Loxodrome - found a mistake on Wolfram MathWorld site?

[kkz23691]

Hello,

Could it be?... We find this claim on Wolfram MathWorld site http://mathworld.wolfram.com/SphericalSpiral.html

The claim is that this curve (given in oblate spheroidal coordinates in the limit where the spheroid is a sphere, measuring the inclination angle $c$ with respect to the xy-plane and making the unusual choice of putting a "minus" sign in z)

${\mathbf r}_x=\cos t \cos c$
${\mathbf r}_y=\sin t \cos c$
${\mathbf r}_z=-\sin c$
where
$c=\tan^{-1}(at)$

is a loxodrome - a curve that lies on the sphere and forms the same angle with all meridians as it winds around the sphere.

Then the angle with respect to all meridians must be the same. This means (take $p$ as an independent parameter)
$\cos\alpha=\frac{\frac{d{\mathbf r}}{dc}\cdot \frac{d{\mathbf r}}{dp}}{|\frac{d{\mathbf r}}{dc}| |\frac{d{\mathbf r}}{dp}|}=\mbox{const}$

It seems that this is not true!
Proof:

Since $c$ is the inclination angle, $\frac{d{\mathbf r}}{dc}$ is a vector tangential to the meridian. Its components are:
$\frac{d{\mathbf r}_x}{dc}=-\cos t \sin c$
$\frac{d{\mathbf r}_y}{dc}=-\sin t \sin c$
$\frac{d{\mathbf r}_z}{dc}=-\cos c$

Now, the sphere is a surface parameterized with 2 parameters $(c,t)$; let the loxodrome be parameterized with one parameter $p$ such that $(c,t)=(c(p), t(p))$. Then

${\mathbf r}_x(p)=\cos t(p) \cos c(p)$
${\mathbf r}_y(p)=\sin t(p) \cos c(p)$
${\mathbf r}_z(p)=-\sin c(p)$

Then the vector tangent to the curve is $\frac{d{\mathbf r}}{dp}$ and its components are

$\frac{d{\mathbf r}_x}{dp}=-(\sin t \cos c) t^{\prime} - (\cos t \sin c) c^{\prime}$
$\frac{d{\mathbf r}_y}{dp}= (\cos t \cos c) t^{\prime} - (\sin t \sin c) c^{\prime}$
$\frac{d{\mathbf r}_z}{dp}=-(\cos c) c^{\prime}$

After taking into account that for any vector ${\mathbf x}$ it is true that $|{\mathbf x}| = \sqrt{{\mathbf x}\cdot{\mathbf x}}$, the norm of this vector is $|\frac{d{\mathbf r}}{dp}| = \sqrt{c^{\prime 2} + t^{\prime 2} \cos^2 c}$ and $|\frac{d{\mathbf r}}{dc}| =1$.

The dot product $\frac{d{\mathbf r}}{dc}\cdot \frac{d{\mathbf r}}{dp}$ turns out to be equal to $c^{\prime}$. Then

$\cos\alpha=\frac{c^{\prime}}{ \sqrt{c^{\prime 2} + t^{\prime 2} \cos^2 c}}$

Now Mathworld claims that $t$ is an independent parameter and $c(t)=\tan^{-1}(at)$. Then $t^{\prime}=\frac{dt}{dt}=1$ and $c^{\prime}=\frac{dc}{dt}=\frac{1}{1+a^2t^2}$.
Plugging in into the expression for $\cos\alpha$ yields

$\cos\alpha=\frac{1}{\sqrt{1+2a^2t^2}}\neq \mbox{const}$

Therefore, the curve given above is not a loxodrome?

 

[Greg Bernhardt]

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

 

[kkz23691]

I pdf-ed the proof and emailed it to mathworld. Hopefully, they will check and let me know.

 

[lavinia]

Your calculations have errors. For instance the formula for c' is wrong.

Notice that cos(c) = 1/(1 + (at)^2)

 

[kkz23691]

Hello Lavinia,
thank you for reviewing this long post!
Here is my reasoning:

$c^{\prime} = \frac{dc}{dt}= \frac{d}{dt}\tan^{-1}(at)= \frac{d}{dt}\arctan(at)=\frac{1} {1+a^2t^2}$

As far as cos(c) is concerned:
Take a right-angled triangle with sides 1, x and hypothenuse equal to $\sqrt{1+x^2}$.
Then the angle over-against x is $\arctan(x)$. The cosine of this same angle is equal to $\frac{1}{\sqrt{1+x^2}}$
Thus,
$\cos(\arctan(x)) = \frac{1}{\sqrt{1+x^2}}$
This means
$\cos(c) = \cos(\arctan(at)) = \frac{1}{\sqrt{1+a^2t^2}}$.

 

[lavinia]

Hello Lavinia,
thank you for reviewing this long post!
Here is my reasoning:

$c^{\prime} = \frac{dc}{dt}= \frac{d}{dt}\tan^{-1}(at)= \frac{d}{dt}\arctan(at)=\frac{1} {1+a^2t^2}$

As far as cos(c) is concerned:
Take a right-angled triangle with sides 1, x and hypothenuse equal to $\sqrt{1+x^2}$.
Then the angle over-against x is $\arctan(x)$. The cosine of this same angle is equal to $\frac{1}{\sqrt{1+x^2}}$
Thus,
$\cos(\arctan(x)) = \frac{1}{\sqrt{1+x^2}}$
This means
$\cos(c) = \cos(\arctan(at)) = \frac{1}{\sqrt{1+a^2t^2}}$.

