如何使Peirce投影算法允许360度旋转?

问题描述 投票:8回答:1

我正在使用this algorithm进行R中的Peirce世界地图投影。我能够做一些精细的地图,例如在函数lambda_0中使用28作为toPeirceQuincuncial参数的值,因为这个角度产生的土地失真较少并且没有重要的岛屿(显然除了南极洲)。该算法使用如下:

# constants
pi<-acos(-1.0)
twopi<-2.0*pi
halfpi<-0.5*pi
degree<-pi / 180
halfSqrt2<-sqrt(2) / 2
quarterpi<-0.25 * pi
mquarterpi<--0.25 * pi
threequarterpi<-0.75 * pi
mthreequarterpi<--0.75 * pi
radian<-180/pi
sqrt2<-sqrt(2)
sqrt8<-2. * sqrt2
halfSqrt3<-sqrt(3) / 2
PeirceQuincuncialScale<-3.7081493546027438 ;# 2*K(1/2)
PeirceQuincuncialLimit<-1.8540746773013719 ;# K(1/2)

ellFaux<-function(cos_phi,sin_phi,k){
  x<-cos_phi * cos_phi
  y<-1.0 - k * k * sin_phi * sin_phi
  z<-1.0
  rf<-ellRF(x,y,z)
  return(sin_phi * rf)
}

ellRF<-function(x,y,z){
  if (x < 0.0 || y < 0.0 || z < 0.0) {
    print("Negative argument to Carlson's ellRF")
    print("ellRF negArgument")
  }
  delx<-1.0; 
  dely<-1.0; 
  delz<-1.0
  while(abs(delx) > 0.0025 || abs(dely) > 0.0025 || abs(delz) > 0.0025) {
    sx<-sqrt(x)
    sy<-sqrt(y)
    sz<-sqrt(z)
    len<-sx * (sy + sz) + sy * sz
    x<-0.25 * (x + len)
    y<-0.25 * (y + len)
    z<-0.25 * (z + len)
    mean<-(x + y + z) / 3.0
    delx<-(mean - x) / mean
    dely<-(mean - y) / mean
    delz<-(mean - z) / mean
  }
  e2<-delx * dely - delz * delz
  e3<-delx * dely * delz
  return((1.0 + (e2 / 24.0 - 0.1 - 3.0 * e3 / 44.0) * e2+ e3 / 14) / sqrt(mean))
}

toPeirceQuincuncial<-function(lambda,phi,lambda_0=20.0){
  # Convert latitude and longitude to radians relative to the
  # central meridian

  lambda<-lambda - lambda_0 + 180
  if (lambda < 0.0 || lambda > 360.0) {
    lambda<-lambda - 360 * floor(lambda / 360)
  }
  lambda<-(lambda - 180) * degree
  phi<-phi * degree

  # Compute the auxiliary quantities 'm' and 'n'. Set 'm' to match
  # the sign of 'lambda' and 'n' to be positive if |lambda| > pi/2

  cos_phiosqrt2<-halfSqrt2 * cos(phi)
  cos_lambda<-cos(lambda)
  sin_lambda<-sin(lambda)
  cos_a<-cos_phiosqrt2 * (sin_lambda + cos_lambda)
  cos_b<-cos_phiosqrt2 * (sin_lambda - cos_lambda)
  sin_a<-sqrt(1.0 - cos_a * cos_a)
  sin_b<-sqrt(1.0 - cos_b * cos_b)
  cos_a_cos_b<-cos_a * cos_b
  sin_a_sin_b<-sin_a * sin_b
  sin2_m<-1.0 + cos_a_cos_b - sin_a_sin_b
  sin2_n<-1.0 - cos_a_cos_b - sin_a_sin_b
  if (sin2_m < 0.0) {
    sin2_m<-0.0
  }
  sin_m<-sqrt(sin2_m)
  if (sin2_m > 1.0) {
    sin2_m<-1.0
  }
  cos_m<-sqrt(1.0 - sin2_m)
  if (sin_lambda < 0.0) {
    sin_m<--sin_m
  }
  if (sin2_n < 0.0) {
    sin2_n<-0.0
  }
  sin_n<-sqrt(sin2_n)
  if (sin2_n > 1.0) {
    sin2_n<-1.0 
  }
  cos_n<-sqrt(1.0 - sin2_n)
  if (cos_lambda > 0.0) {
    sin_n<--sin_n
  }

  # Compute elliptic integrals to map the disc to the square

  x<-ellFaux(cos_m,sin_m,halfSqrt2)
  y<-ellFaux(cos_n,sin_n,halfSqrt2)

