The 9/11 Forum

T_Szamboti wrote:If the bus frame is stiff it can be nearly the same. It all depends on how much deflection there is between the driver and the back of the bus.

Exactly. Rigid body means the points in the body remain a fixed distance under application of force. d'Alembert's principle in action. This is why I specified that both buses (head-on collision, I forgot to mention) crumple a meter or two each. Not rigid. Principle does not hold.

Here the bus drivers get creamed. 'Instant' stop for both of them. Meantime, the buses crumple and the rear ends continue to displace through the crumple distance. The passengers in the back each have the crumple distance in which to come to a stop. The same momentum change occurs (full speed to rest) for these passengers, but the distance over which this occurs is much greater than for the drivers. The average acceleration on the passengers is therefore much less, so perhaps they live.

OneWhiteEye wrote:

T_Szamboti wrote:If the bus frame is stiff it can be nearly the same. It all depends on how much deflection there is between the driver and the back of the bus.

Exactly. Rigid body means the points in the body remain a fixed distance under application of force. d'Alembert's principle in action. This is why I specified that both buses (head-on collision, I forgot to mention) crumple a meter or two each. Not rigid. Principle does not hold.

Here the bus drivers get creamed. 'Instant' stop for both of them. Meantime, the buses crumple and the rear ends continue to displace through the crumple distance. The passengers in the back each have the crumple distance in which to come to a stop. The same momentum change occurs (full speed to rest) for these passengers, but the distance over which this occurs is much greater than for the drivers. The average acceleration on the passengers is therefore much less, so perhaps they live.

I was thinking of a school bus with an engine in front. You are apparently discussing a bus with the driver right at the front, with no benefit from the crumple. There isn't any question that whatever is behind a crumple will receive less shock than that material which is right at the point of impact.

The question is to determine how much deflection there is to see how much shock is transmitted to locations away from the immediate area of impact. It is not insignificant in items which have some level of rigidity like vehicles, buldings, aircraft, etc. In a velocity shock drop test the number of g's can be estimated by the equation G = 2 x drop height/deflection.

I'm not sure if he's made the comment in this thread yet or not (I don't rightly recall, and I don't feel like digging through it at this time), and I can't recall having come across anybody else correct him.

However T_Szamboti has on at least two seperate occasions in two seperate places made the claim that using the differences between successive velocity measurements was using the third derivative, and that you should use a=2d/t^2 instead.

First off, the formula is wrong, and calculates the wrong values for acceleration. I'm fairly sure that the formula that Tony was thinking of is x=(at^2)/2 + (v_0t) +x_0.

Secondly, the formula, as posted requires foreknowledge of whether you're dealing with acceleration or deceleration so that you can assign the appropriate sign to it, unless the acceleration persists long enough to result in a negative net displacement.

As proof below is a table of data generated, with T_Szamboti's acceleration formula.
The data represents 43 seconds in total of data, and is generated by accelerating an object at 9.81m/s/s for 19 seconds, 0m/s/s for 19 seconds, and -57.88 m/s/s.

Code: Select all

t   d   v   a (calc)   a(actual)
0   0   0   0       0
1   4.91   9.81   9.81   9.81
2   19.62   19.62   29.43   9.81
3   44.15   29.43   49.05   9.81
4   78.48   39.24   68.67   9.81
5   122.63   49.05   88.29   9.81
6   176.58   58.86   107.91   9.81
7   240.35   68.67   127.53   9.81
8   313.92   78.48   147.15   9.81
9   397.31   88.29   166.77   9.81
10   490.5   98.1   186.39   9.81
11   593.51   107.91   206.01   9.81
12   706.32   117.72   225.63   9.81
13   828.95   127.53   245.25   9.81
14   961.38   137.34   264.87   9.81
15   1103.63   147.15   284.49   9.81
16   1255.68   156.96   304.11   9.81
17   1417.55   166.77   323.73   9.81
18   1589.22   176.58   343.35   9.81
19   1770.71   186.39   362.97   9.81
20   1957.1   186.39   372.78   0
21   2143.49   186.39   372.78   0
22   2329.88   186.39   372.78   0
23   2516.27   186.39   372.78   0
24   2702.66   186.39   372.78   0
25   2889.05   186.39   372.78   0
26   3075.44   186.39   372.78   0
27   3261.83   186.39   372.78   0
28   3448.22   186.39   372.78   0
29   3634.61   186.39   372.78   0
30   3821      186.39   372.78   0
31   4007.39   186.39   372.78   0
32   4193.78   186.39   372.78   0
33   4380.17   186.39   372.78   0
34   4566.56   186.39   372.78   0
35   4752.95   186.39   372.78   0
36   4939.34   186.39   372.78   0
37   5125.73   186.39   372.78   0
38   5312.12   186.39   372.78   0
39   5498.51   186.39   372.78   0
40   5655.96   128.51   314.9   -57.88
41   5755.53   70.63   199.14   -57.88
42   5797.22   12.75   83.39   -57.88

