I am asking this questions as a trimmed version of my previous question. Now that I have a face looking some position on screen and also gaze coordinates (pitch and yaw) of both the eye. Let us say
Left_Eye = [-0.06222888 -0.06577308]
Right_Eye = [-0.04176027 -0.44416167]
I want to identify the screen coordinates where the person probably may be looking at? Is this possible? Please help!
What you need is:
3D position and direction for each eye
you claim you got it but pitch and yaw are just Euler angles and you need also some reference frame and order of transforms to convert them back into 3D vector. Its better to leave the direction in a vector form (which I suspect you got in the first place). Along with the direction you need th position in 3D in the same coordinate system too...
3D definition of your projection plane
so you need at least start position and 2 basis vectors defining your planar rectangle. Much better is to use 4x4 homogenous transform matrix for this because that allows very easy transform from and in to its local coordinate system...
So I see it like this:
So now its just matter of finding the intersection between rays and plane
P(s) = R0 + s*R
P(t) = L0 + t*L
P(u,v) = P0 + u*U +v*V
Solving this system will lead to acquiring u,v which is also the 2D coordinate inside your plane yo are looking at. Of course because of inaccuracies this will not be solvable algebraicaly. So its better to convert the rays into plane local coordinates and just computing the point on each ray with w=0.0 (making this a simple linear equation with single unknown) and computing average position between one for left eye and the other for right eye (in case they do not align perfectly).
so If R0',R',L0',L' are the converted values in UVW local coordinates then:
R0z' + s*Rz' = 0.0
s = -R0z'/Rz'
// so...
R1 = R0' - R'*R0z'/Rz'
L1 = L0' - L'*L0z'/Lz'
P = 0.5 * (R1 + L1)
Where P is the point you are looking at in the UVW coordinates...
The conversion is done easily according to your notations you either multiply the inverse or direct matrix representing the plane by (R,1),(L,1),(R0,0)(L0,0). The forth coordinate (0,1) just tells if you are transforming vector or point.
Without knowing more about your coordinate systems, data accuracy, and what knowns and unknowns you got is hard to be more specific than this.
If your plane is the camera projection plane than U,V are the x and y axis of the image taken from camera and W is normal to it (direction is just matter of notation).
As you are using camera input which uses a perspective projection I hope your positions and vectors are corrected for it.
as i know, shapely use only cartesian coordinate system. I have two point on the earth with lat and lon coordinates. I need create buffer with 1km radius around this two points and find polygon, where this buffers intersect.
But construstion
buffer = Point(54.4353,65.87343).buffer(0.001) create simple circle, but in projection on the Earth it becomes ellipse, but i need two real circle with 1 km radius.
I think, i need convert my buffers into new projection and then intersect it, but dont now how correct do it.
You need to do what you say. For that, you will need to use a library that handles projections (pyproj is the choice here). There is a similar question in Geodesic buffering in python
import pyproj
from shapely.geometry import MultiPolygon, Polygon, Point
from shapely.ops import transform as sh_transform
from functools import partial
wgs84_globe = pyproj.Proj(proj='latlong', ellps='WGS84')
def point_buff_on_globe(lat, lon, radius):
#First, you build the Azimuthal Equidistant Projection centered in the
# point given by WGS84 lat, lon coordinates
aeqd = pyproj.Proj(proj='aeqd', ellps='WGS84', datum='WGS84',
lat_0=lat, lon_0=lon)
#You then transform the coordinates of that point in that projection
project_coords = pyproj.transform(wgs84_globe, aeqd, lon, lat)
# Build a shapely point with that coordinates and buffer it in the aeqd projection
aeqd_buffer = Point(project_coords).buffer(radius)
# Transform back to WGS84 each coordinate of the aeqd buffer.
# Notice the clever use of sh_transform with partial functor, this is
# something that I learned here in SO. A plain iteration in the coordinates
# will do the job too.
projected_pol = sh_transform(partial(pyproj.transform, aeqd, wgs84_globe),
aeqd_buffer)
return projected_pol
The function point_buff_on_globe will give you a polygon in lat lon that is the result of buffering the given point in the Azimuthal Equidistant Projection centered in that point (the best you can do with your requirements. Two observations:
I don't remember the units of the radius argument. I think is in meters, so if you need a 10 km buffer, you will need to pass it 10e3. But please, check it!
