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Message: <blockquote data-quote="Cleon" data-source="post: 6325323" data-attributes="member: 57383">I did some sums, and in real world terms the impact would be devastating.It's speed of 1,350 feet in a round is 810,000 feet per hour, or roughly 153.4 miles per hour. That's 65.58 metres per second if you prefer SI units.That speed's in the same ballpark as the launch velocity of a long-range bow or siege engine, so we're talking of an impact similar to being hit by a catapult stone that weighs four million tons and is on fire.Kinetic energy is ½ × mass × velocity². With a weight of 4 million short tons that gives KE = ½ × 3,628,738,960 kg × 68.58² metres a second = 4.27 Terajoules.4,270,000,000,000 joules is scary.A normal longbow arrow is about 100 J, give or take.An AP shell from a 16-inch <a href="http://en.wikipedia.org/wiki/16%22/50_caliber_Mark_7_gun" target="_blank">Navy cannon</a> has about 350 Megajoules when leaving the muzzle. You'd need about twelve thousand such shells to match the gammaroid's 8.53 TJ.For another Naval comparison, a <a href="http://en.wikipedia.org/wiki/Nimitz-class_aircraft_carrier" target="_blank">Nimitz-class aircraft carrier</a> would have to be travelling at almost the speed of sound (about 633-648 mph or Mach 0.83-0.85) to have that much KE.However, the gammaroid is so massive compared to most ships and creatures it wouldn't transfer much of that energy to them in an impact - a medium-sized creature would just splat up against it like a fly hitting a speeding sports car.Its speed's the same as the impact velocity of a fall from approximately 787 feet under <a href="http://en.wikipedia.org/wiki/Standard_gravity" target="_blank">standard gravity</a> (9.80665 m/s², or ~32.174 fps²), so relatively speaking it'd be similar to falling onto a mountaintop from that height. Unless it's using Flame Sheath, in which case it'd be like falling onto an erupting volcano.[1,350 feet per round is 225 fps. Divide by g of 32.174 fps² and you get 6.99 s, so it's the distance fallen in about 7 seconds at that acceleration. Acceleration distance from a stationary point equals ½ × acceleration × time², or ½ × 32.174 × 6.99² = 786.7 feet]</blockquote>

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