The column labled a(calc) is the acceleration calculated using T_Szambotti's a=2d/t^2 and a(actual) represents the actual accelerations I applied to generate the displacement data in the first place (which represents total displacement in meters).

I think you should check your calcs.

If I'm reading what you're saying correctly.

at t =2 for example you have d=19.62 therefore as (v_0t) and x_0t both = 0 they drop out of the equation and the solution is, 2*19.62/4 which equals 9.81. not 19.62 as you have in the a(calc) column for t=2.

no_body wrote:I think you should check your calcs.

If I'm reading what you're saying correctly.

at t =2 for example you have d=19.62 therefore as (v_0t) and x_0t both = 0 they drop out of the equation and the solution is, 2*19.62/4 which equals 9.81. not 19.62 as you have in the a(calc) column for t=2.

I was hoping that someone would raise this point.
You're saying that the formula should be applied to the total displacement, rather than the displacement between seconds, however doing so yields an average acceleration between T_t and T_0, because the formula that T_Szambotti is suggesting that we use is only valid on the proviso that the acceleration is constant.

Applying the formula in the manner you're suggesting yields the following results:

Code: Select all

a(Cal-alt)
0
9.81
9.81
9.81
9.81
9.81
9.81
9.81
9.81
9.81
9.81
9.81
9.81
9.81
9.81
9.81
9.81
9.81
9.81
9.81
9.79
9.72
9.63
9.51
9.38
9.24
9.1
8.95
8.8
8.64
8.49
8.34
8.19
8.04
7.9
7.76
7.62
7.49
7.36
7.23
7.07
6.85
6.57

Which are still wrong and still fail to show the deceleration in the data.

(I meant to post this sooner in anticipation of this very point being raised, but was waylaid with RL stuff).

Yes you get the 'average' acceleration doing it with that formula, the change in velocity over time,

Deceleration in which data?

no_body wrote:Yes you get the average acceleration doing it with that formula,

Deceleration in which data?

Deceleration in any data.

The point is that if you're looking for deceleration in (for example) OWE's displacement data, using Tony's formula to calculate the acceleration directly from the displacement data isn't going to work. The only way to do it is to judge the instaneous acceleration from a v-t diagram, which is precisely what he was complaining about people doing.

The point was that using that particular formula only works if the object is initially at rest (hence posting the correct formula, which requires calculation of at least one value for velocity), only provides an average acceleration over the time period being considered, and can not, on its own, tell you if the object is accelerating or deccelerating.

Consider this data:

