Can a Whipple Shield be a viable method of space combat protection?

Question

Like it says on the tin. Problem? Basic configuration is not going to work. If I used several additional layers and all the Outer and Inner layers using carbon nanotubes, whereas the Middle Layer(s) use(es) Kevlar, Titanium or Aluminium, can they still survive incoming 5 kg guided projectiles of doom composed of depleted uranium?

Answer 1

The Whipple shield is a form of spaced armour, causing the incoming projectile to give up part of its energy prior to impacting the main body of the ship. Because the physics of hypervelocity impacts is quite different from things travelling at velocities common to things moving on Earth, the normal form of a Whipple shield is a thin piece of aluminum held mechanically some distance from the ship's hull, rather than a heavy sheet of armour plate (like WWII German tanks often sported towards the end of the conflict).

Since the Whipple Shield is generally for dissipating the energy of micrometeors and small pieces of space debris, the thin shield of aluminum works well. The incoming particles are partially vaporized by the impact and what is left is broken up and dissipated in a cone, so the hull receives many small impacts rather than a large one.

The use of aluminum is more a matter of practicality rather than due to the material properties of the shield. Aluminum is relatively cheap, easy to form into the required shapes and most importantly, very light. With current and foreseeable technology, the most important consideration for spaceflight is the overall mass of the ship, so a Whipple shield shouldn't add too much extra mass to the vehicle.

The projectiles you are describing will have far more energy than a micrometeor or a small piece of debris. If you specify the velocity, the actual impact energy in Joules can be calculated, but it is safe to assume that even at "mere" orbital velocity the slug will have more energy than the binding energy holding the molecules of the shield together. The slug could be badly damaged or even vaporized by the impact, but the hypervelocity pellets of DU will still cause considerable damage to the ship. (Edit to add: At these velocities, it really doesn't matter too much what the slug is made of; 5Kg of kitty litter or wadded up paper towel will have just as much energy. The DU slug will have a smaller cross section, making it harder to get target lock, and the dissipated cone of debris will likely be tighter with DU).

There is a possible and practical method of using the idea of a Whipple shield to protect the ship, known as "Kirklin Mines". As described in Atomic Rockets:

In AV:T are kinetic weapons called "Kirklin mines" (invented by Kirk Spencer). They are dirt cheap chemical fueled anti-missile weapons, specifically anti-Torch missile weapons. The ideas is that they cost a fraction of the price of a missile, yet can scrag it. Using the magic of relative velocity, all they have to do is get in the way (this is why they are used against torch missiles, if the relative velocity isn't large enough the mine might not do enough damage to mission-kill the missile).

In your scenario, the Kirklin mines would be a simple chemical fuelled rockets or coilgun projectiles with an umbrella type structure on the front, launched in the path of the incoming projectiles. The impact should be far enough from the ship that the pieces of destroy slugs are dissipated in a large cone with very few striking the ship (ideally the ship has time to move out of the diameter of the cone and receives no strikes at all). Even if the relative velocity isn't large enough to destroy the incoming slug outright, the energy of the impact will probably cause it to break into several pieces, or cause it to tumble and change its orbit in an unpredictable manner and be more likely to miss the ship.

For more on space weaponry and space warfare, a good starting point is the Atomic Rockets site, specifically here: http://www.projectrho.com/public_html/rocket/spacegunconvent.php, but also look up Rocketpunk Manifesto's posts on space warfare: http://www.rocketpunk-manifesto.com/search?q=space+war

Answer 2

Whipple shield of some sort is definitely most reasonable solution against most hypervolocity projectiles. However, there is broad variaty of designs, and not all Whipple shield have to be necessarily made of thin foil.

The essential principle of Whipple shield is that there is enough space between layers of the shield so that the projectile (or fragment) which impact on one layer expands and therefore spread to large area before it hits next layer. The point is to distribute impact over large area.

This follows the general problem of armor-vs-weapon. Armor have to protect the whole surface of protected object, while the weapon should penetrate it only at one point. That is why armor often require significantly more resources. Considering certain balistic coeffietient in terms of mass/area [kg/m^2], you should always assume thickness of armor is roughly proportional to thickness of projectile. The higher aspect ratio penetrator has the higher penetration or impact depth you can achieve with projectile of given mass and velocity (=>momentum). This is true as well for HEAT, APDFS as for hypervelocity meteorites and projectiles from railgun.

