Does Damage to MH17 indicate or exclude a Particular Buk Launch Location?

Mick West

Administrator
Staff member
If that is the case, the primer would be located at the front end of the warhead, thus producing a backward pointing cone?

AA shows this diagram of a cone from 124 to 68°, with heavy fragmentation at 112 to 68 and light from 124 to 72
https://www.metabunk.org/sk/20150617-054143-zbldf.jpg

For the "heavy" this is symmetrical about the perpendicular +/- 22°, but for "light" it 18° forward, and 34° backwards. This is a bit odd, as you would expect for the "lancet" effect as shown , you would want the heavy fragments backwards. Unless by "heavy" it means "fragmented a lot".

This shows an error in my sim, as I'd just looked at the 56° figure and assumed it was symmetrical. The actual full range is 22° forward to 34° backwards.

This demonstrates the magnitude of this error, with the left side being corrected, and the right side being symmetrical. (plane velocity of zero here)


Another way of looking at this is that the center of the 56 degree field is angled backwards 6°
 
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Ole

Member
For the "heavy" this is symmetrical about the perpendicular +/- 22°, but for "light" it 18° forward, and 34° backwards. This is a bit odd, as you would expect for the "lancet" effect as shown , you would want the heavy fragments backwards. Unless by "heavy" it means "fragmented a lot".

I assume that would be, because the heavier fragments are the slower ones, and thus their resultant speed vector would be less dominated by the speed component induced by the detonation and a little more dominated by the speed of the missile at the instant of detonation.

Indeed some weird assumptions have to be made to derive their dynamic fragment distribution from their static fragment distribution.

There is a saying :"Never go to sea with two chronometers; take one or three." That is because if two chronometers differ you don't know which of them is the correct one, so you need a third chronometer to decide that question. Same thing with their fragment distribution. One of them appears wrong, but which one?
 

mvdb22

Member
I am wondering how, based on the damage, the direction of the missile can be determined. The statement of Almaz the warhead has a scalpel (concentration of major percentage of all fragments in limited area of fragment beam) is false based on the barrel shaped warhead and single primer.

When the missile came from Snizhne the right wing + engine would have damage and not the left. When the missile came from Zaroshens’kye the left wing + engine would have damage.

There are various pics showing parts of one or both engines. The problem is : which engine?
This photo shows damage to the nacelle (outer cover) of one of the engines. Left or right? To make things complicated: damage can also be done by falling debris
 

Rob

Member
The entire purpose of a fragmentation warhead is to inflict the maximum amount of damage at the greatest distance possible for the expected target. All design parameters (number of fragments, ratio of fragment mass to TNT, mass of fragments etc etc) are tuned to achieve that maximum range at which the warhead can destroy its intended target.
That "kill range" appears to be a closely held secret, but for a 9N314M1 warhead it is probably between 10 and 20 meters distance from the warhead.

Either way, the kill range is a crucial parameter, since it determines how many other parts of the system, and specifically the proximity fuse, should operate.

We know that fragments fan out in a forward pointing cone, so you want to hit your target when it appears within that cone but not before it appears within your "kill range".

This means the proximity fuse should look outward in a cone, and the system will need to be tuned so that when the proximity fuse detects a target within the kill range that the warhead fragments will go right through that location. After all, you do NOT want to miss a small target like a drone or the most difficult of all : another missile. So you have to hit the spot that the proximity fuse identified.

In other words, the proximity fuse's sensitivity cone determines WHEN and WHERE the missile will deposit its fragments.

Here is a great article on how a SAM proximity fuse sensitivity cone is achieved :

https://www.feko.info/applications/...-fuse-antenna-for-an-air-defence-missile/view




This guy uses a linear microstrip antenna array at 2.4 GHz, but similar sensitivity cones can be achieved with 80's technology at lower frequency and with coil antennas.

