# Minimum Nuclear Charge Threshold?

1. Mar 30, 2006

### Orion1

Is is possible to shape and configure a nuclear charge to produce only a 700 tonne detonation? (0.7 kilotons, 0.7 kt)

What is the current minimum threshold yield for a nuclear charge detonation based upon current nuclear technology?

Reference:
http://www.breitbart.com/news/2006/03/30/060330162648.wxde5ocl.html [Broken]

Last edited by a moderator: May 2, 2017
2. Mar 31, 2006

### jdsneeder

See the following for a description of the "Davey Crockett" which used the W54 warhead which had a minimum yield of only 10 tons.

http://www.brook.edu/FP/projects/nucwcost/davyc.HTM" [Broken]

Last edited by a moderator: May 2, 2017
3. Apr 1, 2006

### Orion1

Yield:
1 kiloton = 4.184*10^12 J
0.7 kt = 2.929*10^12 J

Typical minimum achievable yield ratio:
600 kt/t

Theoretical tactical thermonuclear device yield mass:

M = 0.7 kt*t/600 kt = 1.167*10^-3 t

M = 1.167*10^-3 t * 1000 kg/t = 1.167 kg

M = 1.167 kg

Yield Energy:
$$E = \text{2.929} \cdot \text{10}^{12} \; \text{J}$$

Granite density:
$$\rho = 2600 \; \text{kg} \cdot \text{m}^{-3}$$

Granite dimensionless constant:
$$c = {\frac {E{t}^{2}}{\rho\,{R}^{5}}}$$

Yield:
$$E = {\frac {c\rho\,{R}^{5}}{{t}^{2}}}$$

$$R = \left( \frac{E t^2}{c \rho} \right)^{\frac{1}{5}$$

Penetrator finite yield strength = target impact pressure:
$$Y_t = \frac{\rho_t v^2}{2}$$ ???

Fatal dose base surge area:
$$A = 3W^{0.6} \; \text{km}^2$$
(W is the explosive energy yield in kilotons of TNT.)
For a typical third-world urban population density of 6000/km^2, estimates imply that a 1 kt weapon would kill 18,000 people.

Vaporized rock cavity radius:
$$R = 2W^{\frac{1}{3}} \; \text{km}$$

$$R = 4W^{\frac{1}{3}} \; \text{km}$$

Reference:
http://en.wikipedia.org/wiki/Nuclear_weapon_yield
http://www.brook.edu/FP/projects/nucwcost/davyc.HTM [Broken]
http://www.aip.org/pt/vol-56/iss-11/p32.html [Broken]

Last edited by a moderator: May 2, 2017
4. Apr 1, 2006

### Astronuc

Staff Emeritus
One would not do a thermonuclear charge for only 0.7 kt.

A simply nuclear charge (fissile) would suffice.

Traditional nuclear warheads were boosted with a fusion (thermonuclear) component to get more bang for the size.

5. Apr 3, 2006

### Morbius

Orion1,

An explosion of 700 tonnes is within the minimum one can do with a nuke. As others
have pointed out the Davy Crocket was 10 tonnes.

The article is in error in saying that this is the first mushroom cloud at NTS since the
halt to atmospheric testing. LLNL has a facility at NTS called BEEF - Big Explosive
Experimental Facility. While the 700 tonnes is greater than other LLNL / LANL
experiments; the smaller yields also give mushroom clouds:

http://www.nv.doe.gov/library/factsheets/DOENV_711.pdf [Broken]

http://www.lanl.gov/news/index.php?fuseaction=nb.story&story_id=3050&nb_date=2002-09-30

See "Area 4" at:

http://www.globalsecurity.org/wmd/facility/nts.htm

As the article states, if one is going to design weapons to attack deeply buried
targets; one has to be able to model the explosion. At some point, one has to tie
the computer simulation back to reality - and this is the way one does that. See:

http://www.llnl.gov/pao/WYOP/Shock_Physics.html [Broken]

Dr. Gregory Greenman
Physicist

Last edited by a moderator: May 2, 2017
6. Apr 3, 2006

### Morbius

Orion1,

That may be true - but if it is your only chance to destroy a nuke before it can be
delivered to a terrorist organization and ultimately to a large city where it could kill
hundereds of thousands or perhaps millions of people - then 18,000 may be the price
one has to pay.

I know it sounds harsh - but if you have a choice of killing 18,000 or allowing the
killing of 1 million; which do you chose? That's the type of brutal calculus one has
to face in a world with the potential for nuclear terrorism.

Dr. Gregory Greenman
Physicist

7. Apr 5, 2006

### Orion1

Taylor Equation...

