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Q: The mathematics of kayaking across a current ( No Answer,   8 Comments )
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

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.
There is no answer at this time.

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 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|
X and Y are unitary vectors
a is the angle (point E - kayak - margin plus 180 degrees)
x and y are coordinates


V = [(p*x)/(x^2 + y^2)^(1/2)]*X + [(p*y)/(x^2 + y^2)^(1/2) + w]*Y


V = dR/dt

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


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:

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)


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.

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