## Thursday, October 18, 2007

### Approximate sunrise and sunset times

This function is not perfect, but it does a reasonable job estimating sunrise and sunset times for my field site. If more accurate data are required, try here. Note, the command to calculate day of year is: strptime(x, "%m/%d/%Y")\$yday+1
``````suncalc<-function(d,Lat=48.1442,Long=-122.7551){
## d is the day of year
## Lat is latitude in decimal degrees
## Long is longitude in decimal degrees (negative == West)

##This method is copied from:
##Teets, D.A. 2003. Predicting sunrise and sunset times.
##  The College Mathematics Journal 34(4):317-321.

## At the default location the estimates of sunrise and sunset are within
## seven minutes of the correct times (http://aa.usno.navy.mil/data/docs/RS_OneYear.php)
## with a mean of 2.4 minutes error.

## Function to convert degrees to radians

R=6378

##Radians between the xy-plane and the ecliptic plane

## Calculate offset of sunrise based on longitude (min)
## If Long is negative, then the mod represents degrees West of
## a standard time meridian, so timing of sunrise and sunset should
timezone = -4*(abs(Long)%%15)*sign(Long)

## The earth's mean distance from the sun (km)
r = 149598000

theta = 2*pi/365.25*(d-80)

z.s = r*sin(theta)*sin(epsilon)
r.p = sqrt(r^2-z.s^2)

t0 = 1440/(2*pi)*acos((R-z.s*sin(L))/(r.p*cos(L)))

that = t0+5

## Adjust "noon" for the fact that the earth's orbit is not circular:
n = 720-10*sin(4*pi*(d-80)/365.25)+8*sin(2*pi*d/365.25)

## now sunrise and sunset are:
sunrise = (n-that+timezone)/60
sunset = (n+that+timezone)/60

return(list("sunrise" = sunrise,"sunset" = sunset))
}

``````

## Sunday, October 14, 2007

### Convert polar coordinates to Cartesian

When I want to calculate the coordinates of a location (e.g., a nest or burrow) based on distance and bearing from a grid point, this function helps me avoid writing down SOH-CAH-TOA every time. Just note that the bearing in this case is from the grid point (known location) to the unknown location.
`polar2cart<-function(x,y,dist,bearing,as.deg=FALSE){  ## Translate Polar coordinates into Cartesian coordinates  ## based on starting location, distance, and bearing  ## as.deg indicates if the bearing is in degrees (T) or radians (F)    if(as.deg){    ##if bearing is in degrees, convert to radians    bearing=bearing*pi/180  }    newx<-x+dist*sin(bearing)  ##X  newy<-y+dist*cos(bearing)  ##Y  return(list("x"=newx,"y"=newy))}##Exampleoldloc=c(0,5) bearing=200 #degreesdist=5newloc<-polar2cart(oldloc[1],oldloc[2],dist,bearing,TRUE)plot(oldloc[1],oldloc[2],xlim=c(-10,10),ylim=c(-10,10))points(newloc\$x,newloc\$y,col="red")`

## Wednesday, October 3, 2007

### Reorder factor levels

Very often, especially when plotting data, I need to reorder the levels of a factor because the default order is alphabetical. There must be many ways of reordering the levels; however, I always forget which package to look in. A direct way of reordering, using standard syntax is as follows:
`## generate datax = factor(sample(letters[1:5],100, replace=TRUE))print(levels(x))  ## This will show the levels of x are "Levels: a b c d e"## To reorder the levels:## note, if x is not a factor use levels(factor(x))x = factor(x,levels(x)[c(4,5,1:3)])print(levels(x))  ## Now "Levels: d e a b c"`
In this case, the level order could be set in the first line; however, if there are many levels (and you don't want to type out all of the levels explicitly), the above method is preferred. Again, if there is a better way to do this, please let me know in the comments.