I am still puzzeling to get a nice equation for the arclength of an hypercycle.

(I asked a similar question (less developed) about a year ago that was never answered, now i am a bit further, i deleted the old question, to prevent duplicates)


  • a standard hyperbolic plane (curvature = -1)

  • two points $S_1$ and $S_2$ a distance $s$ apart.

What is the arclength of the hypercycle $h$ whose axis is a distance $d$ away as function of $s$ and $d$?

How far I came:

  • make the Saccheri Quadrilateral whose summit $S_1S_2$ and whose base is $B_1B_2$ (so the angles $\angle S_1B_1B_2$ and $\angle B_1B_2S_2$ are right base angles) and call the distance $B_1B_2 \ b $.

Then the following equations are valid:

$$ h = b \cosh(d) $$

$$ \sinh \left( \frac {s}{2} \right) = \sinh \left( \frac{b}{2} \right) \cosh(d) $$

But then to remove $b$ out of the equation I get stuck on the rather complicated formula:

$$ h = 2 \operatorname{arsinh} \left( \frac{ \sinh \left( \frac{s}{2} \right) }{\cosh (d)} \right) \cosh (d) $$


$$ s = 2 \operatorname{arsinh} \left( \sinh \left( \frac{h}{2 \cosh (d)} \right) \cosh (d) \right) \cosh (d) $$

Can these formulas be simplified further?

  • $\begingroup$ Originally you wanted $h$ in terms of $s,d.$ Then eliminating $b$ you get to the first additional formula, as in my answer. But your second added formula makes it look like you're going for $s$ in terms of $h,d.$ Which variable do you want in terms of the other two? $\endgroup$ – coffeemath Jan 28 '16 at 16:37
  • $\begingroup$ Yes i am looking for $h$ as function of $s$ and $d$ , just added $s$ as function of $h$ and $d$ for completeness. $\endgroup$ – Willemien Jan 28 '16 at 16:42

$\sinh(x)$ being strictly increasing on the reals, has a unique inverse function. Specific values for the inverse can be found using the definition of $\sinh(x)$ and the quadratic equation, and a logarithm extraction.

You already have $\sinh(b/2)=\sinh(s/2)/\cosh(d)$ in terms of $s,d.$ So after applying inverse sinh and doubling you have $b$ in terms of $s,d,$ which may be then put into $h=b\cosh(d)$ to arrive at $h$ in terms of $s,d.$

Added later: The inverse sinh in terms of log is $$\sinh^{-1}(x)=\log(x+\sqrt{x^2+1}).$$ When used here, it produces a rather involved expression for $h$ in terms of $s,d$ which (as far as I see) doesn't simplify to anything "nice", due to the expression inside the log at the end.

  • $\begingroup$ thanks I changed the question a bit, hope you can help me further $\endgroup$ – Willemien Jan 28 '16 at 12:33
  • $\begingroup$ Since the inverse sinh is in terms of log as noted, I think your formula cannot simplify because of the expression $\sinh(s/2)/(\cosh(d)$ which ends up inside the logarithm, and it cannot be pulled out using log rules. $\endgroup$ – coffeemath Jan 29 '16 at 13:13

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