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March 22

Vectors

You know the position, bearing, and speed of a target. You know your position, and the speed of a missile that you can launch. How do you work out the bearing that the missile should be launched on to hit the target? Does calculus have to be involved? Thanks 92.15.6.157 (talk) 11:31, 22 March 2011 (UTC)[reply]

We just had this question: Wikipedia:Reference_desk/Archives/Miscellaneous/2011_March_15#Question_about_missile_tracking. StuRat (talk) 15:48, 22 March 2011 (UTC)[reply]

No we have not. This is about the mathematics involved, that was something much more general. 92.15.10.228 (talk) 21:49, 22 March 2011 (UTC)[reply]

Did you follow all the links there ? Some of them got deep into the math. StuRat (talk) 04:02, 23 March 2011 (UTC)[reply]

Under the simplifying assumption that the target's velocity is constant and so is the missile's velocity once fired: Let t=0 be the moment when the missile is fired. Write the position of the target as a function of t. Derive from this the distance d(t) between yourself and the target as a function of t. Set d(t) = v_Mt, where v_M is the speed of the missile, and solve for t. After squaring both sides of the equation and simplifying, a quadratic equation in t is left. Its positive solution(s), if any, is the moment the target can be hit by a missile fired at t=0. Compute where the target will be at that time, using your already prepared function. Compute the bearing from you to that position. No calculus. –Henning Makholm (talk) 22:12, 22 March 2011 (UTC)[reply]

What would the formulas involved be please? Isnt it just finding the length of sides of a triangle, although the differing speeds of the target and missile make it more complicated? Thanks 92.15.14.45 (talk) 12:06, 23 March 2011 (UTC)[reply]

The formulas are listed explicitly in my post below. The only difference is, if you insist that your missile travels at an uncontrollable, constant speed, say s, then you have to use the fact that || w || = s. If w = (w₁,w₂,w₃) then s² = w₁² + w₂² + w₃². This will introduce quadratic terms which, as Henning mentioned, means you can have two solutions, one solution, or no solutions. If you solve p + tv = q + tw for w, you see that tw_i = p_i – q_i + tv_i for all 1 ≤ i ≤ 3. Then impose the condition that || w || = s, then you have

t^{2}s^{2}=(p_{1}-q_{1}+tv_{1})^{2}+(p_{2}-q_{2}+tv_{2})^{2}+(p_{3}-q_{3}+tv_{3})^{2}\ \,.

Although, you could assume that you're at the origin, i.e. q₁ = q₂ = q₃ = 0, to simplify. Solve for t, if you can, then put the solution for t back into your solutions for w to give you w, i.e. the required velocity of the missile. But since s is fixed, this just tells you the direction. — Fly by Night (talk) 14:09, 23 March 2011 (UTC)[reply]

Thanks. My maths education stopped when I was 16 years old. Does the above mean that there is no simple formula that tells you what direction to fire the missile in? If I wanted to use it for a computer game, for example. Thanks 92.24.188.210 (talk) 20:53, 24 March 2011 (UTC)[reply]

(edit conflict) The real problem is very, very difficult to solve if you consider a three dimensional world with a curved earth, gravity, wind resistance, rotation of the earth, missile weight changing as fuel is used, etc. To solve this problem you would need to use a lot of calculus. However, if we do a very simple, pool ball model, where everything happens in the plane, there's no friction, etc, then we only need linear algebra. Assume that your target is at point p at time t = 0, and it has a velocity of v; recall that velocity is a vector. After t seconds the target will be at the point p + tv. Assume that you are at point q, if you fire the missile with velocity w then after t seconds the missile will be at the point q + tw. You need to solve the vector equation p + tv = q + tw, which is actually two simultaneous linear equations. We know p, q and v. You could specify a time t and then solve for w. That would give you the velocity, i.e. the speed and direction, that you would need to fire the missile with in order for it to hit the target at time t. This method works not just in the plane, but in n-dimensional space. — Fly by Night (talk) 22:19, 22 March 2011 (UTC)[reply]

As the above responses indicated, in the "real world", you'd better constantly revise you trajectory based on revised data, if you hope to hit the target. This applies whether you use a linear algebra solution to find the initial trajectory, or numerical methods, as I described at the link I provided. StuRat (talk) 04:06, 23 March 2011 (UTC)[reply]

You'd need more than linear algebra; you'd need calculus. The trajectory would be instantly wrong if you used linear algebra. Take a look at our Trajectory article. — Fly by Night (talk) 14:15, 23 March 2011 (UTC)[reply]

I was talking about the "simple pool ball model" you suggested above. But my same comment applies to even the most robust calculus model you can build. You'd still miss the target unless changes in velocity for both objects are taken into account as they occur, resulting in continuous observations, calculations, and adjustments during flight. There are just too many unknowns (like wind) to ignore. StuRat (talk) 15:42, 23 March 2011 (UTC)[reply]

Oh, right. Well you didn't say. Your post implies that you are talking about the "real world". — Fly by Night (talk) 15:46, 23 March 2011 (UTC)[reply]

Well, since it's necessary to continuously recalculate the trajectory in any case, it doesn't particularly matter how simple or complex the initial (and incremental) calculations are. The reduced accuracy of the simpler models is compensated for by them being faster, and thus allowing more iterations per second. An exception might be if the observations or adjustments can only be made infrequently, for some reason. Then it would become important to be as accurate as possible with the calculations at each iteration. StuRat (talk) 16:12, 23 March 2011 (UTC)[reply]

