Feather Fall (normal rate of fall...?)

Hollywood · Mar 2, 2003

The creatures or objects affected fall slowly (though faster than feathers typically do). The rate of falling is instantly changed to a mere 60 feet a round (equivalent to the end of a fall from a few feet), with no damage incurred upon landing while the spell is in effect. However, when the spell duration ceases, a normal rate of fall resumes.

Ok, does anyone know what the normal rate of fall is according to the 3rd Ed. rules?

CRGreathouse · Mar 2, 2003

Hollywood said:
Ok, does anyone know what the normal rate of fall is according to the 3rd Ed. rules?

I believe it's 150 feet for the first round, 300 feet thereafter: DMG 69.

Hollywood · Mar 2, 2003

Re: Re: Feather Fall (normal rate of fall...?)

CRGreathouse said:
I believe it's 150 feet for the first round, 300 feet thereafter: DMG 69.

Mms, thats as good of place to look as any I guess. Thanks.

dcollins · Mar 3, 2003

No, that's specific to creatures who have the capacity for flight but aren't quite up to their "minimum forward speed".

The Dragon magazine adventure "Storm Lord's Keep" by James Wyatt specified a normal falling rate of 670 feet in the first round, 1,150 feet in each succeeding round (p. 79). I personally prefer a straight 500 feet in the first round, 1,000 feet in each later round, based on a skydiving book I saw.

Steverooo · Mar 3, 2003

Formula

d = ViT + 1/2 aT^2

the Distance fallen (d) is equal to the initial velocity (Vi) times the time (T), plus one half the acceleration due to gravity (a), times the time squared... Acceleation (in feet) is 32 feet per second squared, downwards (so negative). What's that in meters? -9.8? The initial velocity is usually zero, so d = 1/2 (-32 ft/sec^2) x T^2. Since a round is six seconds, T^2 is 36 sec^2, so the distance fallen in the first round is:

d = 1/2 (-32 feet/sec^2) x 36 sec^2
d = -16 feet x 36 = -576 feet (or 576 feet straight down).

Thereafter, velocity increases until "terminal velocity" is reached... What TV is is dependent upon the shape of the object (wind resistance, etc.), but IIRC, is around 290 mph...

290 miles/hour x 5,280 feet/mile = 1,531,200 feet/hour
1,531,200 feet/hour x 1 hour/60 min. x 1 min./60 sec. =
425 1/3 feet/sec.

(Heh!) Nope, I must be remembering wrong...

Anyway, if you can get a value for terminal velocity, plug it into the formula, above, as "d", then solve for "T" and divide by six, to turn it into rounds (six seconds). That will give you how many rounds it will take to reach that speed.

What I do, though, is just find the distance to fall, plug it in, and crank out "T", which tells me how many seconds it takes to hit the ground. That's usually what you want to know.

Pax · Mar 3, 2003

Basic physics.

Rounds are about, what ... 6 seconds long, right?

Each second, you fall yoru current speed, plus half the acceleration for thats econd. So (all speeds in feet/second, accelerating under 1 gravity):

1 ... speed 000, accalerate to 032 ... fall 016' ... total fall 016'
2 ... speed 032, accelerate to 064 ... fall 048' ... total fall 064'
3 ... speed 064, accelerate to 096 ... fall 080' ... total fall 144'
4 ... speed 096, accelerate to 128 ... fall 112' ... total fall 256'
5 ... speed 128, accelerate to 160 ... fall 144' ... total fall 400'
6 ... speed 160, accelerate to 192 ... fall 160' ... total fall 560'

Now, somewhere in tehre you reach terminal velocity -- you just won't accelerate any faster. I'm not sure where in there, though.

That would fit the "500' in the first round, 1000' feet every round after that", which DCollins suggests. It's a nice round number, and it's "close enough" to an actual 6-second "rule of thumb" falling rate.

Well, close enough for ME, anyway.

As an aside -- Feather Fall gives you roughly a 10-feet-per-second falling rate. Swift, but not inuriously so. About a slow jog, really (figure it'd manage a 50-yard / 50-meter dash in about ... 15 seconds).

Hollywood · Mar 3, 2003

Ok,

Thanks. I was looking for an official rule on what the "normal falling rate" is.

Btw, the general terminal velocity of a person is ~125mph if you are extended with arms and legs out. About ~200mph in a ball and I think the max. skydiving record is ~300mph if I remember right. This is in practice from sky-diving. So about ~1000' per round, more if the person is tucked, etc.

However, D&D's physics are right anyways, so I'm not particularly worried about the correct physics answer, but rather, as I stated before, the official rule.

As far as I'm concerned, this topic is closed.