I'm not sure if I make myself clear here, but the important part is that the entire system is tuned so that the warhead fragments will hit the location where the proximity fuse sensitivity cone first detects enough metal within the "kill range" distance from the missile path. Which is a RING around the missile path the size of the kill range, and NOT strait ahead.

In the case of the Snizhne approach, that first metal detected was almost certainly (a meter or so into) the nose of MH17, and the warhead indeed delivered the bulk of fragments right at that location.

But in the case of the Zaroshens'kye approach, the first metal that the proximity fuse would have detected would be some part of the right side of the plane, behind the cockpit, in which case it would have blasted that area with fragments.

And that obviously did not happen, so either the launch was not from the Zaroshens'kye direction, or the proximity fuse seriously malfunctioned.
 
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Ole

Member
We know that fragments fan out in a forward pointing cone, so you want to hit your target when it appears within that cone but not before it appears within your "kill range".

Where do we know that from?
 

mvdb22

Member
Almaz Antey presented the slide below. Damage is seen on the left wing and engine. The presentation did not show any photos nor proof which confirms the left wing and engine was damage. Anyone does have such proof?
 

Rob

Member
Where do we know that from?

Simple physics. If you want to hit your target within the set parameters (of damage), you need to wait until it appears within the proximity fuse's trigger range.

Otherwise you set off your warhead too early (not enough fragment density and momentum) or too late (not enough radius).
 

mvdb22

Member
This is an interesting photo. I turned it so the direction of view is roughyl from the front of the nose towards the tail of the B777. This shows the lefthand side of the cockpit. The round silverish area was where the pitot tube was located.



 

Herman Aven

Member
We know that fragments fan out in a forward pointing cone, so you want to hit your target when it appears within that cone but not before it appears within your "kill range".

Instead of "we know" I'd like to see a reference. Something perhaps like the following from: Fundamentals of Naval Weapons Systems United States Naval Academy

Fragment Flight. The fragments of a warhead travel outward in a nearly perpendicular direction to the surface of its casing (for a cylindrical warhead there is a 7- to 10-degree lead angle).. ... The tail and nose spray are frequently referred to separately as the "forty-five degree cone," which is an area of less dense fragmentation. If this payload were to be detonated in flight, the dense side spray would have a slight forward thrust with an increased velocity equal to missile flight velocity.​

There's surely a kill range bit it seems to be more concentrated and perpendicular than many are assuming. But it would make sense, what's the use of an advanced proximity fuze if one is going to spread everything out into a giant cone? It seems more about precision and concentration and modern fragmentation warheads appear to display increasingly "ring" shaped patterns. From the same article:

The latest air target warheads are designed to emit a narrow beam of high-velocity fragments. This type of warhead, called an annular Blast Fragmentation warhead (ABF), has a fragmentation pattern that propagates out in the form of a ring with tremendous destructive potential. A newer type of fragmentation warhead is the Selectively Aimable Warhead (SAW). This "smart" warhead is designed to aim its fragment density at the target. This is accomplished by the fuzing system telling the warhead where the target is located and causing it to detonate so as to maximize the energy density on the target.​

Although I'm not sure if this last bit would apply for these older type warheads. But it's clear that it's a design goal.

In the case of the Snizhne approach, that first metal detected was almost certainly (a meter or so into) the nose of MH17, and the warhead indeed delivered the bulk of fragments right at that location.
Wouldn't be that a waste of a good ring if only the cockpit would be cut? Also you'd expect to see more exit holes right of the cockpit which aren't shown so far (but they might be there).

But in the case of the Zaroshens'kye approach, the first metal that the proximity fuse would have detected would be some part of the right side of the plane, behind the cockpit, in which case it would have blasted that area with fragments. And that obviously did not happen, so either the launch was not from the Zaroshens'kye direction, or the proximity fuse seriously malfunctioned.
That doesn't add up at all. With the shown approach the most signal might have been exactly around the time of detonation. Furthermore it's hard to know exactly the time between fuse activity and impact considering the speeds involved. There's a whole area for tuning available for example knowing the delays there could be an added predictive element to the trigger.
 

jonnyH

Senior Member.
I assume that would be, because the heavier fragments are the slower ones
Wouldn't that be the other way around?

