Saturday, September 26, 2015

Mid-flight Speed - MH370

A mid-flight speed estimate for MH370 was made by Brian Anderson using a time of flight methodology. Brian's method can be found at the link below.


Since it is a slow week, and I can't stand to look at any more Maldives posts without getting upset, I decided to revisit this calculation using a different method than Brian. My method refers to the Doppler diagram below.




At 19:40 the satellite is virtually standing still at the extreme end of the "dither". V_s is essentially zero, and the Doppler shift associated with satellite movement relative to both Perth and the aircraft is negligible. All of the residual Doppler is the result of aircraft motion, D2 - D1 above.

Setting V_s = 0V_p = V_pt, and | R_s - R_p | ~ | R_e - R_p | in the expression for D2 and subtracting D1 results in the following:

D2 - D1 ~ V_pt dot (R_s - R_e) x FL / | R_e - R_p | x C

R_s - R_e  is a vector in the North - South direction of magnitude 1206 km. 

| R_e - R_p | is 36, 738 km (by calculation from satellite to 19:40 range ring)

Using the values above and the values for FL and C we have:

D2 - D1 = (V_pt dot z) x 0.18 (approximation valid near equator)

where V_pt dot z is the scalar component of aircraft tangential (to the earth) velocity in the North - South direction.  

Using a residual Doppler value at 19:40 of 37.8 Hz (D2 - D1), and solving for V_pt dot z yields:

V_pt dot z = 210 m/s = 756 km/hr = 408 knots.

408 knots is the slowest speed that the aircraft could have been traveling to produce a residual Doppler of 37.8 Hz at the 19:40 range ring. That speed would be at a heading of 180 degrees. Headings on either side of 180 degrees would require a higher speed to produce the 37.8 Hz residual Doppler, limited by the aircraft performance envelop.

This value is in general agreement with Brian's calculation. 

Update: 08/05/2016

The graphic below shows the mid-flight speed probability distribution calculated by the DSTG and published in their book "Bayesian Methods..." Figure 5.7. The highest probability speed calculated by the DSTG corresponds exactly with the 408 knot value calculated above.

It should not be inferred that the DSTG result and my result somehow reinforce  each other - although the most probable DSTG speed corresponding to zero BFO error at 180 degrees is identical to my result. The DSTG result is a probability distribution. My result is simply a determination of the minimum speed necessary to produce the 19:40 BFO observation. In fact, I would regard speeds below 400 knots in the above graphic not only to be unlikely, but physically impossible if the 19:40 BFO value is valid. It would require a significant bias drift to account for speeds less than 400 knots at 19:40. A vertical line at 180 degrees in the above figure would depict the probability function the DSTG used for BFO error.

Update: 9/11/16

Given the calculation of speed and heading required at 19:40 it is reasonable to ask what combination of speed and heading, if maintained, would produce an arrival time at 20:40 consistent with the distance to that range ring. The answer is about 425 knots at a heading of about 168 degrees. This answer assumes a late FMT with a latitude at 19:40 of about 8N.

It turns out this heading is directly at the Cocos waypoint. 168 degrees is also consistent with the recent work of Iannello and Godrey which postulates a terminus on the 7th arc around 27S using McMurdo Station, Antartica as a programmed destination.

Update: 9/12/16 

If one carefully extracts the likelihood relative magnitude vs speed in DSTG Figure 5.7 above it is possible to derive the BFO probability density function used by the DSTG in their modeling. This information is displayed graphically in the figure below. The data extraction yields a standard deviation (sigma) of approximately 5Hz. Track maintained at 180 degrees true. Speed is ground speed.