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Message: <blockquote data-quote="dvvega" data-source="post: 2070691" data-attributes="member: 524">At one time I sat down and calculated how fast someone would reach terminal velocity based on a standing start.This was done for a feather fall spell before "immediate" actions were available. The concept was to feather fall someone else falling.Anyways, it worked out that terminal velocity was hit at the start of round 2 given enough falling distance (noting rounds are 6 seconds long).Although terminal velocity varies between 53 m/s to 76 m/s ... 76 m/s is what most physics books quote (it does depend on object mass but let's ignore it for now). This assumes you don't want to be falling and so you've spread yourself out as best as you can.Standard gravity is 9.8 m/s, so to achieve 76 m/s I would need to be travelling for approximately 7 seconds which is just over a round. Using displacement s = ut + 0.5*a*t^2I start at velocity 0 so we are only interested in s = 0.5*a*t^2 and we have a = 9.8 and t keeps growing but in our case 6 seconds is one round and we don't hit terminal until 7 seconds time.So distance in the first round from a standing start = 0.5 * 9.8 * 6^2 = 176.4 meters which converts to 578.74 feet.Second round is a different story after the first 1.7 seconds ... after terminal velocity is reached. So distance fallen until terminal velocity is (continuing from above) 953.15 feet. From that point onwards they constantly fall 76 metres per second or 249 feet. Thus, even if we rounded the extra 1.7 seconds to 2 seconds, they would fall a further 960 feet or almost double what they had fallen already.So end of round 2 sees them at 1910 feet from the starting point.From then on they will indeed fall about 960 per round, however the 500/1000 is a little under quoted. It would be more like 500/1500/1000.Did I just actually write that twaddle!!D</blockquote>

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