Code: Select all

t   s   v   a(calc)   a(actual)
0   0   0      
1   4.91   9.81   9.81   9.81
2   19.62   19.62   9.81   9.81
3   44.15   29.43   9.81   9.81
4   78.48   39.24   9.81   9.81
5   122.63   49.05   9.81   9.81
6   176.58   58.86   9.81   9.81
7   240.35   68.67   9.81   9.81
8   313.92   78.48   9.81   9.81
9   397.31   88.29   9.81   9.81
10   490.5   98.1   9.81   9.81
11   593.51   107.91   9.81   9.81
12   706.32   117.72   9.81   9.81
13   828.95   127.53   9.81   9.81
14   961.38   137.34   9.81   9.81
15   1103.63   147.15   9.81   9.81
16   1255.68   156.96   9.81   9.81
17   1417.55   166.77   9.81   9.81
18   1589.22   176.58   9.81   9.81
19   1770.71   186.39   9.81   9.81
20   1962   196.2   9.81   9.81
21   2163.11   206.01   9.81   9.81
22   2374.02   215.82   9.81   9.81
23   2594.75   225.63   9.81   9.81
24   2825.28   235.44   9.81   9.81
25   3065.63   245.25   9.81   9.81
26   3315.78   255.06   9.81   9.81
27   3575.75   264.87   9.81   9.81
28   3845.52   274.68   9.81   9.81
29   4125.11   284.49   9.81   9.81
30   4414.5   294.3   9.81   9.81
31   4713.71   304.11   9.81   9.81
32   5022.72   313.92   9.81   9.81
33   5341.55   323.73   9.81   9.81
34   5670.18   333.54   9.81   9.81
35   6008.63   343.35   9.81   9.81
36   6356.88   353.16   9.81   9.81
37   6714.95   362.97   9.81   9.81
38   7082.82   372.78   9.81   9.81
39   7460.51   382.59   9.81   9.81
40   7848   392.4   9.81   9.81
41   8088.35   88.29   9.62   -304.11
42   8181.54   98.1   9.28   9.81
43   8284.55   107.91   8.96   9.81
44   8397.36   117.72   8.67   9.81
45   8519.99   127.53   8.41   9.81
46   8652.42   137.34   8.18   9.81
47   8794.67   147.15   7.96   9.81
48   8946.72   156.96   7.77   9.81
49   9108.59   166.77   7.59   9.81
50   9280.26   176.58   7.42   9.81
51   9461.75   186.39   7.28   9.81
52   9653.04   196.2   7.14   9.81
53   9854.15   206.01   7.02   9.81
54   10065.06   215.82   6.9   9.81
55   10285.79   225.63   6.8   9.81
56   10516.32   235.44   6.71   9.81
57   10756.67   245.25   6.62   9.81
58   11006.82   255.06   6.54   9.81
59   11266.79   264.87   6.47   9.81
60   11536.56   274.68   6.41   9.81
61   11816.15   284.49   6.35   9.81
62   12105.54   294.3   6.3   9.81
63   12404.75   304.11   6.25   9.81
64   12713.76   313.92   6.21   9.81
65   13032.59   323.73   6.17   9.81
66   13361.22   333.54   6.13   9.81
67   13699.67   343.35   6.1   9.81
68   14047.92   353.16   6.08   9.81
69   14405.99   362.97   6.05   9.81
70   14773.86   372.78   6.03   9.81
71   15151.55   382.59   6.01   9.81
72   15539.04   392.4   6   9.81
73   15936.35   402.21   5.98   9.81
74   16343.46   412.02   5.97   9.81
75   16760.39   421.83   5.96   9.81
76   17187.12   431.64   5.95   9.81
77   17623.67   441.45   5.94   9.81
78   18070.02   451.26   5.94   9.81
79   18526.19   461.07   5.94   9.81
80   18992.16   470.88   5.94   9.81
81   19467.95   480.69   5.93   9.81
82   19953.54   490.5   5.94   9.81
83   20448.95   500.31   5.94   9.81
84   20954.16   510.12   5.94   9.81
85   21469.19   519.93   5.94   9.81
86   21994.02   529.74   5.95   9.81
87   22528.67   539.55   5.95   9.81
88   23073.12   549.36   5.96   9.81
89   23627.39   559.17   5.97   9.81
90   24191.46   568.98   5.97   9.81
91   24765.35   578.79   5.98   9.81
92   25349.04   588.6   5.99   9.81
93   25942.55   598.41   6   9.81
94   26545.86   608.22   6.01   9.81
95   27158.99   618.03   6.02   9.81
96   27781.92   627.84   6.03   9.81
97   28414.67   637.65   6.04   9.81
98   29057.22   647.46   6.05   9.81
99   29709.59   657.27   6.06   9.81
100   30371.76   667.08   6.07   9.81
101   31043.75   676.89   6.09   9.81
102   31725.54   686.7   6.1   9.81
103   32417.15   696.51   6.11   9.81
104   33118.56   706.32   6.12   9.81
105   33829.79   716.13   6.14   9.81
106   34550.82   725.94   6.15   9.81
107   35281.67   735.75   6.16   9.81
108   36022.32   745.56   6.18   9.81
109   36772.79   755.37   6.19   9.81
110   37533.06   765.18   6.2   9.81
111   38303.15   774.99   6.22   9.81
112   39083.04   784.8   6.23   9.81
113   39872.75   794.61   6.25   9.81
114   40672.26   804.42   6.26   9.81
115   41481.59   814.23   6.27   9.81
116   42300.72   824.04   6.29   9.81
117   43129.67   833.85   6.3   9.81
118   43968.42   843.66   6.32   9.81
119   44816.99   853.47   6.33   9.81
120   45675.36   863.28   6.34   9.81


The acceleration calculated using the 'perscribed' method gives absolutely no indication that there's a 31g jolt of duration 1 second lurking in the displacement data.

