Wednesday, August 20, 2014

The Fall of Gandalf

There are many out there who take issue with the Fall of Gandalf bit from Lord of the Rings: The Fellowship of the Rings. They argue that the Balrog's whip is too long, Gandalf couldn't have fallen the distance, etc. Read the link and you'll see their methodology. The math is fine, but the assumptions are not. Let's go over this with a different set of assumptions and see what it nets us. Did someone say math + LOTR? Let's do this!

So, we know that Gandalf fell for 17 seconds, and that he was head down (which is more aerodynamic than the spread eagle most skydivers perform, which is what the nerds used for the initial equations). This gives a terminal velocity of ~200 mph, or ~90 m/s. We'll find out some stuff with just this.

So, we have a terminal velocity equation for Gandalf of:

This gives us a coefficient of drag of 0.2847 for Gandalf. It will take him 9.2 seconds to reach this velocity. That means, in 17 seconds, we have a distance of:

So Gandalf has fallen less than a kilometer to catch up to the Balrog. This also tells us something about the Balrog's decent, since we know that, adding up the numbers for how long he fell (12 s until the whip, 30 s until Gandalf's fall, 17 s until Gandalf catches up), the Balrog fell 832.5m in 47 s.
Almost there . . . a little more math to go.

Here's where our conjecture comes in, but it'll be a little bit of it and should make sense. The equation, with unknowns, is:
​Now, we don't know what the terminal velocity and the time to it are. However, we can make some educated guesses. First, the terminal velocity is going to be nowhere near Gandalf's (or else Gandalf wouldn't have been able to catch up). Here's a list of some possibilities (vT = terminal velocity, vS = time to get there):

vT    vS
20    5.4
30    21.8
40    35.9

I'm going with 20 m/s because of the wings (feel free to do the math for any of the other possibilities). This also makes sense as the Balrog is made of fire, so he probably doesn't weight that much (otherwise he would essentially crush Gandalf when they sword fight). It could be even less while he's falling initially since he could open up his wings and drift like a parachute; I'll cover that in a sec.

So, assuming a terminal velocity of 20 m/s it would take the Balrog 5.4 s to get there. That makes the distance equation for falling for the initial 12 s:

This would put the Balrog's whip at about 1.5 football fields, which is still ridiculously long; even accounting for some magic stretching which we might expect.

It also means that Gandalf slams into the Balrog with a closing speed of about 150 mph, which would be fatal (or, at least, catastrophic). But these are fairly simplistic calculations for bodies falling. We need to take some other things into account.

The Balrog has wings. While he obviously can't use them to fly, it's silly to assume they do nothing. If he could fly, he would have flown back up and incinerated Gandalf, the Hobbits, et al in a barrage of fiery doom then gone on to find the descendant of Smaug and had an all out, aerial death battle. Now I wish that happened.

I digress. The Balrog had wings, and we can't assume they're useless. If they were to, say, act like a parachute, he would have an almost immediate terminal velocity of ~5 m/s, or about 11 mph. If you watch the clip again, you can see that the Balrog initially falls slowly, then falls rather quickly once he hits Gandalf. It's reasonable to assume that the Balrog flared his wings out in order to get one, last ditch shot at his foe. 5 m/s for 12s is 60m, but that's the distance his feet travelled. He looks to be some 4 m high with arms that are about 2 m, so that gives us 6 m less, give or take. So we're talking a whip length of about 54 m.

The actual fall time is more like 8s, because the initial 4s is more the bridge crumbling (again, watch the clip). So a spread-wing fall could be more on the order of 40m, giving a whip length of some 34 m give or take. Now, this may still be longer than it has a right to be, but the whip did appear out of seemingly nothing, so there's obviously some magic involved. It doesn't take that much suspension of belief to say that the whip could expand from, say, an original length of some 20 m to a bit over 30 m.
I could buy this whip being >30m

Also, once the Balrog saw Gandalf in pursuit, he could have furled his wings for a more aerodynamic profile and thus sped away quicker. Gandalf as well could have splayed out before getting to the Balrog to slow his descent.

