I have two isosurfaces (skull and skin). given point A on skull isosurface, i calculated the normal at point A using "double *pos = pickerCell->GetPickNormal()".
when i print pos, this is what i got: -6.2367, 1.98263, -0.9823
could someone explain to me what these 3 values mean?
I would like to find the intersection point of this normal of point A with the skin isosurface.
Could I use IntersectWithLine() function to do so? If yes, the line in my case would then be the normal? what is the start and end point of the normal?
Or is there a better way of doing ?
As you've found, you need to define the line to intersect with as two points. What is commonly done is to start at the point, P, that you picked (the same point where the normal, v, was computed) and compute two points, A = P + v delta and B = P - v delta where you have to set delta using context (if your model is in a unit cube, delta might be something like .01, where if your model has units of size 1000, delta may be 1, etc.).
Also, I'm not sure why the normal the cell picker returns is not normalized, but I'm assuming that if you normalize it it is the surface normal. I'd call it something other than 'pos' to avoid confusion (as it is a direction, not a position).
Related
After reading several posts about getting the 2D transformation of 2D points from one image to another, estimateRigidTransform() seems to be the recommendation. I'm trying to use it. I modified the source code (to change the RANSAC parameters, because its hardcoded, and the hardcoded parameters are not very good)(the source code for this function is in lkpyramid.cpp). I have read up on how RANSAC works, and am trying to understand the steps in estimateRigidTransform().
// choose random 3 non-complanar points from A & B
...
// additional check for non-complanar vectors
a[0] = pA[idx[0]];
a[1] = pA[idx[1]];
a[2] = pA[idx[2]];
b[0] = pB[idx[0]];
b[1] = pB[idx[1]];
b[2] = pB[idx[2]];
double dax1 = a[1].x - a[0].x, day1 = a[1].y - a[0].y;
double dax2 = a[2].x - a[0].x, day2 = a[2].y - a[0].y;
double dbx1 = b[1].x - b[0].x, dby1 = b[1].y - b[0].y;
double dbx2 = b[2].x - b[0].x, dby2 = b[2].y - b[0].y;
const double eps = 0.01;
if( fabs(dax1*day2 - day1*dax2) < eps*std::sqrt(dax1*dax1+day1*day1)*std::sqrt(dax2*dax2+day2*day2) ||
fabs(dbx1*dby2 - dby1*dbx2) < eps*std::sqrt(dbx1*dbx1+dby1*dby1)*std::sqrt(dbx2*dbx2+dby2*dby2) )
continue;
Is it a typo that it uses non-coplanar vectors? I mean the 2D points are all on the same plane right?
My second question is what is that if condition doing? I know that the left hand side (gives the area of triangle times 2) would be zero or near zero if the points are collinear, and the right hand side is the multiplication of the lengths of 2 sides of the triangle.
Collinearity is preserved in affine transformations (such as the one you are probably estimating), but this transformations also calculate also changes in rotations in point of view (as if you rotated the object in a 3d world). However, these points will be collinear as well, so for the algorithm it may have not a unique solution. Look at the pictures:
imagine selecting 3 center points of each black square in the first row in the first image. Then map it to the same centers in the next image. It may generate a mapping to that solution, but also a mapping to a zoom version of the first one. The same may happen with the third one, just that this time may map to a zoom out version of the first one (without any other change). However if the points are not collinear, for example, 3 corner squares centers, it will find a unique mapping.
I hope this helps you to clarify your doubts. If not, leave a comment
I'm trying to work out the best way to determine whether a point is inside a frustum. I have something working, but not sure whether it is too cumbersome, and perhaps there is a more elegant / efficient way I should be doing this.
Suppose I want to find out whether point 'x' is inside a frustrum:
Once I have the locations of the 8 points of the frustrum (4 near points, four far points), I am calculating the normal for each plane of the frustum based on a triangle made from three of the points. For example (as in the diagram above), for the right side, I am making two vectors from three of the points:
Vector U = FBR - NBR
Vector V = FTR - NBR
Then I am making the cross product between these two vectors, ensuring that the winding order is correct for the normal to be pointing inside the frustum, in this case V x U will give the correct normal.
Right_normal = V x U
Once I have the normal for each plane, I am then checking whether point x is in front of or behind the plane by drawing a vector from x to one of the plane's points:
Vector xNBR = x - NBR
Then I am doing the dot product between this vector and the normal and testing whether the answer is positive, confirming whether point x is the correct side of that plane of the frustrum:
if ( xNBR . Right_normal < 0 )
{
return false;
}
else continue testing x against other planes...
If x is positive for all planes, then it is inside the frustum.
