cv::recoverPose has parameter "triangulatedPoints" as seen in documentation, though math behind it is not documented, even in sources (relevant commit on github).
When I use it, I get this matrix in following form:
[0.06596200907402348, 0.1074107606919504, 0.08120752154556411,
0.07162400555712592, 0.1112415181779849, 0.06479560707001968,
0.06812069103377787, 0.07274771866295617, 0.1036230973846902,
0.07643884790206311, 0.09753859499789987, 0.1050111597547035,
0.08431322508162108, 0.08653721971228882, 0.06607013741719928,
0.1088621999959361, 0.1079215237863785, 0.07874160849424018,
0.07888037486261903, 0.07311940086190356;
-0.3474319603010109, -0.3492386196164926, -0.3592673043398864,
-0.3301695131649525, -0.3398606744869519, -0.3240186574427479,
-0.3302508442361889, -0.3534091474425142, -0.3134288005980755,
-0.3456284001726975, -0.3372514921152191, -0.3229005408417835,
-0.3156005118578394, -0.3545418178651592, -0.3427899760859008,
-0.3552801904337188, -0.3368860879000375, -0.3268499974874541,
-0.3221050630233929, -0.3395139819250934;
-0.9334091581425227, -0.9288726274060354, -0.9277125424980246,
-0.9392374374147775, -0.9318967835907961, -0.941870018271934,
-0.9394698966781299, -0.9306592884695234, -0.9419749503870455,
-0.9332801148509925, -0.9343740431697417, -0.9386198310107222,
-0.9431781968459053, -0.9290466865633286, -0.9351167772249444,
-0.9264105322194914, -0.933362882155191, -0.9398254944757025,
-0.9414486961893244, -0.935785675955617;
-0.0607238817598344, -0.0607532477465341, -0.06067768097603395,
-0.06075467523485482, -0.06073245675798231, -0.06078081616640227,
-0.06074754785132623, -0.0606879948481664, -0.06089198212719162,
-0.06071522666667255, -0.06076842109618678, -0.06083346023742937,
-0.06084805655000008, -0.0606931888685702, -0.06071558440082779,
-0.06073329803512636, -0.06078189449161094, -0.06080195858434526,
-0.06083228813425822, -0.06073695721101467]
e.g. 4x20 matrix (in this case there were 20 points). I want to convert this data to std::vector in order to use it in solvePnP. How to do it, what is the math here? Thanks!
OpenCV offers a triangulatePoints function, which has the same output:
points4D 4xN array of reconstructed points in homogeneous coordinates.
Which indicates that each column is a 3D point in homogeneous coordinate system. However your points looks quite not as I would expect. For instance your first point is:
[0.06596200907402348, -0.3474319603010109, -0.9334091581425227, -0.0607238817598344]
But I would expect the last component to be 1.0 already. You should double check if something is not wrong here. You can always remove the "scaling" of the point by dividing each dimension by the last component:
[ x, y z, w ] = w [x/w, y/w, z/w, 1]
And then use the first three parts for your PnP solution.
I hope this helps
Related
i have shape file (.shp) with data in EPSG:2180.
I have extracted positions like this one:
303553.0249270061 580466.2644065879
Same values i see in equivalent .gml file.
That i am sure values are ok.
But how can i convert it to Latitude and Longitude?
(from comments GPS is WGS 84 (a.k.a. EPSG:4326)).
i see that params about EPSG:2180:
latitude_of_origin=0
central_meridian=19
scale_factor=0.9993
false_easting=500000
false_northing=-5300000
SPHEROID = 6378137,298.257222101
degree=0.0174532925199433
I have tried such simple calculation like
one_degree = 111196.672; //meters
X:= 303553.0249270061;
Y:= 580466.2644065879;
X:= X - false_northing;
Y:= Y - false_easting;
X:= latitude_of_origin + X/one_degree;
Y:= central_meridian + Y/one_degree;
but this show me:
50.3931720629823
19.7236391427847
which is not true. It is near but should be >20.
How this calculation should looks like?
I need it in Delphi application.
I don't know solution for You in Delphi but i found one in js and php.
I found this topic while i was searching "how to transform location units from one system to another" and google EPSG:2180 convert.
