KrisFlyZ 0 #1 January 6, 2004 Quote This one got me thinking. Given: Canopy open at 2,000 FT AGL Canopy forward airspeed of 25 mph Canopy rate of descent of 750 FPM Area that can be reached in 0 and 20-mph winds 0 mph wind: Under canopy for 160 seconds at 36.67 FPS airspeed 5867.20 ft (1.11 miles) in any direction (radius) 3.87 miles^2 the canopy can reach in 0 mph wind 20-mph winds: Under canopy for 160 seconds at 36.67 airspeed and from 7.33 FPS to 66 FPS ground speed 1172.8 ft upwind [holding] 10,560.00 ft downwind [running] 7,509.51 ft crosswind [crosswind] (5862.20ft 9-degrees to the wind line and drift downwind 4693.33ft) If someone would figure the area for 20-mph winds, that would be great. Obviously, 20-mph wind scenario results in a much greater surface area that the canopy can reach. Derek When we tip the cone the circle becomes an ellipse. The area of an ellipse is http://thesaurus.maths.org/dictionary/map/word/2942. The area is actually slightly less .... From the above calculations by Derek, a = 10560 + 1172.8/2 (running distance + holding distance(this is in the canopy's direction when faced into the wind, this number would have to be subtracted when the canopy goes back in the wind or the speed of the canopy is less than the wind speed)/2 = 5866.4 b = distance flown when crabbing 90 degrees to windline. This is same as the radius of the circle in the previous case of no wind. 5867.2 Now this makes the area of the ellipse after converting to Sq. miles Area = 3.14 * (5867.2 * 5866.4 ) divided by 5280 * 5280 = 3.88Sq. Miles. I am saying that this is less because the area of the circle is 3.14 * 5867.2 * 5867.2 before conversion to sq. miles. Interesting !!!! I gave it more thought because it seems impossible. The tip the cone anology makes it seem that the long radius of the ellipse would lengthen when the cone was tipped. However on further thought, it seems like this is a little more complex than making a blanket statement that covers all cases. 750 FPM downward speed might not be realistic for a canopy that only has a forward speed of 25 mph. A change in that to reflect a more floaty canopy would show a considerably different result. Quote Share this post Link to post Share on other sites
Hooknswoop 19 #2 January 6, 2004 Very interesting, 11059.90 sq ft less area the canopy can reach in 20-mph wind vs. 0-wind. So much for obvious . Quote However on further thought, it seems like this is a little more complex than making a blanket statement that covers all cases. Run some numbers and see what you get. I was fooled. Quote 750 FPM downward speed might not be realistic for a canopy that only has a forward speed of 25 mph. I started w/ 1000 FPM, but thought that was too high. 750 FPM gave me an almost 3:1 (2,000 ft down and 5867.2 ft across the ground) glide ratio, which should be right for a 300+ jump F-111 canopy or a 7 cell. Any opinions for say 3 different canopy speeds and descent rates? Derek Quote Share this post Link to post Share on other sites
KrisFlyZ 0 #3 January 6, 2004 Let us go into the world of variables. Canopy starting height = h ft Forward speed of canopy = c ft/s Descent rate of canopy = d ft/s wind speed = w ft/s the area of the ellipse in which the canopy can land is Pi * [(h/d) * c] * {[ (c + w) * (h/d) plus or minus (c - w)* (h/d)]/2} that is Pi* (h/d) * c*c is the same as no wind conditions when the canopy speed exceeds the wind speed. Always!! i.e The wind has no effect on the area in which the canopy can land even though the shape of the area is now an ellipse!! The previous calculation was messed up because of the rounding. I recalculated without rounding and the result is the same. if the wind speed exceeds the canopy speed The formula becomes Pi * (h/d) * c * w and yes this ellipse is bigger in area than the circle. It is obvious afterall!! Quote Share this post Link to post Share on other sites
KrisFlyZ 0 #4 January 9, 2004 QuoteLet us go into the world of variables. Canopy starting height = h ft Forward speed of canopy = c ft/s Descent rate of canopy = d ft/s wind speed = w ft/s the area of the ellipse in which the canopy can land is Pi * [(h/d) * c] * {[ (c + w) * (h/d) plus or minus (c - w)* (h/d)]/2} that is Pi* (h/d) * c*c is the same as no wind conditions when the canopy speed exceeds the wind speed. Always!! i.e The wind has no effect on the area in which the canopy can land even though the shape of the area is now an ellipse!! There is an error in this. The shape of the area in which the canopy will land if wind speed is less than canopy speed will still be a circle. However the center of the circle is shifted by w*(h/d) units in the direction of the wind. In high winds however the shape is an ellipse and the center of the ellipse is shifted the same w* (h/d) units in the direction of the wind. This ofcourse does'nt mean much in the real world. Quote Share this post Link to post Share on other sites
