Guided Missile.

[mfgoode]

Originally Posted by Jacob Visser

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.)

Well your first solutions is way off the mark. We are dealing with missiles and every second counts as you will be aware off if you have some war experience

Surprisingly your other solution (207. .. secs) is exact and correct.

The missile does not compute the point of interception at all neither is it aimed there. Instead it constantly changes course as the plane moves as Drew has explained. It is right on target at all times.

To see this, try stopping the plane with a vector and naturally add the same to the missile. It will take you to your right answer with the appropriate vector diagram.
Best of luck Jakes! c'mon you can do it easily and more convincingly.

Mally

[axn]

Originally Posted by mfgoode

Surprisingly your other solution (207. .. secs) is exact and correct.

No it isn't. That is the _least_ amount of time required, assuming a direct intercept course.

But as per the problem, the missile has to travel along a curved path.

FWIW, my excel calc also tells me 240 secs.

[S485122]

Mally.

Since 207 seconds is the solution where the missile points immediately at the interception point, the real solution where the missile continuously changes course must be longer.
The 240 I got are calculated on the sames basis as the method you propose. The problem if you get 207 seconds is that you do not take small enough time iterations. Do not forget the initial vector of the missile is pointed due north. Next moment the plane has moved a bit east, the missile adapts its vector, and so on. It is only after some time interval that the West-East component of the vector of the missile is bigger than its South-North component.

Let me put it in another way: in 207 seconds the plane travels 1000*207,846/3600 miles = 57.735 miles, in the same time the missile has travelled double this distance: 115.47 miles. Note that 100²+57.735²=13333 and that 115²=13333 (all figures are rounded of which means the numbers are only approximatively equal.). Doesn't this seem like the formula of Pythagoras?

[mfgoode]

Originally Posted by axn1

No it isn't. That is the _least_ amount of time required, assuming a direct intercept course.

But as per the problem, the missile has to travel along a curved path.

FWIW, my excel calc also tells me 240 secs.

Please read my problem again. No where does it say that the missile has to take a curved path It is assumed it takes a curved path but this path is not relevant to the problem. It simply says the missile has the plane in its 'sights' at all times.

I presume that it is not what your exel calc says but how you program it.

It is only natural that a missile should take the least time for an interception.

Men of war dont send missiles on a roller coaster ride! They mean business.

Still I wont twist your arm to believe my solution. From past experience this has only lead to arguments of 'you are wrong; I am right' esp. when Wacky is around and probably still moderator of this thread
.
I have presented a mathematical curiosity and people like Drew have appreciated the depth of it without giving a solution as such.

So axn1 you either take it or leave it and trust your excel calc to work out your solutions. I hope you dont have to sit for an examination as you may be surprised at the result.

Mally

[axn]

Originally Posted by mfgoode

Please read my problem again. No where does it say that the missile has to take a curved path It is assumed it takes a curved path but this path is not relevant to the problem. It simply says the missile has the plane in its 'sights' at all times.

I presume that it is not what your exel calc says but how you program it.

You presumed wrong. The calculation proceeds by slicing up the flight time into discrete time intervals and calculating the coordinates of both the objects. At the end of each time interval, the missile's direction is recomputed (per the requirement that the "missile is constantly pointed at the plane"). The coordinates thus calculated, plots the flight path of the missile, which is most certainly not a straight line.

However, I _think_ I understand where you have gone wrong. The highlighted part of your quote -- but this path is not relevant to the problem -- tells me that you are thinking vectors, velocity and displacement, rather than speed and distance. Let me assure you, the path _is_ very relevant to this problem. Otherwise, as per your argument, if the missile flies due north and then flies due east, it will still take the same time to intercept!!?

Originally Posted by mfgoode

It is only natural that a missile should take the least time for an interception.

Why? I don't see that as obvious at all. The only advantage of "constantly pointing at the target" is that you don't have to precompute any intercept point. But given a predictable course of the target, the optimal path is to compute the intercept point and fly at a straight line to that point.

[Wacky]

Originally Posted by mfgoode

In other words the missile is constantly pointed at the plane.

It simply says the missile has the plane in its 'sights' at all times.

I do not consider these statements to be equivalent.

You dont need calculus to derive the answer.

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.

[S485122]

I was submited this same problem some 20 years ago by my grandfather, its setting was more pacific: it was about a doc running in pursuit of his human :-).

To really solve the problem according to premisses (the direction of travel of the pursuer is always towards the pursued, you need to compute a curvilign integral. I searched "Pursuit Curve" in MathWorld found Pursuit Curve and related links.

Thanks to you axn1 and Whacky for your support :-) I started doubting.

[Maybeso]

Originally Posted by mfgoode

[snip]
The missile does not compute the point of interception at all ...
[snip]
To see this, try stopping the plane with a vector and naturally add the same to the missile.

I'm sorry Mally, but these two statements are contradictory.

"Constantly pointed at the plane" means to only consider current position and ignore any velocity vector - speed or direction. If you add any vector associated with the plane to the missile, then the missile is pointing at the end of the vector, not at the plane. If the vector is '207 seconds long' then the missile has computed the intercept. If the vector is shorter, it is merely an intermediate intercept.

Mally, read Axn1's description of his solution again. The direction of the missile is adjusted at each step by looking at the two positions, ignoring all velocities. Then after the direction is set, the positions are updated with the velocities. This is the proper way to calculate the discrete path of a missile that does not 'lead' its target.

[Maybeso]

Great link Jacob!

[mfgoode]

Originally Posted by axn1

No it isn't. That is the _least_ amount of time required, assuming a direct intercept course.

But as per the problem, the missile has to travel along a curved path.

FWIW, my excel calc also tells me 240 secs.

At times the duration travelled along a curved line between two points is shorter than following a straight line.

Well if you dont agree I advise supplementary reading as below.

http://mathworld.wolfram.com/Brachis...neProblem.html

Mally

[mfgoode]

Originally Posted by Maybeso

I'm sorry Mally, but these two statements are contradictory.

"Constantly pointed at the plane" means to only consider current position and ignore any velocity vector - speed or direction. If you add any vector associated with the plane to the missile, then the missile is pointing at the end of the vector, not at the plane. If the vector is '207 seconds long' then the missile has computed the intercept. If the vector is shorter, it is merely an intermediate intercept.

My dear maybeso! in three dimensions time is not a vector which has magnitude AND direction. Time is unidirectional and taken to move from past to present and we dont know what direction to give it.

Please brush up on your vector theory.

Of course the velocity is pointing at the end of the vector. All you have to do is to get the vertical vector to the plane at rest by completing the rest of the rt.angled triangle.

Mally