I have an angle in degrees, and need to map it to a voltage.
i.e. left: -360 degrees = 0V, straight: 0 degrees = 5V, right: 360 degrees = 10V.
How do you map values in LabVIEW?
I found scaling under the numeric pallet but they seem to be very specific (e.g. for a thermistor) with no generic block.
If the mapping you want is described by a mathematical formula, just implement that calculation using LabVIEW functions. In your case, assuming it's a linear relationship, the formula is voltage = (angle + 360) / 72 .
You could code this in LabVIEW with an Add function followed by a Divide function, or if you think it's clearer written out as a formula then you could use a Formula Node.
If you need to use this calculation in more than one place in your application, make it into a subVI.
Looks like you're trying to read a analog gauge value! Actually it's very simple.
Build a lookup Table
Perform Interpolation followed by thresholding.
Refer the VI Snippet (Just paste this image into your block diagram).
Sample Output:
In my meshing application I will have to specify fix points within a domain. The idea is that, the fix points must also be the element points after the domain is being meshed.
Furthermore, the elements around the fix points should be more dense. The general concept is that for the fix points, there should exist a radius r around those points, such that the mesh size inside r is of different sizes than outside of the r. The mesh sizes inside and outside of the r should be specifiable.
Are these two things doable in CGAL 2D Mesh algorithm?
Using your wording, all the input point of the initial constrained Delaunay triangulation will be fix points, because the 2D mesh generator only insert new points in the triangulation: it never removes any point.
As for the density, you can copy, paste, and modify a criteria class, such as CGAL::Delaunay_mesh_size_criteria_2<CDT> so that the local size upper bound is smaller around the fix points.
Now, the difficulty is how to implement that new size policy. Your criteria class could store a const reference to another Delaunay_triangulation_2, that contains only the fixed points you want. Then, for each triangle query, you can call nearest_vertex and then actually check if the distance between the query point is smaller that the radius bound of your circles. For a triangle, you can either verify that for only its barycenter, or for all three points of the triangle. Then, according to the result of that/those query(s), you can modify the size bound, in the code of your copy of CGAL::Delaunay_mesh_size_criteria_2<CDT>.
Yes, no points will be removed from the triangulation by the mesher.
Note however that if you insert points too close to a constraint this will induce a refinement of the constraint while it is not Gabriel.
How do i calculate point D for triangle 2?
I have the the following coordinates for triangle 1:
a(0,0) b(0,78) c(18,39)
point D is located at (0,39) in triangle 1.
now I change the perspective on my triangle by for example moving coordinate b and c.
the new triangle formed is called triangle 2 with coordinates:
a(0,0) b(11,72) c(37,42)
AS YOU CAN SEE POINT D IS NOT IN THE MIDDLE OF LINE a<-->b BECAUSE OF THE CHANGE IN PERSPECTIVE/SKEW.
How do i calculate point d? I have the coordinates abc of triangle 1 & 2.
Preferably answer in programcode rather than using math signs, since i am not a hero at reading math :)
You need to convert point D to barycentric coordinates using the original triangle coordinates, then convert it back to cartesian coordinates using the modified triangle coordinates.
This looks like a good introduction to triangular barycentric coordinates: http://blogs.msdn.com/b/rezanour/archive/2011/08/07/barycentric-coordinates-and-point-in-triangle-tests.aspx
Also, explicit formulae for converting a point in a triangle to barycentric coordinates are given at the end of the Converting to Barycentric Coordinates section of the Wikipedia article “Barycentric coordinate system”.
I guess there are more ways of calculating a coordinate from one perspective to another.
more on the triangle way is written by culebrón here: Transforming captured co-ordinates into screen co-ordinates
At the same link there is another way by using SVD and calculate an H-matrix which can be used to translate any coordinate from one perspective to another. I am going to use this way because i could solve this way in matlab. Next step in objective-c! i had some trouble calculating the same in objective-c. more on that here: calculate the V from A = USVt in objective-C with SVD from LAPACK in xcode
I would like to know how to solve the triangle way too! i could not figure out what a1 and a2 were in culebron's post: https://stackoverflow.com/a/1690300/1568532 neither the width and height made much sense to me.
Also i would like to know how to calculate the EYE's point of view on a triangle or quadrangle based on 3 or 4 coordinates. if you know the original size of the object.
any ideas on this?
when i search for eye or camera's point of view. there is load of result about photography.
what do i need to use in order to calculate this? maybe some example anyone?
I am working on a program that needs to fit numerous cosine waves in order to determine one of the parameters for the function. The equation that I am using is y = y_0 + Acos((4*pi*L)/x + pi) where L is the value that I am trying to obtain from the best fit line.