I'm not going to do the calculations for all that, but it's entirely reasonable that the scene went down just as we saw it. To say it's physically impossible is to assume, I think, more than one has a right to.

Thursday, February 13, 2014

The Issue with Math

OK, got this sorted and ready to go.  Enjoy!

Mathematicians can do some amazing things, but some have been accused of living too much in the world of theory rather than fact. Here's a neat video:

Basically, it says that:

1-1+1-1+1-1+1-1+1... = ½

Here's the "proof," in a nutshell:

1-1+1-1+1-1+1... = S

1-S = 1-(1-1+1-1+1-1+1-1+1...)

1-(1-1+1-1+1-1+1-1+1...) = 1-1+1-1+1-1+1... = S


1-S = S

1 = 2S

S = ½

Pretty straightforward, right? Plus we know the series oscillates between 0 and 1, so a number of 1/2 makes sense (as it would be the average, as it were).

Not so fast, Mathicus Maximus!
It's a new superhero I'm working on.
The issue here lies in treating S like a natural number; their premise in finding S assumes that S indeed exists and is not simply conceptual. This is a problematic assumption, which I'll demonstrate.

So, let's do us some math wrangling. Buckle up; it's math time!
Please, Hammer, don't infinite Riemann sum them.
It's assumed in the video that one can take S and perform basic arithmetic on it like an algebraic variable. Let's apply some of the same logic and play with the equation some more.

First, let's take 2S. In the video, this is treated like 2*S, thus we can solve:
2S = 1
S = ½

OK, all fair and good. Or is it? Let's do some fandangling with 2S and see what we can see.

S = 1-1+1-1+1-1...
2S = 2(1-1+1-1+1-1...)
2S = 2-2+2-2+2-2...

OK, here's the fun part:
2-2+2-2+2-2... = (1+1)-(1+1)+(1+1)-(1+1)...
(1+1)-(1+1)+(1+1)-(1+1)... = 1+1-1-1+1+1-1-1...
1+1-1-1+1+1-1-1... = (1+1-1-1)+(1+1-1-1)+(1+1-1-1)...
(1+1-1-1)+(1+1-1-1)+(1+1-1-1)... = (1-1+1-1)+(1-1+1-1)+(1-1+1-1)...
(1-1+1-1)+(1-1+1-1)+(1-1+1-1)... = 1-1+1-1+1-1+1-1...

And, by definition:
1-1+1-1+1-1+1-1... = S

2S = S

If this is true (which it is, based on the definition of S and the logic used in the video), then we have:
1-S = S = 2S
1-S = 2S
1 = 3S
S = 

We could get any number of possibilities for S as logic dictates that nS = S (n being 1 or an even integer, at least; haven't tried it with odd integers).

The video treats the equation 1-S = S like we can algebraically manipulate it, but we can't do that and still get real results. This is because they assume that there exists a value S and then go about using that value, but they never actually prove that S exists. If we try the same thing with other abstract concepts/numbers, like 0, we get:

4(0) = 8(0)
4(0)/0 = 8
0/0 = 8/4 = 2
0 = 2(0)

Now to substitute 0=2(0) in our original equation. We still haven't committed any math ills, here.
4(2(0)) = 8(0)
2(0) = 8(0)/4

Now let's substitute again.
2(0) = 8(2(0))/4
0/0 = 16/8 = 2
1 = 2

In this way, we have "proven" that 1 = 2.

Since this is false, one or more of our premises or calculations must be false. We know that it's the division by 0; you can't take 0/0 = 1. The premise from the video commits a similar math sin. Namely, it assumes that S+S = 2S (that S is a number which can be added like that).

However, we have shown that S+S = S, and is in keeping with how the sequence is defined. We cannot simply treat S as a number or we run into the same issue we did above with 0. The defined definition of S will not allow us to simply manipulate it like an algebraic variable. We arrived at the conclusion that 1-S = S through non algebraic means; it was part of the concept of what S is. Similarly, we arrived at the equally valid conclusion that 2S = S. There is obviously more going on here than simple addition or multiplication when we perform arithmetic with it. Therefore, we must either reject the premise that S = ½, or we must accept that S = ¼, or any other value we can manipulate it to give.