Coordinate Transforms
Wayne Trenning
Hi there
can a surveyor confirm for me please that all that is required to
transform data from one 3D coordinate system to another are the 3D
coords (in both grids) of 2 points
Thanx in anticipation
--
Wayne Trenning MAPTEK
Mining Engineer 3D Computer Graphics
Phone: (+61 8) 8379 7333 for Science and Industry
Fax : (+61 8) 8379 7377
http://www.maptek.com.au
Email: wayne.t...@adelaide.maptek.com.au
235 Glen Osmond Rd
Frewville, 5062
South Australia
Mike
Seven degrees:
i.e.
X rotation
Y rotation
Z rotation
X displacement
Y displacement
Z displacement
Scale ratio
Mike Falk,PE,LS
Wayne Trenning <wayne.t...@adelaide.maptek.com.au> wrote in article
<353EE762...@adelaide.maptek.com.au>...
Bryan Williams
You are correct, the transformation matrix can be computed from
two pairs of coordinates. Unfortunately, the parameters obtained from a
"two point" calculation can be slightly incorrect due to rounding
(especially if there are scaling parameters required for one or more axis).
The more pairs of points you can include in a Least Squares solution, the
better the accuracy and validity of the parameters.
Wayne Trenning <wayne.t...@adelaide.maptek.com.au> wrote:
>Hi there
>
>can a surveyor confirm for me please that all that is required to
>transform data from one 3D coordinate system to another are the 3D
>coords (in both grids) of 2 points
>
>Thanx in anticipation
--
Bryan
Outside of a dog, a book is man's best friend.
Inside of a dog, it's too dark to read.
John W Robbins
Wayne Trenning wrote:
> Hi there
>
> can a surveyor confirm for me please that all that is required to
> transform data from one 3D coordinate system to another are the 3D
> coords (in both grids) of 2 points
>
> Thanx in anticipation
>
I think that the previous respondants to this question are half right.
If all one is interested in are the three components of translation and
the three components of rotation, then, yes, two points will uniquely
determine the six-parameter transformation (six known quantities and six
unknown). However, if you should want to include a scale parameter for
the full seven parameter transformation, then you will need a third
point and it becomes a least-squares estimation problem.
Hope this helps - for some background, one might look at
http://www.utexas.edu/depts/grg/gcraft/notes/datum/datum.html
John Robbins
Cliff Mugnier
Wayne Trenning wrote:
>
> Hi there
>
> can a surveyor confirm for me please that all that is required to
> transform data from one 3D coordinate system to another are the 3D
> coords (in both grids) of 2 points
>
> Thanx in anticipation
>
>
> Wayne Trenning MAPTEK
> Mining Engineer 3D Computer Graphics
> Phone: (+61 8) 8379 7333 for Science and Industry
> Fax : (+61 8) 8379 7377
> http://www.maptek.com.au
> Email: wayne.t...@adelaide.maptek.com.au
>
> 235 Glen Osmond Rd
> Frewville, 5062
> South Australia
You need more even in a perfect world. Two 3D points in common between
two coordinate systems gives you a line in 3D space. If there is any
non-parallelism between the two coordinate systems, there exists the
ability to rotate about that line's axis. Three 3D points would "nail
it" if and only if the relation between the two coordinate systems is
perfectly linear.
If you are in an enclosed space, doing optical tooling applications,
that linearity will exist (plant floor to printing press to roller
axis). Two points would give a unique solution, but don't you want a
check on your work?
If you are outside, in the real world, say, with GPS equipment ... it
ain't linear. Only three points can be uncomfortably sparse.
In a mining environment, with ore veins described with strike and dip, I
would not be comfortable with a unique solution if your two coordinate
axes ARE parallel. I would guess that they are not.
Cliff
--
Clifford J. Mugnier (cj...@uno.edu)
The Topographic Engineering Laboratory
Department of Civil and Environmental Engineering
UNIVERSITY OF NEW ORLEANS
New Orleans, Louisiana 70148
Voice and Facsimile: (504) 280-7095
Stan Thomas
Wayne Trenning wrote:
> Hi there
>
> can a surveyor confirm for me please that all that is required to
> transform data from one 3D coordinate system to another are the 3D
> coords (in both grids) of 2 points
>
> Thanx in anticipation
>
>
For a real serious answer you should try an advanced mathematics news
group. However, in the spirit of good sportsmanship I might as well take
my shot an an answer.
