The computation of the atmospheric trajectory, ground track and extrapolated ground level point of a fireball from eyewitness accounts

Jeremy B. Tatum

Note: Some browsers, like Netscape 4.6, will not display the Greek letters in the text correctly. However, they are OK in the equations.

1.   Introduction

Three methods of computing the atmospheric trajectory of a fireball from eyewitness  accounts are described, including the advantages and limitations of each.  The three methods will be referred to as Azimuths First, Plane and Ground, and Intersecting Planes.  It will be concluded that the last of these is far superior to the other two.  Initially it is assumed that there are only two witnesses, which is the minimum number necessary to obtain a trigonometric solution.  Following that will be a section on how to treat the observations from many more than two witnesses.

It is assumed that all of the following conditions are fulfilled:
Each witness is interviewed  1. In situ 
2. By a trained interviewer
3. Within a week of an event
The interviewer 4. Measures the angles indicated by a witness
5. With instruments
Witnesses are distributed 6. On both sides of the fireball, so that some witnesses see the fireball move from left to right, and others see it move from right to left.

   If all of these conditions are not satisfied, a trigonometric solution should not be attempted.  Any results arising from a calculation in which all of these conditions are not satisfied will be pure fiction and the computed atmospheric track and impact point will be grossly in error.   It should also be noted that witness reports (other than purely descriptive reports) from witnesses in moving vehicles should not be used in quantitative calculations.  In order of increasing unreliability of angular reports from moving vehicles are those from cars, boats and aircraft.

    In the calculations described below, a Flat Earth is assumed.  While each method can be modified to include the curvature of Earth, or the height of a witness above sea level, this is not at all justified - and it would be downright unscientific to include such minutiae - for eyewitness accounts, which typically have errors of several degrees or even tens of degrees.  A Flat Earth model assumes that the height of the fireball (typically tens of km) is negligible compared with the radius of Earth (6371 km), and that the height of a witness above sea-level  (at most in Canada, in the unlikely event that a witness was at the summit of Mount Logan and was also interviewed there, 5.95 km) is negligible compared with the typical fireball height of tens of km.  It is doubtful whether a departure from a Flat Earth model is justifiable even from a wide-field photograph, although this will depend on the quality and what landmarks or skymarks are identifiable on the photograph.

2.    Definition of terms

Words such as "the track", "the end-point" are frequently encountered when describing fireballs, yet it is unlikely that any two persons mean the same things by such terms, and unlikely that a speaker and a hearer or a writer and a reader understand the same things by them.

I therefore propose that the following terms be standardized.

3.   Notation

In preparing a calculation, an arbitrary ground-level origin of coordinates is chosen,  conveniently somewhat to the southwest of all witnesses.  Any point relative to this origin can then be represented in the usual way by coordinates (x, y, z), conveniently expressed in km.   I shall assume there are two ground-level witnesses whom, after the manner of Stephen Leacock, I shall refer to as A and B.  Their coordinates are, respectively (xA , yA , 0) and (xB , yB , 0).  That is to say, their heights, relative to the origin, are negligible, compared with the height of the fireball.  As explained in section 2, this assumption is justified within the uncertainties of eyewitness (and perhaps even photographic) records.

Witness A indicates the two points on the sky track, which correspond to two points on the atmospheric trajectory which I call K and L.   Witness B indicates two points on the sky track, which correspond to two points (not necessarily the same two points that were indicated by witness A) on the atmospheric trajectory which I call M and N.

The position of an altazimuth  point on the celestial sphere centred at a witness is to be denoted by the usual spherical coordinates (q , f).  Here q is the zenith distance, or the complement of the altitude.  The azimuth f is measured from the east point of the horizon counterclockwise (northward).  (In the original report from an interviewer, azimuth will probably be given as measured from the north point of the horizon clockwise (eastward); it will be necessary to change this for conformity with the usual mathematical convention for spherical coordinates.)  It is assumed that azimuth is given relative to true geographic coordinates and that any corrections for magnetic declination have been properly made.

