A is a point on the Poincaré disk model of the hyperbolic plane. B is a second point, d hyperbolic distance away from A. The hyperbolic ray AB passes through A at angle θ.

How might one find the coordinates of B, given A, d, and θ?

In other words, how would one answer a question like this:

Beginning at Euclidean position (-.3, .4) on the Poincaré disk, one turns to face Northwest (3π / 2 radians) and travels 3 units. Where does one end up?

Is enough information given?

  • 1
    $\begingroup$ $\theta$: the angle of $AB$ to what other line? $\endgroup$ – zoli Jun 5 '15 at 0:39
  • $\begingroup$ relative to the x axis $\endgroup$ – Nick Barry Jun 5 '15 at 0:42
  • $\begingroup$ What is the angle of $AB$ to the $x$ axis if the two don't intersect within the hyperbolic plane? $\endgroup$ – zoli Jun 5 '15 at 0:44
  • $\begingroup$ then, relative to the line parallel to the x axis (parallel in the Euclidean sense), which passes through A $\endgroup$ – Nick Barry Jun 5 '15 at 0:49
  • $\begingroup$ In other words, the slope of the tangent line of AB at A is equal to tan(θ). $\endgroup$ – Nick Barry Jun 5 '15 at 1:00

Not enough information.

Let $A$ be an arbitrary point on the disk.

Draw the hyperbolic line $l$ through $A$ at with slope $\tan\theta$ at $A$.

Now one can select two points $B$ and $C$ on $l$ so that $d(A,B)=d(A,C)=d$ ($A$ is the midpoint between $B$ and $C$.

  • $\begingroup$ Would one of those points not be indicated for -π/2 < θ < π/2, and the other for π/2 < θ < 3π/2 ? $\endgroup$ – Nick Barry Jun 5 '15 at 1:19
  • $\begingroup$ yes, if you replace "line" with "ray" - but the choice of $A$ remains arbitrary. $\endgroup$ – sds Jun 5 '15 at 1:25
  • $\begingroup$ A good point. I should have said "ray." It's amateur hour in my time zone. $\endgroup$ – Nick Barry Jun 5 '15 at 1:31

Construct a hyperbolic triangle from $A$, $B$, and the center of the unit circle, $C$. Let $a$ be the length of $BC$, $b$ the length of $AC$, and $c$ the length of $AB$. Let $\alpha$, and $\gamma$ be the values of $\angle CAB$ and $\angle ACB$, respectively.

The value of $c$ is given as distance $d$: $$c = d$$

The value of $\alpha$ is determined from the angle of $A$ relative to the unit circle, $\epsilon$; and the angle correspondent to the given direction of travel, $\delta$: $$\alpha = |\pi - |\epsilon - \delta||$$

The value of $b$ can be determined based on the Euclidean distance from the center of the unit circle to $A$, $d_A$: $$b = 2 \operatorname{arctanh}(d_A)$$

The value of $a$, the hyperbolic distance between $B$ and the center of the unit circle, can now be determined by the hyperbolic law of cosines: $$cosh(a) = cosh(b)cosh(c) - sihn(b)sinh(c)\cos(\alpha)$$ $$a = arccosh(cosh(b)cosh(c) - sihn(b)sinh(c)\cos(\alpha))$$

The value of $C$ can be also be determined by the hyperbolic law of cosines: $$\cos(\gamma) = \frac{cosh(a)cosh(b) - cosh(c)}{sinh(a)sinh(b)}$$ $$\gamma = arccos(\frac{cosh(a)cosh(b) - cosh(c)}{sinh(a)sinh(b)})$$

Now, add (or subtract), $\gamma$ from $\epsilon$ to determine the angle of $B$ relative to the unit circle, $\theta$. The Euclidean distance between $B$ and the center of the unit circle, $d_B$, can be determined using the exponential function: $$d_B = \frac{e^a - 1}{e^a + 1}$$

Using this distance, the coordinates $x_B$ and $y_B$ of $B$ can be now be determined using basic trigonometric functions: $$x_B = d_B \cos(\theta)$$ $$y_B = d_B \sin(\theta)$$

The coordinates of destination point $B$ are $(x_B, y_B)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.