A problem with some of the other answers is that they assume that you
are talking linear transformations. I see no evidence from your post
that you are excluding spherical, or cylindrical, or any other 3D
coordinate systems. What all of these systems have in common is that
points in the systems can be uniquely described by the independent
variables. A transformation from one coordinate system to another is a
set of three functions which map coordinates of points in one system to
their coordinates in the other system.
x' = x'(x,y,z), y' = y'(x,y,z), z' = z'(x,y,z).
If these equations are not linear then they can have any number of
parameters. The more parameters, the more points necessary to achieve
the require number of equations to match the number of parameters. Of
course the equations must have unique solutions. I am sure there must be
lot more conditions that these equations must meet (e.g. be continuous,
single valued, one-to-one, onto, ... blah blah ... stuff like that).
If you make the assumption that the equations are linear then the most
general expression is -
x' = a11*x + a12*y + a13*z + c1
y' = a21*x + a22*y + a23*z + c2
z' = a31*x + a32*y + a33*z + c3
This transformation has twelve parameters.. Therefore, you must set up
twelve linearly independent equations to get a unique solution for the
twelve parameters. Each pair of points (x,y,z) -> (x',y',z') will
produce three equations. It appears then that even for a general linear
transformation you will need four pairs of points, not just two. You
can, of course, put additional constraints to reduce the number of
parameters. You could make the origins are the same, for example. This
is the same as saying the (0,0,0) -> (0,0,0). This eliminates three
parameters but it is also just the equivalent of specifying one pair of
the required four points. You could also make the assumption that
stretching is isotropic in all directions. This will knock of a few
parameters too.
Anyway, that's my shot at it.
Stan
Peter B. Legere
Well...... Yes, and No.
(Sorry, couldn't resist)
Peter
Wayne Trenning . wrote in article
Gregory J. Murphy
Wayne Trenning wrote:
>
> Hi there
>
> can a surveyor confirm for me please that all that is required to
> transform data from one 3D coordinate system to another are the 3D
> coords (in both grids) of 2 points
>
> Thanx in anticipation
>
>
> Wayne Trenning MAPTEK
> Mining Engineer 3D Computer Graphics
> Phone: (+61 8) 8379 7333 for Science and Industry
> Fax : (+61 8) 8379 7377
> http://www.maptek.com.au
> Email: wayne.t...@adelaide.maptek.com.au
>
> 235 Glen Osmond Rd
> Frewville, 5062
> South Australia
Hi, Wayne
You will require a least three points for a solution if you assume scale
= 1, a unique solution can be found using two points. It is way better
to use the following for a three-dimensional transformation.
I will assume that what you are look for or, think about, is a straight
forward Three-Dimensional Similarity Transformation with three
Translations, three Rotations, and a Scale change. The basic form of
the equations take the shape of the following:
[x2] = sM [x1] + k
where [x2] = (X,Y,Z) and [x1] = (X',Y',Z'), and k = (Xt,Yt,Zt)
k is translation components, x1 is initial coordinates and x2 is desired
coordinates. M is a lot more involved, however it can be broken down
into rotation components about each axis,
Mx = [1 0 0]
[0 cos b1 sin b1]
[0 -sin b1 cos b1]
My = [cos b2 0 -sin b2]
[0 1 0]
[sin b2 0 cos b2]
Mz = [ cos b3 sin b3 0]
[-sin b3 cos b3 0]
[0 0 1]
where M = Mx*My*Mz
and s is the scale component.
