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Guided Missile.
:rolleyes:
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 :coffee: |
[QUOTE=mfgoode;90699]:rolleyes:
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 :coffee:[/QUOTE] :smile: HINT Not to complicate matters assume the plane is flying at tree top level and so its altitude does not count and is negligible. Mally :coffee: |
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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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 |
Guided missile
:smile:
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 :coffee: |
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)[sup]2[/sup]+P[sup]2[/sup]=M[sup]2[/sup] Since we know that the missile is twice as fast as the plane M=2P This gives 10000+P[sup]2[/sup]=4P[sup]2[/sup] This is equivalent to P[sup]2[/sup]=10000/3 P=100/3[sup]0.5[/sup]=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. |
[QUOTE=mfgoode;90699]:rolleyes:
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 :coffee:[/QUOTE] For what it's worth (puzzle aside), this is a losing pursuit strategy, as the turn rate required of the pursuer increases to infinity as the moment of impact approaches. You don't want to point the nose of the missile at the target. As someone mentioned, you want to fly to the point of interception, instead. But since that point cannot be seen, the goal is to keep the target from moving in the pursuer's field-of-view. This ensures a collision course between the two objects. 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 |
Guided missile
:rolleyes:
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 :coffee: |
[QUOTE=mfgoode;90876]
Please remember this is a mathematical problem and as such the conditions are ideal as Wacky made so 'plain' about the 'plane'. [/QUOTE] Understood, which is why I made the "puzzle aside" comment. I just thought the real-world pursuit problem was so interesting I'd mention a thing or two about it. Drew |
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.) |
misguiding missiles.
[QUOTE=drew;90881]Understood, which is why I made the "puzzle aside" comment. I just thought the real-world pursuit problem was so interesting I'd mention a thing or two about it.
Drew[/QUOTE] I appreciated your analysis in detail, though it did not pertain to the problem directly. Please allow me to add on an aside to that. 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! :wink: Mally :coffee: |
[QUOTE=Jacob Visser;90892]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.)[/QUOTE] :smile: 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 :coffee: |
[QUOTE=mfgoode;90935]Surprisingly your other solution (207. .. secs) is exact and correct.[/quote]
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. |
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 100[sup]2[/sup]+57.735[sup]2[/sup]=13333 and that 115[sup]2[/sup]=13333 (all figures are rounded of which means the numbers are only approximatively equal.). Doesn't this seem like the formula of Pythagoras? |
curved path.
[QUOTE=axn1;90940]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.[/QUOTE] 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 :coffee: |
[QUOTE=mfgoode;90961]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 [B]but this path is not relevant to the problem[/B]. 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. [/quote] 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!!? [QUOTE=mfgoode;90961]It is only natural that a missile should take the least time for an interception.[/quote] 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. |
[QUOTE=mfgoode;90699]In other words the missile is constantly pointed at the plane.[/QUOTE]
[QUOTE]It simply says the missile has the plane in its 'sights' at all times.[/QUOTE] I do not consider these statements to be equivalent. [QUOTE]You dont need calculus to derive the answer.[/QUOTE] 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. |
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 [url=http://mathworld.wolfram.com/PursuitCurve.html]Pursuit Curve[/url] and related links. Thanks to you axn1 and Whacky for your support :-) I started doubting. |
[QUOTE=mfgoode;90935][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. [/QUOTE]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 [B]positions[/B], ignoring all velocities. Then [B]after [/B]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: |
:grin: Great link Jacob!
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Curved Path.
