How long for a kayaker to cross a stream with a current it they aim at the target throughout?

Question

Assume the stream is D miles wide, the paddler paddles over water at p miles/hour, the current is a stready w miles/hour. They aim at the target point straight across on the opposite shore and stay aimed there throughout the crossing. However, the current sweeps them downstream, so they don't take a staight course across. They stay continuously aimed at the original target point, but at an increasing angle. Thus the course is an arc, not a straight line. How long does it take? How far do they paddle over water? Over ground? (Analytical answers preferred over numerical simulations.)

Thanks for the analysis, sgasner. While waiting for answers, I did a similar thing, but avoided using angles, instead using pythagorean identities. That gave me two differential equations, but I don't know how to solve them. Is there even an analytic answer? If not, wouldn't it be just as easy to write a numeric simulation using first principles?

Thanks. --David.

Answer 1

Silly bequalming,

If the angle is 90 degrees, the kayaker has reached the opposite shore. Fortunately these kayakers are tough, and never ever slow down. This really comes down to a ratio, between the kayakers progress moment to moment, and the rate of the current. I think we all know that this is possible, depending on the different rates.

First, some definitions.
T = time
T0 = time at the beginning = 0
Tc = time at completion
T* = some time between T0 and Tc
theta(T) = angle of kayaker, which is dependent on time

and notation:

integral(T*..0) [f] = integral of function f, with upper bound T* and lower bound 0.

lim(T*->Tc) is the limit as T* goes to Tc

Lets pick a hypothetical point half way across (time = T*). The kayaker's distance from the opposite shore is the total distance minus his progress (cos(theta)P), his position downstream is the total effect of the current (W*T) minus what the kayaker has made up (sin(theta)P)..

D - integral(T*..0) [cos(theta(T*))P] = distance to travel
WT* - integral(T*..0)[sin(theta(T*))P] = distance downstream

Now we can form a function of time to describe the angle theta(T) at time T*

tan(theta(T*)) = opp/adj =
D - integral(T*..0) [cos(theta(T*))P] /
WT* - integral(T*..0)[sin(theta(T*))P] =

simplifying:
tan(theta(T*) =
[D - sin(theta(T*))P] / [WT* - (-cos(theta(T*)P - P)]

so

theta(T*) =
arctan{ [D - sin(theta(T*))P] / [WT* - (-cos(theta(T*)P - P)] }

ugly, but manageable.

So now to the real questions, how long to cross? The cross is complete when their distance from the opposite shore is zero, i.e.
we know that

D - integral(Tc..0) [cos(theta(Tc))P] = 0
or
integral(Tc..0) [cos(theta(Tc))P] = D
or
[sin(theta(Tc))P] = D

We just need to solve for Tc, so we try a limit

lim(T*->Tc)[sin(theta(T*))P] = D

sub in our function for theta(T*) to get

lim(T*->Tc)[sin(arctan{ [D - sin(theta(T*))P] /
[WT* - (-cos(theta(T*)P - P)] })P] = D

Numercially i would start with small values of T, and increase them until I get T value for which this is true. That would require values of W, D and P.

Ok, I've wasted an hour. the rest is up to you. no guarantees on the quality, by the way. check my math.

Answer 2

ok.. my guy travels at the speed of light.. so he goes in a straight line... across the stream.. hehe.. so he paddles just the distance across the stream.. and it takes him nearly zero time.. hahaahahaa!!!

ok.. straight line.. is the shortest... when (stream speed / paddling speed) approaches zero.

the problem does NOT state that the kayaker must reach the destination point.. only the other side!!! and since the kayak has a certain width.. it will hit the opposite shore at some angle less than 90 degrees to the straight line across the stream. (if it was 90 degrees.. half of the kayak would be on the shore.)

so... if the kayaker paddles at the SAME speed as the stream... he will stay within a right triangle outlined by the line straight across and one at 45 degrees angle

Ok.. it is late..and i'm braindead.. I'll let you think about it more.. hehe

Answer 3

they never reach the opposite side. the limit increases without bound due to the current.

Oops. I meant, the limit of the angle he has to travel approaches 90º due to the current.
It's like he moves 1/2 the distance every second; that kind of conundrum.

sgasner, I said the limit of the angle. If you create an equation, you will receive a discontinuity at the "solution." You must calculate a limit in order to receive a meaningful answer. SO, my original answer stands: the kayaker does not reach the opposite shore.

Answer 4

Just teach the kayaker to aim upstream! :-)