Least squares is the best method to choose for solving the parameters if
they are unknown. So generally if you are not interested in the
individual components of the transformation just getting from one
othogonal coordinate frame to the other. The final form takes the shape
of the following:
[x2] = M [x1] + k
where M = [a11 a12 a13]
[a21 a22 a23]
[a31 a32 a33]
and k = [ dX ] and x1 = [ X' ] and x2 = [ X ]
[ dY ] [ Y' ] [ Y ]
[ dZ ] [ Z' ] [ Z ]
rewriting the equation, solution takes the form:
U = P * R
[X] [X' Y' Z'1 0 0 0 0 0 0 1 0 0] [a11]
[Y] [0 0 0 X' Y' Z'1 0 0 0 0 1 0] [a12]
[Z]1 [0 0 0 0 0 0 X' Y' Z'1 0 0 1] [a13]
[X] [X' Y' Z'2 0 0 0 0 0 0 1 0 0] [a21]
[Y] [0 0 0 X' Y' Z'2 0 0 0 0 1 0] [a22]
[Z]2 [0 0 0 0 0 0 X' Y' Z'2 0 0 1] [a23]
[X] = [X' Y' Z'3 0 0 0 0 0 0 1 0 0] * [a31]
[Y] [0 0 0 X' Y' Z'3 0 0 0 0 1 0] [a32]
[Z]3 [0 0 0 0 0 0 X' Y' Z'3 0 0 1] [a33]
[X] [X' Y' Z'4 0 0 0 0 0 0 1 0 0] [dX ]
[Y] [0 0 0 X' Y' Z'4 0 0 0 0 1 0] [dY ]
[Z]4 [0 0 0 0 0 0 X' Y' Z'4 0 0 1] [dZ ]
P' = transpose of P, hense
R ={INV [P'P]} * [P'U]
The equation components P and U can continue with more points if
necessary.
Extract the components of M and k and transform your data set.
Hope this helpful!
Greg
TJSongs
Check out www.Surveyors.Com
Klaus Hanke
I cannot confirm this at all. You 'll need 3 of them at least. 2 points
just give an axis in space which still allows rotation around it.
K.Hanke
Eric Collins, CET
for a three parameter (Molodensky) transformation, yes...you compute ECEF
coords and calc the difference...that is your datum shift. This is only
good for regional areas though depending on datums used...
Cliff Mugnier
Eric,
A three-parameter geocentric shift in geocentric coordinates is not in
itself, a Molodensky transform. A true Molodensky transformation
involves the initial subtraction of the geocentric coordinates of the
origin(s) of the classical geodetic datum(s) BEFORE computing a scale
factor. A true Molodensky uses a minimum of four parameters, commonly
seven. The minimum is dX, dY, dZ, AND Scale Factor. The optional three
is Rx,Ry,Rz. The other major 3-, 4-, or 7-parameter transform is the
Bursa-Wolf that does NOT take into account the datum origin, and is more
commonly used when close to the datum origin. The third variant is the
Veis model that uses a different sequential rotation than the other
two. The Veis, I think, uses Euler angles.
What you referred to is the three-parameter Bursa-Wolf transformation.
Eric Collins, CET
Cliff Mugnier wrote:
<sigh> this is a direct quote from my geodesy notes...
Datum Shifts
The most common datum shift methods include those based on interpolation, and
those that require conversion to and from a three dimensional Cartesian
system. NADCON is a datum shift method which is interpolative in nature, and
is used specifically for converting NAD27 coordinates to and from NAD83
coordinates. It can be interpreted geometrically as a surface of minimum
curvature developed from a set of known datum shift values, so that given
specific coordinates, a particular set of datum shift values apply.
Another datum shift method relies on first converting geographic coordinates to
Earth Centered Earth Fixed (or ECEF) coordinates. The ECEF coordinate system is
a 3 dimensional, mutually orthogonal axis Cartesian system, in which the center
of the mass of the Earth is the origin. The Z axis roughly approximates the
polar axis and the positive Y axis intersects the zero meridian.
After coordinates expressed in the first datum are converted to ECEF
coordinates, they may undergo either a translation, or a combination of a
translation , rotation and scale factor before conversion to the second datum.
The former method is called a three parameter, or Molodensky datum shift and
the latter a seven parameter, or Bursa Wolf datum shift. The values of both
three and seven parameter datum shifts are location dependent when used for
coordinate conversion between regional datums, or between world and regional
datums, although one set of seven parameter shift values can be used globally
between two world datums such as WGS84 and WGS 72.
If coordinates in two different datums for the same point are known, three
parameter shift values can be determined which are accurate for the immediate
area, and perhaps a very large area depending on the similarity of the two
datums in question. Seven parameter shift values must be derived by a minimum
of three points in the area of interest, and can often be used over a larger
area than can the three parameter shift values. However, consideration should
be given to the fact that if three or more control points are used, the
relative accuracy of the control becomes embedded into the shift. The same
holds true for a mean three parameter shift developed by the same points.
However, a three parameter shift developed from one station means that any
bias in the control point used will be evenly distributed in the immediate
area resulting in good relative accuracy between shifted points.
Also, implicit in the use of an interpolative datum shift method is the
assumption that the input regional datum contains inherent distortion, so that
a grid of points computed on the datum contains the same relative distortion.