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Lumifer comments on Decoherence is Simple - Less Wrong
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Predictive power. The more accurate a prediction I can make without knowing the value of a given variable, the less important that variable is.
Ugh, imperial measures. Do you mind if I work with a five-centimetre sphere dropped from 60 metres?
A sphere is quite an aerodynamic shape; so I expect, for most masses, that air friction will have a small to negligible impact on the sphere's final velocity. I know that the acceleration due to gravity is 9.8m/s^2, and so I turn to the equations of motion; v^2 = v_0^2+2*a*s (where v_0 is the starting velocity). Starting velocity v_0 is 0, a is 9.8, s is 60m; thus v^2 = (0*0)+(2*9.8*60) = 1176, therefore v = about 34.3m/s. Little slower than that because of air resistance, but probably not too much slower. (You'll also notice that I'm not using the radius of the sphere anywhere in this calculation). It's an approximation, yes, but it's probably fairly accurate... good enough for many, though not all purposes.
Now, if I know the mass but not the shape, it's a lot harder to justify the "ignore air resistance" step...
You're doing the middle-school physics "an object dropped in vacuum" calculation :-) If you want to get a number that takes air resistance into account you need college-level physics.
So, since you've mentioned accuracy, how accurate your 34.3 m/s value is? Can you give me some kind of confidence intervals?
Yes, exactly. Because for many everyday situations, it's close enough.
No, I can't. In order to do that, I would need, first of all, to know how to do the air resistance calculation - I can probably look that up, but it's going to be complicated - and, importantly, some sort of probability distribution for the possible masses of the ball (knowing the radius might help in estimating this).
Of course, the greater the mass of the ball, the more accurate my value is, because the air risistance will have less effect; in the limit, if the ball is a hydrogen balloon, I expect it to float away and never actually hit the ground at all, while in the other limit, if the ball is a tiny black hole, I expect it to smash into the ground at exactly the calculated value (and then keep going).
And thus we get back to the question of what's important in physics equations.
But let's do a numerical example for fun.
Our ball is 5 cm in diameter, so its volume is about 65.5 cm3. Let's make it out of wood, say, bamboo. Its density is about 0.35 g/cm3 so the ball will weigh about 23 g.
Let's calculate its terminal velocity, that is, the speed at which drag exactly balances gravity. The formula is v = sqrt(2mg/(pAC)) where m is mass (0.023 kg) , g is the same old 9.8, p is air density which is about 1.2 kg/m3, A is projected area and since we have a sphere it's 19.6 cm2 or 0.00196 m2, and C is the drag coefficient which for a sphere is 0.47.
v = sqrt( 2 * 0.023 * 9.8 / (1.2 * 0.00196 * 0.47)) = 20.2 m/s
So the terminal velocity of a 5 cm diameter bamboo ball is about 20 m/s. That is quite a way off your estimate of 34.3 and we got there without using things like hollow balls or aerogel :-)
To be fair, a light ball is exactly where my estimate is known to be least accurate. Let's consider, rather, a ball with a density of 1 - one that neither floats nor sinks in water. (Since, in my experience, many things sink in water and many, but not quite as many, things float in it, I think it makes a reasonable guess for the average density of all possible balls). Then you have m=0.0655kg, and thus:
v = sqrt( 2 * 0.0655 * 9.8 / (1.2 * 0.00196 * 0.47)) = 34.0785 m/s
...okay, if it was falling in a vacuum it would have reached that speed, but it's had air resistance all the way down, so it's probably not even close to that. (And it it had been dropped from, say, 240m, then I would have calculated a value of close on 70 m/s, which would have been even more wildly out).
So, I will admit, it turns out that mass is a good deal more important than I had expected - also, air resistance has a larger effect than I had anticipated.