The basic equation for terminal velocity of an object is given by:V = SqRt [ 2*m*g / r*A*C ]where m is the mass of the object in kg; g is acceleration due to gravity = 9.81 m/s/s; r is the air density in kg per cubic m; A is the projected area of the object in square m; C is the coefficient of drag (no dimensions).SqRt denotes the square root, * denotes multiplication and / denotes division.For a long cylindrical rod, from tables, C = 0.8 approximately;A = pi * .004 * .004 = 5E-5 square metre;r = 1.3 kg per cubic m (for cold air);m = .02 kg.These figures give V = 87 m/s = 312 km per hour (195 mile per hour).This is a ball park figure as air density can vary from 1.2 -1.4, and the drag coefficient will likely be affected by the presence of the in-line fins, so C between 0.75 and 0.85 - for simplicity the dart has been approximated as a pure cylindrical rod. Net result, it gives an idea that the dart will hurt if it hits you.It is interesting that tests on the early British 12,000lb Tallboy bomb designed by Sir Barnes Wallis apparently revealed it to reach supersonic terminal velocity and so to wobble off target. But that's for another thread...Barry ,If you do the sums for a person free falling with arms and legs out, the terminal velocity is indeed around 127mph. Taking m = 100kg; A = .5 sq metre; r = 1.3; C = 1 for high resistance body, gives V = 55 m/s or 198 km per hour (124 mile per hour).However, as you will have found when sky diving from 13,500 feet, sticking your head down, tucking your arms by your side and minimising drag, your terminal velocity will probably have touched 200 mph. Again, for m = 100kg, effective area now reduced so that A =.25; drag reduced so C = .8, we have V = 87 m/s = 312 km per hour (195 mile per hour). By coincidence, the same as the aerial dart.