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#1 |
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Bronze Medalist
Jan 2004
Mumbai,India
22×33×19 Posts |
Guided missile. A guided missile base is on the shore 100 miles due south of an enemy plane when it fires a heat seeking missile at the enemy plane which is flying at 1000 m.p.h. due East. In other words the missile is constantly pointed at the plane. The speed of the missile is 2000 m.p.h How far will the plane have travelled when it is intercepted by the missile? How long would it take for interception? Mally
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#2 | |
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Bronze Medalist
Jan 2004
Mumbai,India
22×33×19 Posts |
Quote:
HINTNot to complicate matters assume the plane is flying at tree top level and so its altitude does not count and is negligible. Mally
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#3 |
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Jun 2003
The Texas Hill Country
32×112 Posts |
And further, I think that we should ignore the curvature of the surface and assume that everything is in the plane of the plane. Is that plain?
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#4 |
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Sep 2006
Brussels, Belgium
13·131 Posts |
I did not want to plunge in to the integrals, but via a discrete approach I got 240 s as the interception time. The distance the plane or the missile will have travelled is trivial to compute knowing the speed of each and the time to interception.
Plane 1000*240/3600=66.6666... miles Missile twice that much 133,3333... miles |
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#5 |
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Bronze Medalist
Jan 2004
Mumbai,India
22×33×19 Posts |
240 secs? That a bit too long a duration. Could you please disclose your 'discrete approach'? You dont need calculus to derive the answer and theres no catch in it as such. Mally
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#6 |
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Sep 2006
Brussels, Belgium
13×131 Posts |
I just computed the movement of both the plane (1000 miles/h to the east.) and the misile moving towards the plane at 2000 miles/h. I decomposed the movement in NS and EW direction, calculating the direction via an arctg and this for very small time intervals (about 0,00001 second, lazyness can have you do more work :-).
Even then the result does not seem a lot to me : if plane and missile where traveling in the same direction, one in pursuit of the other it would take 100 / (2000-1000) hours or 360 seconds to interception. In this cas the missile takes a shortcut. The shortest time to intercept would be if the missile pointed directly to the computed point of interception : Let P be the distance flown by the plane, Let M be the distance flown by the missile Because the plane is moving at 90 degrees of the original position from the missile we have (100 miles)2+P2=M2 Since we know that the missile is twice as fast as the plane M=2P This gives 10000+P2=4P2 This is equivalent to P2=10000/3 P=100/30.5=57.735... miles Then the time is 57.735/1000 hours which is equal to aproximately 207.846 seconds. But this is not according to your hypotheses, because to achieve this the missile points towards the expected point of interception. The difference between 240 s and 207,846 does not seem much to me given the different path used. Last fiddled with by S485122 on 2006-11-06 at 18:32 Reason: conclusion |
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#7 | |
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Jun 2005
17E16 Posts |
Quote:
The optimal solution to this type of pursuit is called 'proportional navigation'. The pursuer measures the rate of turn of the line of sight to the target, and turns at a turn rate proportional to (but greater than) the turn rate of the line of sight. The proportionality constant is called the 'navigation constant', or K_nav. This is optimal in ideal circumstances, but imperfect sensors and lags closing the loop on turn rate can make the problem difficult in real-world situations. If you linearize this problem using small angle assumptions, you can deduce that a K_nav of 2 is required to prevent the pursuer's turn rate from increasing as it approaches the target. This represents a constant turn rate toward the target. A K_nav of 3 results in a turn rate of zero at the point of impact, and a K_nav of 4 results in a zero turn rate rate (second derivative) at the point of impact...and so on. It's a fascinating mathematical problem. Drew Last fiddled with by drew on 2006-11-06 at 22:51 |
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#8 |
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Bronze Medalist
Jan 2004
Mumbai,India
22×33×19 Posts |
Drew: I appreciate your analysis of this problem as also Jacobs solution, which for the time being I will remain mum about it, to let others have a go at it.. Please remember this is a mathematical problem and as such the conditions are ideal as Wacky made so 'plain' about the 'plane'. Mind you Jacob I'm not dismissing your solution, but you have given two or more solutions, so which one do you believe is the right one? Mally
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#9 | |
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Jun 2005
2×191 Posts |
Quote:
Drew |
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#10 |
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Sep 2006
Brussels, Belgium
13×131 Posts |
Mally,
I gave only one solution: the first at 240 seconds. The other times I gave where based on other data than that of the problem (both objects travelling in the same direction, one in pursuit of the other gives a 360 interception time and missile pointing directly at the "meeting" place gives 207 odd seconds.) |
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#11 | |
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Bronze Medalist
Jan 2004
Mumbai,India
80416 Posts |
Quote:
Heat seeking missiles have heat sensors perhaps tuned to the jet planes' exhausts. Hence their course is altered continuously pointing to the target's direction and speed. In our last war (Indian ) with our neighbours heat seeking 'Sidewinders' were launched against our jets with devastating effect initially. To 'confuse' their missile guidance system our fighter jets flew in pairs. The result was for the sidewinder to take the resultant path in between them and fly past (in between) without destroying the jets. Simple mathematics put to good use! Mally
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