The derivative of the arctangent(x) is 1/(1+x^2) but you want arctangent(ax). You forgot to use the Chain Rule.

I get cos(c) for the dot product of the derivative of the curve with the unit tangent to the meridian. What do you get?

 

[kkz23691]

The derivative of the arctangent(x) is 1/(1+x^2) but you want arctangent(ax). You forgot to use the Chain Rule.

$a$ is a constant. Only $t$ is a variable. Sorry, I should have said this earlier. In the end, a 3-d curve is to depend on only one variable say $t$: ${\mathbf r}={\mathbf r}(t)$

This would mean that the derivative $\frac{dc}{dt}$ will treat $a$ as a constant.
I see you have found this error in my derivation:

I wrote:
$\frac{d(\arctan(at))}{dt} = \frac{1}{1+a^2t^2}$

while the correct derivative is:
$\frac{d(\arctan(at))}{dt} = \frac{a}{{1+a^2t^2}}$

 

[kkz23691]

I get cos(c) for the dot product of the derivative of the curve with the unit tangent to the meridian. What do you get?

I get $c^{\prime}$. Here is my reasoning:

Since $c$ is the inclination angle, $\frac{d{\mathbf r}}{dc}$ is a vector tangential to the meridian. Its components are:
$\frac{d{\mathbf r}_x}{dc}=-\cos t \sin c$
$\frac{d{\mathbf r}_y}{dc}=-\sin t \sin c$
$\frac{d{\mathbf r}_z}{dc} =−\cos c$

The vector tangent to the curve is $\frac{d{\mathbf r}}{dp}$ and its components are

$\frac{d{\mathbf r}_x}{dp}=-(\sin t \cos c) t^{\prime} - (\cos t \sin c) c^{\prime}$
$\frac{d{\mathbf r}_y}{dp}= (\cos t \cos c) t^{\prime} - (\sin t \sin c) c^{\prime}$
$\frac{d{\mathbf r}_z}{dp}=-(\cos c) c^{\prime}$

Then
$\frac{d{\mathbf r}}{dc}\cdot \frac{d{\mathbf r}}{dp} = c^{\prime}$.

 

[lavinia]

I get $c^{\prime}$. Here is my reasoning:

Since $c$ is the inclination angle, $\frac{d{\mathbf r}}{dc}$ is a vector tangential to the meridian. Its components are:
$\frac{d{\mathbf r}_x}{dc}=-\cos t \sin c$
$\frac{d{\mathbf r}_y}{dc}=-\sin t \sin c$
$\frac{d{\mathbf r}_z}{dc} =−\cos c$

The vector tangent to the curve is $\frac{d{\mathbf r}}{dp}$ and its components are

$\frac{d{\mathbf r}_x}{dp}=-(\sin t \cos c) t^{\prime} - (\cos t \sin c) c^{\prime}$
$\frac{d{\mathbf r}_y}{dp}= (\cos t \cos c) t^{\prime} - (\sin t \sin c) c^{\prime}$
$\frac{d{\mathbf r}_z}{dp}=-(\cos c) c^{\prime}$

Then
$\frac{d{\mathbf r}}{dc}\cdot \frac{d{\mathbf r}}{dp} = c^{\prime}$.

You are correct. I was taking the inner product with a latitude nor a longitude.

 

[kkz23691]

Thank you for your time. I checked and even after correcting the derivative of the $\arctan(ax)$ the cosine of the angle is still not a constant. It looks like this curve is not a loxodrome...

 

[lavinia]

Thank you for your time. I checked and even after correcting the derivative of the $\arctan(ax)$ the cosine of the angle is still not a constant. It looks like this curve is not a loxodrome...

I rechecked your calculations a few times and think they are now correct.

I suspect that the Wolfram article has a typo and c should be the inverse tangent of a^t not a.t.

If you solve for the curve in the unit disk that makes a constant angle with the radial lines emanating from the origin, you get a logarithmic spiral. Project this spiral into the unit sphere via stereographic projection. This will be a curve that makes a constant angle with the meridians,this because the radial lines project to the meridians and stereographic projection is a conformal map.

I got the curve,

( sech(Kt)cos(t), sech(Kt) sin(t), -tanh(Kt)).

where K is a constant that depends on the cosine of the angle of intersection of the radial lines with the logarithmic spiral.

This curve in spherical coordinates is

( cos(f(Kt))cos(t), cos(f(Kt))sin(t),-sin(f(Kt)))

where the function,f, is the inverse sin of tanh(Kt).

I found an identity says that f can be expressed as the arcsine of an exponential. Take a look at the Wikipedia article on the Gudermannian function for this identity.

 

[kkz23691]

Hello Lavinia,
all of this makes sense. The curve you have derived is also found here:
http://en.wikipedia.org/wiki/Rhumb_line#Mathematical_definition
and matches your result.

Thank you for your time!