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Q: The mathematics of kayaking across a current ( No Answer,   8 Comments )
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 Subject: The mathematics of kayaking across a current Category: Science > Math Asked by: djlewis9-ga List Price: \$30.00 Posted: 14 Jun 2006 18:58 PDT Expires: 14 Jul 2006 18:58 PDT Question ID: 738250
 ```A kayaker wants to cross a wide river that runs from south to north. The river is exactly D nautical miles wide from shore to shore at all points, and the current is the same at all points and times. (Yes, this is a very dull river; to compensate, picture beautiful mountains in the background.) The current runs at a steady w knots (nautical miles / hour) from south to north throughout the crossing. The kayaker starts out at a point on the eastern shore and heads for a point on the western shore exactly E nautical miles downstream (E > 0, north) or upstream (E < 0, south). If E = 0, she is heading straight across. Not realizing there are other strategies, the kayaker, who paddles at a steady rate of p knots relative to the water, simply points her boat at the target on the other shore and paddles forward. The current, of course, sweeps her a bit north as she paddles, but she is unaware of this and constantly adjusts her boat unconsciously to keep aiming at and paddling toward the target. Obviously, therefore, her course is an arc that heads north-west-ish at first and the curves around to approach the target heading south-west-ish. Question -- How long does it take the kayaker to cross the river and how far does she paddle relative to the water and relative to the ground, as three functions of D, E, p and w? For extra credit, what are the functions when it's a tidal race instead of a river, and the current w varies over time as a parameterized sinusoidal function? (Tricky -- the current could even reverse during the crossing.) You may assume that the kayak is a mathematical point and the current simply exerts a steady south-to-north motion on it. But I will ask later if it makes a difference to assume the kayak is 18 feet long and that when the kayaker is paddling any direction relative to the current other than straight with it, the current also exerts a sideways force on the kayak. Or, you may answer now and I'll give a generous tip.```
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 Subject: Re: The mathematics of kayaking across a current From: myoarin-ga on 15 Jun 2006 04:37 PDT
 ```Using her system, if w, the speed of the current, is greater than p, her paddling speed, it would seem that she would be swept past the point E distance from her start and would never be able to reach her goal, since once she was level with it, she could never make headway against the current. Indeed, if w is greater than p by a defined proportion, one could calculate that if E were less that "x", she could not reach the opposite shore by paddling straight west across the river - perhaps by paddling southwest right from the start. I think the relationship of w to p must be defined in some way to allow a solution.```
 Subject: Re: The mathematics of kayaking across a current From: frde-ga on 15 Jun 2006 05:08 PDT
 ```|Obviously, therefore, her course is an arc that heads north-west-ish at first and the curves around to approach the target heading south-west-ish.| Not so obvious, she might have lined up two trees - or have a GPS gizmo```
 Subject: Re: The mathematics of kayaking across a current From: bcattwood-ga on 15 Jun 2006 06:51 PDT
 ```It is not necessary to know the relationship between w and p to write the governing equations. If the kayaker can't make it, then the solution to the equations will simply be unphysical, i.e. negative or infinite time. I don't have time to go into the whole solution, but will say that you need to write a system (two) of differential equations for the x and y components of the kayakers velocity. The x component (dx/dt) will simply be the x component of the velocty vector p pointing from the kayakers instantaneous position (x,y) to the destination (D,E), while the y component (dy/dt) will be the y component of the same vector plus w. Have fun!```
 Subject: Re: The mathematics of kayaking across a current From: djlewis9-ga on 15 Jun 2006 12:11 PDT
 ```Yep -- two differential equations. I got that far, and even started to work out the geometry. But (a) I wasn't sure I had it right; (b) I thought I could get someone else would do it for me. ;-))) But if you are saying that there is no analytical solution (is there?), then I may just as well simulate the whole thing from first principles, right? I can do that pretty easily. As for the comment about lining up two trees -- that is called a "range" in navigation, though it's unlikely to work very well with two trees; it will need a tree and a more distant object like a mountain, or a buoy and a tree, etc. But a range is fundamentally a way to make a straight course, and the mathemtics of that are very simple -- a right triangle with sides D and w*t, and hypoteneuse p*t. My question was what if she doesn't use a range, but simply keeps her boat pointed straight at the target. That makes the course an arc, and takes longer and more effort. In fact, my ultimate goal is to quantify the extra effort. Thanks. --David.```
