From: robert@Xenon.Stanford.EDU (Robert Kennedy) Subject: Request for help: Problem in Calculus of Variations Date: 15 May 2000 19:15:43 GMT Newsgroups: sci.math.research Summary: [missing] I race motorcycles and I'm not afraid of math and physics, but I don't have the background to solve a racing-inspired problem I've managed to come up with. It would seem to be the sort of problem that suggests itself to any mathematician or physicist who thinks about racing, so I'm sure it must be well studied. Only trouble is, I haven't managed to find anything about it in textbooks. I'm trying to understand how I would set up a system of differential equations (that I will probably need to solve numerically -- that part is no problem) that will describe the optimum path through a corner on a race track. I believe this is must be a fairly standard problem in the calculus of variations, but I don't know how to apply CoV in this sort of situation where there are forces that can't be modeled as potential fields. Also, there are side conditions that aren't present in the sort of CoV problems I know how to solve. I want to make a number of simplifying assumptions that I'll be able to generalize away once I understand the basic techniques I'll need. I'm just looking for someone who can show me the basic techniques (or possibly refer me to a textbook presentation). Let's assume a race track lying completely in the x-y plane, and that my vehicle has an arbitrarily powerful motor and brakes, but is bound by the limits of traction. So if I let: t be time (which I'll conveniently use as my independent parameter); c(t) be cornering acceleration (acceleration parallel to the track surface and perpendicular to the path of travel); a(t) be braking or engine acceleration (parallel to the path of travel, signed appropriately of course); then we have that c(t)^2 + a(t)^2 <= 1 (available traction is not exceeded) It will be easy to see from my problem formulation below that we might as well suppose equality in that condition; the optimum path will use all the available traction resources at every point. Some more definitions and notation: v(t) is speed of travel; (x(t), y(t)) is position in an orthogonal coordinate system in the plane of the track; h(t) is direction of travel expressed as an angle counterclockwise from the x axis. Relationships and constraints: dx/dt = v(t) cos(h(t)) (obvious) dy/dt = v(t) sin(h(t)) (obvious) dv/dt = a(t) (obvious) dh/dt = c(t)/v(t) (cornering changes direction of travel) c(t)^2 + a(t)^2 <= 1 (no more than available traction is used) So far, we've made no assumptions about the shape of the track except that it's flat. Now let's assume the simplest possible turn: There are two parallel walls, say at x = -1 and x = 1 that I have to stay between, and there is a cone set up midway between the walls, at (0,0). I begin very far away, say at (-1+epsilon, -100), and my goal is to get to (1-epsilon, -100) as quickly as possible, going around the cone in the process, and not violating the traction constraint. Since we're free to choose my initial and final speeds and directions, it's clear (with some hand-waving) that the optimum path will satisfy the traction constraint with equality at every point. I think this specification is enough to yield uniqueness of a solution, but I have no idea how to find the path. If it would be helpful to add some reasonable additional conditions, I'm willing, even though they will change the problem slightly. An example condition that I can imagine might be helpful is an initial condition saying that the cornering acceleration is zero when first start out. This must be a classic problem, and surely there's a well written explication of its solution somewhere... Or maybe not, I guess. Any and all help/pointers/whatever appreciated. Thanks! -- Robert Kennedy ============================================================================== From: billryan22@aol.com (BillRyan22) Subject: Re: puzzled pilot problem Date: 04 Oct 2000 14:10:38 GMT Newsgroups: sci.math RIV You are Kaoru Iwamoto, a Japanese kamikaze pilot during the second world >war. Barely escaping from the enemy fighters you are somewhere in an >enemy-controlled zone of the pasific. Your radar tells you there are a >hundred ships in the area but you have no idea which are friends, foes >or civillians. > To see what kinds of ships you're dealing with you have to visit them >one by one, for your goal is to sink the most threatening ship and one >can attack only once... > You can of course just visit all hundred posibillities and then revisit >the biggest threat, but your fuel is uncertain and more fighters might >turn up. > Of course it's even worse to waste your explosive power - enough to >sink a battleship or carrier - on something like a light transport or >some yacht... > So what's your best strategy? and what are the chances you sunk the >most threatening ship? * * * * * You have presented a rather challenging problem in an interesting way. I agree with Robert Israel’s recent post that more definition is desirable for its resolution. I will assume the following for discussion purposes: (1) The 100 ships are aligned in a column, exactly one mile apart. The pilot is now at ship #1. They all look alike on the radar screen. (2) The pilot has a consistent probability of p of flying one more mile. without being shot down. He can see enemy planes on his radar screen and can accurately determine p. (3) The plane has just enough fuel to go 100 miles. (4) The pilot has no knowledge of navel warships. For all he knows there may well be 200,000 ton battleships or carriers with 3000 foot long decks. (5) Knowing nothing about ships, he equates “threatening” with “big” and he is able to accurately determine the size of the ships he sees. (6) He has no prior knowledge of the proportion of civilian ships. There are too few of them to create a concern about sampling error. * * * * * Let's say that at some point the biggest ship so far (BSF) is #a. He decides to fly to #x =>a. As x increases, so does his probability of sinking the biggest ship in the group provided he can get to x and then back to the BSF =< x. For a given a, the probability of making this round trip is p^[2(x-a)] subject to fuel limitations. His expected probability of success in sinking the biggest in the group is therefore f(x) = x*p^(2x-2a). . He wants to optimize f(x). Taking the derivative and equating to 0 yields an o[optimum result when x=-1/2ln(p). Note that a has no influence on the result. (Since x must be an integer, some rounding is required.) He will not be affected by fuel limitations so long as his optimum x is =< 51, 1.e., when p < -e-(1/102) or about .9902. He can make the 99 mile round trip from #1 to #51 and back to #1 if that is the biggest =<#51. If p<.9902. we can develop the probability of success beginning by summarizing the geometric series: p^(x-1) +... + p^(2x-2) Each ship has an equal chance of being biggest. This can be expressed as: [1/p][ p^x + ...+ p^(2x-1)] This sums to S = [1/p][p^x - p^2x]/[1-p] Since x ~ 1/2ln(p), p^x ~ e-.5 and p^2x ~ e^-1. This makes: S ~ [e^-.5 - e^-1]/[p(1-p) ~ .239/[p(1-p)] The average is S/x. The probability of success = [Average (A)] times [probability that biggest in group is = 51 (p>.9902) except to note that the pilot should not fly beyond (101 + BSF)/2 so that he can return to the BSF at the last point possible. He should fly only as far as #61 if BSF = #1. He should attempt to fly back to #1 rather than fly farther and risk having to make a round trip to a BSF he may find beyond #51. Bill Ryan