I have one start point x0,y0,z0 and two angles. Theta(polar(0-180)) and Phi(azimuth (0-360)). And now I need to find direction of vector of reflected line 3d space box with parameters (4,5,6)
I tried to calculate using geometry formula and use cos and sin, adjacent, opposite. But it gave me wrong answer for my further cycle calculation I suppose I need some equation to write. Some kind of i, j, k. I could not calculate slope as I don’t have enough input.
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i have two concentric circles and three points are given for each circle that are on circumference.
I need a optimized method to check if a given random point exist inbetween these circles or not.
You can compute (x²+y²), x, y, 1 for each point. The last entry is simply the constant one. Put these terms for four given points into a matrix and compute its determinant. The determinant will be zero if the points are cocircular. Otherwise the sign will tell you which point is on which side with respect to the circle defined by the other three. Use a simple example to check which sign corresponds to which direction. Be prepared for the fact that the three circle-defining points being oriented in a clockwise or counter-clockwise orientation will affect this sign, too.
Computing a 4×4 determinant can be done horribly inefficiently, too. I'd suggest you compute all the 2×2 minors from the first two rows, and all the 2×2 minors from the last two, then you can combine them to form the full determinant. See this Math SE post for details. If you need further mathematical help (as opposed to programming help), you might find more suitable answers there.
Nothe that the above works for each circle independently. Check whether the point is inside the one, then check whether it is outside the other. It does not make use of the fact that the circles are assumed to be cocircular.
I struggle with the orientation of an object I am moving along a hermite curve.
I figured out how to move it at constant speed at also have the tangent of my curve, which would be the forward vector of the moving object. My problem is: How do I know the up and right vector? The easiest way would be to start at a given rotation and then step through the curve always taking the last rotation as a reference for the next one, like in this reference:
Camera movement along a splinepdf
But this would result in an uncontrollable rotation at the end of the spline. What I am trying to do is to create an algorithm which gives you the correct orientation at any point of the curve, without stepping through it. Ideally it would use the orientation of the two controlpoints for the current segment as a reference.
I thought of using some kind of pre-calculated data, which is created from the two orientations of the controlpoints and the current curve segments form, but didn't manage to come up with a solution.
I would be happy to get any answers or just ideas how to approach this problem.
Let C(t) be the camera trajectory, with tangent vector T(t). The tangent vector controls the pitch and the yaw. What you are missing is roll control.
Define an auxiliary trajectory D(t) that "parallels" C(t) and use the vector CD(t). The up vector is given by U(t)=T(t) /\ CD(t) (normalized), and the right vector by U(t) /\ T(t) (normalized).
OK i came up with a solution using frenet frames. I define an orientation for each of my control points, then i calculate a number of points along the spline for each segment. Each points orientation is then calculated using the previous points orientation. The orientatin of the first point equals the orientation of the control point.
Here is a very nice description of the procedure.
After calculating each points orientation, you can interpolate them so the last points orientation matches the orientation of the next controlpoint.
Given A: a point, B: A point known to exist on a plane P, C: the normal of plane P. Can I determine if A lies on P by the result of a dot product between (A - B) and C being zero? (or within a certain level of precision, I'll probably use 0.0001f)
I might just be missing some obvious mathematical flaw but this seems to be much simpler and faster than transforming the point to a triangle's coordinate space a.la the answer to Check if a point is inside a plane segment
So secondly I guess; if this is a valid check, would it be computationally faster than using matrix transformations if all I want is to see if the point is on the plane? (and not whether it lies within a polygon on said plane, I'll probably keep using matrix transforms for that)
You are correct that B lies on the plane through A and with normal P if and only if dotProduct(A-B,P) = 0.
To estimate speed for this sort of thing, you can pretty much just count the multiplications. This formula just has three multiplications, so its going to be faster than pretty much anything to do with matrices.
The above answers are closer to the proof, but not sufficient. It should be intuitive that using just two vectors is insufficient because for one, point P can be above the plane and a vertical line drawn from it to the plane would still generate a zero dot product with any single vector lying on the plane, just as it would for a point P on the plane. The necessary and sufficient condition is that if two vectors can be found on the plane then the actual plane is represented unambiguously by the cross product of the two vectors i.e.
w=uxv. By definition, w is the area vector, which is always perpendicular to the plane.
