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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?

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# Loxodrome - found a mistake on Wolfram MathWorld site?

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