Fall Speeds

[Kashell]

The player's guide says under the Feather Fall spell that one's fall speed is reduced to 60 ft. per round...but what is the normal fall speed? I can't find it in the player's guide, DM's manuel, or anything else for the life of me...

I would appriciate it.

 

[Staffan]

Not sure what the terminal velocity of a falling body is, but ignoring air resistance, falling gives you an accelleration of roughly 10 m/s^2. That is, 10 meters per second, per second - after one second, you're falling at 10 m/s, after another, it's at 20 m/s, 30 m/s, and so on. For those of you using archaic units, I think that's 32 feet/second/second.

Using the wonders of math, we get:
distance = accelleration*time^2/2
time^2 = 2*distance/accelleration
time = square root of (2*distance/accelleration)

So falling 200 ft would take: sqrt(2*200/32) = sqrt (400/32) = ~3.5 seconds.

 

[Doomhawk]

I believe that somewhere in the DMG, it states that you fall 150 feet in the first round of falling, and 300 feet each round thereafter. I could be wrong, though.

 

[shilsen]

I believe that somewhere in the DMG, it states that you fall 150 feet in the first round of falling, and 300 feet each round thereafter. I could be wrong, though.

That's listed on pg. 20 of the DMG under the "Movement in Three Dimensions" section and specifically referring to what happens when a flying creature fails to maintain its minimum forward speed and falls.

 

[Piratecat]

Yup. It's horribly inaccurate compared to real life, but it does okay for a game approximation.

 

[Vrecknidj]

Staffan is right, in that you fall at about 32 feet per second, per second. In other words, by the end of the first second, you're falling at a speed of 32 feet per second. In that first second, you actually move 16 feet. In the second second, your speed increases so that by the end of it, you're moving at a speed of 64 feet per second, and in the second second, you move 48 more feet, so now you're 64 feet from where you were. By the end of the third second, your speed is now 96 feet per second, and you've fallen a total of 144 feet. By the end of the fourth second, your speed is now 128 feet per second, and you've fallen a total of 256 feet. By the end of the fifth second, your speed is 160 feet per second, and you've fallen a total of 400 feet. By the end of the sixth second, your speed is 192 feet per second, and you've fallen a total of 576 feet.

That's a bit more than the 150 feet the DMG states. It ignores air resistance, but, unless there's a really strong wind blowing, air resistance is going to be negligible anyway.

The shortened version of the formula I used (physics buffs can check me on this) is that the distance you fall, if you started from not falling, is 16*t*t, where t is the number of seconds you've fallen. The speed at the end of each second (with the same assumptions) is 32*t. I know there's a terminal velocity, but I don't think you reach it in the first six seconds.

Dave

 

[Len]

It only takes about 10 seconds to reach terminal velocity, which is around 180-200 ft/s for a skydiver (ie. a person wearing a pack). (Sources: 1, 2, 3)

A good D&D approximation would be 500 feet in the first round and 1000 feet in each subsequent round.

 

[Kashell]

Hmm...I wonder...

 

[dcollins]

The figure in the DMG is for a normally-flying creature (with wings, feathers, etc.) who is somewhat damaged, in a not-quite-controlled descent: a fairly slow rate of 150 ft. the first round, 300 ft. each subsequent round.

The D&D 3E adventure from Dungeon #93, "The Storm Lord's Keep", had the official rate for normally-falling PCs (p. 79): 670 feet in the first round, 1,150 feet each subsequent round.

As Len said, what I personally use is 500 feet in the first round, 1,000 feet each subsequent round, which is very close to the figures in the book "Skydiving".