Oswald took 10.2 seconds to fire all three shots.

[Charles Collins]

I was responding to your statement that I was wrong when I said "The first cannot be reconciled with Altgens photo at z255.".  The "first" refers to Myers' web page which has W001 3.66 seconds before the head shot which puts W015 at z256-257.  I didn't say his "second" could not be reconciled with Altgens.  I just said it required the Cabell car to go from not being visible in Altgens to having completed the turn onto Elm in .76 seconds.
It is a bit difficult to stop Tyler's animation at the right spot.  He uses the 255.1 frame in the cover to his manual. I wasn't sure what frame you were using. There is not much difference, I agree. Besides, there is also some uncertainty in exactly when Altgens took his #6 photo - it could be z254-z256 based on Jackie's hand on JFK's right sleeve and the car-right flag flutter. I would suggest that z254-z255 is probably more accurate.
Myers' work appears to be, from an animation point of view, well done.  But the problems arise because it is not possible to check his work, being based on a model that he keeps secret and will not share.  He presents it as highly accurate but doesn't show any error range.  This is not the way a scientific analysis should be presented.  I would be interested to know how his model compares to the Knott Laboratory model created by using actual measurement data obtained by a laser scan of Dealey Plaza.

He presents it as highly accurate but doesn't show any error range.

On page 28 of the pdf Myers qualifies his error range for this project:

In synchronizing the amateur films for this project, an error ratio of plus-or-minus (+/-) one
frame (approximately one tenth of a second) was deemed acceptable, and therefore considered
synchronous.

Also, he does qualify the following on page 30 of the PDF:

Each of the nine amateur films used in this analysis was prepared from original sources.103

103. With the exception of the F.M. “Mark” Bell, and John Martin, Jr., films, which used the best available sources.

[Andrew Mason]

He presents it as highly accurate but doesn't show any error range.

On page 28 of the pdf Myers qualifies his error range for this project:

In synchronizing the amateur films for this project, an error ratio of plus-or-minus (+/-) one
frame (approximately one tenth of a second) was deemed acceptable, and therefore considered
synchronous.

Also, he does qualify the following on page 30 of the PDF:

Each of the nine amateur films used in this analysis was prepared from original sources.103

103. With the exception of the F.M. “Mark” Bell, and John Martin, Jr., films, which used the best available sources.

His "error ratio" is the lowest possible amount of error. He couldn't very well say to within less than a frame.   The cameras were not all operating at the same speed or making exposures at the same time. 

That is not a margin of error with respect to the measurements he uses to reach his conclusions: ie. positions of people and vehicles, speeds of vehicles, angles of vehicles and rates of change of angles, limitations of film resolution, uncertainties of film speed etc. 

In order to determine a margin of error you need to follow accepted rules. 

Example: You make a measurement 10 different times.  You then use an appropriate statistical method for determining the most likely value and the error range. Assuming the samples formed a normal distribution about a mean value one would use the average of those 10 readings and work out the standard deviation (the square root of the sum of the squares of the difference between each value and the mean).  The margin of error would be two standard deviations, which is the range within which 95% of the values occur.  That would then be the error range for that measurement. 

Of course, that assumes that you are not basing the measurements on assumptions that introduce further uncertainty. If you are making the measurements based on values that themselves have a margin of error, then you need to factor those error margins into your work.  And you need to show how you calculated these things by providing the data so that the reader can check your work.

[Charles Collins]

His "error ratio" is the lowest possible amount of error. He couldn't very well say to within less than a frame.   The cameras were not all operating at the same speed or making exposures at the same time. 

That is not a margin of error with respect to the measurements he uses to reach his conclusions: ie. positions of people and vehicles, speeds of vehicles, angles of vehicles and rates of change of angles, limitations of film resolution, uncertainties of film speed etc. 

In order to determine a margin of error you need to follow accepted rules. 

Example: You make a measurement 10 different times.  You then use an appropriate statistical method for determining the most likely value and the error range. Assuming the samples formed a normal distribution about a mean value one would use the average of those 10 readings and work out the standard deviation (the square root of the sum of the squares of the difference between each value and the mean).  The margin of error would be two standard deviations, which is the range within which 95% of the values occur.  That would then be the error range for that measurement. 

Of course, that assumes that you are not basing the measurements on assumptions that introduce further uncertainty. If you are making the measurements based on values that themselves have a margin of error, then you need to factor those error margins into your work.  And you need to show how you calculated these things by providing the data so that the reader can check your work.

I have a 10-shot string.  I have measured the velocity for each shot with a chronograph.  The velocities (fps) are: 834, 832, 828, 828, 832, 829, 833, 827, 829, 829. The standard deviation is indicated to be 2.3 by the chronograph software program. The average is indicated to be 830. It is actually 830.1 if done manually. The spread is indicated to be 7.

Given the above please show us how to calculate the margin for error. And be sure to show the calculations so we can check your work. Thanks…

[Andrew Mason]

I have a 10-shot string.  I have measured the velocity for each shot with a chronograph.  The velocities (fps) are: 834, 832, 828, 828, 832, 829, 833, 827, 829, 829. The standard deviation is indicated to be 2.3 by the chronograph software program. The average is indicated to be 830. It is actually 830.1 if done manually. The spread is indicated to be 7.

Given the above please show us how to calculate the margin for error. And be sure to show the calculations so we can check your work. Thanks…

The variance (mean square of the difference of values from the mean) is 5.29 so the standard deviation is 2.3 (square root of the variance).

[Charles Collins]

The variance (mean square of the difference of values from the mean) is 5.29 so the standard deviation is 2.3 (square root of the variance).

I don’t think that standard deviation is the same as margin for error. I am also struggling to understand how any of this applies to the accuracy of synchronizing the various films. If I remember correctly, you stated that Mark Tyler’s work is supposed to be with a margin of error of seven frames. Did Mark Tyler provide all the data and calculations demonstration that you are demanding of Dale Myers’ work?