I know that it is possible to do this correctly by hand for each set of data, but what is the best way to automate this process? I am currently reading in the data from text files, and running a loop with the initial paramiters changing until I have an array of paramater values that have an amplitude similar to the data, then I check the percent difference between points on the center peak and two end peaks to try to pick the best one. It in consistently picking lower values than what I get when fitting by hand (almost exactly one phase off). So is there a way to improve this method, or another method that works better?
Edit: My LabVIEW version has a cos fitting VI which is what I am using, the problem is when I try to automate the fitting by changing the initial parameters using a loop, I cant figure out how to get the program to pick the same best fit line as a human would pick.
Why not just use a Fast Fourier Transform? This should be way faster than fitting a cosine. In the result vector of complex numbers look for the largest peak of in the totals. You're given frequency (position in the FFT result vector), amplitude and phase.
You can evaluate the goodness of the fit by computing the difference between fitting curve and your data. A VI does this in the "Advanced curve fitting" palette. Then all you have to do is pick up the best fit.
My question is fairly simple. I have two tetrahedra, each with a current position, a linear speed in space, an angular velocity and a center of mass (center of rotation, actually).
Having this data, I am trying to find a (fast) algorithm which would precisely determine (1) whether they would collide at some point in time, and if it is the case, (2) after how much time they collided and (3) the point of collision.
Most people would solve this by doing triangle-triangle collision detection, but this would waste a few CPU cycles on redundant operations such as checking the same edge of one tetrahedron against the same edge of the other tetrahedron upon checking up different triangles. This only means I'll optimize things a bit. Nothing to worry about.
The problem is that I am not aware of any public CCD (continuous collision detection) triangle-triangle algorithm which takes self-rotation in account.
Therefore, I need an algorithm which would be inputted the following data:
vertex data for three triangles
position and center of rotation/mass
linear velocity and angular velocity
And would output the following:
Whether there is a collision
After how much time the collision occurred
In which point in space the collision occurred
Thanks in advance for your help.
The commonly used discrete collision detection would check the triangles of each shape for collision, over successive discrete points in time. While straightforward to compute, it could miss a fast moving object hitting another one, due to the collision happening between discrete points in time tested.
Continuous collision detection would first compute the volumes traced by each triangle over an infinity of time. For a triangle moving at constant speed and without rotation, this volume could look like a triangular prism. CCD would then check for collision between the volumes, and finally trace back if and at what time the triangles actually shared the same space.
When angular velocity is introduced, the volume traced by each triangle no longer looks like a prism. It might look more like the shape of a screw, like a strand of DNA, or some other non-trivial shapes you might get by rotating a triangle around some arbitrary axis while dragging it linearly. Computing the shape of such volume is no easy feat.
One approach might first compute the sphere that contains an entire tetrahedron when it is rotating at the given angular velocity vector, if it was not moving linearly. You can compute a rotation circle for each vertex, and derive the sphere from that. Given a sphere, we can now approximate the extruded CCD volume as a cylinder with the radius of the sphere and progressing along the linear velocity vector. Finding collisions of such cylinders gets us a first approximation for an area to search for collisions in.
A second, complementary approach might attempt to approximate the actual volume traced by each triangle by breaking it down into small, almost-prismatic sub-volumes. It would take the triangle positions at two increments of time, and add surfaces generated by tracing the triangle vertices at those moments. It's an approximation because it connects a straight line rather than an actual curve. For the approximation to avoid gross errors, the duration between each successive moments needs to be short enough such that the triangle only completes a small fraction of a rotation. The duration can be derived from the angular velocity.
The second approach creates many more polygons! You can use the first approach to limit the search volume, and then use the second to get higher precision.
If you're solving this for a game engine, you might find the precision of above sufficient (I would still shudder at the computational cost). If, rather, you're writing a CAD program or working on your thesis, you might find it less than satisfying. In the latter case, you might want to refine the second approach, perhaps by a better geometric description of the volume occupied by a turning, moving triangle -- when limited to a small turn angle.
I have spent quite a lot of time wondering about geometry problems like this one, and it seems like accurate solutions, despite their simple statements, are way too complicated to be practical, even for analogous 2D cases.
But intuitively I see that such solutions do exist when you consider linear translation velocities and linear angular velocities. Don't think you'll find the answer on the web or in any book because what we're talking about here are special, yet complex, cases. An iterative solution is probably what you want anyway -- the rest of the world is satisfied with those, so why shouldn't you be?
If you were trying to collide non-rotating tetrahedra, I'd suggest a taking the Minkowski sum and performing a ray check, but that won't work with rotation.
The best I can come up with is to perform swept-sphere collision using their bounding spheres to give you a range of times to check using bisection or what-have-you.