The lighter bits are going to decelerate much faster than the heavier fragments. You would expect to be able to throw a rock much further/faster than a handful of sand for instance.
 

Mick West

Administrator
Staff member
Simple physics. If you want to hit your target within the set parameters (of damage), you need to wait until it appears within the proximity fuse's trigger range.

Otherwise you set off your warhead too early (not enough fragment density and momentum) or too late (not enough radius).

Sure, but AA get around this by saying:
  • The missile explodes 3-5 meters beyond the detection point.
  • The missile was designed to take out the heavily armored A-10 (or similar)
  • The missile had a very stong perpendicular component to the frag energy after accounting for the speed of the missile..
I've not found them explicitly saying the first point in the video though, not giving any real details on the fuse. And the last point has several problems discussed above.
 

Mick West

Administrator
Staff member
Wouldn't that be the other way around?

The lighter bits are going to decelerate much faster than the heavier fragments. You would expect to be able to throw a rock much further/faster than a handful of sand for instance.

It's the initial speed that's more relevant here. You can throw a light rock much farther than a heavy rock.
 

Mick West

Administrator
Staff member
Instead of "we know" I'd like to see a reference. Something perhaps like the following from: Fundamentals of Naval Weapons Systems United States Naval Academy

Fragment Flight. The fragments of a warhead travel outward in a nearly perpendicular direction to the surface of its casing (for a cylindrical warhead there is a 7- to 10-degree lead angle).. ... The tail and nose spray are frequently referred to separately as the "forty-five degree cone," which is an area of less dense fragmentation. If this payload were to be detonated in flight, the dense side spray would have a slight forward thrust with an increased velocity equal to missile flight velocity.

"The fragments of a warhead travel outward in a nearly perpendicular direction to the surface of its casing" - does no sound like AA's 56° spread

"Slight forward thrust" does not tally with a 1000 m/s missile and a 2400 m/s frag velocity.
 

jonnyH

Senior Member.
You can throw a light rock much farther than a heavy rock.

Yes, but I can throw my 175g Ultrastar further than my 160g Sky-styler.

Isn't the shrapnel all made of the same stuff, just broken into different sized fragments? If thats the case "heavy" implies large and "light", small. Hence the rocks vs sand analogy. But then I suppose this did all happen in too short a period of time considering the distances and the speeds for it to make much difference.
 

Mick West

Administrator
Staff member
Yes, but I can throw my 175g Ultrastar further than my 160g Sky-styler.

Isn't the shrapnel all made of the same stuff, just broken into different sized fragments? If thats the case "heavy" implies large and "light", small. Hence the rocks vs sand analogy. But then I suppose this did all happen in too short a period of time considering the distances and the speeds for it to make much difference.

To be clear there's two overlapping things here.

Firstly there's the force of the explosion. Force = mass * acceleration, so a = F/m, so the larger mass the lower acceleration, and the slower the initial speed. (surface area comes into effect too, but I'm simplifying here). So larger fragments will start moving slower in the explosion.

Secondly there's air resistance, which is a tad more complex, but again the larger the mass, the less deceleration it has. So larger objects will slow down slower.

So there's an overlap. You can tell how things start out (larger objects are slower) and how they end up (larger objects are faster), but you can't say at the point of impact with the target where you are on that overlap.
 

