the Moon, and the extra distance due to the curvature of the Earth will become less and less significant compared to the greater and greater distances or ??height?? away from the Earth. As an observer recedes into outer space, the distance to the horizon continue to increase until it far exceeds even that of the entire circumference of the Earth, let alone just a small extra arc of the curvature. The observer could be lightseconds, lightminutes, or light hours away, while the entire circumference of Earth is just only 0.133 light second round, if light could whip around the Earth. .
But for low heights, such as from a small hill, a tall building, a low flying plane, or from a watch tower, the extra arc distance along the ground to the horizon would always be greater than that of a straight line vision from above.
Actually the visual horizon is slightly further away than the calculated visual horizon, due to the slight refraction of light rays due to the atmospheric density. There may be also the effect of mirage playing tricks on the eyes, lifting an invisible distant city above the horizon.
1 statute mile = 5,280 feet = 1,609 meters
Geographical Locations of Cities:
Location Longitude (E) Latitude (N)
Penang 100 0 15 50 25
Phuket 98 0 22 80 0
Kuala Lumpur 1010 41 30 9
Singapore 1030 51 10 17
Distances between Neighboring Cities
Using the above geographical coordinates, and applying spherical trigonometry or the Haversine formula, the shortest distances along the curvature of the Earth between Malaysia and neighboring countries are:
Kuala Lumpur  Penang: 297.9 km = 185.1 statute miles = 160.8 nautical miles
Kuala Lumpur ?C Singapore: 318 km = 198 miles = 172 nautical miles
Penang  Phuket (South Thailand): 354.64 km = 220.36 miles = 191.49 nautical miles
The Question Now:
Having calculated out the distances along the Great Circle of some of the towns and cities between Malaysia, Thailand and Singapore, and we have shown that are within the theoretical distances that can be seen from an aircraft flying 10,000 metres above the ground, can we actually see Penang and Singapore from Kuala Lumpur, or Phuket above Penang from a plane flying at 10,000 metres over these areas? We presume there is no blanket of cloud cover, and that the weather is perfect with visibility up to infinity?
Frankly, I have never been able to locate where the horizon is whenever I fly. All I saw were just sheets, and sheets of thick white clouds below, and they stretched as far away as I could see. I seek your valuable experience and your expert comments.
Thank you Captain.
JB Lim
Malaysia
Hi Dr JB Lim,
Thank you for the explanation on how to calculate the visibility of an object on the ground from a particular height. Yes, pilots have been using the same formula (First Equation) to work out the the lineofsight distance, not so much to look at any particular object, but to find out the range when they could start to receive a VHF radio transmission.
You see, VHF (Very High Frequency) radio transmissions travel in a straight line and could not bend or follow the curvature of the Earth like the HF (High Frequency) transmission could.
The latest weather of a particular airport is provided by the ATIS (Aerodrome Terminal Information Service) transmistted on VHF. When I am flying at 35,000 feet, I know that at around 220 nautical miles, I would be able to receive the latest weather from the ATIS of an aerodrome.
In this sense, the formula you used to calculate the line of sight distance is useful. As to actually see a city from 220 nautical miles at 35,000 feet, it is not always possible to do so even if the visibility is up to infinity (due to traces of haze or the effect of oblique visibility).
Perhaps, at night, one may be able to see the lights of the cities on a good day when the weather is perfect, but otherwise, just like you, I would see sheets and sheets of thick white clouds below me most of the times! However, if the cities were in the range of around 100 nautical miles, I probably could see and recognise them in the horizon (crosschecking with the radar on board if needed!)
