How far would you fall in one round assuming you started falling this round, and were high enough that you did not hit anything? Would this follow the math, or is there a rule written somewhere I'm missing?
Keep in mind that human terminal velocity peaks around 53-56m/s... about 170-183 ft/sec... or about 1100 ft/round. It should take, at 32ft/sec (which is close enough for this), it's about 5.7 sec...
So, since terminal velocity is hit during turn 1... half that is the distance covered in turn 1...
So turn 1 should be to about 550 feet, and each turn thereafter another 1100 feet.
Now, peak recorded speed was about 125mph in random posture, or about 210mph in bullet posture... for about 308 ft/sec...
so, if the character's in bullet mode...
So, a little physics: falling like a ragdoll: 550' in round 1, 1100' each round after.
For simplicity, make it 500 and 1000 in ragdoll, and 500, 1500, and 1800 for diving.
Also note the 4E DMG says 500 feet in round 1, which is about 50' short, but could be covered by not being in ragdoll, but parasol, or by a 28' per second per second gravity.
This is from the Dungeon Master's Guide p48 under CrashingDDI also p209 in the Rules Compendium (thanks @Iszi).
The math more or less follows the DMG.
According to the rules, a creature falls 100 squares (500 feet) in the first (six second) round.
According to the laws of physics, a creature falls y = 0.5 g t2 feet, where g is the acceleration due to gravity (32 ft/s2 on Earth) and t is the number of seconds in free fall. Substituting, you get 0.5 * (32 * 62), or 576 feet. Not too far off.
Fun with Fantasy Physics
If you assume the 500 feet is a law and not an approximation, you can figure out the gravitational acceleration constant for the D&D world! That is, you can solve for g in the equation. If y = 0.5 g t2, and y=500 and t=6, then 500 = 0.5 g * 62.
So in the D&D world, g=27.78 ft/s2, not 32 ft/s2 as it does on Earth. This means that it takes longer for a creature to reach terminal velocity in the fantasy world...
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