Ole

Member
I'm basing it on their slides.
They say, at 14:00

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.
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Actualy they don't say this depicts the fragment distribution in the frame of reference of the warhead.
In the static plane ...
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might try to convey that this is the fragment distribution in the frame of reference of the plane.
Just tinking loud: Many things make more sense from this point of view. Assuming the fragment distribution in the frame of reference of the warhead actually is as drawn by
This was my back of fag packet calculation:BUK Vp.jpg
then the fragment distribution in the frame of reference of the plane would be tilted/shifted forward. The size of that shift is a function of the relative speed between missile and plane. In a tail chase scenario of a fast jet that relative speed would be ~500 m/s in a head on scenario to an SR 71 or to a missile that relative speed would be >2000m/s. Wikipedia has the maximum target speed for the BUK 9M38 as Mach 4 (~1300m/s).

The 56° sector would define the sector which the fragment cone can shift in (in the frame of reference of the plane) between these two extreme scenarios.
That would explain why the slower (red line) fragments are shifted more to the front than the faster fragments.

It would also explain why the the difference in the shift is bigger on the front limit than on the stern limit. On the front limit the longitudinal velocity component is more dominated by the relative speed between missile and target than on the rear limit. This relative speed is independent of the detonation induced speed, thus the difference in fragment speed due to differing acceleration by the detonation matters less on the front limit.

From this point of view slide 11 depicts the sector into which the fragments can be ejected from the warhead depending on the relative speed between missile and target. A simplistic fusing logic has to make sure that the target always is within this sector when fusing.
 

Mick West

Administrator
Staff member
Actualy they don't say this depicts the fragment distribution in the frame of reference of the warhead.
In the static plane ...
might try to convey that this is the fragment distribution in the frame of reference of the plane.
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They say that the second diagram accounts for the speed of the missile and the aircraft. So that's in the frame of reference of the plane.

The first diagram was clearly a static non-moving frame of reference - determined from stationary tests.

See my previous post:

I'm basing it on their slides.
They say, at 14:00

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.
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So there they describe an even distribution of velocities.

[15:07]
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.
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Then they show it modified to account for the speed of the missile and aircraft.

[15:37]
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.
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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.)

You can't tell how much the offset is unless you know what the velocities are. I'm using their figures of a resultant velocity of 2,400 m/s, and missile speed 1000 m/s. My sim draws the resultant cone from the frame of reference of the aircraft.
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David Coulter

Senior Member.
The missing bit here is the deceleration of high explosive shrapnel. I agree with Mick's vector geometry, but it would be good to understand the energy vs distance calculations. I am not sure that there would be any penetration energy on the opposite side of the aircraft as claimed in the OP. Distance ^2 like gravity?
 

Ole

Member
As said before, just thinking aloud:

They say that the second diagram accounts for the speed of the missile and the aircraft. So that's in the frame of reference of the plane.

The second diagram then would take into account the speed of the missile and the specific speed of the aircraft in this scenario (head on to a target flying at 250 m/s), whereas the previous slide is valid for all possible speeds of a target.
 

Ole

Member
The entire purpose of a fragmentation warhead is to inflict the maximum amount of damage at the greatest distance possible for the expected target. All design parameters (number of fragments, ratio of fragment mass to TNT, mass of fragments etc etc) are tuned to achieve that maximum range at which the warhead can destroy its intended target.

The smaller the opening angle of the fragmentation cone, the smaller the hitting probability, or the better the missile has to "aim" at the target. So it might also be a design goal to have the opening angle of the fragmentation cone as large as possible (i.e. as perpendicular as possible) for most intercepting scenarios.
 

Mick West

Administrator
Staff member
The missing bit here is the deceleration of high explosive shrapnel. I agree with Mick's vector geometry, but it would be good to understand the energy vs distance calculations. I am not sure that there would be any penetration energy on the opposite side of the aircraft as claimed in the OP. Distance ^2 like gravity?

Bit more complicated. And you are goign to get vastly more energy loss in punching through metal than in traveling through three meters of air. Consider a bullet, it't pretty lethal at 1m, 10m and 50m.