(For the benefit of astronomers it might be mentioned here that in geophysics the term "magnetic declination" is used to describe the difference between true and magnetic north; this clearly is quite unrelated to the word "declination" as used in spherical astronomy.)

The directions to the four points indicated by the two witnesses are then given, in obvious notation, by

4.   Azimuths First

In this method, witness A indicates the directions to two points K and L in the usual manner, and  witness B indicates the directions to the same two points K and L.   Let K' and L' be the points at ground level vertically below K and L, which are on the atmospheric trajectory.   That is to say, K' and L' are on the ground track.  The distances of K' and L' from the two witnesses are then easily found simply by making a scale drawing as illustrated in figure 1, in which the full lines represent the azimuths of K' and L' relative to A and B, and the dashed line is the ground track.

Figure 1

This graphical method of locating K' and L' is so simple that many may not wish to do it by trigonometrical calculation.   However, others will wish to programme the calculation for automatic computation, and I therefore give here the coordinates of K' and L' in terms of the measured data.:



and for L', merely substitute L and L' for K and K' in these equations.

The distances a, b, c, d indicated in figure 1 are:


and similarly in an obvious manner for the others.

The x- and y-coordinates of K and L on the atmospheric trajectory are, of course, the same as those of K' and L', and their z-coordinates (heights) are


and     (5)

The two ways of calculating each of these serves as a check against arithmetic mistakes.

Assuming that K is higher than L (if it isn't the meteoroid is rising, not falling), the coordinates of the extrapolated ground level point O are



This point is, of course, on the ground track.  The impact point, also on the ground track, will fall short of this point by an amount that cannot be determined from geometric considerations alone.

While this method is simple and straightforward it has one very severe limitation, which renders its use invalid except in most unlikely circumstances, namely, it is essential that the two witnesses indicate the same two points on the atmospheric trajectory.  This might, on rare occasion, be possible, if there were two well-defined flares unambiguously marking two specific and unmistakable points on the atmospheric trajectory and the investigator is certain that each witness has indicated these same two points.   In practice it is almost never the case that two witnesses indicate the same two points.  The problem is infinitely compounded if there are many more than two witnesses. If the above analysis is carried out on the erroneous assumption that the two witnesses have indicated the same two points, the computed ground track will be totally erroneous and not even approximately correct.

The situation is illustrated in figure 2.  The dashed line is the true ground track.  K' and L' are the ground projections of the two points indicated by witness A, and M' and N' are the ground projections of the two points indicated by witness B.  The dash-dotted line is the erroneous ground track.  In the example (by no means an extreme one) of figure 2, the angle between the correct and erroneous ground tracks is 58 degrees.  In figure 2 we indicate the points K" and L", which are the quite erroneously-deduced positions of K' and L'.  We see further that the distances a, b, c, d, which should correctly be, respectively, AK', AL', BM', BN', will be erroneously taken to be AK", AL", BK", BL".  The heights deduced from these erroneous values will be absurdly wrong and the deduced extrapolated ground level point is likely to be many hundreds or even more than a thousand kilometres from the correct point and nowhere near the correct ground track.

Figure 2

5.   Plane and Ground

Witness A, situated at the point (xA , yA ), indicates the directions to two points, K and L, described by spherical coordinates ( qAK , fAK ) , ( qAL , fAL ).  The atmospheric trajectory of the fireball must lie in the plane that includes the witness and the vectors from the witness to K and L.  The equation to this plane is


This can be written in the form

PAx  +  QAy  +   RAz  +  1  =  0     (9)

and it intersects the ground in the line (not the ground track!)

Px +  QAy  +  1 =  0.    (10)

Similarly the plane defined by witness B intersects the ground in a line

 PBx  +  QBy   +  1 =  0.       (11)

The solution of equations (10) and (11) gives the coordinates of the extrapolated ground-level point.