 Subject: Re: The mathematics of kayaking across a current From: youreh-ga on 20 Jun 2006 18:13 PDT
 ```Is this a homework problem? That is, can we give an answer and stay on good moral ground? It seems a lot of questions recently on answers.google sound an awful lot like homework questions...```
 Subject: Re: The mathematics of kayaking across a current From: goldenvoucher-ga on 21 Jun 2006 04:14 PDT
 ```It is probably best to define a referential with origin E and velocity 0. The velocity vector (V) is the sum of P and W (vectors). W = w*Y P = cos(a)*p*X + sin(a)*p*Y cos(a) = -|x*X|/|x*X + y*Y| sin(a) = -|y*Y|/|x*X + y*Y| where X and Y are unitary vectors a is the angle (point E - kayak - margin plus 180 degrees) x and y are coordinates then V = [(p*x)/(x^2 + y^2)^(1/2)]*X + [(p*y)/(x^2 + y^2)^(1/2) + w]*Y now V = dR/dt where R is the position vector (R = X*x + Y*y) I got the following differential equations: [-(p*x) / (x^2+y^2)^(1/2)] * dt = dx [-(p*y) / (x^2+y^2)^(1/2) + w] * dt = dy The ugly analitical solution is the function you are looking for. I think it is best to make a simulation in a program (e.g. Dynamics Solver). ---- If the current changes according to a sinusoidal function: define l as periodicity, and m as the initial phase. You can then plug w*sin(l*t) in the above equation. [-(p*x) / (x^2+y^2)^(1/2)] * dt = dx [-(p*y) / (x^2+y^2)^(1/2) + w*sin(l*t + m)] * dt = dy w becomes maximum velocity of current. Here's a simulation I ran: http://www.angelfire.com/planet/maquiavel/photos/w.JPG ---- What do you think?```
 Subject: Re: The mathematics of kayaking across a current From: berkeleychocolate-ga on 21 Jun 2006 10:42 PDT
 ```The good news is that the time T = [ p*sqrt(D^2 + E^2) ? E*w]/(p^2 ? w^2) . The bad news is that I have a very complicated proof of this. Since the answer is fairly simple, I suspect there is a simpler proof. Here?s a sketch of the hard way: Note: For naturalness of the coordinate system, I assumed the girl started at the western shore and went northeast. Let r (vector) = (x,y). Let s (vector) = r ? (0,w*t) ? (D,E). Note: t is time. Then ds/dt = dr/dt ? (0,w). The assumptions are that ds/dt has a constant length p and is parallel to (D,E) ? r, which is ?s ? (0,w*t). So also ds/dt is parallel to s + (0,w*t). Since ds/dt has a constant length p, we can write ds/dt = p(cos(theta), sin(theta)). Note that theta is the angle from the position (x,y) to the final destination (D,E), where she starts at (0,0) and (D,0) is on the positive x-axis. Let I stand for the integral sign. Integrating the previous equation, we get s = (I p*cos(theta) dt, I p*sin(theta)) - (D,E). Two vectors (x,y) and (u,v) are parallel when y/x = v/u. Applying this to the two parallel vectors above, we get tan(theta) = [ I p*sin(theta) dt + w*t ?E ]/[ I p*cos(theta) dt ? D ]. After considerable manipulation of these two equations, one arrives at a second order differential equation: d^2 theta/d(theta)^2 + ( 2*tan(theta) + (p/w)*sec(theta)) * (d(theta)/dt)^2 = 0. Using an integrating factor trick, one can solve this equation: Letting b = a/w, we get: [tan(theta) + sec(theta)]^b * [b*sec(theta) ? tan(theta)]/(b^2 ? 1) = m*t + n, where m and n are related to the initial conditions. (You can verify this by substituting this equation into the differential equation.) These can be found since we know the initial value x=y=0 and initial direction (pointing towards (D,E) ). Now knowing m and n, we can find d(theta)/dt by differentiating both sides of the above equation. Knowing d theta/dt, we can go back and calculate x = I p*cos(theta) dt. It turns out that x = (D/K)*[ K ?( tan(theta) + sec(theta) )^b], where K = [tan(theta0) + sec(theta0) ]^b and theta0 is the angle theta at t=0. Finally it?s a simple thing to set x = D and solve for the time, T. It turns out that thetafinal = -pi/2, which means at the last instant the girl is aiming the canoe directly backwards. The integral for the total distance does not seem to be manageable. Anyone for a simpler proof?```
 Subject: Re: The mathematics of kayaking across a current From: berkeleychocolate-ga on 23 Jun 2006 15:39 PDT
 ```A few additional comments to my earlier ones. Note: a above should be p and canoe should be kayak. Here's a solution to the path, parameterized in terms of theta which goes from a start of arctan(E/D) to -pi/2 (backwards). Here t is time, x and y are the coordinates at time t. It is useful to define: b = p/w theta0 = arctan(E/D) K = ( tan(theta0)+sec(theta0) )^b L = b*sec(theta0)-tan(theta0) M = b*tan(theta0)-sec(theta0) Then: t = (D/(K*w*(b^2-1))*[ K*L - ( tan(theta )+ sec(theta) )^b * ( b*sec(theta) - tan(theta) ] x = (D/K)* [ K - ( tan(theta) + sec(theta) )^b ] y = w*t - (D*b/K)* [ (tan(theta) + sec(theta) )^b * ( b*tan(theta) - sec(theta)) -K*M ] One can graph the path by letting theta vary in increments from theta0 to -pi/2 and solving for t, x and y. I did this for p=5,w=2,d=30,e=25 and got a pretty graph roughly resembling an apostrophe.```