There's some math on this here:
http://fas.org/man/dod-101/navy/docs/es310/warheads/Warheads.htm

Reduction in Velocity with Range

As soon as the fragments are thrown outward from the casing, their velocity will begin to drop due to wind resistance (drag). The drag force is given by:

Drag = ½ rv2 Cd A

where: r = The density of air. Normally 1.2 Kg/m3.
V = The fragment velocity.
Cd = The coefficient of drag. Depends on the shape of the fragment and to some extent, the velocity.
A = The cross-sectional area of the fragment.

We can solve the equations of motion for the projectile and get the fragment's velocity as a function of the distance traveled:





where s = the range, and v0 is the initial fragment velocity.



Example- find the fragment velocity 100 m from the detonation of a M61 hand grenade, given:

v0 = 2150 m/s
A = 1 cm2
Cd = 0.5
m = 2 g

We use the default value for the density of air. This gives a velocity of

v(at 100 m) = (2150 m/s) e-(1.2 x 0.5 x 0.0001 x 100)/(2 x 0.002)
v = 480 m/s
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If you take that equations and plot a graph based on distance, you get:


Not a hugely significant difference over a few meters, and this is for a grenade, with lighter pieces, which will slow down quicker.
 
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Mick West

Administrator
Staff member
The second diagram then would take into account the speed of the missile and the specific speed of the aircraft in this scenario (head on to a target flying at 250 m/s), whereas the previous slide is valid for all possible speeds of a target.

Or equally invalid :) If you don't account for the speed then you don't know where the fragments go.

The speed of the aircraft is only 1/4 the speed of the missile, and 1/10 the speed of the fragments, so it does not have a huge impact on the dispersion pattern.
 

Ole

Member
Or equally invalid :) If you don't account for the speed then you don't know where the fragments go.

The speed of the aircraft is only 1/4 the speed of the missile, and 1/10 the speed of the fragments, so it does not have a huge impact on the dispersion pattern.

Yet the missile is designed for targets at a relative speeds between ~500 m/s and >2000 m/s, so from the designer's point of view the relative speed has a huge impact on the dispersion pattern. When designing the dispersion pattern (for warhead at rest) and the antenna diagram of the fuse, they have to cover this whole range of possible dispersion patterns (for plane at rest).

Russian tech is said to be unsophisticated and robust, so maybe they just chose a dispersion pattern that is perpendicular in the average case of relative speeds, and one or two sideways and sidways-forward looking antenna diagramms


To achieve maximum effectiveness of combat equipment at different shooting conditions, the inclination angle of the antenna diagram has to be a function of the vector of the approach speed of the missile to the target, i.e it is a function of the missile speed, and of the approach angle between missile and target.

If the inclination angle of the antenna diagramm relative to the longitudinal axis of the missile
is assumed to be constant, then the adaptation of the fuse of the warhead only works for some average conditions between missile and target.

If the conditions differ from those anticipated (eg. of the assumed target speed, the distance to the intercepting point, the altitude and heading parameters, etc.) the effectiveness of the missile and consequently the probability of destruction of the target is lowered.

The airspeeds of modern air targets vary within wide limits. In order to adapt fuse and warhead throughout the whole speed range, not only one but several fixed angles of the antenna diagram of the fuse are required. Which of the directional patterns is to be used, will be determined according to the intercept conditions between target and missile. '
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from https://static.allmystery.de/upics/45674f_S12.jpg
via http://www.i2ocr.com/free-online-german-ocr and google translate with corrections by me.
(in the meantime I figured out that the original source is this: http://www.zvab.com/Schießen-Fla-Raketen-F-Neupokojew/198485491/buch)
 

Rob

Member
A few questions:

-1 spreading pattern:
Could the pattern be (more or less) compared to a Gaussian distribution (or "normal distribution)?
normaal verdeling six sigma sigmaniveau.jpg
if spreadingpattern is indeed comparable with a Gaussian distribution, would it be possible to include this (roughly) in the vectormodel?

Very good suggestion. Thanks !
Does the density distribution follow a Gaussian distribution ?
I am not a warhead designer, but from a physics point of view that makes some sense.
After all, there is a general direction (outward) but also random factor involved when fragments leave the warhead.

