Rough guess on first strike capacity in 1990.

A Minuteman III intercontinental ballistic missile successfully launches at 1 a.m. Nov. 5 from Vandenberg Air Force Base, Calif. The missile was configured with a National Nuclear Security Administration test assembly in which a single unarmed re-entry vehicle traveled approximately 4,190 miles to their pre-determined targets near the Kwajalein Atoll in the Marshall Islands. (U.S. Air Force photo/Joe Davila)

Did the Soviet Union, that now dead Evil Empire, have a first strike capability against US nuclear forces in 1990?

Did the US have first strike capability?

Let’s ponder those questions.


It is important to add a disclosure at this point on the source of data. Every piece of information going into this series of posts is obtained from public sources. I have not used anything that is classified to develop this discussion. Again, everything is from an open, public source.

In a funny way, that means you can ignore everything I say because any specific piece of information may or may not be correct. On the other hand, that means I can freely talk about all this information and speculate in any way I wish to speculate.

Finally keep in mind I am uneducated in a lot of the things discussed here, so there is a distinct possibility I grossly misunderstood something or omitted a massively critical factor. In fact, I am quite willing to admit it is even possible I don’t have any idea what I’m talking about.

Range of information in public sources

This series of posts on nuclear weapons has accumulated data on the American and Soviet nuclear inventories at several points in time: 1980, 1983, 1990, and 2020.

Only point for which I have found circular error probable (CEP) data for all the ICBMs is 1990.

The same resource which provided that information, Physics and Nuclear Arms Today (Readings from Physics Today), also contained a graph which moves us forward on considering first strike capability.

This particular graph plotted out each weapon in terms of expected pounds per square inch (psi) of overpressure at the target given the weapon’s CEP and yield.

A Minuteman missile silo, according to the article, is constructed to withstand 2000 psi. That amount or more will damage the silo so much that the missile inside cannot be launched. In a very dark, macabre, humorous way the graph also contains a notation at which point the target would have at least 4 m of dirt over it. In addition to damaging the silo so badly that it could not be used for launch, there would also be at least 12 feet of dirt piled on top of the silo.

That graph provides us the ability to determine which of the American and Soviet ICBMs and SLBMs would have accuracy and yield sufficient to on average kill a silo hardened to US standards. I will combine that information with the weapon inventory. Will also assume that Soviet silos are hardened to the same strength.

This graph shows the SLBMs would not be able to kill an ICBM silo, so I will leave those out of the discussion.

Next step is to keep in mind the CEP concept means half the missiles will hit closer and half the missiles will hit further away. I will make the simplifying assumption that 50% of the missiles would kill the target and 50% would be a miss.

That in turn allows calculating how many missiles it would take to have a very high probability of killing one silo. Dividing the number of ICBMs by opposing targets shows how many shots can be taken against each silo.

So, let’s do some calculations.

Calculation of first strike capability

Previous analysis showed that all of the American ICBMs would not only knock out a silo but also throw at least 4 m of dirt on top. None of the SLBM’s would take out a land-based silo.

Following table shows 2,450 reentry vehicles (RV) are available which could destroy a silo in the American inventory. There are 1,000 launchers.

US ICBM recap  launcher  total warhead  total mt  total Equiv mt
damage on 2,000# psi silo
no            –           –          –          –
yes            –           –          –          –
4 m       1,000       2,450      2,452     1,375
 —  —  —  —
total       1,000       2,450      2,452     1,375


There is a wider range of effectiveness in the Soviet inventory. The SLBMs are, um, not effective against a land-based silo so I will inventory the ICBM in terms of effectiveness. (That is an amusing understatement because one of their SLBMs had a CEP of 13 kilometers, or 13,000 meters.)

I calculated there are 6,197 reentry vehicles which could destroy a silo. There are 1,451 ICBM launchers.

Soviet ICBM recap  launcher  total warhead  total mt  total Equiv mt
damage on 2,000# psi silo
no          460         460         436        443
yes          168         852         306        412
4 m          823       5,345       2,786     3,461
 —  —  —  —
total       1,451       6,657       3,528     4,316


Pulling the information together shows how many reentry vehicles have a more than 50% probability of destroying a target and the number of silos on the opposing side.

If we assume each side were to target all of its ICBMs against the opposing ICBMs this gives us the number of RVs per target, in other words how many RVs could be launched against each silo.

US attack Soviets Soviets attack US
yes              –            852
4 meter         2,450         5,345
>50% kill         2,450         6,197
ICBM target         1,451         1,000
ratio             1.7             6.2


We can then calculate the average probability of one RV destroying one silo. The probability is 50% because the concept of CEP means half the missiles will hit closer and half further away. Since there will be multiple RVs per target, we can calculate the probability of two RVs destroying a silo, which is 75%. The probabilities can be extended. Here’s my calculation:

US attack Soviets Soviets attack US
probability of kill
1 50.0% 50.0%
1.7 69.2%
2 75.0% 75.0%
3 87.5%
4 93.8%
5 96.9%
6 98.4%


My massive string of assumptions and calculations show that with the US having an average of 1.7 ICBM RVs available per target there would be a 69% probability of destroying an individual target.

The far larger Soviet inventory would allow targeting six RVs against each silo which will give a 98% probability of destroying an individual target.


If we then apply the average probability of killing one target against the number of targets, we can calculate the number of silos expected to be destroyed.

We can subtract that from the number of targets to calculate the expected number of surviving silos.

Since the success of individual RVs is an average, it would not be possible to determine which specific silos would be destroyed. So I calculated the average number of RVs per launcher.

We can combine the number of silos that would be expected to survive with the average number of RVs per launcher to make a completely wild guess on the number of RVs that might be expected to survive.

US attack Soviets Soviets attack US
probability 69.2% 98.4%
targets         1,451         1,000
expect kill         1,004            984
expect silo survive            447              16
ave RV/launcher           6.25           2.45
expect RV survive         2,791              38



The result of my extremely long list of disclosed assumptions suggest that something in the range of 2,800 Soviet RVs might be expected to survive a first strike attack from the US.

In contrast, with my list of assumptions, there might have been something in the range of 40 American RVs expected to survive a first strike attack from the Soviet Union.

It looks to me like the Soviet Union did in fact have a first strike capability against the United States in 1990.

It looks to me like the United States did not have a first strike capability against the Soviet Union in 1990.


What do you think of my analysis? My conclusion?

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