It would be possible to write expressions for the coordinates of the ground-level point explicitly in terms of the original data, and the investigator might like to try to do this, though it is more likely that the entire procedure will be carried out numerically, step-by-step.

This method is simple and it has the great advantage over the Azimuths First method that it does not require that the two witnesses indicate the same two points.   While it does quickly produce the extrapolated ground-level point, it is to be remembered that the impact point falls on the ground track short of the extrapolated ground-level point, and the method has not determined the ground track, so that we do not know in what direction to search starting from the extrapolated ground-level point.

6.   Intersecting Planes

The Intersecting Planes method is similar to the Plane and Ground Method, except that it completes the calculation and determines not only the extrapolated ground-level point, but also the ground track, atmospheric trajectory, heights, and true angle of descent.  In effect, the Plane and Ground method is just a small part of the complete Intersecting Planes method.  Neither method requires that the two witnesses indicate the same two points on the atmospheric trajectory.  Thus, as previously described, witness A indicates points K and L, and witness B indicates points M and N.

As described in section 5, each witness and the directions to the indicated points describes a plane, the two planes being given by

PAx  +  QAy  +   RAz  +  1  =  0         (12)

and    PBx  +  Qy +   RBz  +  1  =  0.   (13)

Equation (12) is just equation (9) repeated, and the method of setting it up is described in section 5.  The atmospheric trajectory is in each of these planes; that is to say the atmospheric trajectory is the intersection of the two planes and indeed equations (12) and (13) are the equations to the atmospheric trajectory.   (For those who may not be conversant with algebraic geometry in three dimensions, a line in three dimensional space is always represented by two equations.)

As in section 5, the extrapolated ground-level point can be found by putting z = 0 in each of these two equations, to form equations (10) and (11), and solving these equations for x and y.  If instead of putting z = 0 we eliminate z between equations (12) and (13), we shall obtain a linear equation in x and y , and this equation is the equation to the ground track, which will include both the extrapolated ground-level point and the impact point.  As pointed out earlier, geometric considerations alone cannot determine the distance between these two points.   What is certain, at least, is that the impact point is between the extrapolated ground-level point and the more forward of L' or N', which are the ground-level points vertically below the second points for the two witnesses.
For any point (x , y) on the ground track, the corresponding height z on the atmospheric trajectory can be found from either (or, preferably, both, as an arithmetic check) of equations (12) or (13).  If Dx , Dy , Dz are the differences between the coordinates of any two points on the atmospheric trajectory, the true angle of descent y below the horizontal is given by


7.   Pre-encounter orbit

In order to determine the pre-encounter orbit (i.e. the six orbital elements) of the meteoroid, it is necessary to know the velocity of the meteoroid at a specified point before it is slowed down by the atmosphere.  That is to say, it is necessary to know the three components of its heliocentric velocity and the three components of its heliocentric position vector (corrected for both zenith attraction and aberration) before it is slowed down by impact with the atmosphere.  As soon as a fireball is visible and incandescent, its very incandescence indicates that its speed has already been reduced enormously.  It is not possible to determine, even approximately, the speed through the atmosphere, and still less the pre-encounter speed, from eye-witness accounts alone.  Any orbits so claimed are therefore entirely spurious.

The method of computing the pre-encounter orbit presuming that the position and velocity vectors before atmospheric entry are known, is outlined in a separate MIAC document by the author.

8.   Combination of Observations

I have hitherto described how to determine the atmospheric trajectory and related quantities from two witnesses, which is the absolute minimum number of witnesses necessary to determine the trajectory.  (In a separate document in the MIAC archives I have explained in detail why it is impossible to determine, even approximately, the direction of motion of a fireball from observations made from a single station, in spite of regular claims to have done so).
In this section I make some suggestions as to how best to combine observations from many witnesses scattered over a wide area. It is assumed that all conditions described in section 1 have been satisfied.