Then, for the point P in question, constructing a third vector s from either u or v should be tested against w by the dot product, s.t.
w.s=|w||s|cos(90)=0 implies that the point P lies on the plane described by w, which is in turn described by vectors u and v.
I'm trying to find a way to calculate the intersection between two arcs.
I need to use this to determine how much of an Arc is visually on the right half of a circle, and how much on the left.
I though about creating an arc of the right half, and intersect that with the actual arc.
But it takes me wayyy to much time to solve this, so I thought about asking here - someone must have done it before.
Edit:
I'm sorry the previous illustration was provided when my head was too heavy after crunching angles. I'll try to explain again:
In this link you can see that I cut the arc in the middle to two halves, the right part of the Arc contains 135 degrees, and the left part has 90.
This Arc starts at -180 and ends at 45. (or starts at 180 and ends at 405 if normalized).
I have managed to create this code in order to calculate the amount of arc degrees contained in the right part, and in the left part:
f1 = (angle2>270.0f?270.0f:angle2) - (angle1<90.0f?90.0f:angle1);
if (f1 < 0.0f) f1 = 0.0f;
f2 = (angle2>640.0f?640.0f:angle2) - (angle1<450.0f?450.0f:angle1);
if (f2 < 0.0f) f2 = 0.0f;
f3 = (angle2>90.0f?90.0f:angle2) - angle1;
if (f3<0.0f) f3=0.0f;
f4 = (angle2>450.0f?450.0f:angle2) - (angle1<270.0f?270.0f:angle1);
if (f4<0.0f) f4=0.0f;
It works great after normalizing the angles to be non-negative, but starting below 360 of course.
Then f1 + f2 gives me the sum of the left half, and f3 + f4 gives me the sum of the right half.
It also does not consider a case when the arc is defined as more than 360, which may be an "error" case.
BUT, this seems like more of a "workaround", and not a correct mathematical solution.
I'm looking for a more elegant solution, which should be based on "intersection" between two arc (because math has no "sides", its not visual";
Thanks!!
I think this works, but I haven't tested it thoroughly. You have 2 arcs and each arc has a start angle and a stop angle. I'll work this in degrees measured clockwise from north, as you have done, but it will be just as easy to work in radians measured anti-clockwise from east as the mathematicians do.
First 'normalise' your arcs, that is, reduce all the angles in them to lie in [0,360), so take out multiples of 360deg and make all the angles +ve. Make sure that the stop angle of each arc lies to clockwise of the start angle.
Next, choose the start angle of one of your arcs, it doesn't matter which. Sort all the angles you have (4 of them) into numerical order. If any of the angles are numerically smaller than the start angle you have chosen, add 360deg to them.
Re-sort the angles into increasing numerical order. Your chosen start angle will be the first element in the new list. From the start angle you already chose, what is the next angle in the list ?
1) If it is the stop angle of the same arc then either there is no overlap or this arc is entirely contained within the other arc. Make a note and find the next angle. If the next angle is the start angle of the other arc there is no overlap and you can stop; if it is the stop angle of the other arc then the overlap contains the whole of the first arc. Stop
2) If it is the start angle of the other arc, then the overlap begins at that angle. Make a note of this angle. The next angle your sweep encounters has to be a stop angle and the overlap ends there. Stop.
3) If it is the stop angle of the other arc then the overlap comprises the angle between the start angle of the first arc and this angle. Stop.
This isn't particularly elegant and relies on ifs rather more than I generally like but it should work and be relatively easy to translate into your favourite programming language.
And look, no trigonometry at all !
EDIT
Here's a more 'mathematical' approach since you seem to feel the need.
For an angle theta in (-pi,pi] the hyperbolic sine function (often called sinh) maps the angle to an interval on the real line in the interval (approximately) (-11.5,11.5]. Unlike arcsin and arccos the inverse of this function is also single-valued on the same interval. Follow these steps:
1) If an arc includes 0 break it into 2 arcs, (start,0) and (0,stop). You now have 2, 3 or 4 intervals on the real line.
2) Compute the intersection of those intervals and transform back from linear measurement into angular measurement. You now have the intersection of the two arcs.
This test can be resumed with a one-line test. Even if a good answer is already posted, let me present mine.
