Wasserfall's PK against Allied Aircraft if deployed?

[Wanderer]

A lot of Nazi Wank Scenarios have the German Wasserfall become this massive Aircraft killer downing allied bombers left and right and when I asked the creators of these wank scenarios what the PK(probability of a kill) percentage was, I get blank looks.

So out of morbid curiosity I am asking the more knowledgeable members of the board what Wasserfall's PK would have been against Allied Bombers after they get past the massive ECM/ECCM of the bomber stream?

For those who don't know what Wasserfall is, here you go.

Also, the Wasserfall despite the articles claims was not connected in any way to U.S. development of SAMs which was done independently of any Nazi projects. Project Nike began in 1944 when the War Department demanded a new air defense system to combat the new jet aircraft, as existing gun-based systems proved largely incapable of dealing with the speeds and altitudes at which jet aircraft operated. Two proposals were accepted. Bell Labs offered Project Nike. A much longer-ranged collision-course system was developed by General Electric, named Project Thumper, eventually delivering the BOMARC missile.

[MKSheppard]

I'd say it's probably 0.25 or perhaps 0.15. Which means to down a single aircraft, you'd need about 4 to 7 wasserfall. Of course, this is a massive improvement over the 3,343 shells fired from an 88mm Flak Cannon to down a heavy bomber.

In “The Wages of Destruction: The Making and Breaking of the Nazi Economy” it's estimated that a single 105mm Howitzer round required 66 pounds of steel per shell. Since a 105mm HE round weighed only 32 lbs; this meant that it took twice as much steel to make a shell as it weighed. Since a 88 shell weighed 20 lbs, it probably took 41.2 pounds of steel to make an 88mm shell.

So those 3,343 shells mean it takes the equivalent of about 68 tons of steel to down each bomber; steel that could be used elsewhere.....

[Coyote]

So, it may not have improved the kill ratio, but it would have improved the economics behind the kill ratio...

[Wanderer]

I'd say it's probably 0.25 or perhaps 0.15. Which means to down a single aircraft, you'd need about 4 to 7 wasserfall.

The .25 is for the B-17/B-24 and .15 for the B-29, I assume?

[MKSheppard]

The .25 is for the B-17/B-24 and .15 for the B-29, I assume?

No, it's a range of best case (0.25) to worst case (0.15) for the B-17/B-24.

[Wanderer]

The .25 is for the B-17/B-24 and .15 for the B-29, I assume?

No, it's a range of best case (0.25) to worst case (0.15) for the B-17/B-24.

Thanks for clearing that up Shep.

[MKSheppard]

Honestly, the issue is really fuzzy; because the early guided weapons of this era were easily decoyed; for example, the Germans scored quite a few notable successes with their Fritz-X guided bombs, but when the US Navy deployed electronic jammers to their ships, the effectiveness of Fritz-X fell off dramatically.

[Sea Skimmer]

It is a very complicated issue, made worse by the fact that the Germans did work on several different guidance systems.

I think a 25% hit rate is unlikely, even those anti ship glide bombs only managed a slightly higher success rate and that’s aiming for a much bigger target moving at 1/10th the speed in 2-D only. I like 5-10% is probably the best they would have managed, before jamming dropped this away to less then 1%. As for cost effectiveness, the missile weighed no less then 3,700kg and had a liquid fuel rocket motor, it would have been very expensive to build, estimated at 10,000 reichsmarks per unit assuming a high rate of mass production. A V-2 for comparison cost about 50,000 reichsmarks with mass production.

However something to keep in mind is that the missile had a huge warhead, and if and when a missile did hit a bomber formation it would have the potential to damage or destroy multiple tightly packed aircraft. This might force the USAAF to simply abandon daylight raids over Germany until it could equip aircraft with jammers. The RAF bombing at night would be largely unaffected, the guidance system is too crude to hit anything at night with enough reliability to be worth firing.

[Stuart]

It is a very complicated issue, made worse by the fact that the Germans did work on several different guidance systems.