[QUOTE=axn1;90940]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.[/QUOTE] :rolleyes: 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. [url]http://mathworld.wolfram.com/BrachistochroneProblem.html[/url] Mally :coffee: |
[QUOTE=Maybeso;90974]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. :maybeso:[/QUOTE] :rolleyes: 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 :coffee: |
Grand Father
[QUOTE=Jacob Visser;90973]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 [url=http://mathworld.wolfram.com/PursuitCurve.html]Pursuit Curve[/url] and related links. Thanks to you axn1 and Whacky for your support :-) I started doubting.[/QUOTE] :mad: I would have given you a no nonsense solution which has no argument as to the path. I have given you a hint in my post to maybeso and since I dont think ANY of you mentioned deserve to know the simple but rigorous solution I will refrain once more from presenting it. When I say 'once more' this is not the only solution or theory of mine which will make waves in future which I have actually withdrawn from this forum for lack of support from the members. It awaits publication in a prestigious journal very soon to be refereed by IMPARTIAL math'cians. Amen. As I told axn1 to either take it or leave it. To be sure there are others who will do the needful. Mally :coffee: |
[QUOTE=mfgoode;91003]At times the duration travelled along a curved line between two points is shorter than following a straight line.[/QUOTE]
That is true ONLY when the speed is affected by an acceleration such as that caused by gravity. Without acceleration, the time of travel is directly proportional to the distance and a straight line is always the shortest path between two points. |
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Mally,
You agree with me that initially, the missile points due North towards the plane? (Or even if it was pointing in another direction, just after departure it points towards the plane, thus due North.) Do you agree with me that the plane after 36 seconds has travelled 10 miles due East? Do you agree with me that at that moment the missile is 20 miles from its departure point? Do you agree with me that at that moment the missile is pointing at a point 10 miles east of the departure place of the plane? Do you agree with me that the plane after another 36 seconds has travelled 10 more miles due East? Do you agree with me that at that moment the missile is pointing at a point 20 miles east of the departure place of the plane? Do you agree that since the place the missile is pointing to is changing, that it is not travelling on a parallel to that of the plane (contrary to hypotheses) its direction of travel is changing and thus its path is not a straight line? I have made a drawing. On the left the solution most of the posters are proposing. On the right the solution you are proposing. This last solution implies that from the start the plane travels North-East-North and this while the plane is due North. |
Rigorous? As in mathematically rigorous? Or real world rigorous?
In the real world, the missile takes some amount of time to accelerate to its cruise speed. I don't see mention above of this acceleration time. I presume the missile will be launched from the ship at some angle. How far will it travel up before modifying its course to be parallel to the ground? Fusion |
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Hi
I am woting for the 240 sec. solution - or rather 239.87 sec. Since this is a matematical problem the turnrate can be infinit high during the last part of the interception. No problem there. The orginal question stated clearly that: [QUOTE]In other words the missile is constantly pointed at the plane.[/QUOTE] So no onboard computer to calculate the point of interception. I did a stepwise calculation in Berekely Madonna: [QUOTE]METHOD RK4 STARTTIME = 0 STOPTIME=275 DT = 0.001 Speed_plane = 1000/60/60 Speed_missile = 2000/60/60 init plane_x = 0 init plane_y = 100 Next plane_y = plane_y Next plane_x = plane_x+speed_plane*dt Angel = if (plane_x-missile_x) > 0 then arctan((plane_y-missile_y)/(plane_x-missile_x)) else Pi/2 init missile_x = 0 init missile_y = 0 Next missile_x = missile_x + speed_missile*dt*cos(angel) Next missile_y = missile_y + speed_missile*dt*sin(angel) Limit missile_x <= plane_x Limit missile_y <= plane_y[/QUOTE] The missile starts at origin (0,0) and the plane at (0,100). X-coordinat is east and Y-coordinat is north. Confirmed hit after 239.87 sec, the plain travels 66.63 miles + fall to the ground. :grin: Se the graph intercept for coordinats of missile and plane vs time Se the graph phase plot for plane_y and missile_y vs plane_x - this is the flightpath seen from above. -Eivind |
[QUOTE=Fusion_power;91027]Rigorous? As in mathematically rigorous? Or real world rigorous?
In the real world, the missile takes some amount of time to accelerate to its cruise speed. I don't see mention above of this acceleration time. I presume the missile will be launched from the ship at some angle. How far will it travel up before modifying its course to be parallel to the ground? Fusion[/QUOTE] Please provide: 1) Altitude difference between missile and plane. 2) Launch angel in Z-coordinat 3) Function or tabel for acceleration of the missile This is easy to incorporate into the model. -Eivind -Eivind |
[QUOTE=Jacob Visser;91024]Mally,
You agree with me that initially, the missile points due North towards the plane? (Or even if it was pointing in another direction, just after departure it points towards the plane, thus due North.) Do you agree with me that the plane after 36 seconds has travelled 10 miles due East? Do you agree with me that at that moment the missile is 20 miles from its departure point? Do you agree with me that at that moment the missile is pointing at a point 10 miles east of the departure place of the plane? Do you agree with me that the plane after another 36 seconds has travelled 10 more miles due East? Do you agree with me that at that moment the missile is pointing at a point 20 miles east of the departure place of the plane? Do you agree that since the place the missile is pointing to is changing, that it is not travelling on a parallel to that of the plane (contrary to hypotheses) its direction of travel is changing and thus its path is not a straight line? I have made a drawing. On the left the solution most of the posters are proposing. On the right the solution you are proposing. This last solution implies that from the start the plane travels North-East-North and this while the plane is due North.[/QUOTE] Jacob, don't waste your time. This is Mally's typical M.O. I've discovered after several such discussions with Mally that he cannot be reasoned wiith. It doesn't matter if there is a rational explanation for the proper solution to the problem, he'll stick with *his* answer no matter what arguments you present to convince him of the correct answer. Since he presented the problem himself, I think he believes his answer must be the right one irrespective of any logic or mathematics. I have no doubt that the problem, as posed to him, has the answer Mally presented, but I believe he added the "In other words the missile is constantly pointed at the plane" comment in an attempt to clarify the problem not realizing that it intruduced an assumption that yields a vastly different answer. But he won't admit it. In any case, I corroborate your answer. We all know what the proper solution is, so don't lose any sleep if you cannot convince Mally. It's a lost cause. I've linked below another thread in which Mally could not be reasoned with. That was my first encounter with Mally and it's put a bad taste in my mouth ever since. You can also see how rude he was to me right off the bat, despite having the wrong solution. There are more examples in this forum of the same type of behavior but I cannot be bothered to find them right now. Hopefully, this can convince you that you're fighting a losing battle: [url]http://mersenneforum.org/showthread.php?t=5068[/url] Drew |
Howlers!