Beware of using this with radius to wide or points that are to far away from each other. Projections work well when the points are near to the point you are centering the projection.
Here is my situation:
The company I work for are using some third party software that plots the location of devices on a wifi network. The software requires you to upload a floor plan of the area (this is just a 2000 x 1187 px jpeg). The software spits out xy coordinates for every device location which is visualised over the floor plan.
I have been asked to take these xy coordinates and display them using the Google maps API, so therefore need to somehow convert them to latitude and longitude.
The third party software uses the pixel dimensions of the jpeg as reference to calculating the xy coordinate, according to the limited documentation.
I have attempted to use Python and the pyproj module to make the conversion (there are over half a million coordinates to convert), but the results just seem to plot the lat long in roughly the same position every time.
Here is what I have so far:
projection = Proj("+proj=lcc +lat_1=53.466082 +lat_2=53.464971 +lat_0=53.465482 +lon_0=-2.338944 +x_0=0 +y_0=0 +no_defs +a=6378137 +rf=298.257222101 +to_meter=1 +k_0=200")
wgs84 = Proj(init='epsg:4326')
lat = xCoordinate
lon = yCoordinate
xx, yy = transform(projection, wgs84, lat, lon)
latLong = {'lat':xx, 'long':yy}
return latLong
I admittedly have very little experience with mapping etc and have very little time to get my head around the topic.
I hope this is enough information to go on. If anyone can help in pointing me in the right direction then I would be eternally grateful.
Cheers.
I have a point (x,y,z) in 3d that I would like to rotate. First I would like to rotate the point around another point (0,0,0) 360 degrees. Then I would like to change the plane that the point rotates in by 1 degree and repeat. I have been looking at the rotation_matrix function in http://www.lfd.uci.edu/~gohlke/code/transformations.py.html , however it seems as if the rotation only goes around the x,y or z axis rather than an arbitrary angle. Does anyone know how to accomplish this?
Rotating around the x, y, and z pretty much is the only way to do it. Rotating a plane is exactly like rotating around an axis. Check out my scratch project for the math.
EDIT: Just found out that I need to convert latitude, longitude and elevation of a location on earth to J2000 coordinates and nothing to do with ra/dec or the moon. Sorry for this. Your answers did give me a lot of insights. Please see the edited question below.
Question: how do i convert latitude, longitude and elevation to J2000 coordinates (XYZ). Is there a conversion present in ephem? I checked the docs but I couldnt find something I need (or mightve overlooked something due to my lack of knowledge in this field). Thanks
***************** OLD (Disregard) ******************
I have the moon position in Right Ascension (RA) and Declination (Dec) and I want to convert them into X Y Z coordinates. Is there a built-in PyEphem function for this? Also, what is the math behind it? Thanks.
EDIT: I am using the J2000 coordinate system (which is equatorial i think, this is my first time working with astronomy). I have the distance to moon available. The ra/dec values are already in J2000 (equatorial) coordinates.
X points North
Y points West
Z points towards the sky
Best answer:
It has just come to my attention that, in June 2011, the Naval Observatory released a Python interface to the powerful NOVAS reference software with which the highest-precision astronomical computations are performed:
http://aa.usno.navy.mil/software/novas/novas_py/novaspy_intro.php
With this library you can get the answer you are seeking, at far higher precision than PyEphem has ever offered:
from novas import compat as novas
jd_tt = novas.julian_date(2012, 9, 8, 12.00)
delta_t = 66.603 # from http://maia.usno.navy.mil/ser7/deltat.preds
lat = 42.3583 # positive is north
lon = -71.0603 # negative is west
observer = novas.make_observer_on_surface(lat, lon, 0, 0, 0)
print novas.geo_posvel(jd_tt, delta_t, observer)
On my machine this gives the answer:
((-3.5081406460494928e-06, 3.135277713623258e-05, 2.858239912567112e-05),
(-0.00019753847060164693, -2.2329942271278055e-05, 2.4885824275734915e-07))
Try it yourself and see if this gives you the kind of results that you need!
Newer answer:
It appears that the answer is “no” — PyEphem, to my surprise, gives no easy way to get the answer to the question "where, in x, y, z coordinates, is (say) Boston at time t ?”