  # Reflect the Southern Hemisphere outward

  if(phi < 0) {
    if (lambda < mthreequarterpi) {
      y<-PeirceQuincuncialScale - y
    } else if (lambda < mquarterpi) {
      x<--PeirceQuincuncialScale - x
    } else if (lambda < quarterpi) {
      y<--PeirceQuincuncialScale - y
    } else if (lambda < threequarterpi) {
      x<-PeirceQuincuncialScale - x
    } else {
      y<-PeirceQuincuncialScale - y
    }
  }

  # Rotate the square by 45 degrees to fit the screen better

  X<-(x - y) * halfSqrt2
  Y<-(x + y) * halfSqrt2
  res<-list(X,Y)
  return(res)
}

library(rgdal)
ang <- 28
p <- readOGR('~/R/shp','TM_WORLD_BORDERS-0.3') # read world shapefile
for (p1 in 1:length(p@polygons)) {
  for (p2 in 1:length(p@polygons[[p1]]@Polygons)) {
    for (p3 in 1:nrow(p@polygons[[p1]]@Polygons[[p2]]@coords)) {
      pos <- toPeirceQuincuncial(p@polygons[[p1]]@Polygons[[p2]]@coords[p3,1],
                                 p@polygons[[p1]]@Polygons[[p2]]@coords[p3,2],ang)
      p@polygons[[p1]]@Polygons[[p2]]@coords[p3,1] <- pos[[1]][1]
      p@polygons[[p1]]@Polygons[[p2]]@coords[p3,2] <- pos[[2]][1]
    }
  }
}
z <- toPeirceQuincuncial(0,-90,ang)[[1]][1]
p@bbox[,1] <- -z
p@bbox[,2] <- z
# plotting the map
par(mar=c(0,0,0,0),bg='#a7cdf2',xaxs='i',yaxs='i')
plot(p,col='gray',lwd=.5)
for (lon in 15*1:24) { # meridians
  pos <- 0
  posAnt <- 0
  for (lat in -90:90) {
    if (length(pos) == 2) {
      posAnt <- pos
    }
    pos <- toPeirceQuincuncial(lon,lat,ang)
    if (length(posAnt) == 2) {
      segments(pos[[1]][1],pos[[2]][1],posAnt[[1]][1],posAnt[[2]][1],col='white',lwd=.5)
    }
  }
}
lats <- 15*1:5
posS <- matrix(0,length(lats),2)
posST <- 0
pos0 <- 0
posN <- matrix(0,length(lats),2)
posNT <- 0
for (lon in 0:360) {
  posAntS <- posS
  posAntST <- posST
  posAnt0 <- pos0
  posAntN <- posN
  posAntNT <- posNT
  pos0 <- unlist(toPeirceQuincuncial(lon,0,ang))
  posST <- unlist(toPeirceQuincuncial(lon,-23.4368,ang))
  posNT <- unlist(toPeirceQuincuncial(lon,23.4368,ang))
  for (i in 1:length(lats)) {
    posS[i,] <- unlist(toPeirceQuincuncial(lon,-lats[i],ang))
    posN[i,] <- unlist(toPeirceQuincuncial(lon,lats[i],ang))
  }
  if (lon > 0) {
    segments(pos0[1],pos0[2],posAnt0[1],posAnt0[2],col='red',lwd=1)
    segments(posNT[1],posNT[2],posAntNT[1],posAntNT[2],col='yellow')
    for (i in 1:length(lats)) {
      segments(posN[i,1],posN[i,2],posAntN[i,1],posAntN[i,2],col='white',lwd=.5)
    }
    if (!(lon %in% round(90*(0:3+.5)+ang))) {
      for (i in 1:length(lats)) {
        segments(posS[i,1],posS[i,2],posAntS[i,1],posAntS[i,2],col='white',lwd=.5)
      }
      segments(posST[1],posST[2],posAntST[1],posAntST[2],col='yellow')
    } else {
      for (i in 1:length(lats)) {
        posS[i,] <- unlist(toPeirceQuincuncial(lon-0.001,-lats[i],ang))
        segments(posS[i,1],posS[i,2],posAntS[i,1],posAntS[i,2],col='white',lwd=.5)
        posS[i,] <- unlist(toPeirceQuincuncial(lon,-lats[i],ang))
      }
      posST <- unlist(toPeirceQuincuncial(lon-0.001,-23.4368,ang))
      segments(posST[1],posST[2],posAntST[1],posAntST[2],col='yellow')
      posST <- unlist(toPeirceQuincuncial(lon,-23.4368,ang))
    }
  }
}

