My question refers to Gauss's heliotrope - an optical device for projecting sunlight to a distance by a reflecting mirror, that was used in geodetic surveys in order to mark reference points. The problem is (and Gauss was aware of this problem) that since the index of refraction changes with the alttitude above sea, than according to snell's law the rays of light are not straight but rather curved. So my question is how did geodesits make a practical use of hte heliotrope? and did Gauss originally face the problem?
$\begingroup$
$\endgroup$
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
5
In geodesy, one usually looks horizontally, or almost horizontally, so the ray travels in the same layer of the atmosphere, where the temperature and pressure are almost constant, thus refraction is negligible.
Sun only serves as the source of light in this case, so the refraction of the ray coming from the Sun to the devise does not matter.
Refraction is not negligible for astronomic observations (which were also used in geodesy, for the measurements in which the Sun position is relevant), to take it into account, the tables of refraction are used, but the true refraction may depend on the state of the atmosphere and be different from the tables value. This is one of the major difficulties with astronomical measurements. (Gauss certainly was aware of this problem, it is well known since the antiquity, and the most precise tables of refraction in Gauss's time are due to his friend Bessel).
There are several possible uses of the heliotrope, so one has to specify the exact setting to discuss whether refraction is relevant or not in this setting.
answered Dec 29, 2017 at 14:25
-
1$\begingroup$ i think saying that for light that is reflected horizontally refraction is neglible is a mistake, since even if a ray of light is parallel to the ground than still the curvature of earth (and thus the curvature of athmospheric layers) plays role in "bending" of the light ray. I read somewhere that Gauss did some work on the bending of ilght rays in the athmosphere but i don't remenber where. $\endgroup$– user2554Commented Dec 29, 2017 at 14:35
-
$\begingroup$ Geodesic measurements are made on relatively small distances (direct visibility) so the curvature of Earth has negligible effect on refraction at these distances. Gauss indeed did some work on refraction. But it is relevant for astronomical refraction, not for a ray from one mountain top to another. $\endgroup$ Commented Dec 29, 2017 at 14:42
-
1$\begingroup$ I voted your answer simply because i respect the effort you made. But i read that typical distances in a geodetic survey are of the order of tens of kilometers, and i don't think these are neglible distances relative to earth radius (6400 km) . For the purpose of using a heliotrope even an angular deviation of one degree may make the heliotrope station not visible. $\endgroup$– user2554Commented Dec 29, 2017 at 14:52
-
-
$\begingroup$ @user2554: I am afraid you don't understand what we are talking about. "One degree" is not an error, this is a blunder! Astronomical refraction usually does not exceed few minutes. Terrestrial refraction is of the magnitude of few seconds! The Gauss heliotrope accuracy (of reading the scale) was probably round 20". That's why plus or minus 5" could not be relevant. $\endgroup$ Commented Dec 30, 2017 at 2:06
$\begingroup$
$\endgroup$
In 1825, Gauss introduced the coefficient of refraction k, which represents a way to quantify terrestrial refraction. He defined it as the ratio of the radius of the earth R and radius of curvature r of light ray (the line of sight). He found, on the basis of reciprocal vertical angle measurements near hannover, an average value of 0.13 for this coefficient.
Now, in geodesy there is a mathematical trick of dealing with the curvature of line of sight - it's called effective earth radius convention, and it's very useful in calculating horizons seen from an heightened point. It's defined mathematicaly as:
$$ R_{eff} = \frac {{R_{earth}}}{{1 - k}}$$
By replacing earth with an imaginary earth with greater radius, one can make calculations from a new reference system. In order to direct the heliotrope ray to a distant observer, one has to calculate angles on the basis of the horizontal distance x between two points A and B, their heights $h_1$ and $h_2$, and the effective earth radius $R_{eff}$.
$\begingroup$
$\endgroup$
2
(Cannot comment, so answer to the comment of Alexandre Eremenko here. However, this also contains an answer to the original question — if one takes it literally, not “in spirit”.)
According to a text of 1875, the mean error of Gauss’ survey was 0.59″. So an error of 5″ would be a killer. (No, I have no clue how one would be able to average enough theodolite measurements to get down to 0.59″. I would expect one needs thousands of them…)
Also: I have no idea why precision of heliotropes enters the equation. With Sun’s angular size of 0.5°, you do not need precision much more than this. I imagine the process as a heliotrope at one end, and a theodolite at the other end. It is the precision of theodolite which matters, right?
-
$\begingroup$ Please do not post comments as answers. Either wait until your reputation is higher or upgrade this to an alternative answer. $\endgroup$– ConifoldCommented Mar 29, 2018 at 20:17
-
$\begingroup$ Done. (In fact, it already was.) $\endgroup$ Commented Mar 30, 2018 at 22:56