Let assume that the first arc is A:(a0,a1) and the second arc is B:(b0,b1). I assume that the angle values are unique, i.e. in the range [0°,360°[, [0,2*pi[ or ]-pi,pi] (the range itself is not important, we will see why). I will take the range ]-pi,pi] as the range of all angles.
To explain in details the approach, I first design a test for interval intersection in R. Thus, we have here a1>=a0 and b1>=b0. Following the same notations for real intervals, I compute the following quantity:
S = (b0-a1)*(b1-a0)
If S>0, the two segments are not overlapping, else their intersection is not empty. It is indeed easy to see why this formula works. If S>0, we have two cases:
b0>a1 implies that b1>a0, so there is no intersection: a0=<a1<b0=<b1.
b1<a0 implies that b0<b1, so there is no intersection: b0=<b1<a0=<a1.
So we have a single mathematical expression which performs well in R.
Now I expand it over the circular domain ]-pi,pi]. The hypotheses a0<a1 and b0<b1 are not true anymore: for example, an arc can go from pi/2 to -pi/2, it is the left hemicircle.
So I compute the following quantity:
S = (b0-a1)*(b1-a0)*H(a1-a0)*H(b1-b0)
where H is the step function defined by H(x)=-1 if x<0 else H(x)=1
Again, if S>0, there is no intersection between the arcs A and B. There are 16 cases to explore, and I will not do this here ... but it is easy to make them on a sheet :).
Remark: The value of S is not important, just the signs of the terms. The beauty of this formula is that it is independant from the range you have taken. Also, you can rewrite it as a logical test:
T := (b0>a1)^(b1>a0)^(a1>=a0)^(b1>=b0)
where ^ is logical XOR
EDIT
Alas, there is an obvious failure case in this formula ... So I correct it here. I realize that htere is a case where the intersection of the two arcs can be two arcs, for example when -pi<a0<b1<b0<a1<pi.
The solution to correct this is to introduce a second test: if the sum of the angles is above 2*pi, the arcs intersect for sure.
So the formula turns out to be:
T := (a1+b1-a0-b0+2*pi*((b1<b0)+(a1<a0))<2*pi) | ((b0>a1)^(b1>a0)^(a1>=a0)^(b1>=b0))
Ok, it is way less elegant than the previous one, but it is now correct.
I'm working in OpenCV but I don't think there is a function for this. I can find a function for finding affine transformations, but affine transformations include scaling, and I only want to consider rotation + translation.
Imagine I have two sets of points in 2d - let's say each set has exactly 50 points.
E.g. set A = {x1, y1, x2, y2, ... , x50, y50}
set B = {x1', y1', x2', y2', ... , x50', y50'}
I want to find the rotation and translation combination that gets closest to mapping set A onto set B. I guess I would define "closest" as minimises the average distance between points in A and corresponding points in B. I.e., minimises the average distance between (x1, y1) and (x1', y1'), etc.
I guess I could use brute force testing all possible translations and rotations but this would be extremely inefficient. Does anyone know a simpler way?
Thanks!
This problem has a very elegant solution in terms of singular value decomposition of the proximity matrix (distances between pairs of points). The name of this is the orthogonal Procrustes problem, after the Greek legend about a fellow who offered travellers a bed that would fit anyone.
The solution comes from finding the nearest orthogonal matrix to a given (not necessarily orthogonal) matrix.
The way I would do it in Excel is to make a couple columns representing the points.
Cells representing rotation/translation of a set (no need to rotate and translate both of them).
Then columns representing those same points rotated/translated.
Then another column for the distance between the points of the rotated/translated points.
Then a cell of the sum of the distances between points.
Finally, use Solver to optimize the rotation and translation cells.
If you fix some rotation you can get an answer using ternary search. Run search in x and for every tested x run it in y to get the best value. This will give you the correct answer since the function (sum of corresponding distances) is convex (this can be proved through observing that restriction of the function to any line is a one-dimensional convex function; and the last is a standard fact: the sum of several convex functions is convex).
Instead of brute force over the angle I can propose such a method based on the ternary search. Choose some not very large step S. Compute the target function for every angle in (0, S, 2S,...). Then, if S is small enough, we can exclude some of segments (iS, (i + 1)S) from consideration. Namely ones with relatively large values of function with angles iS and (i + 1)S. Being implemented carefully this can give an answer and can do it faster than brute force.