I am getting GPS information from a device like this
052340.000,A
32.46275,N
75.310415,E
I know N is for north and E for east but what A is representing?
Looking at the value, and some of the other comments, it is unlikely to be an altitude in meters. If this has been extracted from a GPGLL NMEA sentance, the value is time of fix, e.g. 05:23:40, as per the following
$GPGLL
Geographic Position, Latitude / Longitude and time.
eg2. $GPGLL,4916.45,N,12311.12,W,225444,A
4916.46,N Latitude 49 deg. 16.45 min. North
12311.12,W Longitude 123 deg. 11.12 min. West
225444 Fix taken at 22:54:44 UTC
A Data valid
eg3. $GPGLL,5133.81,N,00042.25,W*75
1 2 3 4 5
1 5133.81 Current latitude
2 N North/South
3 00042.25 Current longitude
4 W East/West
5 *75 checksum
$--GLL,lll.ll,a,yyyyy.yy,a,hhmmss.ss,A
llll.ll = Latitude of position
a = N or S
yyyyy.yy = Longitude of position
a = E or W
hhmmss.ss = UTC of position
A = status: A = valid data
Altitude. it would depend on the device as to what units it is. from the number shown in your example i would doubt it is meters, unless you are in an aeroplane.
more info here
Related
I have a dataset with various longitude and latitude datapoints. I would like to develop a centroid or "average" of those longitude and latitude coordinates in bigquery at a specific level of granularity.
Example of Current Data:
ID
LONG
LAT
101
-71.23403
42.01979
101
-91.469621
44.867211
102
78.8952716
38.4022661
102
80.8518668
35.3152386
Desired Output (output centroid is made up)
ID
CENTROID_LONG
CENTROID_LAT
101
-71.23403
42.01979
102
-91.469621
44.867211
Where the values above are aggregated to a centroid lat and long number.
Consider below
select id, st_union_agg(st_geogpoint(long, lat)) points,
st_centroid(st_union_agg(st_geogpoint(long, lat))) centroid
from your_table
group by id
if applied to sample data in your question - output is
which is visualized as
I have lat/long data of two animals tracked in Western Australia and I'd like to find their home ranges using adehabitatHR.
library(sp)
library(rgdal)
library(raster)
library(adehabitatHR)
library(sf)
quolls<-read.csv("quolls.csv")
head(quolls)
Latitude Longitude animal_ID
1 -22.62271 117.1247 1
2 -22.62286 117.1246 1
3 -22.62192 117.1223 1
4 -22.62021 117.1224 1
5 -22.61989 117.1244 1
6 -22.62022 117.1260 1
But the home range estimates of each animal are obviously too small.
I think the EPSG must be wrong but after a very long time looking I still can't find the right one.
Can anyone point me in the right direction please?
# make a SpatialPoints dataframe without a CRS
quolls2 <- quolls
quoll.latlong<-data.frame(x=quolls2$Longitude,y=quolls2$Latitude)
coordinates(quolls2) <- quoll.latlong
# add crs
proj4string(quolls2) <- CRS(SRS_string = "EPSG:4283")
mcp<-mcp(quolls2[,7],percent=95,unout = c("ha"))
mcp
Home range for animal 1 is 1.217428e-08 and animal 2 is 6.253689e-08.
And likewise with kernel density estimation;
quoll_ud <- adehabitatHR::kernelUD(quolls2[7],grid = 450)
quoll_hr <- adehabitatHR::getverticeshr(quoll_ud, 99)
print(quoll_hr)
which estimates animal 1 at 2.36917592701502e-08 and animal 2 at 1.16018636413173e-07.
Just stumbled across the answer.. it's EPSG 28350.
I got it to work in the end by abandoning the raw lats and longs and instead importing a shapefile I had of the animal data with st_read.
Then st_transform to 28350.