[QUOTE=Wacky;91020]That is true ONLY when the speed is affected by an acceleration such as that caused by gravity. Without acceleration, the time of travel is directly proportional to the distance and a straight line is always the shortest path between two points.[/QUOTE]
:grin: Well Richard Wackerbath old boy you must be joking. You couldn't be serious as I cant believe it. What howlers! Please brush the cobwebs from your brain Regards my old friend, Mally :coffee: |
Rigorous!?
[QUOTE=Fusion_power;91027]Rigorous? As in mathematically rigorous? Or real world rigorous?
In the real world, the missile takes some amount of time to accelerate to its cruise speed. I don't see mention above of this acceleration time. I presume the missile will be launched from the ship at some angle. How far will it travel up before modifying its course to be parallel to the ground? Fusion[/QUOTE] :furious: I have emphasized its mathematical and not a physical problem. This is strictly not a real world problem. Ship? from off the shore? The vertical component is 100/sq.rt.3. It travels vertically as high as the plane no higher and horizontally as far as the plane travels dumkoff ! Mally :coffee: |
Step wise ?
[QUOTE=Eivind;91028]Hi
I am woting for the 240 sec. solution - or rather 239.87 sec. Since this is a matematical problem the turnrate can be infinit high during the last part of the interception. No problem there. The original question stated clearly that: So no onboard computer to calculate the point of interception. I did a stepwise calculation in Berekely Madonna: The missile starts at origin (0,0) and the plane at (0,100). X-coordinat is east and Y-coordinat is north. Confirmed hit after 239.87 sec, the plain travels 66.63 miles + fall to the ground. :grin: Se the graph intercept for coordinats of missile and plane vs time Se the graph phase plot for plane_y and missile_y vs plane_x - this is the flightpath seen from above. -Eivind[/QUOTE] :cool: Forget about step wise calculations Elvind. Thats okay for computer programmers but not in pure math. Use an Integral if you can which is the ideal summation. If you are at Berkeley I would expect your English to be better. As it is, its atrocious! Being from the School of Hard Knocks I believe in nipping things in the bud. Sorry! Mally :coffee: |
Good advice!
[QUOTE=drew;91030]Jacob, don't waste your time. This is Mally's typical M.O. I've discovered after several such discussions with Mally that he cannot be reasoned wiith. It doesn't matter if there is a rational explanation for the proper solution to the problem, he'll stick with *his* answer no matter what arguments you present to convince him of the correct answer. Since he presented the problem himself, I think he believes his answer must be the right one irrespective of any logic or mathematics.