This problem name is:
Transform a point coordinate from one map projection to another
So solution is to make mathematical transformation and maby other calculations. You can Check:
Proj4 Library
with API Interfaces: C, C++, Python, Java, Ruby
My solution was to use library Proj4js.
Maby try to open source's on github of this library and analize function transform and with luck You will find answer ; )
I used php version of this package => Proj4jsphp
My problem was how to transform Coordinates in EPSG:2180 => GPS is WGS 84 (a.k.a. EPSG:4326).
Later i found out there are different names for this systems:
EPSG:2180 responds to Poland CS92
EPSG:4326 responds to WGS84 -> where WGS84 is the most popular system (lat, lon).
I am surprising to notice that it is somehow difficult to obtain a correct fit of interaction function from gam().
To be more specific, I want to estimate an additive function:
y=m_1(x)+m_2(z)+m_{12}(x,z)+u,
where m_1(x)=x^2, m_2(z)=z^2,m_{12}(x,z)=xz. The following code generate this model:
test1 <- function(x,z,sx=1,sz=1) {
#--m1(x) function
m.x<-x^2
m.x<-m.x-mean(m.x)
#--m2(z) function
m.z<-z^2
m.z<-m.z-mean(m.z)
#--m12(x,z) function
m.xz<-x*z
m.xz<-m.xz-mean(m.xz)
m<-m.x+m.z+m.xz
return(list(m=m,m.x=m.x,m.z=m.z,m.xz=m.xz))
}
n <- 1000
a=0
b=2
x <- runif(n,a,b)/20
z <- runif(n,a,b)
u <- rnorm(n,0,0.5)
model<-test1(x,z)
y <- model$m + u
So I use gam() by fitting the model as
b3 <- gam(y~ ti(x) + ti(z) + ti(x,z))
vis.gam(b3);title("tensor anova")
#---extracting basis matrix
B.f3<-model.matrix.gam(b3)
#---extracting series estimator
b3.hat<-b3$coefficients
Question: when I plot the estimated function by gam()above against its true function, I end up with
par(mfrow=c(1,3))
#---m1(x)
B.x<-B.f3[,c(2:5)]
b.x.hat<-b3.hat[c(2:5)]
plot(x,B.x%*%b.x.hat)
points(x,model$m.x,col='red')
legend('topleft',c('Estimate','True'),lty=c(1,1),col=c('black','red'))
#---m2(z)
B.z<-B.f3[,c(6:9)]
b.z.hat<-b3.hat[c(6:9)]
plot(z,B.z%*%b.z.hat)
points(z,model$m.z,col='red')
legend('topleft',c('Estimate','True'),lty=c(1,1),col=c('black','red'))
#---m12(x,z)
B.xz<-B.f3[,-c(1:9)]
b.xz.hat<-b3.hat[-c(1:9)]
plot(x,B.xz%*%b.xz.hat)
points(x,model$m.xz,col='red')
legend('topleft',c('Estimate','True'),lty=c(1,1),col=c('black','red'))
However, the function estimate of m_1(x) is largely different from x^2, and the interaction function estimate m_{12}(x,z) is also largely different from xz defined in test1 above. The results are the same if I use predict(b3).
I really can't figure it out. Can anybody help me out by explaining why the results end up with this? Greatly appreciate it!
First, the problem of the above issue is not due to the package, of course. It is closely related to the identification conditions of the smooth functions. One common practice is to impose the assumptions that E(mj(.))=0 for all individual function j=1,...,d, and E(m_ij(x_i,x_j)|x_i)=E(m_ij(x_i,x_j)|x_j)=0 for i not equal to j. Those conditions require one to employ centered basis function in series estimator, which has been done already in GAM package. However, in my case above, function m(x,z)=x*z defined in test1 does not satisfy the above identification assumptions, since the integral of x*z with respect to either x or z is not zero when x and z have range from zero to two.
Furthermore, series estimator allows the individual and interaction function to be identified if one impose m(0)=0 or m(0,x_j)=m(x_i,0)=0. This can be readily achieved if we center the basis function around zero. I have tried both cases, and they work well whenever DGP satisfies the identification conditions.