So this seems to work, but I'm just wondering whether I'm doing this in a stupid way. I didn't even know what 'cross product' meant until yesterday, so it's all rather new to me and I might be doing something rather silly.
To adapt the approach you have taken, rather than change it totally, you can make use of the fact that 2 of the pairs of planes are parallel. Create only one normal for that pair of planes. You already have the test for the point being "in front" of one of the planes, but assuming you know the depth of the frustum, you can use the same distance to test the point against the other parallel face.
double distancePastFrontPlane = xNBR . Right_normal;
if (distancePastFrontPlane < 0 )
{
// point is in front of front plane
return false;
if(distancePastFrontPlane > depthFaceRtoFaceL)
{
// point is behind back plane
return false;
}
}
If you have multiple points to test against the same frustum you can benefit because you only calculate the frustum depth once (per pair of parallel planes).
i'm new to opencv & it's developing. I have x,y,z coordinates ( 0.00949334694383068, -0.3999829847985352, 0.8449078780010854) by using the given coordinates how could i find the direction.
for an example
input one : x,y,z = 0.00949334694383068, -0.3999829847985352, 0.8449078780010854
input two : x,y,z = 0.01603883281370779, 0.6066595296580494, 0.5342810849038371
At the finally i want to compare input_one direction and input_two direction. Any help is appreciated
This is called vector math.
"Coordinates" are a special kind of vector, relative to some origin (in your case x=0,y=0,z=0). For that reason, the difference x1-x2, y1-y2, z1-z2 is a vector from point 2 to point 1. The inverse x2-x1, y2-y1, z2-z1 is a vector from point 1 to point 2.
The direction of a vector is usually defined by ignoring its length, or alternatively by setting its length to one. So we need to first define the length, which is L = √(x*x + y*y + z*z). We can define the vector x/L, y/L, z/L which points in the same direction as x,y,z but with length one.
Finally, to compare two directions we can calculate the inner product of those two directions: x1/L1 * x2/L2 + y1/L1 * y2/L2 + z1/L1 * z2/L2. If that's one, they point in the same direction. If it's 0, they're orthogonal. If it's -1, they point in opposite directions.
As you can see, the vector 0,0,0 has length 0 and no direction. That can complicate things a bit.
In OpenCV: class Vec. The length function is called norm(v) and the inner product is called v1.mul(v2)
What you're trying to do is called calculating the azimuth. If you're interested in doing this for navigational or geographic purposes and need a thorough understanding of this, you can start here:
http://mobile.codeguru.com/cpp/cpp/algorithms/article.php/c5115/Geographic-Distance-and-Azimuth-Calculations.htm
Otherwise you could look for a library for calculating the azimuth bases on 3d co-ordinates
I would like to implement the Ramer–Douglas–Peucker_algorithm in C++.
The pseudo code looks like this:
function DouglasPeucker(PointList[], epsilon)
//Find the point with the maximum distance
dmax = 0
index = 0
for i = 2 to (length(PointList) - 1)
d = OrthogonalDistance(PointList[i], Line(PointList[1], PointList[end]))
if d > dmax
index = i
dmax = d
end
end
//If max distance is greater than epsilon, recursively simplify
if dmax >= epsilon
//Recursive call
recResults1[] = DouglasPeucker(PointList[1...index], epsilon)
recResults2[] = DouglasPeucker(PointList[index...end], epsilon)
// Build the result list
ResultList[] = {recResults1[1...end-1] recResults2[1...end]}
else
ResultList[] = {PointList[1], PointList[end]}
end
//Return the result
return ResultList[]
end
Here is my understanding so far.
It is a recursive function taking in an array of points and a distance threshold.
Then it iterates through the current points to find the point with the maximum distance.
I got a bit lost with the Orthographical Distance function. How do I compute this? I've never seen a distance function take a line segment as a parameter.
I think aside from this I should be alright, I will just use std::vectors for the arrays. I think I will use std::copy and then push or pop according to what the algorithm says.
Thanks
The OrthogonalDistance is shown in this picture:
So it's the distance from your point and the point on the line which is the projection of that point on the line.
The distance from a point to a line is usally something like this:
(source: fauser.edu)
where x0 and y0 are the coordinates of the external point and a, b, c are the coefficient of the equation of your line.
That's what I remember from school, long time ago.
The orthogonal distance from a point P to a line L is defined by the distance between point P and point P2, where P2 is the orthogonal projection of P on line L.
How you compute this value depends on the dimension of the space you work on, but if it's 2D, you should be able to figure that out by drawing an example on a piece of paper !
Look at the topcoder tutorial and the
double linePointDist(point A, point B, point C, bool isSegment);
method it it what your looking for.