This is a surprise because “libastro”, the library behind PyEphem, of course has to compute this internally in order to figure out where other objects are relative to an observer. It seems to do so in two places. In parallax.c it defines ta_par() which talks only about angles on the outside, but on its inside you can see that it temporarily computes the x, y, z of the observer. You can even see the important constant 298.257 hidden inside there, which measures how flat the earth is, since it is not a perfect sphere.
The other place is in earthsat.c which looks like a completely different code base from the rest of “libastro”, and so it duplicates some of the logic. Its EarthFlat constant of 298.25 is a bit less precise, but is doing the same job. And its function, GetSitPosition(), actually exposes x,y,z coordinates instead of keeping them hidden. But it is declared static so there is no way to call in to this useful function from outside!
So for the moment, PyEphem gives you no way to compute your x,y,z directly. But it does provide one important piece of information: the current sidereal time, which you will (I think) be able to use to figure how far around the earth Boston (or wherever) has traveled by time t, which will be important in figuring out your coordinates.
I will see if I can work up a quick solution in Python that combines the hour angle from PyEphem with some explicit trigonometry to get you an answer. But, for the moment, no: PyEphem does not expose this information directly, sadly enough; I will put it on the list of things for a future version!
Older answer, from when the question was about the x,y,z position of the Moon:
PyEphem does not, alas, have built-in functions for converting from the polar coordinates used in amateur astronomy to the x/y/z coordinates which will let you map out how objects are distributed in space around the Earth. But the conversion is easy to do yourself:
import ephem
import math
m = ephem.Mars('2012/8/1')
print m.ra, m.dec
x = math.cos(m.dec) * math.cos(m.ra)
y = math.cos(m.dec) * math.sin(m.ra)
z = math.sin(m.dec)
print x, y, z
print 'sanity check: vector length =', math.sqrt(x*x + y*y + z*z)
The output of this script is:
12:58:51.20 -6:24:05.6
-0.961178016954 -0.252399543786 -0.111495695074
sanity check: vector length = 1.0
The position of Mars for the random date that I used here are quite reasonable values: an RA that is almost one hour more than halfway around the great circle (since 12h would be exactly halfway), and a declination that pushes the position a bit south. Thus the x, y, and z that we get out: the z is a slightly negative number since -6° is indeed south of the equator, and x and y are both negative since going 13h around a 24h circle puts you down in the negative/negative quadrant of a normal unit circle.
Note that although J2000 has a north and south — so that we can truthfully say that the slightly negative z is a southward direction — it does not have an east and west, since the earth turning below it is constantly swinging east and west in all directions. Instead, RA measures from “the first point of Ares” which is the direction in which the sun lies during the spring Equinox. So x and y are not east or west; they are coordinates pointing out into the solar system on a fixed axis defined by the direction that the Earth sits in every Spring.
This x y z vector I have created is a “unit vector” — a little vector that has the magnitude 1.0, as I verified in the script to make sure I had the formulae correct. If you were computing x y and z coordinates for objects whose distance from the earth you knew, then you could get a real vector — whose magnitude were distances, instead of fractions of 1 — by multiplying each of the three x y and z by the distance to the object.
Does that help you out? From your description — and your question about east and west — I could not tell if you wanted RA and dec turned into x y z or whether you are actually wanting the azimuth and altitude converted (but the math is the same either way). That would look something like:
x = math.cos(m.alt) * math.cos(m.az)
y = math.cos(m.alt) * math.sin(m.az)
z = math.sin(m.alt)
What are you trying to accomplish with these coordinates? That could help us make sure that we are giving them to you in a useful format.
I suggest you look at the PyEphem documentation for coordinate conversion.
Basically, PyEphem only deals with three coordinate systems - equatorial, ecliptic, and galactic - each defined by two angles and an epoch (adjustment offset for polar progression).
Depending on what your coordinate scheme looks like, you should be able to use trigonometry to convert to it if you also have the object's distance.
Edit: your "X-Y-Z" coordinates seem to be lefthanded ecliptic coordinates.
from ephem import Equatorial, Ecliptic, degree
def convert_equatorial_to_XYZ(ra, dec, dist=1.0, epoch='2000'):
"""
Given
ra right ascension (in hours)
dec declination (in degrees)
dist distance (optional, defaults to 1.0)
epoch epoch (optional, assumes J2000)
Return
degrees North, degrees West, distance
"""
eq = Equatorial(ra, dec, epoch=epoch)
ec = Ecliptic(eq)
return ec.lat/degree, 360.0 - ec.lon/degree, dist