I'm trying to iterate over 2 parameters to get two splines for each pair. The code:
y_arr:[0.2487,0.40323333333333,0.55776666666667,0.7123]$
str_h_arr:[-0.8,-1.0,-1.2,-1.4]$
z_points:[0,0.1225,0.245,0.3675,0.49,0.6125,0.735,0.8575,0.98,1.1025,1.225,1.3475,1.47,
1.5925,1.715,1.8375,1.96,2.0825,2.205,2.26625,2.3275,2.3765,2.401,2.4255,2.43775,
2.4451,2.448775,2.45]$
length(a)$
length(b)$
load(interpol)$
for y_k:1 thru length(a) do (
for h_k:1 thru length(b) do (
y:y_arr[y_k],
str_h:str_h_arr[h_k],
bot_startpoints: [[-2.45,0],[0,y],[2.45,0]],
top_startpoints: [[-2.45,str_h_min],[0,y+str_h],[2.45,str_h_min]],
spline: cspline(bot_startpoints),
bot(x):=''spline,
print(bot(0))
)
);
//Part with top spline is skipped.
For all iterations output is now the same: 0.7123
What I want to get is two splines like in picture
Members of y_arr are y values in x=0, str_h_arr: height between splines in x=0.
So bot(0) should give me all values from y_arr.
If i don't use loop and just give this block values of y_k and h_k, it's working properly.
Can anybody point me to where I'm (or Maxima is) wrong with using loop with cspline?
The problem is that quote-quote (two single quotes, '') is applied only once, when it is read in input; it is not applied every time the expression in evaluated in the loop.
Looks like you need only to evaluate the spline at x = 0 and nothing else. So I'll suggest ev(spline, x=0) to evaluate it. You can also construct a lambda expression and evaluate that.
Here is the program after I've revised it as described above. Also, it is simpler and clearer to write for y in y_arr do (...) rather than making use of an explicit index for y_arr.
y_arr:[0.2487,0.40323333333333,0.55776666666667,0.7123]$
str_h_arr:[-0.8,-1.0,-1.2,-1.4]$
z_points:[0,0.1225,0.245,0.3675,0.49,0.6125,0.735,0.8575,0.98,1.1025,1.225,1.3475,1.47,
1.5925,1.715,1.8375,1.96,2.0825,2.205,2.26625,2.3275,2.3765,2.401,2.4255,2.43775,
2.4451,2.448775,2.45]$
load(interpol)$
for y in y_arr do (
for str_h in str_h_arr do (
bot_startpoints: [[-2.45,0],[0,y],[2.45,0]],
top_startpoints: [[-2.45,str_h_min],[0,y+str_h],[2.45,str_h_min]],
spline: cspline(bot_startpoints),
print (ev (spline, x=0))));
This is the output I get:
0.2487
0.2487
0.2487
0.2487
0.40323333333333
0.40323333333333
0.40323333333333
0.40323333333333
0.55776666666667
0.55776666666667
0.55776666666667
0.55776666666667
0.7123
0.7123
0.7123
0.7123
I am working on my project that uses elgamal elliptic curve.
I know when the elgamal ec encrypt by following steps
Represent the message m as a point M in E(Fp).
Select k ∈R [1,n−1].
Compute C1 = kP.
Compute C2 = M +kQ.
Return(C1,C2).
Where Q is the intended recipient’s public key, P is base point.
My qusetion at number one. how represent m as a point. Is point represent one character or represent group of characters.
There's no obvious way to attach m to points in E(Fp). However, you can use variant algorithm of ElGamal such as Menezes-Vanstone Elliptic curve cryptosystem to encode a message
in a point, a good reference here(P.31).
As for java code, I suggest you do some work, and post another question on SO when you encounter a problem you really can't solve by yourself.
DISCLAIMER: This question is only for those who have access to the econometrics toolbox in Matlab.
The Situation: I would like to use Matlab to simulate N observations from an ARIMA(p, d, q) model using the econometrics toolbox. What's the difficulty? I would like the innovations to be simulated with deterministic, time-varying variance.