A brief description of the math required can be found here. Just realize that you can swap the word "Orthogonal" for "Perpendicular" when dealing with 2D. The site linked has instructions for lines defined by two points as well as lines defined by slope-intercept form.
The short version is reproduced here:
If the line is represented in slope intercept from: ax + by + c = 0, and the point is represented by x0, y0, then the function that will give the Orthogonal distance is:
abs(a*x0 + b*y0 + c)/sqrt(a*a + b*b)
It's not clear to me whether you want the distance of the point to the (infinite) line through the two points, or the distance to the line segment defined by the points, but I suspect its the latter.
Consider the somewhat contrived example of the points (0,0) (1,0) and (10, t) where t is small. The distance of (10,t) from the line through the first two points (ie the x axis) is t, while the distance (10,t) from the line segment with end points (0,0) and (1,0) is hypot(9,t) ~ 9. So if you were using distance to the line, there is a danger that the
algorithm above would not split at (10,t).
The method mentioned by jethro above handles both lines and line segments.
I have a parametric curve, say two vectors of doubles where the parameter is the index, and I have to calculate the angle of the tangent to this curve at any given point (index).
Any suggestion or link about how to do that?
Thanks.
Here's a short formula, equivalent (I think) to pau.estalella's answer:
m[i] = (y[i+1] - y[i-1]) / (x[i+1] - x[i-1])
this approximates, reasonably well, the slope at the point (x[i], y[i]).
Your question mentions the "angle of the tangent". The tangent line, having slope m[i], makes angle arctangent(m[i]) with the positive x axis. If this is what you're after, you might use the two-argument arctangent, if it's available:
angle[i] = atan2(y[i+1] - y[i-1], x[i+1] - x[i-1])
this will work correctly, even when x[i+1] == x[i-1].
I suggest you check out the Wikipedia article on numerical differentiation for a start. Before you go much further than that, decide what purposes you want the tangent for and decide whether or not you need to try more complex schemes than the simple ones in the article.
The first problem you run into is to even define the tangent in one of the vertexes of the curve. Consider e.g. that you have the two arrays:
x = { 1.0, 2.0, 2.0 };
y = { 1.0, 1.0, 2.0 };
Then at the second vertex you have a 90-degree change of direction of the line. In that place the tangent isn't even defined mathematically.
Answer to gregseth's comment below
I guess in your example the "tangent" at the second point would be the line parallel to (P0,P2) passing through P1... which kind of give me the answer : for any point of index N the parallel to (N-1, N+1) passing through N. Would that be a not-too-bad approximation?
It depends on what you are using it for. Consider for example:
x = { 1.0, 2.0, 2.0 };
y = { 1.0, 1000000, 1000000 };
That is basically an L shape with a very high vertical line. In your suggestion it would give you an almost vertical tangent. Is that what you want, or do you rather want a 45-degree tangent in that case? It also depends on your input data how you sould define it.
One solution is get the two vectors connection to the vertex, normalize them and then use your algorithm. That way you would get a 45-degree tangent in the above example.
Compute the first derivative: dy/dx. That gives you the tangent.
The tangent to a smooth curve at a point P is the parametric straight line P + tV, where V is the derivative of the curve with respect to "the parameter". But here the parameter is just the index of an array, and numerical differentiation is a difficult problem, hence to approximate the tangent I would use (weighted) least squares approximation.
In other words, choose three or five points of the curve around your point of interest P (i.e. P[i-2], P[i-1], P[i], P[i+1], and P[i+2], if P==P[i]), and approximate them with a straight line, in the least squares sense. The more weight you assign to the middle point P, the more close the line will be to P; on the other hand, the more weight you assign to the extremal points, the more "tangent" the straight line will be, that is, the more nicely it will approximate you curve in the neighborhood of P.
For example, with respect to the following points:
x = [-1, 0, 1]
y = [ 0, 1, 0]
for which the tangent is not defined (as in Anders Abel's answer),
this approach should yield a horizontal straight line close to the point (0,1).
You can try to compute the tangent of an interpolating curve that passes through the given points (I'm thinking of a cubic spline, which is pretty easy to derive) or compute the tangent directly from the data points.
You can find a rough approximation of the derivative in the following manner
Let a curve C pass through points p1,p2 and p3. At point p2 you have two possible tangents: t1=p2-p1 and t2=p3-p2. You can combine them by simply computing their average: 0.5*(t1+t2)
or you can combine them according to their lengths (or their reciprocal 1/length)
Remember to normalize the resulting tangent.
In order to compute the angle between the tangent and the curve, remember that the dot product of two unit vectors gives the cosine of the angle between them. Take the resulting tangent t and the unit vector v2=|p3-p2|, and acos(dot(t,v2)) gives the angle you need.