Typically the distribution of witnesses will be somewhat clumpy.  There are likely to be several witnesses within two or three km of each other in each town or village visited by the investigators.  (I have used the plural here - it order to obtain the necessary measurements from a wide area within a week, it is necessary to have a team of trained investigators.)  In addition there will be a few isolated witnesses from rural areas.  Each group of witnesses (whether of several people in a town, or a "group" of one individual in the countryside) must be counted as a single observation site and an average sky track for all the witnesses in that group must be determined.  The differences in apparent sky tracks reported by individuals within a few km of each other are not trigonometrical differences and must not be treated as such., but differences in recollection.

The altitudes and azimuths of the first and last points for each witness within a group can be plotted on graph paper and joined to show the several indicated sky tracks.  There will be appreciable variation among them, and, if one of more are obviously completely and unrealistically different from the general trend of the rest, they should probably at this stage be discarded.

The next stage is to draw a "least-squares sky track" through the remaining (undiscarded) first and second points for all the witnesses within the group.  The least-squares calculation is best carried out by a mathematically and statistically trained investigator.  Most "packaged" least-squares programs are, technically, least-squares "regressions of y upon x", and such a regression is entirely inappropriate here.  It should be assumed that the random errors in altitude and in azimuth are equally likely, so what has to be minimized is the sum of the squares of the perpendicular (not vertical) distances of each point from the computed regression line, or, more correctly, since we are dealing with the celestial sphere, the sum of the squares of the great circle distances of each point from the computed regression line.

This least-squares sky track then forms the average sky track for all the members within that (closely-spaced) group.  When combining this group with others, the group can be given a weight proportional to the square root of the number of witnesses in the group.  (It is my practice - after having done numerous interviews - not to attempt to assign a weight to individual witnesses based on a "gut feeling" of how reliable a particular witness was.  I  give zero weight to any witness in a moving vehicle, whether car, boat or aircraft, and perhaps to any witness who had obviously not seen the same thing as everyone else; but I have found no rational way to quantify my "gut feeling" - probably quite unfairly wrong - about the reliability of individual witnesses.)

We then have to "pair" each group on one side of the trajectory with every group in turn on the other side.  Groups on the same side of the trajectory must not  be paired with each other.  In the latter case the angles between the intersecting planes are small and the errors in determining the line of intersection are hopelessly large - an error of a degree in the angle of intersection being easily capable of shifting the ground track by thousands of kilometres or even placing the atmospheric trajectory beneath the ground.

If there are N1 groups on one side of the trajectory and N2 on the other, the number of pairings will be N1N2, so you will calculate N1N2 extrapolated ground-level points and N1N2 ground tracks.  A least-squares ground track from all of these, and a "centre of gravity" extrapolated ground-level point can then be calculated, giving a weight proportional to the geometric mean of the weights you have given to each of the two groups.
Having determined the "centre of gravity" extrapolated ground-level point, you then go there to search not for  meteorites but for additional witnesses.

9.   Concluding remarks

After a fireball, it takes one week to interview witnesses.   This is, in practice, a major effort, and it may well be objected that there are so many witnesses that it is impossible to interview them within a week.  That may well be so, but eyewitness accounts after a week are unreliable and should not be used in the calculations.  If photographs have been taken, however, the subsequent measurements can be made at leisure.  The calculation is long and must be done carefully, with a combination of the investigator's judgment - e.g. whether to reject particular witnesses, or how many groups to form, or how to pair the groups - and automatic computer calculation.  You cannot expect merely to put the raw observations into a computer program as data and wait for an instant unique solution. Even after all the data have been received from all the field interviewers (and it may take some time for them all to reach you) you should expect to take several days until you are entirely satisfied with your solution.  "Instant" solutions may give fame to their author, but are highly suspect.

10 January 2002 -- Copyright J Tatum