[QUOTE=drew;92754]Any missile must operate under certain assumptions regarding the trajectory of the target that won't hold steady because a maneuvering target is unpredictable. It's reasonable to chase the target vehicles current inertial state, so computing the intercept vector (using the visual feedback I described earlier) is a good approach even for a sophisticated missile.

Against a maneuvering target, it needs to be prepared to continuously re-evaluate this interception as the target maneuvers.

Drew[/QUOTE]

Agreed, it has to keep re-evaluating based on new information as it comes in.

Knowing the target's current direction and speed at each point of evaluation is useful information in making a decision of the direction to intercept that the missile should take right now. If either changes, the missile needs to adapt, but such things are common sense ;-)

Your decision on missile heading may be modified by whether you believe the target will have spotted your missile (eg based on radar range). Thus if the target is still heading east it likely will continue to do so until the missile will eventually be spotted. I would recommend some initial missile direction as to move towards some point between the current location of the target and the point where the missile would intercept provided the target maintained speed and direction. This compromise does not put you too far away to the east if things change, but sends you generally in the right direction until you're spotted on radar and the game changes ;-)

Better is to send two missiles on divergent courses so that one is closing in whichever way the target goes.

Realistic problem
 

[QUOTE=Peter Nelson;92751]Mally I don't think this is a very realistic problem

I have worked on imaging systems ie cameras and image processing electronics, which are used for realtime analysis onboard "smart" missiles. Rather than just look at the heat of a vapour trail (where your two plane technique could avoid being hit) they actually LOOK and would IDENTIFY BOTH targets and parts of targets like a wing. They would not be fooled by such tricks so the Indian airforce had better have some new evasive techniques or be shot down in flames.[/QUOTE]
:smile:
Peter: you are absolutely right and up to date in missile technology:
Drew:. so are you.
My problem was an academic one and not suited practically to meet modern state of the art technology.
The Keely brothers (both AI's) had no problem with vapour trails in the last war or heat seeking missiles and were decorated for bravery We found that it was the man behind the machine that counts eventually
Yes we are taking evasive action and possiby switching to stealth fighters and bombers AFSAIK once we master the technology.
Regards,
Mally :coffee:

Calculus solution.
 

[QUOTE=Wacky;90966]

I had wondered HOW you were able to compute your answer. I could not figure out how to do so without the use of calculus.

It is obvious that your interpretation of the constraints is different from that of the rest of us.[/QUOTE]

:smile:

Guided missile problem with Calculus solution.

We all are well versed in the problem by now so I wont repeat it again.
Please refer to the left thumbnail of Jacob Visser as a guide.
The solution has been kindly forwarded to me by Prof. V.S. Mokashi of Nagpur.

Taking the general case, let the constant magnitudes of the velocities be Va and Vb of A and B resp. be v and u (v >u) and let L be the initial distance AB between them.
At instant ‘t’ let A be at Ai on the curve and B at Bi along the axis eastward. Let Eta be the angle made by Va with the easterly direction
Resolving the velocities along Ai ,Bi, the components of velocity of A relative to B taken along Ai, Bi will be (v – u cos Eta) which means that the distance between them is diminishing at the varying rate of (v – u cos Eta) per unit time.

The initial distance between A and B being L if T is the time taken by A to meet B we shall have
S (abbrv. For Integral) we have S = (v – u cos Eta).dt – L = 0
S having limits from 0 to T.

Hence vT – u S cos Eta.dt = L ----------------(1)

Also at the instant of meeting, equating the distance travelled eastwards by both A and B is S vcos Eta.dt = uT or v S cos Eta dt

Hence S cos Eta. dt = uT/v -------------------(2)

Here both Eta and t are variables and Eta as a f(t) being unknown the integral cannot per se be evaluated. However for the sake of the solution of the problem this is not needed to be done.

Putting the value of the integral (2) in (1) we get

vT - u(uT/v) = L

or T = vL/(v^2 – u^2)

By data v = 2u

Therefore T = 2uL/(4u^2 – u^2) = 2 L/3u

Substituting we get T = 2*100/3*1000 =2/30 = 0.066666* hrs.

Converting into secs we get 239. 9999* secs. As obtained by you all.

Mally :coffee: