Speed of the butterfly effect Consider the simplest abstract model of a double pendulum. What is the maximum possible amount of change in the initial conditions so that the trajectory doesn't visibly change in the first minute, i.e. the maximum change in the angle is less than 1 degrees? I'm not looking for an exact figure, I just want to know the order. Is it in order of $10^{-3}$ degrees, or $10^{-15}$ or even less?
 A: The hallmark of chaos is exponential divergence of nearby trajectories. This means that if you start the system in state $\pmb{x}+\pmb{\epsilon}$ the divergence between the unperturbed trajectory $\pmb{f}_0(t)$ and the perturbed one $\pmb{f}_\epsilon(t)$ will grow exponentially as $\epsilon\rightarrow 0$: $||\pmb{f}_0(t)-\pmb{f}_\epsilon(t)||\propto e^{\lambda t}$ where $\lambda$ is the Lyapunov exponent. In fact, one can speak of several Lyapunov exponents describing the divergence of trajectories in different directions, but it is the largest exponent that describes how mixing the system is.  
So we can restate the question like: given a chaotic system with largest exponent $\lambda$, after time $T$, what is the initial deviation $\epsilon$ that produces a given deviation $\Delta$? The above gives $\Delta \approx \epsilon e^{\lambda T}$, or $$\epsilon \approx \Delta e^{-\lambda T}.$$
The problem is finding the Lyapunov exponent of the double pendulum. The exponent is energy dependent: if you start with a nearly vertical pendulum it will just be a perturbed non-chaotic pendulum, but with enough energy it will be chaotic. And of course, it depends on the length and mass of the pendulum and how you measure time. Plus, the exponents are averages over long time which means that for a particular trajectory the deviations behave differently. In short, published values rarely replicate. Anyway, this paper gives 7.5 or 7.9/s depending on whether it is simulation or measurement. This paper gives 0.179/s. This one 0.02-0.04. This one 0.41/s. This question has about 0.27. So I think any answer between 0.02 and 10 is possible for some configuration. 
Anyway, if we try with $\lambda=0.1, 1,$ and $10$, $T=60$ and $\Delta=1^\circ$
 I get $\epsilon=0.0024, 8.756\times10^{-27}$ and $2.65\times 10^{-261}$ degrees, respectively. The really small exponent 0.02 gives a modest 0.30 degree perturbation (for modest values of $\lambda T$ one can Taylor-expand the exponential and make a linear estimate).
In short, this is somewhat of a "how long is a piece of string?" question. The key thing to note is that it really matters how wildly the pendulum is swinging: if $\lambda T>1$ nearly any deviation becomes macroscopic. 
A: You would consider that the error would accumulate linearly here.
If you have 60 oscillations in the minute, and the divergence wanted as 1 degree, then it would travel through 60*360 degrees, making a divergence of 21600 degrees.  This would mean that an angular velocity of the order of 10 seconds, would not show for the first minute, and one of 2 seconds in the first 5 minutes.
Thus it's of the order of 10^-4 degree.
