I don't think they explicitly said that. Why do you see that suggestion there?Interesting to compare the resultant vector distribution with what the Russians presented:
View attachment 13373
Theirs seems to suggest that the sideways fragments are traveling the fastest.
In the diagram.I don't think they explicitly said that. Why do you see that suggestion there?
Question : Could you possibly display the angle that the missile is pointing at ?
That makes it easier to set incoming missile angle accurately.
Page 21 has a map with the last FDR point indicated, which match with GPS coordinates 48.127997°N, 38.526789°E.External Quote:The data on the Flight Data Recorder indicated that the aircraft was flying at FL330, on a constant displayed heading of 115° and at a constant speed of 293 kts computed airspeed (groundspeed 494 kts, equals 915 kilometers per hour
Correct me if I am wrong but from what I have read it appears that the missile's trajectory in the horizontal plane is not going to be straight. The missile uses a proportional navigation system which would put the missile onto a straight line intercept IF both the missile and the plane are travelling at constant speeds. In the presentation we are told that MH17 was maintaining a constant speed, but the missile started at 0 and went supersonic. This means that the path of the missile is going to curve slightly as it adjusts its trajectory in order to maintain a constant angle of line of sight to the target as it accelerates.
Can we consider the claim that the missile would travel to the 777 using exotic angles, curves or route modifications to be unfounded?
Which leaves little doubt where that "secret" report from last month originated from...
Interesting to compare the resultant vector distribution with what the Russians presented:
View attachment 13373
Theirs seems to suggest that the sideways fragments are traveling the fastest. If they all start out the same then this is wrong. Those segment are propelled backwards relative to missile, so they end up going slower (as seem in mine).
Imagine you are in a car at 90 mph throwing rocks at street signs. If you throw forward, your 30 mph throw is combined with the speed of the car, and it hits the sign at 120 mph. You can throw backwards and still hit it, but you'll only hit it at 60 mph. Same kind of thing here.
So it looks to me like their diagram is kind of from the point of view of the missile and not the plane. So either I'm missing something, or that's crazy wrong.
I fail to see the benefit of optimising the system for a situation where the target maintains constant speed and trajectory. This may be effective if it was designed to take down commercial aircraft but one might expect military jets to attempt some sort of evasive action more often than not. There seems little point in making sure the missile flies in a straight line toward the anticipated intercept immediately from launch as it seems probable that the target will change course.But wasn't the whole point of the BUK that it had all those radar and targeting systems? That would mean the BUK software very likely could calculate, as default, the shortest route to estimated point of impact using the vector, target speed and direction but of course also knowing its own acceleration. This would save the most fuel, increase the range and decrease the time to impact and limit reaction times. Quite important if one tries to get to fighter jets or other missiles.
As @Rob says, on the basis of what we know about the movement of MH17, there is nothing to indicate that the missile needed to resort to "exotic angles" in order to find its target, but gentle curves and route modifications can't be entirely ruled out and may impact the accuracy of the extrapolation by a few degrees.Can we consider the claim that the missile would travel to the 777 using exotic angles, curves or route modifications to be unfounded?
It's a bit tricky to assert your own vector distribution as equally authoritative overruling the one supplied by the manufacturer. What's the exact base for that apart from very basic assumptions? The dispersion cloud surely has certain distinct properties which can only be described by hard evidence from testing and specs. If you want to doubt the manufacturers version, that's possible, but without more solid evidence it remains extremely fanciful.
External Quote:In the static plane the splitter area is two leaves of 56°, approximately. The green arrows show areas of light fraction and the red shows heavy fraction. I'd like to draw your attention to the fact that before installation of this warhead to the missile, the designers carried out dozens of, hundreds, tests of warheads, separately from the missile in assembly with the missile in it, we can have the proofs that not less than 96% of all hitting segments are distributed in this two sectors, or other 4% can have the tolerance of 2 to 3 degrees.
External Quote:We took into, during the tests, the speed of the missile and the plane were taken into account. The form of the fragment cloud is changing with speed, and two striking fronts are build. These red lines are lighter splinters with higher speed and the violet line shows the front of "I" shaped heavy fraction.
External Quote:The main peculiarity of rocket 9M38M is the special area which is called a "lancet", or the killing lancet, which is perpendicular [inaudible] basically the area of concentration of more than 40% of the all splinter mass, and one half of the whole kinetic energy.
General question for Mick - can we play around with your model or will that mess up what you have done?
I'm basing it on their slides.
They say, at 14:00
External Quote:In the static plane the splitter area is two leaves of 56°, approximately. The green arrows show areas of light fraction and the red shows heavy fraction. I'd like to draw your attention to the fact that before installation of this warhead to the missile, the designers carried out dozens of, hundreds, tests of warheads, separately from the missile in assembly with the missile in it, we can have the proofs that not less than 96% of all hitting segments are distributed in this two sectors, or other 4% can have the tolerance of 2 to 3 degrees.
So there they describe an even distribution of velocities.