I'm not sure that simpler methods being iterated would always result in a strike. The linear algebra model would be immediately wrong. Gravity (and its non-homogeneity), friction (especially close to the speed of sound), air pressure, wind velocity, the aerodynamic properties of the missile itself, etc., will all have massive effects on the trajectory. Then there's all that to deal with with the target and the possibility of it changing direction. An, often substantial, adjustment would be needed following each and every iteration. Would it be possible in terms of aerodynamic law and general Newtonian mechanics, to get the missile to make the adjustments quickly enough without the target getting away, the missile becoming aerodynamically unstable, or the missile running out of fuel? I think, in terms of the linear algebra model, no. I think there's a point where the model can become so bad that you'd find it hard to hit anything; even with arbitrary processing speeds. More sophisticated mathematics and engineering are required. — Fly by Night (talk) 18:22, 23 March 2011 (UTC)[reply]

Surface-to-air missiles have been around for a long time (development started at the end of WW2). I doubt if the early ones would have had the processing power to do complex calculations, yet they still managed to occasionally hit the target. StuRat (talk) 04:40, 24 March 2011 (UTC)[reply]

That's true, and I didn't claim otherwise. In fact it proves my point that they would have used Calculus (which has been around since the 1600's). At the end of the second world war, programmable computers had only just been invented and were the size of entire rooms (see Colossus). Rockets didn't contain computers, and flight could not be corrected or changed; except for the inclusion of a gyroscope (see, for example, V-1 and V-2). The ballistics equations are derived with calculus and are what have been used throughout the ages, from cannon ball, to rifle shot, all the way through to anti-aircraft guns and ICBMs. — Fly by Night (talk) 15:26, 24 March 2011 (UTC)[reply]

And those are OK for a fixed target within a limited range (or if you use a nuke so accuracy doesn't much matter), but hitting a moving target (the subject of this question) without in-flight adjustments is darn near impossible. StuRat (talk) 21:13, 24 March 2011 (UTC)[reply]

I think you'll find that the accuracy of ICBMs is vital, given that the destructive radius of most ICBMs is measured in dozens of kilometres, see for example this graphic from the 1945 nuclear attacks on Japan). Given the distance they travel, their accuracy needs to be within fractions of one percent. — Fly by Night (talk) 00:15, 25 March 2011 (UTC)[reply]

Right, but kilometers is pathetic compared with what a rocket that continuously adjusts it's trajectory can do, and is completely unacceptable for most target/warhead combos. StuRat (talk) 04:15, 25 March 2011 (UTC)[reply]

Min/max value of a vector equation

I believe this should be very simple, but I simply cannot get it to work. I would like to know how to solve this problem. I want to know the min/max possible values of the vector equation:

{\frac {A\cdot B}{\left\Vert A\right\Vert ^{2}+\left\Vert B\right\Vert ^{2}-A\cdot B}}

.

I attempted to simplify it by changing the equation to:

{\frac {\left\Vert A\right\Vert \left\Vert B\right\Vert \cos \Theta }{\left\Vert A\right\Vert ^{2}+\left\Vert B\right\Vert ^{2}-\left\Vert A\right\Vert \left\Vert B\right\Vert \cos \Theta }}

I know that the min-max values of $\left\Vert A\right\Vert$ and $\left\Vert B\right\Vert$ are 0 and infinity. For $\cos \Theta$ , min-max is -1 and 1. I can also see that when all values in sets A and B are zero, the result is undefined. Then, I hit the limit of my ability to simplify further to get a hard min or max limit. I don't know if I didn't learn this in set theory or if it wasn't taught 20 years ago. I can only go with a gut assumption that the min is around -0.5 and the max is around 1. -- kainaw ™ 16:03, 22 March 2011 (UTC)[reply]

Just look at the case of A=B, which reduces it to a scalar problem. The function is unbounded, having neither a maximum not a minimum. Sławomir Biały (talk) 16:17, 22 March 2011 (UTC)[reply]

I am very sorry. I forgot to square the denominator terms. Setting A=B, I can show that the min is -0.333 and the max is 1. I cannot prove that for all cases of A and B. -- kainaw ™ 16:32, 22 March 2011 (UTC)[reply]

That changes things. Let x=|A|/|B|. Then

{\frac {A\cdot B}{\left\Vert A\right\Vert ^{2}+\left\Vert B\right\Vert ^{2}-A\cdot B}}={\frac {\cos \theta }{x+1/x-\cos \theta }}.

The denominator is bounded from zero, and the function tends to zero uniformly as x tends to infinity, so this function has a maximum and minimum at its critical points. These are

(1,0),(-1,\pi ),(1,\pi ),(-1,0)

extended periodically in theta, giving a max of 1 and min of -1/3. Sławomir Biały (talk) 17:03, 22 March 2011 (UTC)[reply]

Thanks. Now I can work on Tanimoto coefficient. It implies that the result is -1 to 1 like vector cosine, but my gut feeling was that it couldn't be -1. So, it need to be reworded to remove that implication. -- kainaw ™ 17:15, 22 March 2011 (UTC)[reply]