Taylor equation:
$$c = {\frac {E{t}^{2}}{\rho \,{R}^{5}}}$$

How is the time component of taylors equation solved without an actual detonation?

$$t = \sqrt{\frac{c \rho R^5}{E}}$$

The reference lists 't' as the 'time since the blast'. Is this simply a time measurement of how long it takes the 'shock wave' to reach a point at a range from the detonation?

If shock waves from a nuclear detonation travel at acoustical velocity through the medium of travel, then the 'time since the blast' should be?:

$$t = \frac{R_r}{v_s}$$
$$t$$ - 'time since the blast'
$$R_r$$ - range from detonation
$$v_s$$ - acoustical velocity through medium

Integrating acoustical velocity with taylor equation?:
$$t = \sqrt{\frac{c \rho R_b^5}{E}} = \frac{R_r}{v_s}$$

Taylor equations?:
$$c = \left( \frac{R_r}{v_s} \right)^2 \frac{E}{\rho R_b^5}$$

$$c = \frac{E}{\rho v_s^2} \left( \frac{R_r^2}{R_b^5} \right)$$

$$R = \left( \frac{E t^2}{c \rho} \right)^{\frac{1}{5}$$

For a given kiloton detonation inside solid granite, would the 'vaporized rock cavity radius' be equivalent to the 'blast radius'?

blast radius = vaporized rock cavity radius:
$$R = \left( \frac{E t^2}{c \rho} \right)^{\frac{1}{5}} = 2W^{\frac{1}{3}} \; \text{km}$$

Reference:
http://en.wikipedia.org/wiki/Nuclear_weapon_yield

Last edited: Apr 5, 2006
8. Apr 5, 2006

### Astronuc

Staff Emeritus
I believe the answer is no. There can be additional deformation in the solid material, that is the solid ground will be displaced but not vaprized.

The amount of material vaporized will be less than the amount displaced.

The blast radius will depend in the density and viscosity (friction) of the medium in which the blast occurs.

The situation becomes more complicated if there is considerable variation in mechanical/physical properties of the medium, e.g. a composite medium with high and lower density components. The blast wave (energy) takes the path of least resistance.

9. Apr 14, 2006

### Orion1

What is the minimum nuclear charge threshold for a thermonuclear detonation?

.7 kt is the minimum thermonuclear charge threshold?

Teller blast cavity radius:
$$R = C \frac{Y^{\frac{1}{3}}}{(\rho h)^{\frac{1}{4}}}$$

C = 57.70 - 60.48 Granite
Y = kt (kilotons)
rho = 2.7 Mg*m^-3 (Megagrams per cubic meter)
h = depth m (meters)

Thermonuclear threshold blast cavity radius: (.7 kt)
$$R = 60.48 \frac{.7^{\frac{1}{3}}}{(2.7 h)^{\frac{1}{4}}}$$
$$R(h) = \frac{41.893}{h^{\frac{1}{4}}}$$

$$\boxed{R(100) = 13.248 \; \text{m}}$$

A .7 kt thermonuclear charge at a depth of 100 meters inside solid granite will produce a blast cavity radius equivalent to 13.248 meters.

kiloton blast cavity radius:
$$R(h) = \frac{60.48}{(2.7 \cdot h)^{\frac{1}{4}}} = \frac{47.181}{h^{\frac{1}{4}}}$$

$$R(h) = \frac{47.181}{h^{\frac{1}{4}}}$$
$$\boxed{R(100) = 19.125 \; \text{m}}}$$

A 1 kt thermonuclear charge at a depth of 100 meters inside solid granite will produce a blast cavity radius equivalent to 19.125 meters.

1 kt in tuff at 122 m can produce a crater with a radius of 20.2 m

Penetrator finite yield strength = target impact pressure:
$$Y_t = \frac{\rho_t v^2}{2}$$

Penetrator velocity threshold: (.7 kt)
$$v_t(Y_t) = \sqrt{\frac{2 Y_t}{\rho_t}} = \sqrt{\frac{2(.7 \cdot 4.184 \cdot 10^{12} \; \text{J})}{2700 \; \text{kg} \cdot \text{m}^{-3}}$$
$$\boxed{v_t(.7 \; \text{kt}) = 46.577 \; \text{km} \cdot {s}^{-1}}$$

Penetrator velocity threshold: (1 kt)
$$v_t(Y_t) = \sqrt{\frac{2 Y_t}{\rho_t}} = \sqrt{\frac{2(4.184 \cdot 10^{12} \; \text{J})}{2700 \; \text{kg} \cdot \text{m}^{-3}}$$
$$\boxed{v_t(1 \; \text{kt}) = 55.671 \; \text{km} \cdot {s}^{-1}}$$

Reference:
http://nuclearweaponarchive.org/Library/Effects/UndergroundEffects.html

Last edited: Apr 14, 2006