That’s even better then. If it’s converging on 0.9, the minimum time at which they’ll meet is distance/0.9.
I think I added a sentence to the end of my post after you responded though.
A is, by choosing the minimum angle of turn he possibly can, deciding the formula for the outside of a circle he wants to traverse. Whether he chooses to go left or right by that amount makes no difference, it merely creates a mirror image that doesn’t change anything.
B is sitting inside that circle, traveling towards the edge. B is traveling at the same speed as A and is traveling on a line that curves more sharply, by definition. And by definition, a sharper turn creates a shorter segment. Two ships traveling at the same speed, but one traveling a longer line, the ship traveling the shorter path will overtake the one traveling the longer path.
As I read it, there’s no stipulation that A must travel in a circle. We’re just told that they can’t keep travelling in a straight line. They can take any other kind of path they like, as long as they travel through at a constant speed.
No, SaintCad is solving the intercept problem. One of the legs of the isosceles triangle is drawn along A’s trajectory; B chooses his trajectory along BC, with C along A’s trajectory such that AC=BC, and thus since A and B are moving at equal speeds they will reach C at the same time. (So the lengths of the legs are a velocity times a time, and the units are happy.)
Of course this is only possible when the angle BAC<90°, as he says. If not, then B cannot choose a course that will intercept A at finite time.
So if A is forced to choose a new heading once a minute, but is allowed to choose a heading that always has this angle >=90°, then B will never catch up to A. If A is forced to change course by at least some minimum angle, then B will approach A arbitrarily closely, however.
HOWEVER, if the ship being chased is traveling 90 degrees or more to the base, the chaser will never catch it since no isosceles triangle can be formed. Therefore, as long as the chased ship is always traveling along an acute angle to the segment between the two ships despite the twists and turns, it will always be caught. Consequently, if it maintains a right angle or obtuse angle heading relative to the segment as it steers, it will never be caught.
Sure you can. They’re called vectors.
The other two sides are the distances the ships have traveled so all of the units are the same unit.
You see, for one boat to catch the other one, they must be at the same point at the same time. Distance equals rate x time (customarily abbreviated as d=rt) we can rewrite it as t = d/r and since both boats have equal speeds, if they have travel equal distances, we can show they have traveled an equal amount of time.
Since they are now at the third point of the triangle and have traveled an equal distance (that is called an isosceles triangle), we conclude that they are at the same point at the same time.
See, not misguided at all.
Oh, thanks, I see now what they were doing. My apologies for my misinterpretation. But they’ve only shown “If A and B are both forced to remain on a straight path, can B intercept A? No, not if A’s path is at suitable angle to the vector between their starting positions.” But that’s not the relevant question. There’s nothing forcing A and B to move in a straight path until the interception point.
That is, the distance between A’s starting point and the intercept point may be different than the distance between B’s starting point and the intercept point; for example, one might move around in a circle and come back to where it started, while the other just moves in a straight line. Or myriad other possibilities. So there’s no reason to assume that the intercept point will form an isosceles triangle with the starting locations of A and B.
Since they must constantly turn and their wisest turn is the minimum, it will be a constant rate of turn. That defines a circle. If they alternate going left-right-left-right, then they won’t make a circle if you plot their course, but so far as the math is concerned it’s still a circle.
Say I draw a circle and then draw a spiral starting at the center and rotating clockwise to meet the edge. Then I say that instead of ship A going in a circle, it goes 90 degrees around a circle and then turns around and goes back over it’s same course, repeating this back and forth. My “spiral”, matching this, will do a near about face at every 90 degree interval, creating something like a wave shape, but in the end result, the distance it travels before hitting the edge will not change.
A) As I read it, they don’t have to constantly turn, just turn infinitely often. E.g., they can trace out a piecewise linear path if they like, as long as they eventually turn off of any particular direction they turn on to
B) Even if they must constantly turn, they could start by turning at an angle of 1 degree per second, then later turning at an angle of 0.1 deg/sec, then later turning at an angle of 0.01 deg/sec, and so on, this being even “wiser” than sticking to some constant rate of turn; this, of course, would not trace out a circle.
This is true, but again doesn’t make any difference to the math. It just makes it an n-gon arranged on a circle and less easily visualized.
Why do that rather than starting with the smallest most infinitesimal, indivisible angle and keeping to it? That’s the only way to maximize the distance between the two ships. If you want them to turn at 1/infinity degrees, then go for it, but I’ll wait for you to figure out exactly what angle that would be.