The only guidance system that was practical for Wasserfall was manual command to line of sight (MCLOS). Essentially the operator held the target in view with a pair of binoculars and used a joystick to keep the missile in the line between those binoculars and the target. This has quite a few problems, not least of which is that the missile is on a very fuel-inefficient course. There's also a capture problem in that the operator has to get the missile within his binocular's field of vision very quickly after launch; if he fails to do so, then the missile goes ballistic and crashes.

I think a 25% hit rate is unlikely, even those anti ship glide bombs only managed a slightly higher success rate and that’s aiming for a much bigger target moving at 1/10th the speed in 2-D only. I like 5-10% is probably the best they would have managed, before jamming dropped this away to less then 1%.

25 percent is way out of the ballpark. To get an idea of the effectiveness of MCLOS, anti-tank missiles using this system got, at best, 10 percent accuracy (rarely they got 25 percent when fired against stationary targets) and a hit rate of 1 - 2 percent was more common. If we average that out, we can give MCLOS anti-tank missiles a pk of around 0.05. Your logic about target speed and the 3-D environment is spot-on; also the range factor is greater for aircraft and the missile speed is much greater for teh Wasserfall - this makes capture and control much harder (by the way, note that Wasserfall cannot be used at night or in bad weather). So, I would expect the pk of the missile against aircraft to be at least an order of magnitude less against aircraft - so we have a pk of 0.005 (a hit rate of 0.5 percent). That's very optimistic.

The radio link to Wasserfall is extremely susceptible to jamming. Just blasting out white noise across the whole frequency spectrum will do it. Other tricks applicable include dropping high-intensity magnesium flares to obscure the target and to blind the operators (think of looking at the sun through binoculurs. ) I'd say that countermeasures would cut the effectiveness of Wasserfall by at least another order of magnitude - and probably much more but let's stay with that.

So, the final Pk against aircraft equipped with defensive countermeasures would be 0.0005, a hit rate of 0.05 percent, one hit for every 2,000 missiles fired. The Germans could do maths, I'd guess that's why they cancelled the missile (they wouldn't have the anti-tank missile experience to darw on but they had been playing with teh guidance system so they knew it was a non-runner

However something to keep in mind is that the missile had a huge warhead, and if and when a missile did hit a bomber formation it would have the potential to damage or destroy multiple tightly packed aircraft. This might force the USAAF to simply abandon daylight raids over Germany until it could equip aircraft with jammers.

The warhead on Wasserfall was 306 kilograms, 684 pounds. Now remeber, destructive power is proportional to the cube root of explosive power (because the blast dissipates in three dimensions). The explosive weight of an 88mm shell was around 7 pounds (the whole shell weighed 20 - 23 pounds). So, allowing for the cube effect, the destructive radius of a Wasserfall is around 4.7 times greater that for an 88mm shell. That's bad but not critical. The destructive radius of an 88mm shell is (IIRC) around 20 feet max so that would give the Wasserfall a destructive radius of around 94 feet. B-17s even when boxed up had more than 100 feet between them so the prospect of multiple kills is remote.*****

By the way, the rate of fire of an 88 maxed out at around 20 rounds per minute. That means the shells it could put into the air in that time would have the following cumulative radii

One Wasserfall, radius of destruction = 27,763 square feet
20 88mm shells radius of destruction = 25,136 square feet

That gives a single Wasserfall a 10.5 percent edge over a single 88mm gun firing for one minute. Hardly an impressive performance edge and cost-efficiency is appalling (the 88mm can go on to fire 20 more rounds, the Wasserfall is GONE).

***** I checked the mathematics of this to confirm it.

The lethal blast radius of a 25 kiloton anti-aircraft warhead is 1,000 yards. 25 kilotons is the equivalent of 25 million kilograms of TNT. The explosive load of an 88mm shell is around 3 kilograms so the 25 kiloton warhead is 8.33 million times greater in explosive power than that of an 88 mm shell. the cube root of 8.33 million is around 204. The lethal radius of an 88mm is 20 feet so 20 x 204 = 4080 feet or 1,360 yards which is close enough for government work. Actually I suspect the lethal radius of an 88mm is closer to 15 feet than 20 which makes the maths work out almost perfectly.