Then as mcp accepts only SpatialPoints, I converted the object with
as(obj, "Spatial").
i am new here, I am using MERRA monthly solar radiation data. I want to convert w/M^2 to MJ/m^2
I am bit confused, how to convert solar radiation monthly average data W/m^2 to MJ/m^2
so far i understood by reading different sources,
Firstly i have to convert w/m^2 to kw/m^2
after kw/m^2 to mj/m^2 .......
Am i doing correctly
Just i am taking one instance:
For may month i have value 294 w/m^2
So 294 * 0.001 = 0.294 kw/m^2
0.294 * 24 (kw to kwh (m^/day)) = 7.056 kwh/m^2/day
7.056 * 3.6 (kwh to mj) = 25.40 mj/day
i am confused i am doing right or wrong .
Not sure why you would take the kWh step in between.
Your panels do 294 Watt per m², i.e. 294 Joule per sec per m². So that's 24*60*60 * 294 = 25401600 Joule per m² per day, or 25.4016 MJ per m² per day.
So if:
1 W/m2 = 1 J/m2 s
Then:
294 W/m2 = 294 J/m2 s
if you want it in days then:
1 day = 60s * 60min *24h = 86400s
294 J/m2 s x 86000s/1day = 25284000 J/m2 day
25284000 J/m2 day x 1MJ/1000000J = 25.284 MJ/m2 day
all together:
294 W/m2 = 294/(1000000/86400) = 25.4016 MJ/m2 day
A watt is the unit of power and Joules are the units of energy, they are related by time. 1 watt is 1 Joule per second 1W = 1 J/s. So the extension of that equation is that 1J = 1w x 1second. 1J = 1Ws. A loose analogy is if you say Litre is a unit of volume and L/S is a unit of flow. So your calculation needs to consider how long you are gathering the solar energy. So the number of Joules, if the sunlight shines at 90degrees to the solar panel for 1 hour is 294W/m2 x 3600s and would give ~1 x 10^7 joules per square metre. Of course as the inclination [the angle of light] varies away from 90 degrees, this will cause the effective power and hence the energy absorbed to drop, as a function of the sine of the angle to the sun. 90 degrees gives a sine of 1 and is full power.
We are investigating migrating a prototype into SQL Server (azure).
We have LineStrings that also have M values. What we would like to do is given another M value find out what its geographical location is.
To aid your visualisation, here is a real-world example:
I have a linestring that represents a flight path. Because the flight goes up and down the distance the plane has actually moved is not the same as the total length of the linestring. We have calibrated M values as a part of the linestring but need to be able to plot on it where a given event occurred. All we know about this event is its M value.
SET #g = geometry::STGeomFromText('LINESTRING(1 0 NULL 0, 2 2 NULL 5, 1 4 NULL 9, 3 6 NULL 15)', 0);
Given something like the above, what is the lat and long of a point with an M value of 8?
This should be an equivalent postgis's ST_LocateAlong
The M value is not a time, but a distance. It should be understood that this distance is arbitrary and does not directly relate to the length of the line and is calibrated against known points. This is due to the set being based on historic data that is in no way accurate by today's standards.
*Note I am not sure if I have Nulled the Z or M value. The extra parameter we are considering here is the M only.
I'm writting a NMEA sentences parser, and couldn't find ant documentation about GPRMA sentence except that it is: "Recommended minimum specific Loran-C data". does anyone know what is the meaning of this sentence?
does the longitude and latitude in it refer to the gps device current location?
thanks
From the very handy guide at http://aprs.gids.nl/nmea/#rma:
eg. $GPRMA,A,llll.ll,N,lllll.ll,W,,,ss.s,ccc,vv.v,W*hh
1 = Data status
2 = Latitude
3 = N/S
4 = longitude
5 = W/E
6 = not used
7 = not used
8 = Speed over ground in knots
9 = Course over ground
10 = Variation
11 = Direction of variation E/W
12 = Checksum
Now, LORAN data is not GPS. It is similar, but an old standard of ground stations that were used to find positions. So to specifically answer your question, no, this is not GPS data. If you want GPS data, you will need $GPRMC.