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Mike wrote: 
		
	
		
		
		
		
		
	
		 
		
	
	
	Steven P. McNicoll wrote: "Mike" wrote in message . .. Using trigonometry, I get ~ 785ft 2.5 miles out from the touch down zone, so your method is pretty accurate. Here's my calculation: Assuming: Distance = 15,000 ft Slope: 3 degrees Height = Distance * sin(Slope) = 785.04 ft. A 3 degree glidepath descends 318 feet per nautical mile. 318 x 2.5 = 795. Sorry, the original calculation was based on bad data. 15,000 feet is not 2.5nm as stated in the original post. 1nm = 6,076ft 2.5nm = 15,190ft Elevation = 15,190 * sin(3-degrees) = 795 ft Sorry for the double post. Last send just "hung" so I resent thinking it didn't send the first time. -- Mike  | 
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