Interesting is that IF the density distribution is Gaussian, that we can determine the angle of the "lancet" that AA is talking about based on their own info.

normaal verdeling six sigma sigmaniveau.jpg

Almaz Antey state that 96% of the fragments are launched within a 56 deg cone. 96 % spans 4 sigma, or 14 deg per sigma.

They also state that 42 % of the fragments end up in the "lancet". 42 % is about 1.6 sigma (anyone want to do a more accurate computation ?), which implies that, if the fragment distribution is Gaussian, Almaz Antey's "lancet" is something like 22 degree wide.
 

Rob

Member
Yet the missile is designed for targets at a relative speeds between ~500 m/s and >2000 m/s, so from the designer's point of view the relative speed has a huge impact on the dispersion pattern. When designing the dispersion pattern (for warhead at rest) and the antenna diagram of the fuse, they have to cover this whole range of possible dispersion patterns (for plane at rest).

That does not make any sense to me from an engineering perspective.

If you design a fragmentation warhead, you want to restrict the angle of dispersion as much as possible, to achieve maximum damage at maximum distance.

Which has NOTHING to with tuning the system so that the missile fragments hit the first metal detected by the proximity fuse, under different approach velocities.

Two entirely different problems.
 
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Ole

Member
Two entirely different problems.

Well, the problem of hitting the target at all and hitting it with the maximum destructive energie are somehow intertwined. If you don't solve the first problem you don't need to bother solving the second.
 

Rob

Member
streubereich.jpg

If that is the case, the primer would be located at the front end of the warhead, thus producing a backward pointing cone?

Ole, could it be that you have that primer located incorrectly ?
After all, there does not seem to be any engineering or physics reason to blast fragments backward.
You would just be wasting kinetic energy.

To stick with Mick's analogy, if you want to hit something out of a fast moving train, you will hit it harder if you throw the stone forward, rather than backward.
 
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Rob

Member
Well, the problem of hitting the target at all and hitting it with the maximum destructive energie are somehow intertwined. If you don't solve the first problem you don't need to bother solving the second.

It appears that the engineers from Almaz Antey solved both problems satisfactory within the engineering parameters set.

Otherwise, 298 innocent people would not have died on July 17.
 
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Rob

Member
Bit more complicated. And you are goign to get vastly more energy loss in punching through metal than in traveling through three meters of air. Consider a bullet, it't pretty lethal at 1m, 10m and 50m.

There's some math on this here:
http://fas.org/man/dod-101/navy/docs/es310/warheads/Warheads.htm

Reduction in Velocity with Range

As soon as the fragments are thrown outward from the casing, their velocity will begin to drop due to wind resistance (drag). The drag force is given by:

Drag = ½ rv2 Cd A

where: r = The density of air. Normally 1.2 Kg/m3.
V = The fragment velocity.
Cd = The coefficient of drag. Depends on the shape of the fragment and to some extent, the velocity.
A = The cross-sectional area of the fragment.

We can solve the equations of motion for the projectile and get the fragment's velocity as a function of the distance traveled:





where s = the range, and v0 is the initial fragment velocity.



Example- find the fragment velocity 100 m from the detonation of a M61 hand grenade, given:

v0 = 2150 m/s
A = 1 cm2
Cd = 0.5
m = 2 g

We use the default value for the density of air. This gives a velocity of

v(at 100 m) = (2150 m/s) e-(1.2 x 0.5 x 0.0001 x 100)/(2 x 0.002)
v = 480 m/s
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If you take that equations and plot a graph based on distance, you get:


Not a hugely significant difference over a few meters, and this is for a grenade, with lighter pieces, which will slow down quicker.

