Puzzling with curvature This is a post inspired by Gauss Bonnet theorem validation with hyperbolic circles
I was thinking also about curvature and how to measure it 
And was puzzling with the following :
Suppose some critters land on a hyperbolic plane (a surface with a constant negative curvature) and they want to measure the curvature of it.
(critters is a nod to an old question about critters on a hyperbolic plane :)
I was thinking on how they could measure the curvature and I found one the following
 way:
The critters have brought  their own measure m (think meters/ miles/yards /feet)  and construct an isosceles right triangle with the top angle as the right angle 
The legs of this triangle are both length $a m$ and the hypotenuse has length $b m$ , 
(In Euclidean geometry  the base angles are $45^o$ and the length of the hypotenuse is always $bm = \sqrt {2} a m$ 
)
In hyperbolic geometry the base angles are $< 45^o$ and length of the hypotenuse is between $\sqrt {2} a m $ and $ 2 a m $
Even better from this construction  you can estimate  what the hyperbolic absolute lengths  of $a m$ and $b m $ are.
$a m$ and $b m$ are  also related by $\cosh(b m) =  \cosh(a m) \cosh(a m) =\cosh^2(a m)$ 
So for every $\frac{b m}{a m} = \frac{b}{a } $ there is only one pair of absolute lengths   $l_h = am$ and $h_h = b m$
And from this $m$ and the curvature can be estimated .
To make it all a bit clear in a table: ( I am still myself puzzeling with this ) 
 b/a   -> l     -> h
 1.43  0.372    0.532
 1.44  0.479    0.690
 1.45  0.567    0.822
 1.5   0.919    1,378
 1.6   1.523    2.438
 1.7   2.224    3.816

So for example: if the by de critters measured lengths  are $a m= 2m$ and $bm = 3m$
then the lengths in  absolute hyperbolic lengths are 0.919 and 1,378 meaning $m=\frac{0.919}{2}=\frac{1,378}{3}= 0.459$ 
But then: 
What is the curvature ? Is it $ -m$ ,$ -m^2$,  $\frac{1}{-m}$ or $\frac{1}{-m^{2}} $?
Is there a more direct way to calculate $l_h$ and  $h_h$ from a/b? (Instead of estimating it by looking it up in a table) 
I like this way of calculating curvature (it is quite simple just a simple triangle an measuring of lenghts ) are there an even more simple ones ?
 A: Did not fully understood what geodetic information the critters get, however, am attempting a basic  answer hoping it would serve as background for possible discussion / insight with 3D hyperbolic triangles constructed by pseudospherical trigonometry  measurements.
Given: $\,C= \pi/2,a=b$
To Find:
A relation between $(a,A)$ for hyperbolic isosceles right triangles to find other angles and sides.
Sinh and Cosh Laws in pseudospherical Trigonometry
(The term used from Roberto Bonola's Non-Euclidean Geometry text-book pp 137)
Let Gauss curvature be $ -1/T^2.$
By hyperbolic Law of (Sinhs)
$$ \frac{\sin A}{\sinh (a/T)}=\left(\frac{\sin B}{\sinh (b/T)}\right)=\frac{\sin C}{\sinh (c/T)}=\frac{1}{\sinh (a/T)}$$
$$ \frac{\sin A}{\sinh (a/T)}=\frac{1}{\sinh (c/T) }   \tag1$$
From hyperbolic Law of  Cosines (Coshs)  triangle sides with a right angle can be generalized to "hyperbolic Pythagoras " theorem:
$$\cosh(c/T) = \cosh(a/T)  \cosh(b/T) =  \cosh^{2}(a/T) \tag2$$
Combine (1) and (2) and simplify to get a relation  for isosceles hyperbolic right triangles as
$$\sin A \sqrt{ 1+ \cosh^{2} a/T }=1 \tag3$$
The graph  of above trigonometric relation.. $ (a,A) $ plotted on $(x,y)$ axes respectively.. and  representative 3D projection of  triangles  made on  bisected sides  $a=(1,2) $  is shown red and blue triangles. Also
$$\sin A \sqrt{ 1+ \cosh \, c/T }=1 \tag4$$

Eliminate $A$ from (3) and (4) side/hypotenuse ratio is found variable for right hyperbolic isosceles requirement but you are trying to fix it up .. so cannot understand how that could be done.
If $ a=1,c=\sqrt2 $ are given, dividing (3) and (4) numerical solution $ T \approx 7300,K=-1/T^2 \approx \frac{-1}{7300^2}$  the Gauss curvature is seen to be very low, near to a flat plane.

c=Sqrt[2.]; a=1. ;Plot[{Cosh[c/tt]/Cosh[a/tt]^2}-1., {tt,6500.,
8000.}]


Numerical data
Clairaut's constant for all geodesics shown
$$r_i \sin \psi_i= 1/4 $$
$$ K=-1/T^2,\, (a,b,c)= (2,2,3.38021); \quad (a_{small},b_{small},c_{small})=(1,1,1.51337);\quad (A,B,C)= (0.249795,0.249795,\pi/2);\quad(for \ both) $$
When three arcs are plotted with these lengths and surface angles, a closed isosceles triangle forms checking accuracy of calculation.
An interesting take on this is about Integral curvature $\int\int K dA$ is same for all right isosceles  triangles same  $A$ when $a$ is partitioned in anyway.. as the pseudospherical deficit is same for the red and blue side triangles given above or for any other triangle constructed in a similar way.It evaluates to $ I.C.= (\pi/2-2A)$.
The set of hypotenuses are not parallel in any sense.
In response to your question in comments that I still cannot understand.. for all  prescribed ratios of leg length $a$ to  hypotenuse $c..$  by virtue of Equation (2) we can  have only one trivial triangle solution $ a=c=0.$
In conclusion, ratio $c/a$ is a fixed number. In the last $\cosh$ graph $c/T,a/T$ relation is not  a straight line through the origin. To me it appears the  "ratio" of side/hypotenuse restriction is perhaps our hang-up from Euclidean trigonometry concept of angle definition. One has to abandon it .. large nonlinearities of non-Euclidean geometries demand it. Even in spherical trigonometry a construction in with different $c/a$ ratios produces different integral curvatures.
EDIT1:
The conclusion was not comprehensive enough.
We should be able to plot $K$ as a function of hypotenuse to side ratio $c/a$. The plot should include flat Euclidean special case as well as spherical geometry  if possible. The following does precisely that. Dividing (3) by (4)
$$ \cosh (c \sqrt{|K|})/ \cosh ^2 (a \sqrt{|K|})  = 1 ,\quad \cos (c \sqrt{|K|})/ \cos  ^2 (a \sqrt{|K|})  = 1  \tag5 $$

The hyperbolic triangle hypotenuse/side ratio$c/a$ should be $\sqrt 2 $  when $K \approx 0$ for flatter triangles and $c/a=2$ when the sides are deep and curved...
You cannot choose a ratio like 1.2 or 3.0 outside this interval.
That solves the puzzle perhaps more than earlier ..
