# Transform 1860s Australian (English?) map coordinates to modern ones

I hope this is the right place. I found an 1862 reference to a very specific land location, not from where I live in Melbourne, Australia.

However, when I entered the latitude and longitude into Google Earth it placed the point in the sea. I wasn't really surprised, but am now looking for clues as to how I could transform the two values to 'modern' ones. (eg UTM/WGS84). This is not my area, but I'm ok with maths, Excel, Python etc.

I'd assume it would mean first finding out what the common 1860s Australian (therefore British) projections/datums would have been.

These are the values:

``````'Longitude E. 9h 39m 39s.3, Latitude S 38 52' 6".6'
``````

He was an Astronomer so, I assume, was giving the Longitude as a Right Ascension...and we multiply (the decimal longitude) by 15

• I'm getting a lon, lat of 144.91375, -38.8685 putting those coords in Bass Strait directly south of Melb. Where is the location supposed to be? Jul 10, 2020 at 2:59
• Maybe the degrees part of latitude is 37 instead of 38? Because S37° 52' 6.6", E144° 54' 49.5" falls just 200 meters from Williamstown Timeball Tower in Melbourne, erected in 1852, and the timeball served as a marine chronometer adjustment method to allow sailors to determine accurate time and longitudes. This sounds like it was a significant benchmark back then. Jul 10, 2020 at 19:01

I would hazard a guess that the latitude is off by 1 degree, and the coordinates may refer to Williamstown Observatory, because using 37 instead of 38 degrees, the coordinates would be S37° 52' 6.6", E144° 54' 49.5", a mere 200 meters away from the Williamstown Timeball Tower (37°52′00.6″S 144°54′45.7″E) (seen on the right in the picture below). This site had an observatory named Williamstown Observatory that was used until 1863 before the astronomical instruments were moved to Melbourne Observatory.

The observatory had transit instruments allowing to determine accurate timing by observing stars, similar to the Greenwich Observatory in the UK. It also served as a standard meridian for the early Geodetic Survey of Victoria (1858-1872)

Here are some interesting links for further reading on the Williamstown Observatory:

Indeed, the coordinates seem to fall a bit offshore when we use the modern WGS84. This isn't surprising, because back then they determined the geographic coordinates of Williamstown Observatory astronomically, which typically can differ from modern values by hundreds of meters, because of different sources of errors in the astronomical method. For example, the deflection of the vertical being around 7"S, 4"W (estimated from EGM2008 geoid), even with perfect instruments and chronometers, they would measure the coordinates to be 7"S, 4"W of the modern coordinate in WGS84.

Since Williamstown served as an important reference benchmark for laying out the Geodetic Survey of Victoria, the rest of the survey follows the then determined coordinate and is "off" by a certain amount when compared to WGS84. Also, the use of a different ellispoid (Clarke 1858) to compute coordinates away from Williamstown also contributes to differences in geographic coordinates when you compare to WGS84.

This old 1874 document seems to report the coordinates of Williamstown Observatory as S37° 52' 7", E144° 54' 42". I don't know if this is the reference coordinate they used, but if we knew them for sure, then it might be possible to calculate an approximate mathematical transformation from that datum to WGS84. Ideally, we would need a reference azimuth, or other reference points, to determine a transformation. There is a drawback to the mathematical approach though; positions derived from surveys back then weren't quite as consistent as today, so transformation errors are still to be expected, but for a relatively small region (eg a city), it could give an interesting approximation. Maybe a transformation already exists, but I haven't found any.

However, if the coordinates you are talking about were determined purely astronomically, then I'm afraid the margin of error of these instruments is too large to determine a consistent transformation.

• Fascinating discussion and theory, thank you. Yes it's the Williamstown Observatory. Could the differences in the Latitude be explained by the older spheroid etc? (Not my area). Interesting that something similar happened to me with GPS. But isn't "S37° 52' 6.6", E144° 54' 49.5" still in the sea? Jul 11, 2020 at 9:52
• See my edits in the answer where I elaborate a bit more on a potential approximate solution for the transformation. Jul 11, 2020 at 15:31
• Thank you again. I have a vague memory that there were TWO observatories here in Williamstown; one for a short time. I'll do some work on this. Old mate gives such an apparently precise Longitude, but that's not the same as being accurate Jul 11, 2020 at 23:00
• I've found some contemporary values for the Melbourne Observatory and found a similar situation. Very interesting. dwids.wordpress.com/2020/07/12/… Jul 12, 2020 at 8:28
• Interesting finds! But the corresponding decimal latitude for 37° 49' 53" is -37.831388, not -37.89805556. That's why your point was so far away from the observatory. Jul 12, 2020 at 15:04

An Australian National Grid page says:

Prior to 1966, Australian maps used a transverse mercator projection, in yards, with the Clarke 1858 ellipsoid. This was called the Australian National Grid (ANG).

Where in the World… Geodetic Datum Mapping Issues suggests the same:

As an example, in Australia, there are 3 main datums in use: AGD66, AGD84 and the latest one – GDA94. Each of these datums is supposed to reference particular spheroids – AGD datums use ANS (Australian National Spheroid) and GDA uses GRS1980. However, an issue that often occurred in pre-1970 wells and surveys, was the use of an older spheroid – the Clarke 1858 Spheroid.