Straight path has nothing to do with it. As long as the two ships are on a heading that meet as the sides of an isosceles triangle, they are getting closer to each other and eventually they will meet. Do me a favor, draw a segment 2" long and start to make a triangle with 2 segments 1" long 60 degrees to the base. Now connect these endpoints. This segment is shorter than the base. Now make rays 70 degrees to the top base of the trapezoid. Even though they changed direction, it is still an acute angle and they will meet . . . so what does changing direction have to do with the conditions I set forth?
Now if your argument is that A might follow a curve, I would argue that at each instantaneous point, it can be treated as following a straight line along the tangent (like in calculus). As long as A’s path, as curvy as it is, never has it on a heading 90 or more degrees from the segment AB, it will be caught. Likewise, if A never curves back to have an acute heading to the segment AB, A will never be caught.
If it goes back and forth between acute, right and obtuse, the question is unanswerable unless you can describe the exact path A is taking.
No, the problem as posed is underdetermined, which of course leads to lots of people solving different problems and confusing each other. It’s stated that B responds instantaneously to A’s change of heading, but it’s not specified how B reacts. Two obvious choices are (1) B sets an intercept course, if possible (as implicit in Mangetout’s diagram, assumed by SaintCad and dracoi, etc.) and (2) B plots a direct pursuit course, always heading straight toward A (sailor assumes this).
It’s also left unstated how often and how much A’s course changes (continuously? at some constant interval? … etc.). The cases where A’s course is piecewise linear are pretty easy to deal with. Whether the requirement that A change course “every so often” is meant to imply a piecewise linear trajectory I can’t say.
The optimum strategy for B depends on A’s behavior; if B knows that A will maintain a course for at least some period of time, then it makes sense for B to set an intercept course; otherwise this may not make sense, and the optimal chioice may lie somewhere between the intercept course and the pursuit course.
Well, that comes back down to assumptions. If A only changes course “every so often” then maybe B can assume A’s course to be piecewise linear, in which case setting an intercept course makes sense if A and B are sufficiently close. Of course if B can intercept A at C then B can intercept A at any point beyond C as well, but the construction given is for the fastest intercept assuming constant velocities.
Because there may not be a smallest, most infinitesimal, indivisible angle. Indeed, presumably, on a most straightforward mathematical idealization of the question, there isn’t.
I was simply noting that reasoning about the impossibility of constructing isosceles triangles subject to certain conditions proves nothing of any relevance, until one has established that the intercept point must form an isosceles triangle with the starting positions of A and B. And there’s no reason it must. The intercept point may be further from the starting point of one ship than it is from the starting point of the other ship, even if both ships keep moving at the same speed. The total distance, in the odometer sense, covered by a ship over the entire course of its journey need not equal the actual distance from its starting point to its ending point. It’s only the odometer sense in which the two ships must cover the same distance from start to intercept; the actual distances from their starting point to the intercept may be wildly different.
No, as I said, that depends on your assumptions. You appear to be assuming that A is traveling in a circle. SaintCad is assuming that A travels on a piecewise linear course. If B can reasonably assume that A’s trajectory is known for the next (say) one minute, then B can aim ahead of A to meet him at the intercept point C, as SaintCad has shown. It is true that for the intercept course B does not initally approach A as rapidly as B could. The point C is chosen to minimize the time to intercept, assuming A maintains a constant-velocity trajectory.
It seems to me that the B will get increasingly closer to A, but never catch him. The first part is easy to see. For the second part, I see that every time B gets closer, A just has to change course 90° to B’s course. This creates a series of increasingly smaller triangles, but there will always be a triangle. Granted, this goes for our hypothetical ships, which have no dimension.
Reading your post again I agree. You are not arguing about an optimal trajectory (least time to intercept) but just saying that as long as A’s heading is less than 90° from the direction to B, B can choose a course to meet A; and if A’s heading is 90° or greater, then B cannot. No argument there.
When you said that “The chaser wants to take a heading…” I understood you to be talking about finding an optimal trajectory; this depends on B’s model for A’s behavior.
This doesn’t follow. B’s spiral could end up getting asymptotically close to the circle without B ever actually catching up, and I think that that’s what actually happens. What I find is that if A is moving around the unit circle with unit speed and B starts from the origin at time 0, then at time t>>1 the distance between B and the origin is approximately 1-2/t[sup]2[/sup], and the distance between A and B is approximately 2/t.