WaPo: China’s carrier-killer missile

[Doktor]

So, since about 1/3 of the carriers are docked at almost all of the time, or should be... the best way to sink one would be to send a salvos to the yards and wait for the other 2/3 to come to you. At least you killed 4 carriers and your military.

[Blademaster]

Here's USN's response to the DF-21.

Breaking China's DF-21D missile kill chain: US expert | Missile ThreatMissile Threat

Breaking the Kill Chain

It seems that to counter the DF-21, the USN will kill the missile's kill chain first before putting in the carrier in harm's way first.

[cdude]

too many technical questions and no demonstrated Chinese capability to fire that many missiles at one target at one time. let alone at a defended and moving target.

I think the whole thread is based on the assumption that China had the capability to do certain things.

With the capability to fire multiple missiles on one ship in a coordinated way, "to move twice the width of the beam of the ship" is not going to help that much.

[Captain Worley]

@Weaponww, a warhead big enough to threaten a carier and moving at ballistic speeds is an [kinetic] energy machine to rival said carriers reactor, around 9 billion joules... You are not going to easily or radically change course to X with something that has 663 million foot pounds of force going in direction Y. You claim .5 degrees is an easy change, I strongly disagree unless you can show how a warhead can apply over 3 million foot pounds of force almost instantly.

.5 mass x velocity squared= physics is a bitch.

To pick nits, foot pounds is a unit of work, .5mv^2 calculates kinetic energy, and pound(f) is a unit of force.

Also, warheads wouldn't need as much steering input as the shuttle because the have much less mass and rotational inertia to overcome. The US spent a lot of money into steerable warheads in the eighties and early nineties...I'm really not sure what the state of the art is now as I've been out of the loop for a couple of decades.

[Josh]

ASAT potential

Satellite guidance wont matter much for aiming IF the US deemed the "carrier killer" a threat. Especially against weaponizing something like this. No more satellite for aiming.

China deeply suspicious of X-37B - Technology & science - Space - Space.com | NBC News

There are a couple other unclassified systems that would probably easier to use for LEO satellite engagements than the X-37. The missile defense system and MIRCL come to mind. SM-3 may have altitude restrictions that don't make it truly effective for normal engagements, but the GMI or whatever its called now would have plenty of KE and most high angled recon orbits would take you over the site eventually. There might be classified systems as well; there were a number of experiments with micro satellites carrying out close maneuvers around each other that seemed a little iffy on other utilities, except maybe opposing satellite inspection. The PRC also maneuvered a pair of its satellites dangerously close together, in addition to their more blatant ASAT test.

[Officer of Engineers]

Know what? Optical solution will not work for the DF-21D. You need at least two pictures taken at least one second apart to determine vector. That's 3 seconds out of a 8 second window gone.

[cdude]

Know what? Optical solution will not work for the DF-21D. You need at least two pictures taken at least one second apart to determine vector. That's 3 seconds out of a 8 second window gone.

Use multiple missiles in a coordinated attack, just kill the whole block. Aint need no pictures.

[Doktor]

Use multiple missiles in a coordinated attack, just kill the whole block. Aint need no pictures.

Check the hit (not kill) ratio from coastal batteries during WW2 vs capitol ships. Might give you an idea if your plan is workable. That's with eyes on the ship.

[Officer of Engineers]

Use multiple missiles in a coordinated attack, just kill the whole block. Aint need no pictures.

Your block is 200 miles radius.

[cdude]

Check the hit (not kill) ratio from coastal batteries during WW2 vs capitol ships. Might give you an idea if your plan is workable. That's with eyes on the ship.

I was thinking of multiple DF-21's in a coordinated attack. It would give you a better chance of hitting a moving ship, that's all.

[cdude]

Your block is 200 miles radius.

I'm assuming the PLA knows where the ship is and tracking it. Otherwise, why bother.

[Doktor]

I'm assuming the PLA knows where the ship is and tracking it. Otherwise, why bother.

The tracking part is tricky. It has been discussed in depth so far.