(So yes Tony, calculating the acceleration that way will show no jolt - even where a jolt exists, because calculating the acceleration that way is largely incapable of showing you a Jolt).

BTW the average acceleration of this data is 0.7g.

Yes but there is no deceleration in the displacement data which ever way you do it.

Taking velocity deltas over the displacement data only shows acceleration. Well in the displacement data I've got anyhow. I've not seen the OWE data, the data I've recorded is for every frame of video over the first 44m of the collapse and the velocity delta calculation shows no deceleration occurs.

Taking a difference of velocity between two points in time that results in a positive number when divided by delta t doesn't necessarily mean you're decelerating, it depends on what your acceleration was before you did the measurement. And so as Tony says what you are then effectively doing is the third derivative, the rate of change in acceleration, not acceleration. If however the positive number you calculate results in the acceleration changing from negative values to positive, then you are decelerating, but this would be seen in the acceleration curve anyway, so you only need the acceleration curve in the first place to see that there is no deceleration. That is if the instantaneous acceleration is enough to make acceleration positive only then are you decelerating.

I also think it would be better if you used the actual displacement data rather than running your program for 120 secs at 1 second intervals (If I'm reading your calculation correctly).

no_body wrote:Yes but there is no deceleration in the displacement data which ever way you do it.

If you haven't looked at or seen OneWhiteEye's data, then how can you honestly make this statement?

no_body wrote:Taking velocity deltas over the displacement data only shows acceleration. Well in the displacement data I've got anyhow. I've not seen the OWE data, the data I've recorded is for every frame of video over the first 44m of the collapse and the velocity delta calculation shows no deceleration occurs.

Technical notes on video motion analysis

no_body wrote:Taking a difference of velocity between two points in time that results in a positive number when divided by delta t doesn't necessarily mean you're decelerating, it depends on what your acceleration was before you did the measurement.

In the context of the post that you're replying to, this makes little or no sense.
I'm telling you that the second data set that I posted is for an object free falling for 2 minutes with a 31g deceleration introduced for a duration of 1 second, at the 41 second mark. I'm also telling you that applying your method for calculating the acceleration doesn't show the jolt, even though I know it's there because I put it there.

Both of these sets of displacement data are data sets that I know have decceleration and jolts in them, because I put them there, and neither of them show the decceleration or jolts calculating the acceleration using your method.

no_body wrote:And so as Tony says what you are then effectively doing is the third derivative, the rate of change in acceleration, not acceleration. If however the positive number you calculate results in the acceleration changing from negative values to positive, then you are decelerating, but this would be seen in the acceleration curve anyway, so you only need the acceleration curve in the first place to see that there is no deceleration. That is if the instantaneous acceleration is enough to make acceleration positive only then are you decelerating.

Maybe i'm just having a bad day, but this doesn't make a whole lot of sense to me either.
Look, I have a data set that represents the displacement in each time interval.
I then calculate the velocity that represents by calculating the change in displacement with respect to the change in time, this then gives me the average velocity between the time intervals, or the rate of change in displacement.
If I then use those consecutive points in the resultant v-t diagram to calculate the acceleration, i'm not calculating the rate of change of the acceleration, i'm calculating the change in the rate of change in displacement across those three points, which, given that constant acceleration give a parabolic curve in a displacement-time graph, and you need at least 3 points to define a curve...

But, if you don't believe me, do a dimensional analysis for yourself.
Displacement = Pixels.
Velocity = (pixels-pixels)/(frames-frames) = pixels/frame
Acceleration = ΔVelocity/ΔTime = V2-V1/t2-t1 = pixelsperframe - pixelsperframe/(frames-frames) = Pixels per frame per frame.

no_body wrote:I also think it would be better if you used the actual displacement data rather than running your program for 120 secs at 1 second intervals (If I'm reading your calculation correctly).

