Following the information of the destruction of an enemy (NATO) bunker buried 200-300 meters in the vicinity of Lvov by a Kinzhal missile strike, some claimed that the Kinzhal does not have this possibility. So I contacted one of our French readers, an eminent mathematician:
Boris Karpov 
Chroniques de Russie
«
As a high-level scientist, could you explain, concretely, the consequences of the arrival of a "Kinzhal" hypersonic missile on a target, if it does not contain any explosives.
Kinetic energy, and destructive power. The main question being "Can he destroy a concrete bunker located XXX meters underground", by varying XXX and the nature of the "earth": Basalt, concrete etc.
This is because following information that we (among others) have disseminated about a Kinzhal shooting having pulverized an underground bunker and liquidating a few hundred NATO officers, some claim that it is impossible. I am well placed to know that the info is correct but a mathematical/physical demonstration would be ideal! »
Here is the answer of Jean-Marie ARNAUDIES
Note: The ^ sign means “power”. 10^5 = 10 to the power of 5 = 100000 (The number 1 followed by 5 zeros)
Here is a very first approach comparing to the Hiroshima atomic bomb
The energy E_H deployed by the Hiroshima atomic bomb is estimated at 50 terajoules (abbreviated: TJ). So E_H=50 x 10^12 Joules= 5 x 10^(13) Joules
Let's calculate the Kinzhal's kinetic energy. This missile weighs 4000 kg. Its cruising speed can be estimated at a little over Mach10 (low estimate in my opinion). But let's count Mach 10 to play devil's advocate, that is to say to lead to a reduction of the energies involved.
We immediately arrive at the following kinetic energy, in Joules: the abstract formula is (1/2) M x V^2, where M designates the mass and V the speed. The mass will be estimated in kg and the speed in m/s (= meter per second). We thus find that the kinetic energy E_K of the Kinzhal at its full speed is given, in Joules, by
E_K = 1/2 x 4000 x (10 x 344)^2 Joules = 2,36672 x 10^(10) Joules.
Now we have to estimate the pressures per square meter, in the case of Kinzhal and in the case of the Hiroshima bomb.
The energy deployed on its impact by the Kinzhal is easy to estimate: the entire force of the missile is concentrated in at most one square meter. Indeed, the area of a section of the Kinzhal perpendicular to its axis is significantly less than 1 square meter.
So we arrive at an energy applied by the Kinzhal on a single square meter at its impact with the ground equal to or greater than 2,36672 x 10^(10) Joules.
But the energy of the Hiroshima A-bomb was exerted on well over a square meter. This energy has charred everything on a surface corresponding to a disk with a radius of 600 m. The bomb did not dig a big hole, as the initial explosion was mostly airborne, although it was near the ground. 600 m radius is 1130976 square meters. To satisfy the devil, let's only count 10^6 square meters. At the supposed center of the explosion, the temperature was around 6000 degrees Celsius and at 600 meters it was still around 1300 degrees Celsius. If we reduce the area to 400 m distance from the center, we see that the temperature at 400 m from the center was greater than 1600 degrees Celsius over an area of more than 500 square meters.
Suppose that all the energy of the Hiroshima bomb was exerted on these 500 square meters. Within this perimeter, very few metals were able to resist complete melting (melting temperature of iron =000 degrees Celsius). We can estimate that the pressure was exerted on average on this perimeter, which will give an extremely pessimistic reduction.
So each square meter of this area from 0 to 400 meters from the "center" of the explosion will have endured at worst a Hiroshima A-bomb energy greater than or equal to (5 x 10^(13)) Joules divided by 500 000, or 10^8 Joules, which gives us the following energy, noted P_H:
P_H= 10^8 Joules.
It follows that the energy E_K undergone by the impact of the Kinzhal on its contact with the ground is
2,366672 x 10^(10) Joules
over a square meter, so E_K is more than TWO HUNDRED TIMES the energy per square meter of the main center of the Hiroshima explosion.
There is also a big difference between the A-bomb of Hiroshima and the Kinzhal: the devastation of Hisroshima was essentially superficial, because the bomb did not explode following an impact with the ground, while the Kinzhal with its pointed nose exerted all of its kinetic energy in one direction, carving its way from the surface to its depths. So the devastation suffered by the effect of the Kinzhal was several hundred times stronger on the square meter of its point of impact than the devastation on each square meter of the hypercentre of the effects of the Hiroshima A-bomb.
This means that the force of the Kinzhal at its contact with the ground has been INSANE, without any measure with that produced by the A-bomb of Hiroshima on each square meter of its hypercenter. At a depth of two or three hundred meters, the earth's crust is extremely soft in the face of energies per square meter of this magnitude. This strength of Kinzhal in contact with the ground is beyond comprehension!
This assessment is verified by comparing with the soil cores taken in oil exploration. We dig holes up to 3000 m deep without needing a Kinzhal! So whether it's granite or dense, soft marl, for a Kinzhal, digging a hole 200 or 300 m deep is a joke!
In this study, the brevity of which you will kindly excuse, I took no account of a possible charge (nuclear or not) carried by the Kinzhal. If such a load is added to it, upon impact with the ground, this load will DECREASE the maximum depth of the casing created by the Kinzhal, because this load will exert its effects in all directions, thus considerably enlarging the initial opening of the hole dug by the missile, which will decrease the force of the impact per square meter.
The margin seems so huge that this remark should allow the military to refine the effect of the Kinzhal: if you are looking for a deep target, say more than 300 meters deep, you have to adjust the explosive charge carried by the Kinzhall so that it only explodes when the Kinzhall has reached the desired depth. For a very deep target, you don't even need a charge! on the other hand, for a shallow target, say less than 100 m deep, the Kinzhal's explosive charge can be set so that it explodes a few hundredths of a second after the missile hits the ground.
This very brief and very rough study convinces me that as long as the Russians are ahead of the USA in very high speed missiles, the Kinzhals with or without nuclear charges and their descendants will defeat all imaginable US armaments, atomic, even thermonuclear or not. If I had power, I would ask my scientists to do their utmost to increase this enormous speed. For example, if we go to 20 Mach instead of 10 Mach, the energy in contact with the ground will be multiplied by 4 and if we arrive at Mach30, then the energy in contact with the ground will be almost 10 times greater, which shows that then no underground shelter will be possible for the enemy (the advantage of the Kinzhal will then be of the order of thousands of times more force in contact with the ground than by nuclear or thermonuclear armament). It therefore seems to me vital for Russia that it deepen and increase this advance to the maximum and above all, that it does everything necessary to prevent treason from helping the USA to also manufacture Kinzhal-type missiles.
Jean-Marie ARNAUDIES
I think this definitively closes the discussion on the "impossibility" of the Kinzhal to destroy a bunker buried 200-300 meters away
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