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Bar-headed Goose (Anser indicus), Family
Anatidae |
Credit & Copyright:
Dr. Bruce G. Marcot
Explanation: This week, the main star is one of the highest-fliers in the world: a Bar-headed Goose ... vying for the altitude title of all birddom. Often touted to be the highest flier, amazingly Bar-headed Geese can surmount the ridgelines and peaks of the mighty Himalayas. Research suggests that Bar-headed Geese follow valleys through their migration path across the Himalayas, staying mostly below 18,000 feet elevation. But a recent study found that occasional Bar-headed Geese will reach nearly 21,120 feet (that's an altitude of 4 miles above sea level, or 6,437 meters). Their physiology provides more efficient red blood cells and greater blood flow with more capillaries, than your average honker. Other
anecdotal sightings by mountain climbers near the top of Makalu have
reported Bar-headed Geese flying far overhead, thus reaching a reported
altitude of 29,500 feet (nearly 5.6 miles or 8,991 meters above sea
level!). Tested in wind
tunnels, Bar-headed Geese have been found to be able to sustain
hyperventilation of super-thin air supplies.
Ruppell's Griffon Vultures have been reported to fly as high as 36,000 feet. In one report from Côte d'Ivoire in east Africa, a Ruppell's Griffon Vulture apparently collided with an aircraft at 37,000 feet (that's 7 miles or about 11,300 meter straight up).
Consider that at sea level (elevation zero), we experience what is called one bar of atmospheric pressure. One bar is equivalent to about 14.7 psi (pounds per square inch). Laying on the beach at sea level, nearly all of the atmosphere is above you, save for those rare places such as Death Valley that dip below sea level, and for various caves and mines that might also extend below sea level. So what is the air pressure in, say, a small single-engine plane flying at an altitude of 5,000 feet AMSL (above mean sea level)? This is told by the simple formula for calculating air pressure: where
p = the resulting atmospheric pressure measured in bars; The result of all this is that the higher you fly, the lower is the atmospheric pressure ... and it lowers by a negative exponential function, meaning that the pressure drops less rapidly the higher you go. But it still drops. If our hypothetical plane is flying at 5,000 feet (= 947 meters or 0.947 km), then p = (1)(e-(0.947/7)) = 0.87 bars. Remember, at sea level, the atmospheric pressure is (by definition) 1 bar. So the plane is flying at an altitude where the air has 87% of the pressure as compared to sea level. Another way to think about this is that the plane is flying above 100 - 87 = 13% of the Earth's atmosphere, by pressure.
OK, so what about these high avian fliers? The Bar-headed Goose, flying above Makalu at its record of approximately 29,500 feet or 8,991 meters or 8.991 km, is flying where air pressure is at p = (1)(e-(8.991/7)) = 0.28 bars. Where it flies, it is above 100 - 28 = 72% of the Earth's atmosphere (again, by pressure)! And the Ruppell's Griffon Vulture? If a plane actually struck one at 11,300 meters (11.3 km) altitude, then the vulture was probably panting for thin oxygen at p = (1)(e-(11.3/7)) = 0.20 bars ... above 80% of the Earth's atmosphere (by pressure)!
So THIS then is how high a bird can fly!
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Next week's picture:
Khasi Pitcher Plant by the Roadside
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