I have no doubt that the problem, as posed to him, has the answer Mally presented, but I believe he added the "In other words the missile is constantly pointed at the plane" comment in an attempt to clarify the problem not realizing that it intruduced an assumption that yields a vastly different answer. But he won't admit it. In any case, I corroborate your answer. We all know what the proper solution is, so don't lose any sleep if you cannot convince Mally. It's a lost cause. I've linked below another thread in which Mally could not be reasoned with. That was my first encounter with Mally and it's put a bad taste in my mouth ever since. You can also see how rude he was to me right off the bat, despite having the wrong solution. There are more examples in this forum of the same type of behavior but I cannot be bothered to find them right now. Hopefully, this can convince you that you're fighting a losing battle: [url]http://mersenneforum.org/showthread.php?t=5068[/url] Drew[/QUOTE] :yucky: Etu Brute (with an accent). What may I ask is the right answer?? Its the case of 'once bitten twice shy' for Drew who this time flew! I'm truly sorry for you Jacob. You are the only one taking this problem seriously. Maybe I will PM you with the actual solution tho' Ive dropped hints here and there. Please dont be fooled by these self made geniasses and pull out of the thread as you will be misled into a maze of confusion. So Drew you fished out a thread to prove your point. Is Revenge that sweet ?? Mally :coffee: |
[QUOTE=mfgoode;91036]What may I ask is the right answer??[/quote]
To the question you asked (as posed) the true answer is the one *everyone* in this thread except you has agreed upon...239.87 seconds. [quote]So Drew you fished out a thread to prove your point. Is Revenge that sweet ??[/QUOTE] No need to fish...I remember that thread well. It's not often I'm greeted so rudely in a mathematical forum. Honestly, I don't think anyone is waiting to see your solution...we all know what a 30-60-90 triangle looks like. Drew |
[QUOTE=drew;91039]To the question you asked (as posed) the true answer is the one *everyone* in this thread except you has agreed upon...239.87 seconds.[/QUOTE]
Excuse me I stand by the 240 seconds. One gets a bit less if one takes bigger steps, to get 240 seconds I computed 2[sup]21[/sup] steps. One gets the minimum if one takes only one step: a little bit less than 208 seconds. I played around with different speeds and came to the conclusion that the time to interception must be a simple equation involving the different speeds and the initial distance, but I haven't found which at the moment. Because of al this fuss about the rejection of the solution I proposed I remembered that evening with my grandfather long ago. A good side effect :-) Mally, I find you a bit rude towards other posters. Did you look at the diagram I posted ? If you send me your solution please take the time to answer the questions I asked you. And please forgive my typos and incorrect use of the language :-) |
[QUOTE=mfgoode;91034]:cool:
If you are at Berkeley I would expect your English to be better. As it is, its atrocious! Being from the School of Hard Knocks I believe in nipping things in the bud. Sorry! Mally :coffee:[/QUOTE] When in a corner Mally's standard tactic is to turn nasty. Because he seems to be wrong more often than most, it is a regular occurrence. It is a shame because he has interesting, if idiosyncratic, things to say. I hope that Eivind is not put off by being the latest object of Mally's insults. In any case, I am certain Eivind's English is much better than Mally's German or Latin. "Dumkoff", Mally, is spelt Dummkopf [post 31]. "Etu" [post 33] is two words - Et tu. It means "even you" and is an expression of dismay a supposed friend's treachery. I am still trying to understand how it fits in the context in which Mally used it. |
[QUOTE=Jacob Visser;91059]Excuse me I stand by the 240 seconds.[/QUOTE]
You're right. I took Elvind at his word because his solution appeared more precise, but now that I've worked it out myself, I agree with you. The answer is exactly 240 seconds. Drew |
You know Mally, there is something to be said for always being polite to others.
I remind myself regularly that mathematicians as a group are usually socially inept. You have carried that trait to a new low. I am now convinced that Bob Silverman is an example of a fine human being by comparison. Maybe you need to study the difference between discrete and discreet. Better yet, remind me someday what a fine upstanding Christian you are:sick:.....after you have read Matthew 7:12 and Galatians 6:10. Last, but not least, post your answer to this problem so you can either be vindicated or else proven a fool. Fusion |
Hi guys
Drew and Jacob you are right - i rechecked the output, and 240 sec. is correct. Due to roundup (Glyphosate) the missile north value changed to 100 at 239.87 sec, but the east values of the plane and missile align at 240 sec. My iterations was *only* 2,7*10^6. So please add 0.13 sec. to the previous post :wink: 99.94: Don't worry I am not easy scared. Atleast i know my limitations :lol: And yes - even though English nor German is my native language, I speek both quite good. Zweimal hefe weiss bitte. Mally: No I don't study or work at Berkeley - nor do I ever intent to do so. I am a chemical engineer working with verry nasty chemicals like PCl3 and SOCl2. Please explain why a 4. order Runge-Kutta numerical integration is inaccurate to solve the problem. I would be glad to see the error difference between this and an algebra solution. -Eivind |
Error admitted.