Question 1) Can I do this using the in-built matlab simulate function without altering it myself? As near as I can tell, this is not possible. From my reading of the docs, the innovations can either be specified to have a constant variance (ie same variance for each innovation), or be specified to be stochastically time-varying (eg a GARCH model), but they cannot be deterministically time-varying, where I, the user, choose their values (except in the trivial constant case).
Question 2) If the answer to question 1 is "No", then does anyone see any reason why I can't edit the simulate function from the econometrics toolbox as follows:
a) Alter the preamble such that the function won't throw an error if the Variance field in the input model is set to a numeric vector instead of a numeric scalar.
b) Alter line 310 of simulate from:
E(:,(maxPQ + 1:end)) = Z * sqrt(variance);
to
E(:,(maxPQ + 1:end)) = (ones(NumPath, 1) * sqrt(variance)) .* Z;
where NumPath is the number of paths to be simulated, and it can be assumed that I've included an error trap to ensure that the (input) deterministic variance path stored in variance is of the right length (ie equal to the number of observations to be simulated per path).
Any help would be most appreciated. Apologies if the question seems basic, I just haven't ever edited one of Mathwork's own functions before and didn't want to do something foolish.
UPDATE (2012-10-18): I'm confident that the code edit I've suggested above is valid, and I'm mostly confident that it won't break anything else. However it turns out that implementing the solution is not trivial due to file permissions. I'm currently talking with Mathworks about the best way to achieve my goal. I'll post the results here once I have them.
It's been a week and a half with no answer, so I think I'm probably okay to post my own answer at this point.
In response to my question 1), no, I have not found anyway to do this with the built-in matlab functions.
In response to my question 2), yes, what I have posted will work. However, it was a little more involved than I imagined due to matlab file permissions. Here is a step-by-step guide:
i) Somewhere in your matlab path, create the directory #arima_Custom.
ii) In the command window, type edit arima. Copy the text of this file into a new m file and save it in the directory #arima_Custom with the filename arima_Custom.m.
iii) Locate the econometrics toolbox on your machine. Once found, look for the directory #arima in the toolbox. This directory will probably be located (on a Linux machine) at something like $MATLAB_ROOT/toolbox/econ/econ/#arima (on my machine, $MATLAB_ROOT is at /usr/local/Matlab/R2012b). Copy the contents of #arima to #arima_Custom, except do NOT copy the file arima.m.
iv) Open arima_Custom for editing, ie edit arima_Custom. In this file change line 1 from:
classdef (Sealed) arima < internal.econ.LagIndexableTimeSeries
to
classdef (Sealed) arima_Custom < internal.econ.LagIndexableTimeSeries
Next, change line 406 from:
function OBJ = arima(varargin)
to
function OBJ = arima_Custom(varargin)
Now, change line 993 from:
if isa(OBJ.Variance, 'double') && (OBJ.Variance <= 0)
to
if isa(OBJ.Variance, 'double') && (sum(OBJ.Variance <= 0) > 0)
v) Open the simulate.m located in #arima_Custom for editing (we copied it there in step iii). It is probably best to open this file by navigating to it manually in the Current Folder window, to ensure the correct simulate.m is opened. In this file, alter line 310 from:
E(:,(maxPQ + 1:end)) = Z * sqrt(variance);
to
%Check that the input variance is of the right length (if it isn't scalar)
if isscalar(variance) == 0
if size(variance, 2) ~= 1
error('Deterministic variance must be a column vector');
end
if size(variance, 1) ~= numObs
error('Deterministic variance vector is incorrect length relative to number of observations');
end
else
variance = variance(ones(numObs, 1));
end
%Scale innovations using deterministic variance
E(:,(maxPQ + 1:end)) = sqrt(ones(numPaths, 1) * variance') .* Z;
And we're done!
You should now be able to simulate with deterministically time-varying variance using the arima_Custom class, for example (for an ARIMA(0,1,0)):
ARIMAModel = arima_Custom('D', 1, 'Variance', ScalarVariance, 'Constant', 0);
ARIMAModel.Variance = TimeVaryingVarianceVector;
[X, e, VarianceVector] = simulate(ARIMAModel, NumObs, 'numPaths', NumPaths);
Further, you should also still be able to use matlab's original arima class, since we didn't alter it.