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
Related
I have an NxN grid with 2 points, the source and destination. I need to move step by step from the source to the destination (which is also moving). How do I determine what the next point is to move to?
One way is to assess all 8 points and see which yields the lowest distance using an Euclidian distance. However, I was hoping there is a cool (mathematical) trick which will yield more elegant results.
Your question statement allows moving diagonally, which is faster (since it's moving both horizontally and vertically in a single step): this solution will always do that unless it has the same x or y coordinate as the target.
using Position = pair<int,int>;
Position move(Position const ¤t, Position const &target) {
// horizontal and vertical distances
const int dx = target.first - current.first;
const int dy = target.second - current.second;
// horizontal and vertical steps [-1,+1]
const int sx = dx ? dx/abs(dx) : 0;
const int sy = dy ? dy/abs(dy) : 0;
return { current.first + sx, current.second + sy };
}
I'm not sure if this counts as a cool mathematical trick though, it just depends on knowing that:
dx = target.x-current.x is positive if you should move in the positive x-direction, negative if you should go in the negative direction, and zero if you should go straight up/down
dx/abs(dx) keeps the sign and removes the magnitude, so it's always one of -1,0,+1 (avoiding however division by zero)
I suppose that answer to your question is Bresenham's line algorithm. It allows to build sequence of integer points between start and end points in your grid. Anyway you can adapt ideas from it to your problem
For more information see https://www.cs.helsinki.fi/group/goa/mallinnus/lines/bresenh.html
I would simply use some vector math, take dest minus source as a vector, and then calculate the angle between that vector and some reference vector, e.g. <1, 0>, with standard methods.
Then you can simply divide the circle in 8 (or 4 if your prefer) sections and determine in which section your vector lies from the angle you obtained.
See euclidean space for how to calculate the angle between two vectors.
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).
This is quite complicated to explain, so I will do my best, sorry if there is anything I missed out, let me know and I will rectify it.
My question is, I have been tasked to draw this shape,
(source: learnersdictionary.com)
This is to be done using C++ to write code that will calculate the points on this shape.
Important details.
User Input - Centre Point (X, Y), number of points to be shown, Font Size (influences radius)
Output - List of co-ordinates on the shape.
The overall aim once I have the points is to put them into a graph on Excel and it will hopefully draw it for me, at the user inputted size!
I know that the maximum Radius is 165mm and the minimum is 35mm. I have decided that my base Font Size shall be 20. I then did some thinking and came up with the equation.
Radius = (Chosen Font Size/20)*130. This is just an estimation, I realise it probably not right, but I thought it could work at least as a template.
I then decided that I should create two different circles, with two different centre points, then link them together to create the shape. I thought that the INSIDE line will have to have a larger Radius and a centre point further along the X-Axis (Y staying constant), as then it could cut into the outside line.
So I defined 2nd Centre point as (X+4, Y). (Again, just estimation, thought it doesn't really matter how far apart they are).
I then decided Radius 2 = (Chosen Font Size/20)*165 (max radius)
So, I have my 2 Radii, and two centre points.
Now to calculate the points on the circles, I am really struggling. I decided the best way to do it would be to create an increment (here is template)
for(int i=0; i<=n; i++) //where 'n' is users chosen number of points
{
//Equation for X point
//Equation for Y Point
cout<<"("<<X<<","<<Y<<")"<<endl;
}
Now, for the life of me, I cannot figure out an equation to calculate the points. I have found equations that involve angles, but as I do not have any, I'm struggling.
I am, in essence, trying to calculate Point 'P' here, except all the way round the circle.
(source: tutorvista.com)
Another point I am thinking may be a problem is imposing limits on the values calculated to only display the values that are on the shape.? Not sure how to chose limits exactly other than to make the outside line a full Half Circle so I have a maximum radius?
So. Does anyone have any hints/tips/links they can share with me on how to proceed exactly?
Thanks again, any problems with the question, sorry will do my best to rectify if you let me know.
Cheers
UPDATE;
R1 = (Font/20)*130;
R2 = (Font/20)*165;
for(X1=0; X1<=n; X1++)
{
Y1 = ((2*Y)+(pow(((4*((pow((X1-X), 2)))+(pow(R1, 2)))), 0.5)))/2;
Y2 = ((2*Y)-(pow(((4*((pow((X1-X), 2)))+(pow(R1, 2)))), 0.5)))/2;
cout<<"("<<X1<<","<<Y1<<")";
cout<<"("<<X1<<","<<Y2<<")";
}
Opinion?
As per Code-Guru's comments on the question, the inner circle looks more like a half circle than the outer. Use the equation in Code-Guru's answer to calculate the points for the inner circle. Then, have a look at this question for how to calculate the radius of a circle which intersects your circle, given the distance (which you can set arbitrarily) and the points of intersection (which you know, because it's a half circle). From this you can draw the outer arc for any given distance, and all you need to do is vary the distance until you produce a shape that you're happy with.
This question may help you to apply Code-Guru's equation.
The equation of a circle is
(x - h)^2 + (y - k)^2 = r^2
With a little bit of algebra, you can iterate x over the range from h to h+r incrementing by some appropriate delta and calculate the two corresponding values of y. This will draw a complete circle.
The next step is to find the x-coordinate for the intersection of the two circles (assuming that the moon shape is defined by two appropriate circles). Again, some algebra and a pencil and paper will help.
More details:
To draw a circle without using polar coordinates and trig, you can do something like this:
for x in h-r to h+r increment by delta
calculate both y coordinates
To calculate the y-coordinates, you need to solve the equation of a circle for y. The easiest way to do this is to transform it into a quadratic equation of the form A*y^2+B*y+C=0 and use the quadratic equation:
(x - h)^2 + (y - k)^2 = r^2
(x - h)^2 + (y - k)^2 - r^2 = 0
(y^2 - 2*k*y + k^2) + (x - h)^2 - r^2 = 0
y^2 - 2*k*y + (k^2 + (x - h)^2 - r^2) = 0
So we have
A = 1
B = -2*k
C = k^2 + (x - h)^2 - r^2
Now plug these into the quadratic equation and chug out the two y-values for each x value in the for loop. (Most likely, you will want to do the calculations in a separate function -- or functions.)
As you can see this is pretty messy. Doing this with trigonometry and angles will be much cleaner.
More thoughts:
Even though there are no angles in the user input described in the question, there is no intrinsic reason why you cannot use them during calculations (unless you have a specific requirement otherwise, say because your teacher told you not to). With that said, using polar coordinates makes this much easier. For a complete circle you can do something like this:
for theta = 0 to 2*PI increment by delta
x = r * cos(theta)
y = r * sin(theta)
To draw an arc, rather than a full circle, you simply change the limits for theta in the for loop. For example, the left-half of the circle goes from PI/2 to 3*PI/2.
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).
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.