So the point is to spread the projectile over the large area. You can achieve that by various means, e.g. by heating it by laser (it may vaporize and expand), but the projectile is fast which means:

• you have short time to transfer required energy for expansion ($E_{EXP}$) => need high power source $P_{EXP} = E_{EXP}/t= E_{EXP} L/ v_{IN}$, where $L$ is the distance from which you start to irradiate and
• the velocity of expansion $v_{EXP}=\sqrt{2E_{EXP}/m}$ must sufficient. The expansion cone with angle $\alpha$ can be calculated as $tan\alpha = v_{EXP}/v_{IN}$ whre $v_{IN}$ is incident velocity of projectile.

=> you can see that power of the defensive laser depend super-linearly on incoming projectile velocity $~v_{IN}^{3/2}$.

Much better option how to heat incoming projectile fast enough it to heat it using its own energy by impact on spaced armor - that is the principle of Whipple shield.

By impact between projectile of mass $m_1$ on element of armor of mass $m_2$ you gain energy according to conservation of momentum $\Delta E = m_1 v_{IN}^2/2 - (m_1+m_2)(m_1/(m_1+m_2)v_{IN})^2/2 = (m_1(m_1+m_2)-m_1)/((m_1+m_2)) v_{IN})^2 = m_1m_2/(2(m_1+m_2)) v_{IN}^2 $

In other words $\Delta E = m_1m_2/(m_1+m_2) E_{IN}$ Considering that ratio $f_M = m_2/m_1$ is proportional to balistic coefficient of projectile and area density (thus thickness) of the armor, the thickness of the armor must be sufficient.

$\Delta E = f_M/(1+f_M) E_{IN} = 1-1/(1+f_M) E_{IN} $ You see that you need to make the ratio as high as possible in order to achieve sufficient expansion velocity $v_{EXP}$ (or expansion angle $\alpha$ ).

So there is fundamental trade-off for Whipple shield:

• Either you make the thickness of each layer high (so that you transform most of forward velocity of the projectil $v_{IN}$ into sideway expansion velocity $v_{EXP}$)
• Or you make the separation between layers high, so that you are fine with narrow expansion angle $\alpha$
  • yet, in this case still the thickness of first layer (thus energy converted to heat) must be sufficient to evaporate the projectile.

Answer 3

The relevant factor here is about how much slug can be disintegrated by how much armor. First let's look at the material properties of the weapon you are facing off against. For this I will use the US made M829A3 DU anti-tank round as a base line, then scale things to fit your scenario. The M829A3 fires a DU penetrator rod that is 10kg in mass, 800 mm long, and 25mm in diameter. Its front 100mm are made of hardened steel, and the remainder is DU. Maintaining this aspect ratio for a 5kg projectile, you get a rod that is about 19.4mm in diameter and 622mm long with a 78mm steel tip. Since this is for a space battle where projectile velocity is vital to landing hits, let's assume you are firing using a railgun or light gas gun at a velocity in the 3-7 km/s range.

Using this handy calculator http://www.longrods.ch/perfcalc.php we can estimate that this projectile can penetrate ~760-829 mm of armor grade steel whereas micro-meteors can only penetrate about .01-4mm of the same armor.

But total penetration is not the only factor here. The reason wiffle shields work is because micro-meteors are tiny and not super hard or thermal resistant, thus they vaporize very quickly on contact with pretty thin armor. Even though such a projectile can penetrate up to 4mm when moving at speeds of 20kps, they can be melted and break up from impact speeds of ~100-1000mps depending on what it is made out of; so, even a .8mm steel shield can break up the absolute nastiest of micro-meteors only needing to stop 1/20th of the projectile's total kinetic energy.

In contrast, DU penetrators are often tipped with hardened steel tips that are designed to not break up. Even if they broke up as easily as meteors (which they don't), it would still take several centimeter thick steel plates to get a similar effect.

Your carbon nanotubes (or more likely graphene) are much harder to find exact scientific data on, but they should perform a good bit better than steel. For the impact to generate enough heat to melt the steel tip before the tip melts or penetrates the carbon armor, would require about 2-3 cm of carbon.

That said, that much graphene would probably offer enough resistance to shatter and deflect the projectile making it not a whipple shield any more. Also, it is about 60-90 million layers thick making the cost to manufacture such an armor WAY outweigh the benefits.

In short, the answer is probably no, wiffle armor is going to be too heavy to make an effective deterrent to a 5kg DU penetrator slug unless you are looking at massive capital ships (at which point they are probably facing something bigger than 5kg slugs). Something more like a tank's reactive armor (which is designed around stopping this sort of weapon) is going to be much more cost and mass efficient. Thats said, spaceships are not tanks. In reality, there probably is not any armor that is actually light enough that you would want to use it on a spaceship to stop that kind of shell. Instead, it makes more since for your ship to focus on factors such as maneuverability, narrow cross-sections, kiting, and compartmentalization to increase survivability against such a weapon. See: What shape of a ship would be most effective in real life space combat? for more on this subject.