*NEW – May 28, 2022*

The other day, Roscosmos demonstrated what kind of crater the “Sarmat” warhead left after falling on the Kura test site (in Kamchatka):

The crater is about 8 meters deep. The diameter is 20 meters. The volume is about 1000 cubic meters.

Let me remind you that this “pothole” was left by an empty warhead, without a nuclear charge. Such a warhead weighs 1 ton. Its speed in the atmosphere is Mach 15, that is, under 5 km/s. This is 6 times faster than a Kalashnikov bullet.

The energy of motion of such a body is equal to 12.5 billion Joules. A **50-ton tank**, accelerated to 3600 km/h, will have approximately the same energy. Just imagine the force of its impact.

Now we also take into account that one “Sarmat” carries **10 such warheads at once**. Their total weight is 10 tons. And the total energy exceeds 120,000 Megajoules. One will need to accelerate 300,000 passenger cars to 100 km/h to get the same kinetic energy…

Overall, the numbers are impressive. And yet, we counted them for the “Sarmat”, which carries empty blocks. Without a nuclear charge.

With a charge, everything turns out differently. Each of the 10 atomic warheads carries the energy equivalent to 750 kilotons of TNT – 50 times more than the Hiroshima bomb.

10 warheads, respectively, carry **500 Hiroshimas**, or 7.5 million tons of TNT equivalent.

## What kind of crater will such an explosion leave?

Or, more precisely, explosions, because the warheads are separated and do not hit the same point.

The size of the craters, of course, depends on a bunch of parameters. First of all, on **blast heights**.

Thus, in the case of a real nuclear strike, the warheads explode high enough above the ground (so the damaging effect is much greater), so the crater itself may not form at all.

But even with an explosion on the surface of the earth, the size of the crater depends, say, on the landscape and geological rocks. In sand and clay, the crater is larger. In rocky terrain – less.

To get a rough idea of the destructive power of a “Sarmat” warhead, let’s imagine that it explodes on the surface of the earth in an ordinary non-rocky area.

For comparison, the Hiroshima bomb, if detonated in such conditions, would leave a “crater” **with a diameter of 90-100 meters** and about 10-15 meters deep.

It may seem that if one “Sarmat” warhead is 50 times more powerful, then the crater will also be 50 times larger. But no.

- To increase the diameter of the crater by only 2 times, the power of the explosion must be increased by 8 times (that is, 2 cubed – 2x2x2=8).
- To increase the crater by 3 times, the power of the explosion must be increased by 27 times (3x3x3=27)

Etc. That is, the dependence of the diameter of the crater on the force of the explosion – **cubic** (so to speak).

I won’t bother you with calculations and will say that if we increase the power of the explosion by 50 times (from Hiroshima to “Sarmat”) then the crater will grow by 3.7 times (the cubic root of 50).

And if after the Hiroshima bomb, the crater could be under 100 meters, then after one “Sarmat” warhead – **370 meters**. Well, roughly speaking, 400.

In depth, it will be about 7-10 times smaller than the diameter. This is 40-60 meters.

Here is a comparison of such a crater with a football field:

Well, or here it is – the Pentagon. The General Staff of the entire American Army:

By a strange coincidence, its dimensions just fit into a circle with an approximate diameter of 400 meters:

And here it the “Titanic” in a “Sarmat” crater. Also purely for comparison:

To some, the crater may seem small. But I remind you that this is only one crater. The** blast wave** itself spreads and destroys everything over a much larger area.

In addition, the blast wave is only one of the damaging factors of a nuclear explosion (it is also worth adding penetrating radiation, radioactive contamination and light radiation).

In short, where a 750-kiloton warhead falls, no one will find it small. And in total there are 10 of them in “Sarmat”. And 50 “Sarmats” will be delivered to the troops.

Total – **500 such craters** with all the concomitant damaging factors. Well, this is not counting “Bulava”, “Yars”, “Topol” and other Russian sweets…

To those who were wondering how to calculate the size of the crater from the explosion of any bomb, I can share one simple formula.

It is, of course, quite approximate, but “plus or minus” with the help of it you can find out the dimensions of the craters. The appearance of the formula is attached as a photo, but it looks like this:

“Crater diameter = 38 times the cubic root of the explosion power in kilotons”

For example, if a bomb has a power of 8 kilotons, then it will leave a crater with a diameter of about 76 metres (the cubic root of 8 is 2, and 2 times 38 will be 76).

It is quite interesting to use this formula to calculate the diameter of a crater after the explosion of a very super-bomb that our sofa warriors are all threatening to send to the United States (100 megatons, that is, 100,000 kilotons).

According to the formula, the diameter is about 1,800 meters. Some countries of the world would fit there entirely.

*I explain with my fingers*

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