Good catch, Dr. G, NIST's fit is probably worthless. what follows is not exactly what you say in the other thread, but definitely on the same wavelength.
They're wrong.
NIST's choice of function to fit, as you say, is without physical meaning. I can forgive them on this point as their intent is apparently interpolation rather than explanation. However, that doesn't relieve them of the burden of doing everything else right. If they've interpolated their data correctly, and it looks like they have, then either their data is bogus, or...
They've been bitten by the old t(0) problem.
I've plotted their position function (scaled to meters) against freefall. The NIST y(t) curve is a solid teal line, and
g is solid red. Larger versions to follow in next post.
As you can see, freefall greatly outpaces their curve, maybe because their curve languishes for the better part of a second. The dotted red line represents freefall but with a t(0) at +0.7 seconds. This about splits the difference from about 3.5 seconds on, but freefall still initially outpaces the NIST curve noticeably for about the first two seconds.
Now, I'm going to add the last data I posted after reworking einsteen's smear. There are problems with this data because I mistakenly grabbed the cubic bezier
control points instead of the
anchor points - that's why it's so jaggy - but, hey, it's good enough for this. These data points are shown as magenta dots with a connecting line:
Almost indistinguishable from freefall starting at 0.7 seconds! But still very different from NIST. Can this be fixed? If you look at the dotted gray lines that have been added, it shows that the y values are equal at about 3 seconds and 3.7 seconds, respectively. How about I add 0.7 seconds to the time on my data?
Pretty good match from 3 seconds on, but my data is even steeper than theirs! Way faster than
g! How can that be when I can get a decent fit at less than
g even with a quadratic? I've added a blue line for this fit of y(t) = 0.112 - 4.808*t + 4.532*t^2:
It is a really good fit over most of the data but, as noted elsewhere, second degree fits do not capture the beginning well. The constant acceleration from this fit is about 9 m/s^2 but the initial velocity is about -5 m/s starting from a meter higher than the building stood! This is not real!
If the first two data points are removed from this set (the first is t0 defined as t0 = (0,0) and the time is arbitrary, anyway) and 0.7 seconds subtracted from the time, this is what the data looks like:
Of course, it looks like
g, that was already established above when the freefall curve was translated 0.7 seconds the other way. But there are two important things to note. While it looks like
g, it isn't necessarily so. The best second degree fit,
which is an excellent fit all the way around, shows a constant acceleration of about 9 m/s^2. The other thing is, when my curve was translated to better match NIST's curve, it looked steeper than theirs, except we see that it isn't from this different perspective.
Hmmm....
Let's chop 1.2 seconds off the front of the NIST y(t) function!
That's more like it. This is one of your points, Dr. G. Down in the mud of initiation, where a +/- 1 pixel accuracy counts for a lot more than in the later stages, it doesn't look like a parabola but, from t=2 forward, it parallels free fall pretty well. I'm not responsible for the funky choice of function, so I can't be blamed for how it looks in the first half second. Had NIST chosen a quadratic to fit their data, my sampling of their interpolation would look exactly like a parabola!
Had they chosen a later t(0), they may have opted for a quadratic fit after all, since they wouldn't have that long slow creep to try to fit with one function. As it is, a quadratic fits their exponential frankenstein fairly well at an acceleration of 9 m/s^2 - exactly what my data says. A comparison coming up, along with links to large versions of the graphs above.