Here's an outline of a closed-form mathematical approach. Each element of this will be easy to express individually, and the final combination of these would be a closed form expression if one could ever write it out:
1) The equation of motion for each point of the tetrahedra is fairly simple in it's own coordinate system. The motion of the center of mass (CM) will just move smoothly along a straight line and the corner points will rotate around an axis through the CM, assumed to be the z-axis here, so the equation for each corner point (parameterized by time, t) is p = vt + x + r(sin(wt+s)i + cos(wt + s)j ), where v is the vector velocity of the center of mass; r is the radius of the projection onto the x-y plane; i, j, and k are the x, y and z unit vectors; and x and s account for the starting position and phase of rotation at t=0.
2) Note that each object has it's own coordinate system to easily represent the motion, but to compare them you'll need to rotate each into a common coordinate system, which may as well be the coordinate system of the screen. (Note though that the different coordinate systems are fixed in space and not traveling with the tetrahedra.) So determine the rotation matrices and apply them to each trajectory (i.e. the points and CM of each of the tetrahedra).
3) Now you have an equation for each trajectory all within the same coordinate system and you need to find the times of the intersections. This can be found by testing whether any of the line segments from the points to the CM of a tetrahedron intersects the any of the triangles of another. This also has a closed-form expression, as can be found here.
Layering these steps will make for terribly ugly equations, but it wouldn't be hard to solve them computationally (although with the rotation of the tetrahedra you need to be sure not to get stuck in a local minimum). Another option might be to plug it into something like Mathematica to do the cranking for you. (Not all problems have easy answers.)
Sorry I'm not a math boff and have no idea what the correct terminology is. Hope my poor terms don't hide my meaning too much.
Pick some arbitrary timestep.
Compute the bounds of each shape in two dimensions perpendicular to the axis it is moving on for the timestep.
For a timestep:
If the shaft of those bounds for any two objects intersect, half timestep and start recurse in.
A kind of binary search of increasingly fine precision to discover the point at which a finite intersection occurs.
Your problem can be cast into a linear programming problem and solved exactly.
First, suppose (p0,p1,p2,p3) are the vertexes at time t0, and (q0,q1,q2,q3) are the vertexes at time t1 for the first tetrahedron, then in 4d space-time, they fill the following 4d closed volume
V = { (r,t) | (r,t) = a0 (p0,t0) + … + a3 (p3,t0) + b0 (q0,t1) + … + b3 (q3,t1) }
Here the a0...a3 and b0…b3 parameters are in the interval [0,1] and sum to 1:
a0+a1+a2+a3+b0+b1+b2+b3=1
The second tetrahedron is similarly a convex polygon (add a ‘ to everything above to define V’ the 4d volume for that moving tetrahedron.
Now the intersection of two convex polygon is a convex polygon. The first time this happens would satisfy the following linear programming problem:
If (p0,p1,p2,p3) moves to (q0,q1,q2,q3)
and (p0’,p1’,p2’,p3’) moves to (q0’,q1’,q2’,q3’)
then the first time of intersection happens at points/times (r,t):
Minimize t0*(a0+a1+a2+a3)+t1*(b0+b1+b2+b3) subject to
0 <= ak <=1, 0<=bk <=1, 0 <= ak’ <=1, 0<=bk’ <=1, k=0..4
a0*(p0,t0) + … + a3*(p3,t0) + b0*(q0,t1) + … + b3*(q3,t1)
= a0’*(p0’,t0) + … + a3’*(p3’,t0) + b0’*(q0’,t1) + … + b3’*(q3’,t1)
The last is actually 4 equations, one for each dimension of (r,t).
This is a total of 20 linear constraints of the 16 values ak,bk,ak', and bk'.
If there is a solution, then
(r,t)= a0*(p0,t0) + … + a3*(p3,t0) + b0*(q0,t1) + … + b3*(q3,t1)
Is a point of first intersection. Otherwise they do not intersect.
Thought about this in the past but lost interest... The best way to go about solving it would be to abstract out one object.
Make a coordinate system where the first tetrahedron is the center (barycentric coords or a skewed system with one point as the origin) and abstract out the rotation by making the other tetrahedron rotate around the center. This should give you parametric equations if you make the rotation times time.
Add the movement of the center of mass towards the first and its spin and you have a set of equations for movement relative to the first (distance).
Solve for t where the distance equals zero.
Obviously with this method the more effects you add (like wind resistance) the messier the equations get buts its still probably the simplest (almost every other collision technique uses this method of abstraction). The biggest problem is if you add any effects that have feedback with no analytical solution the whole equation becomes unsolvable.
Note: If you go the route of of a skewed system watch out for pitfalls with distance. You must be in the right octant! This method favors vectors and quaternions though, while the barycentric coords favors matrices. So pick whichever your system uses most effectively.