On Tue, Apr 1, 2014 at 6:20 PM, Steven D'Aprano <st...@pearwood.info> wrote: > On Tue, 01 Apr 2014 16:01:40 +1100, Chris Angelico wrote: > > [...] >>> The scenario you describe has (effectively) infinite rate-of-change-of- >>> acceleration, often called "jerk". (A jerk is a rapid change in >>> acceleration.) Human comfort is (within reasonable limits) more >>> affected by jerk than acceleration. The passengers will feel three >>> quite distinctive jerks, one when the brakes are first applied (which >>> is probably reasonable), then one at 1s, then again at 2s. That's not >>> comfortable by any stretch of the imagination. >> >> It actually is a smooth increase in deceleration, but I'm operating the >> simulator on a 1s period, so it's actually an average across the first >> second, and an average across the next second... > > Hmmm. A 1-second resolution doesn't really sound too great to me. > > Suppose the deceleration increases linearly from 0 to 0.85 m/s over two > seconds. Averaging it in the way you suggested, we get a total distance > of: > > 2*u - 0.5125 > > (see my previous post for details) where u is measured in metres per > second. Multiply by seconds to get the units right. For a Japanese bullet > train where u = 320 km/hr that corresponds to > > py> 2*88.888889 - 0.5125 > 177.26527800000002 > > metres. > > Now let's do it properly! Given our assumption that the deceleration is > linear, the jerk will be constant: > > j = Δa/Δt > = (0.85 - 0)/2 > = 0.425 m/s^3 > > Let's start integrating! > > py> from sympy import integrate > py> from sympy.abc import t > py> j = '0.425' > py> a = integrate(j, t) > py> v = 88.888889 - integrate(a, t) > py> s = integrate(v, (t, 0, 2)) > py> s > 177.211111333333 > > compared to 177.26527800000002 calculated the rough way. That's not bad, > only about 5cm off! Effectively, your rough calculation was accurate to > one decimal place.
5cm when we're dealing with track distances measured in meters and usually hundreds of them? Sounds fine to me! > Of course, if the equation for acceleration was more complex, the > approximation may not be anywhere near as good. And that's the bit I can't know. I've added a comment to the code and will toss it back to my brother - hopefully he'll either know the answer or know where to look it up / who to ask. > [...] >>> This becomes a simple question for the four standard equations of >>> motion: >>> >>> (1) v = u + at >>> (2) s = 1/2(u + v)t >>> (3) s = ut + 1/2(at^2) >>> (4) v^2 = u^2 + 2as >>> >>> Only (1) and (3) are needed. >> >> Okay, what's u here? Heh. > > They're *standard* equations of motion. Weren't you paying attention > through the, oh, three years of high school where they teach this? :-P They're certainly the standard equations, but I learned that: d = v₀t + at²/2 which is the same as you had, except for the names: d for distance, v₀ (V zero, that's a subscript zero if encodings or fonts kill you here) for initial velocity (ie velocity at time zero - vₜ (that's a subscript t, U+209C, but it isn't working in this font) is velocity at time t, which is v₀+at), and a and t are the same as you have. > s = displacement (distance) > t = time > u = initial velocity > v = final velocity > a = acceleration I have to say, the one-letter variables are a lot easier to work with. Guess I learned them the hard way, heh. Although remembering that v is velocity is easier than remembering which of u and v is initial and which is final. >> Fair enough. :) That's why I asked. Is it naughty enough to break into >> two statements, or is it better to combine it into a single multiple >> assignment? > > Since they are unrelated assignments, of two different variables, they > should be written as two separate assignments, not one using sequence > unpacking. Change made. Yeah, that looks a bit cleaner. Although I would prefer a simple formula to what I have there, and I'm still not perfectly sure it's accurate. Advice appreciated. :) ChrisA -- https://mail.python.org/mailman/listinfo/python-list