The exponential drop off of velocity (and thus exponential drop off of kinetic energy) over distance of the fragments is interesting, since not only the blast cone will change, but also would allow us to estimate how the speed of fragments changes (and thus how their angle of movement changes) after they penetrated the hull (and thus if they can still "exit" the hull on the opposite side of the plane), if we would know the mass (and material) of the fragments...

Anyone ?
 
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The missing bit here is the deceleration of high explosive shrapnel. I agree with Mick's vector geometry, but it would be good to understand the energy vs distance calculations. I am not sure that there would be any penetration energy on the opposite side of the aircraft as claimed in the OP. Distance ^2 like gravity?
There have been reports of exit holes in the floor of the cockpit. If true they might require much the same energy
 
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Mick West

Administrator
Staff member
The exponential drop off of velocity (and thus exponential drop off of kinetic energy) over distance of the fragments is interesting, since not only the blast cone will change, but also would allow us to estimate how the speed of fragments changes (and thus how their angle of movement changes) after they penetrated the hull (and thus if they can still "exit" the hull on the opposite side of the plane), if we would know the mass (and material) of the fragments...

Anyone ?

Not really, because we are not talking about a huge change over the distance involved. Saying "exponential drop-off" is a bit misleading, as things can drop off "exponentially", but not significantly in a the time of interest, and in fact actally be linear. The AA scenario is in 1-6 meters range for the cockpit damage, and the Snizhen scenario is about 4-9 meters.

For a hand grenade over that range there's an approximately 15% essentially linear drop-off.


For a heavier fragmentation warhead, the drop-off would be considerably less. It's just not of any significance given the range of uncertainties we have here.
 
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Ole

Member
Ole, could it be that you have that primer located incorrectly ?
Actualy I would be glad if somebody with better Russian than mine can deny or confirm. But anyhow we can't be shure if that graphic depicts the warhead at question.

After all, there does not seem to be any engineering or physics reason to blast fragments backward.
You would just be wasting kinetic energy.

To stick with Mick's analogy, if you want to hit something out of a fast moving train, you will hit it harder if you throw the stone forward, rather than backward.

It is not the entire purpose of a fragmentation warhead, to inflict the maximum amount of damage at the greatest distance possible for the expected target, by putting the maximum kinetic energy into it. If there is a trade-off between damage inflicted and probabilty of hitting the target, then a smart engineer might settle for a sufficent amount of damage?

IMHO there are two different questions involved in this thread:
1. Does AA try to deceive the public?
2. Would they talk utter nonsens to achieve that?

A reasonable starting point could be:
1. I don't know.
2. That doesn't make sense.
 

Mick West

Administrator
Staff member
IMHO there are two different questions involved in this thread:
1. Does AA try to deceive the public?
2. Would they talk utter nonsens to achieve that?

No, the questions are
1. Can we determine form the available evidence where the Buk was fired from?
2. If not, then can we rule out anywhere?

AA present some claims of evidence that they say rules out Snizhne and instead indicated Zaroshens’kye. Hence we are looking at those claims of evidence to see if they hold up.

Several problems have emerged - most specifically the conflict between the static and dynamic fragment distribution diagrams that cast doubt on the accuracy of AA's interpretation.

So that's what needs discussing.
 

Mick West

Administrator
Staff member
I have renamed this thread: "Does Damage to MH17 indicate or exclude a Particular Buk Launch Location?" from "Was MH17 downed by a buk fired from Zaroshens’kye", to more accurately reflect the discussion in the thread.
 

Ole

Member
Several problems have emerged - most specifically the conflict between the static and dynamic fragment distribution diagrams that cast doubt on the accuracy of AA's interpretation.

The explanation they give of the "static" fragment distribution ends up beeing very ambiguous (maybe caused by the translation). So we have to second-guess what they want to tell us with their "static" fragment distribution. We have done that in two different ways. One way creates a conflict, for the other way I don't see one yet, other than that the fragment distribution is subobtimal energywise while probably optimizing the spacial distribution of the fragments (and that's not a conflict that would make their assesment doubtable).
 

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