I had the idea that the ship of that size can't be hard to track. Apparently.... :pari:

[Officer of Engineers]

I'm assuming the PLA knows where the ship is and tracking it. Otherwise, why bother.

This has been our entire discussion on this thread.

[weaponww]

@Weaponww, a warhead big enough to threaten a carier and moving at ballistic speeds is an [kinetic] energy machine to rival said carriers reactor, around 9 billion joules... You are not going to easily or radically change course to X with something that has 663 million foot pounds of force going in direction Y. You claim .5 degrees is an easy change, I strongly disagree unless you can show how a warhead can apply over 3 million foot pounds of force almost instantly.

.5 mass x velocity squared= physics is a bitch.

your are using kinetic energy which is the energy release upon impact due to motion. it should be Force, specifically angular force or commonly known as centripetal force, The component of force acting on a body in curvilinear motion that is directed toward the center of curvature or axis of rotation. So...

if an object is traveling at 6000m/s, and has turn rate of 0.3degree/s(.28 from the previous post, assume it can turn that much for 1sec). then if the object continue the same speed, direction and turn rate, it will eventually make a circle, 360degree.
360d/.3d/s=1200s to make a full circle.
1200s*6000m/s=7200000m, this is the Circumference of the circle
7200000m/(2*pi)=radius of the circle= 1145900m, lets called it r
Now the Physic part. for the angular force exert on a object traveling anularly
F=mv^2/r the m lets pick from 100kg-1000kg.
F=100kg*(6000m/s)^2/1145900m=3141.6kg m/s^2=3141.6 Newton.
so what is this mean, lets do the G-force conversion
1g=100kg*9.8m/s^2=980 for 100kg mass 1g=1000kg*9.8m/s^2=9800 for 1000kg
so
3141/980=3.2G force apply to the object. if plug in 1000kg as the mass into mv2/r and the 1g force, the outcome G-force is the same.
if plug in .5 degree/s G-force is between 5-6G.
I'm pretty sure even a human can handle 3-6G for 1 sec. cause thats amount of force exert on the missile for 1sec, and the missile only need to change .3 degree at 40km for its new path.

if you still doubt the result, goto any university ask the physic professor on this!!

[weaponww]

And this is demonstrated where?. Like the example I gave you. The Space Shuttle starts final approach 145 miles out. You don't change the trajectory of a high speed, high mass object as easy as moving the barrel of your gun.


No its not, and this is the part that you are not grasping. Its not like moving the rifle and adjusting your aim. There the projectile is traveling at 0.0 MPS. This is like trying to Jeddi Mind trick the bullet into altering its course. The bullet is rapidly going from your 1Km range to the 20 meter range as you are trying to break its inherent want to travel down the original path. And while you are doing that you are eating distance and increasing the angle of your solution.

Your not aiming a rifle from above, your steering a bullet that is happily flying down a high speed stabilized path that it will resist you trying to change.

And while we are figuring out this ballistic mission, don't forget to throw in the effects on the flight path (and your attempt to change it) of such things as air density,air temp, rotation of the earth, gravity, ballistic and drag coefficient, and at the target, sea state. Those are just the things that come to mind

the angle between 3 pts and distance is relative. what ppl say about 3d angle is the angle relative to X-axis(lets call ax), angle relative to y-axis(ay), and angle relative to z-axis(az).

do this experiment, holding a waterbottle in one hand(the bottle position is b), and use your index finger point at it(your hand position is a). then move the bottle updown,side in any direction, your new bottle position at X time is c, now point your finger at C.
thus you establish a triangle base on a b c. and the relation of the angle is base on the length/distance of the side. this angle has ax and ay component in it.
you can move your hand or bottle in any direction, but the angle relationship between these point can be calculate via geometry/trigonometry. thus its the relative position between these points matter, and its distance.

and you can ask a good math/physic teacher about it.
if you want get into deeper as how to find the relations between your angle of change vs ax ay az use rotation matrix
Rotation matrix - Wikipedia, the free encyclopedia

if the angle of change relative to a b c pts is not known, but you know the ax ay az, then you can find the angle of change.