la.vallen wrote:

(In reply to comment #0)

Example: for Earth, ±1' ≈ ±1.8km (±1 arcminute); ±1" ≈ ±31m (1 arcsecond)...
What does a precision of ±0.01° represent in reality?

Well, it partly depends on where you are. Here at 63N is ±1' less than at 30N.

Indeed the "linear precision" is a function of the latitude at that point. Here are the details of the calculation, using an ellipsoid rather than a sphere for more precision.

Given
Req = Radius of Globe at equator (Earth=63781370m)
Rpo = Radius of Globe at pole (Earth=63567523m)
lat = (angular) latitude of point
Pan = angular precision (in degrees)

Sought value
Pln = linear precision

Compute
Tlat = eccentric anomaly

= arctan( (Req/Rpo) * tan(lat) )

Rlat = Radius of Globe at latitude of point

= Req * cos(Tlat)

Plat = Perimeter of Globe at latitude of point

= 2*π*Rlat

Pln = Plat * (Pan/360)

Thus linear precision is
Pln = (Pan/180)*(π*Req*cos(Tlat))

Ex:
lat 0° 30° 50° 70° 85°
Tlat 0 30.08 50.09 70.06 85.016
±0.1° ±11.1km ±9.63km ±7.14km ±3.80km ±967m

±1'  ±1.85km  ±1.61km  ±1.19km   ±633m   ±161m
±1"   ±30.9m   ±26.8m   ±19.8m   ±10.5m  ±2.69m

Dupe of T89218?

Almost dupes, yes. But note that T89218 is only about special values and this is about all values.

Same problem with both suggestions: Conversion to meters depends on the location and is therefor hard to do, if not impossible (e.g. what should happen if I edit the coordinates, should the precision change accordingly or not?). It's fine if external tools allow using meters, but adding this to Wikibase' editing UI would be more confusing than helpful, I'm afraid.