[15:07]
External Quote:We took into, during the tests, the speed of the missile and the plane were taken into account. The form of the fragment cloud is changing with speed, and two striking fronts are build. These red lines are lighter splinters with higher speed and the violet line shows the front of "I" shaped heavy fraction.
Then they show it modified to account for the speed of the missile and aircraft.
[15:37]
External Quote:The main peculiarity of rocket 9M38M is the special area which is called a "lancet", or the killing lancet, which is perpendicular [inaudible] basically the area of concentration of more than 40% of the all splinter mass, and one half of the whole kinetic energy.
So how do they get this weird velocity distribution? And how do they get some of the fragments traveling backwards? What does the static velocity field look like, and why did they present it as even?
It may well be that the "lancet" design calls for higher speed ejecta in the center of the pattern, however they do not show this in the "static" diagram, and they only say the next diagram is accounting for the speed on the missile (and the plane.)
They do say that it took a lot of testing to get it so that a small area (the cockpit) took a big hit.So I see no _technical_ reason to distrust the manufacturer when he implies a non-uniform velocity distribution for his warhead.
Actually I haven't found any place yet, where Almaz states, that the velocity distribution within the two leaves of 56° is uniform. The most extreme form of a non-uniform velocity distribution of ejecta would be a shaped charge: https://en.wikipedia.org/wiki/Shaped_charge
So I see no _technical_ reason to distrust the manufacturer when he implies a non-uniform velocity distribution for his warhead.
With a static warhead. And they very clearly say it's a 56 degree spread of all the shrapnel, and only 42% is in the "scalpel" area. At no point do they suggest non-uniform initial velocity.They do say that it took a lot of testing to get it so that a small area (the cockpit) took a big hit.
If the maximum speed they claim is 2400 m/s then shouldn't the distance between the missile and the edge of the red shaded bit be taken to represent 2400 m/s as its the longest distance travelled. If you use that to calculate the length of a 1000 m/s vector it ends up being about twice as long as in your picture and makes the static velocity profile point entirely backwards.Hence, for the "scalpel" to be at the angle they claim, it would have to be traveling at 5.9x the speed of the missile, or 5900 m/s, with a resultant velocity of just slightly less. And yet they only claim 2400 m/s
Actually when they say that 42% of the splinter mass and more than 50% of the kinetic energy is contained in the "lancet", they suggest/imply a non-uniform velocity distribution. That is because with a strictly uniform velocity distribution the percentage of kinetic energy and the percentage of splinter mass in a given sector would have to be equal.The diagram seems to show an initial uniform distribution of velocity. They never suggest at that point it's not uniform.
The length of this offset is the speed of the missile relative to that particular representation of velocity vectors. The speed of the missile is assumed to be 1000 m/s, so the length of each of those thin black lines represents 1000 m/s. We can now calculate the spread of velocities by the length of the adjusted vectors. (and you can verify this yourself simply by measuring them on-screen)
Hence, for the "scalpel" to be at the angle they claim, it would have to be traveling at 5.9x the speed of the missile, or 5900 m/s, with a resultant velocity of just slightly less. And yet they only claim 2400 m/s
1000 m/s is the speed of the missile. It's just the distance between the two top angles of the before and after vectors in their diagram for the red curve.How did you work out the length of your 1000 m/s vector?
Assuming a non-uniform velocity distribution, the blue vector with caption 2200 m/s would imho represent one of the slower splinters and would have a velocity far below 2200 m/s.
If the maximum speed they claim is 2400 m/s then shouldn't the distance between the missile and the edge of the red shaded bit be taken to represent 2400 m/s as its the longest distance travelled. If you use that to calculate the length of a 1000 m/s vector it ends up being about twice as long as in your picture and makes the static velocity profile point entirely backwards.
How did you work out the length of your 1000 m/s vector? I've missed something somewhere haven't I, am poised to kick myself.
if we assume the velocity to be lowest at the limit of the fragment sector (red line in original graphic), then the 1000 m/s vector is OK. The 2200 m/s should be more like ~1/3 (by eyesight) of the vmax (2400 m/s). vmax is represented roughly by a vector from the center along the black line to the edge of the red shaded bit.
See above, the 2200 m/s is necessary for their diagram to work. If it were slower, then the entire cone would get a lot more of its velocity (proportionally) from the missile, and be at a much steeper forward angle.
For this change in angle of the forward limit of the spread, it MUST be 2.2x the velocity of the missile, so it must be 2200 m/s if the missile velocity is 1000 m/s
That video shows a BUK + warhead exploding before burning its fuel, so we (or at least I) don't know if that explosion is augmented by exploding fuel.A video of a BUK warhead exploding (prematurely) can be found here: http://www.ntv.ru/novosti/141322/video/
To get a "lancet" with 2400 m/s that is perpendicular with a warhead moving at 1000 m/s the lancet would have to point to the stern with a static warhead. That angle (to the stern) would be sin^1 (1000/2400) ~ 25°. So in this slide the lancet with static warhead would have to point to ~115°