:sad:
Before we continue with this interesting thread (to me at least) I would like to offer my sincere apologies to all those I sent rude and overconfident replies at several stages of its development. These are to Wacky, Jacob Visser, Drew (post # 29 hit the nail on the head) axn1, Maybeso, Fusion Power (Matt. 7:12 : Gal. 6 :10 ) , Eivind (zweimal...), 99.94 ('wrong more often than right'). Having said that all I can say to you great guys is ' All I know is that I was blind but now I can see right' I have re-examined the problem, which Drew guessed right that I had doctored it, and suggest we revert back to the original Jacob Visser's 'Grandfather's problem' which actually speaks of two men as referred too instead of plane and missile. Let the distance be 100 metres and the speeds x and 2x metres /sec. in a direction that always faces where B is at the moment. A is south of B. Well its much the same but eliminates gravitational effects the missile is subjected too since we are talking on differences in decimal points. Yes Eivind 4 order Runge-Kutta numerical integration gives as good an approximation as the actual integral generally which at present I have not derived myself However I am happy that this problem generated more depth than what I anticipated and has amused and entertained one and all judging by the replies. Once again I am sorry I pressed the attack button, though I learnt in Martial arts that 'attack is the best from of defence' Regards to one and all, Mally :coffee: |
Thanks Mally !
For the anecdote the "doc" in the problem posed by my grandfather was a dog (one more typo on my part.) In the mean time, I did some calculations with different starting distances, speeds... It appears that the time is directly proportional to the initial distance. Then the differences in time are inversely proportionnal to the differences in speed but something else as well : if the dogs speed is an epsilon above the humans speed the factor is half that of the case where the dogs speed tends to inifinity. T the time to interception, D the initial distance, V[sub]P[/sub] the speed of the dog, V[sub]H[/sub] the speed of the human and X some other factor depending on the relative speeds. T=X*D/(V[sub]P[/sub]-V[sub]H[/sub]) Where the value for X for V[sub]P[/sub] almost equal to V[sub]H[/sub] is half the value for X for V[sub]P[/sub] many times bigger than V[sub]H[/sub]. (It is not an arcsinus...) But I got no further. The solution of the problem in MathWorld dit not enlighten me :-( |
[QUOTE=mfgoode;91136]:sad:
Before we continue with this interesting thread (to me at least) I would like to offer my sincere apologies to all those I sent rude and overconfident replies at several stages of its development. These are to Wacky, Jacob Visser, Drew (post # 29 hit the nail on the head) axn1, Maybeso, Fusion Power (Matt. 7:12 : Gal. 6 :10 ) , Eivind (zweimal...), 99.94 ('wrong more often than right').[/quote] Thanks, Mally. I know it's difficult to swallow your pride and admit when you're wrong, and I appreciate it. Hopefully we can be more cordial with each other in future discussions. [quote]...in a direction that always faces where B is at the moment.[/quote] I'm afraid you've made the same error again, however. The above statement is the crux of everyone's disagreement. Your solution (~108 seconds) necessitates a direct path to the point of interception. When you move towards the target's current location, you follow a curved path for which it takes longer to intercept the target (140 seconds). The acceleration of gravity isn't the issue as you said the target flies low as to eliminate the necessity to climb. [quote]Once again I am sorry I pressed the attack button, though I learnt in Martial arts that 'attack is the best from of defence'[/quote] For what it's worth, take people's corrections as constructive criticism, so there's no reason to get defensive. They are not personal attacks, only attempts to convince everyone of the true answer. No good mathematician will tolerate errors in logic without attempting to set things straight in a constructive manner. Drew |
Mally I don't think this is a very realistic problem.
The missile and plane speeds are unlikely to remain constant, for example because constant thrust would have more effect on the craft as it gradually burns away its fuel reserves. Some of that fuel might be expended to indirectly offset the gravitational tendency to fall to earth, and to overcome air friction and wind and precipitation. These factors mean there is not a linear relationship between fuel burn and speed. Similar considerations apply at missile launch as to initial velocity of leaving the base. Secondly, a plane detecting a missile launch would likely take evasive maneuvers. If it chose to head north away from the missile, the missile could well run out of fuel ie beyond its design range and crash harmlessly. If the plane was on its way to bomb a city, it might choose an alternative city to bomb as a secondary objective, making initial assumptions about its course redundant. Any attempt to initially compute an intercept vector assumes the plane will maintain its direction and/or speed which is unlikely the case if it sees a missile coming. I believe the missile would take a curved path, from the information given. These are also not very sophisticated missiles. 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=Peter Nelson;92751]Any attempt to initially compute an intercept vector assumes the plane will maintain its direction and/or speed which is unlikely the case if it sees a missile coming.[/QUOTE]
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=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: |
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