Geeky LOTR falling Gandalf question

[Gnarlo]

Was watching the Two Towers today, and got to pondering the scene with Gandalf and the Balrog falling into Durin's Deep. I started thinking about the question of how deep the Deep is, since Gimli states in the book that no one had ever plumbed its depths... The two of them fall in the movie for 70 seconds or so from the time that Gandalf lets go/ slips from the bridge; figuring the rate of acceleration and a terminal velocity of around 120 miles / hr (got that from several sky diving sites as an average, Gandalf could fall faster at the beginning when he was head down, but the pair of them bounce of the sides at several points slowing them down I'd imagine), I figure they fell roughly 10,000 feet in those 70 seconds... Any one got a better number or want to check the math?

Yes, these are the things that keep me awake at night

 

[KenM]

Then only thing about the scene that i'm wondering about is Gandalf's sword. Since it is not as heavy as Gandalf or the Balrog, it should fall at a faster rate then either of them. Gandalf gets to sword before balrog.

 

[DanMcS]

Then only thing about the scene that i'm wondering about is Gandalf's sword. Since it is not as heavy as Gandalf or the Balrog, it should fall at a faster rate then either of them. Gandalf gets to sword before balrog.

Weight makes no difference as to how fast you fall.

 

[Kid Charlemagne]

Weight makes no difference as to how fast you fall.

You must not have seen Rosencrantz and Guildenstern Are Dead...



Seriously, though, air resistance would be the factor here, and both Gandalf and the Balrog would produce more air resistance than the sword. I assumed that Gandalf was using a little subtle mojo to direct/impel himself in the direction and speed he required.

 

[kingpaul]

*puts on geek hat*
OK, the kinematic equations we are looking for are:

vf = vi + a*t
vf^2 = vi^2 + 2*a*d

which yields:

(vi + a*t)^2 = vi^2 + 2*a*d

Solving for d:

d = (vi^2 + 2*vi*a*t + a^2*t^2) / (vi^2 + 2*a)

vi = 0
t = 70 {your initial guess}
a = 32.2 ft/s^2

d = (32.2^2*70^2) / (2*32.2) = (1036.84*4900)/64.4 = 78,890 ft.

Now, none of that takes into account terminal velocity. Assuming your vt = 120 mph = 176 ft/s.

So taking the 2nd kinematic:

176^2 = 0 + 2*32.2*d
d = 30967/64.4 = ~481 ft

Now, how long did it take him to achieve that velocity?

d = vi*t + 0.5*a*t^2
481 = 0 + 0.5*32.2*t^2
t = sqrt(481/16.1) = ~5.47 s

So, that leaves ~64.53 s left to fall at a constant velocity.

d = ((vi + vf)/2) * t
d = 176 * 64.53 = 11,357.28

So, total distance traversed would be 11,838.28 ft.

 

[Welverin]

Of course all of this assumes that the entire fall was shown.

Of course the whole thing would have worked better and made more sense had the Balrog whipped and dragged Gandalf down immediately like he should have, but nooo.

 

[Black Omega]

I assumed that Gandalf was using a little subtle mojo to direct/impel himself in the direction and speed he required.

Considering how much magic is simply subtle little things like the clouds parting to show the moon at the right time, I assumed that was Gandolf just giving himself a subtle little push to make sure he caught his sword.