Peirce projection with lambda_0 = 28°

lambda_0的不同价值观,我发现我显然无法选择任何我想要的价值。似乎该功能只能用我认为它的一半可能性。

Peirce projection with 9 different lambda_0 values

数字表示lambda_0的值。如您所见,北美在80°和200°之间从右向左顺时针移动,并在240°和40°之间再次开始相同的移动。

如何更改算法以允许我想要的任何角度(例如,北美指向上方)?

r map-projections
1个回答
5
投票

我做了一个简单的黑客“解决”了这个问题。首先,我发现哪些角度是其他角度的重复:在46到225度之间。然后,对于这些角度,我只需要翻转shapefile的边界框:

z <- toPeirceQuincuncial(0,-90,ang)[[1]][1]
if (ang > 45 & ang < 226) {
  p@bbox[,1] <- z
  p@bbox[,2] <- -z
} else {
  p@bbox[,1] <- -z
  p@bbox[,2] <- z
}

其他的事情:我改进了toPeirceQuincuncial的R代码,因为它是通过坐标转换坐标。由于R是一种矢量语言,我将其改编为一次转换一组坐标,这使得代码非常快。现在lambdaphi可能都是矢量(当然大小相同),而lambda_0仍然是一个数字。

toPeirceQuincuncial2<-function(lambda,phi,lambda_0=20.0){
  # Convert latitude and longitude to radians relative to the
  # central meridian

  lambda<-lambda - lambda_0 + 180
  lambda[which(lambda<0.0 | lambda>360.0)] <-
    lambda[which(lambda<0.0 | lambda>360.0)] - 360*floor(lambda[which(lambda<0.0 | lambda>360.0)]/360)
  lambda<-(lambda - 180) * degree
  phi<-phi * degree

  # Compute the auxiliary quantities 'm' and 'n'. Set 'm' to match
  # the sign of 'lambda' and 'n' to be positive if |lambda| > pi/2

  cos_phiosqrt2<-halfSqrt2 * cos(phi)
  cos_lambda<-cos(lambda)
  sin_lambda<-sin(lambda)
  cos_a<-cos_phiosqrt2 * (sin_lambda + cos_lambda)
  cos_b<-cos_phiosqrt2 * (sin_lambda - cos_lambda)
  sin_a<-sqrt(1.0 - cos_a * cos_a)
  sin_b<-sqrt(1.0 - cos_b * cos_b)
  cos_a_cos_b<-cos_a * cos_b
  sin_a_sin_b<-sin_a * sin_b
  sin2_m<-1.0 + cos_a_cos_b - sin_a_sin_b
  sin2_n<-1.0 - cos_a_cos_b - sin_a_sin_b
  sin2_m[which(sin2_m < 0.0)] <- 0.0
  sin_m<-sqrt(sin2_m)
  sin2_m[which(sin2_m > 1.0)] <- 1.0
  cos_m<-sqrt(1.0 - sin2_m)
  sin_m[which(sin_lambda < 0.0)] <- -sin_m[which(sin_lambda < 0.0)]
  sin2_n[which(sin2_n < 0.0)] <- 0.0
  sin_n<-sqrt(sin2_n)
  sin2_n[which(sin2_n > 1.0)] <- 1.0
  cos_n<-sqrt(1.0 - sin2_n)
  sin_n[which(cos_lambda > 0.0)] <- -sin_n[which(cos_lambda > 0.0)]

  # Compute elliptic integrals to map the disc to the square

  x<-ellFaux(cos_m,sin_m,halfSqrt2)
  y<-ellFaux(cos_n,sin_n,halfSqrt2)

  # Reflect the Southern Hemisphere outward

  y[which(phi<0 & lambda<mthreequarterpi)] <- PeirceQuincuncialScale - y[which(phi<0 & lambda<mthreequarterpi)]
  x[which(phi<0 & lambda>=mthreequarterpi & lambda<mquarterpi)] <- -PeirceQuincuncialScale - x[which(phi<0 & lambda>=mthreequarterpi & lambda<mquarterpi)]
  y[which(phi<0 & lambda>=mquarterpi & lambda<quarterpi)] <- -PeirceQuincuncialScale - y[which(phi<0 & lambda>=mquarterpi & lambda<quarterpi)]
  x[which(phi<0 & lambda>=quarterpi & lambda<threequarterpi)] <- PeirceQuincuncialScale - x[which(phi<0 & lambda>=quarterpi & lambda<threequarterpi)]
  y[which(phi<0 & lambda>=threequarterpi)] <- PeirceQuincuncialScale - y[which(phi<0 & lambda>=threequarterpi)]

  # Rotate the square by 45 degrees to fit the screen better

  X<-(x - y) * halfSqrt2
  Y<-(x + y) * halfSqrt2
  res<-list(X,Y)
  return(res)
}
© www.soinside.com 2019 - 2024. All rights reserved.