Do you really?
Then you haven't understood the point of posting the TEST DATA, which is ACTUAL DISPLACEMENT of a hypothetical object undergoing various acceleration regimes, and I'm wasting my time.

The point of the test data you're complaining about is to illustrate that even in data where a jolt exists (and I know it exists, because I explicitly and deliberately engineered the test displacement data to include one) calculating the acceleration using 2d/t^2 CAN NOT ever show the jolt for what it is, or it's actual magnitude, and CAN NOT show deceleration.

In fact, including a 31g jolt makes my acceleration data behave in exactly the same way as it does in one of the graphs in your essay.

The original means of calculating velocity in the Missing Jolt was discussed here and subsequently revised in the paper. It was a monotonic increasing function so incapable of showing deceleration. Adjacent points would be best but the current method is a big improvement.

OneWhiteEye wrote:The original means of calculating velocity in the Missing Jolt was discussed here and subsequently revised in the paper. It was a monotonic increasing function so incapable of showing deceleration. Adjacent points would be best but the current method is a big improvement.

Awesome Sauce (although, we were debating the calculation of acceleration).

Using symmetric differentiation of distance fallen vs. time (Dn+1-Dn-1)/(Tn+1-Tn-1) is the first derivative, used to find velocity. This method would show a deceleration, if there was one.

If the distance fallen is shorter in a given increment of time, than it was in the previous equivalent increment of time, then a velocity loss and hence a deceleration would have occurred. Velocity is still positive overall, unless there is a bounce.

The only thing needed to show whether or not deceleration occurred is whether or not a velocity loss occurred and that only requires the use of the first derivative.

Differentiating again to find acceleration would be the second derivative, and it isn't necessary to show whether or not deceleration occurred, so I don't understand why anyone is even talking about the third derivative.

So what is up with these little jolts?

Do they exist or not? Trippy, OWE, what are your current opinions?


On transmission of shock through a medium T Szamboti wrote:

I was thinking of a school bus with an engine in front. You are apparently discussing a bus with the driver right at the front, with no benefit from the crumple. There isn't any question that whatever is behind a crumple will receive less shock than that material which is right at the point of impact.

The question is to determine how much deflection there is to see how much shock is transmitted to locations away from the immediate area of impact. It is not insignificant in items which have some level of rigidity like vehicles, buldings, aircraft, etc.

Imagine if the shock is administered along some base spandrel plate of a large sheet of perimeter columns moving downwards. Or the shock is given by slamming the outer connections of flooring attached to the moving perimeter sheet just inside the sheet.

We would expect the shock to be transmitted to the top of the moving spandrel plate without much loss.

Unlike a bus, large unbuckled stiff perimeter sheets may show more detailed movement from jolts along their upper edges (the roofline).


The upper block perimeter moves down, spandrels hammer the floor connections one by one and we see these little jolts along the roofline. Maybe?

Major_Tom wrote:So what is up with these little jolts?

Do they exist or not? Trippy, OWE, what are your current opinions?

Purple. Or should I say 42.

If I were to give an honest opinion I would say that the potential remains for a jolt of some kind 'lurking' in the data. It's possible to fit a reduced jolt to Szamboti & Mc Queen's data, however their data lacks sufficient resolution to determine if it's genuine or not, however DBB has made the point that OWE's higher resolution data shows no indication of a jolt at that particular location, and Szamboti has never adressed the issue (that I can recall).

What I have at the moment is a spreadsheet with femr2, OWE, and Szamboti & McQueens data on it, and a prediction of what would happen if a freely falling object was to be subjected to 6g of upwards force for 900ms every 3.6m of total displacement. I can honestly the say the result looks nothing like what I had expected based on posts in this forum, and predictions made in papers.

What I would appreciate is if femr2, OWE, and Szamboti could work out between them some common feature for synching their data streams up so that I can work with the 4 data sets (and any others produced) and actually compare apples with apples.

The other thing I would like to see (and DBB might be the best bet here) is the displacement predicted by the current model (is it BV&L?) I've seen DBB make several posts that suggest that he has access to at least partial data along this lines, and it would be great if this could be synched up with one of the sets of measured data.

I would also consider using nobody's data, if he's willing to make his raw numeric data available.