Here is my code specific to rotation things, given by pseudo-code.
var rad = (targettransform.rotationEuler.y) * Math.PI / 180;
var difVec = targettransform.pos - othertransform.pos;
//the difVec.z which means radius
this._targetPos.z -= difVec.z * (1 - Math.cos(rad));
this._targetPos.x += difVec.z * Math.sin(rad);
this.transform.position = this._targetPos;
this.transform.lookAt(targettransform.position, Utils.UnitY);
The above pseudo-code is nothing more than just rotating, but I don't understand what 1 - Math.cos(rad) does, it could be some derivation relative to trigonometric. Would someone explain that math in detail.
Edit
Sorry, my mistake, the main purpose of the code snippet is to offset "_targetpos" by vec3(difVec.z * Math.sin(rad),0,this._targetPos.z) on x-z plane. I manage to rewrite the pseudocode to make sense. It seems that the code snippets only implement the function of keeping relative distance and look at target.
1 - cos(Θ)= 2 sin²(Θ/2) and sin(Θ) = 2 sin(Θ/2) cos(Θ/2), so that you are multiplying the scalar δ by 2 sin(Θ/2), and by a vector in the direction -Θ/2, which is added to the target position.
We can see such terms (1-Cos(Fi)) and (Sin(Fi)) in rotation formula around arbitrary axis by Rodrigues
But question does not provide enough details to be sure what that code really does.
Related
I'm making an image using python. But the Lambertian shading does not work.
At first the image saved like this.
enter image description here
But when I reversed the normal vector of sphere, the image saved like this.enter image description here
This is my shading code.
v = -m*ray
if s == 'Sphere':
n = view.viewPoint - list[idx].c - v
n = -n / np.sqrt(np.sum(n*n))
for i in light:
l_i = v + i.position - view.viewPoint
l_i = l_i / np.sqrt(np.sum(l_i * l_i))
x = list[idx].s.d[0] * i.intensity[0] * max(np.dot(l_i, n), 0)
y = list[idx].s.d[1] * i.intensity[1] * max(np.dot(l_i, n), 0)
z = list[idx].s.d[2] * i.intensity[2] * max(np.dot(l_i, n), 0)
list is sphere's list and idx is the number of the closest sphere.
I'd be grateful if anyone could help me. I have been doing this for a week
You have not stated what you think is wrong.
Where is the light in relation to the spheres in the first image? Is it above and slightly behind them? If so - the image looks correct.
Assuming the statements above are correct, then the second image looks correct. The reason the light is on the bottom of the spheres is because the normal is now pointing "in" so the dot() product sign will be opposite to that in the first image.
Note that in your example code, it doesn't look like you have any shadow ray treatment. In other words - all objects will be lit as if all other objects are transparent. No objects will cast shadows on to other objects. This also explains why you can see the bottom of the spheres when the light is coming from the top. If you had proper shadow rays, then it wouldn't actually matter which way the normal is pointing (I would remove the max() functions at that point).
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I'm working with SVR, and using this resource. Erverything is super clear, with epsilon intensive loss function (from figure). Prediction comes with tube, to cover most training sample, and generalize bounds, using support vectors.
Then we have this explanation. This can be described by introducing (non-negative) slack variables , to measure the deviation of training samples outside -insensitive zone. I understand this error, outside tube, but don't know, how we can use this in optimization. Could somebody explain this?
In local source. I'm trying to achieve very simple optimization solution, without libraries. This what I have for loss function.
import numpy as np
# Kernel func, linear by default
def hypothesis(x, weight, k=None):
k = k if k else lambda z : z
k_x = np.vectorize(k)(x)
return np.dot(k_x, np.transpose(weight))
.......
import math
def boundary_loss(x, y, weight, epsilon):
prediction = hypothesis(x, weight)
scatter = np.absolute(
np.transpose(y) - prediction)
bound = lambda z: z \
if z >= epsilon else 0
return np.sum(np.vectorize(bound)(scatter))
First, let's look at the objective function. The first term, 1/2 * w^2 (wish this site had LaTeX support but this will suffice) correlates with the margin of the SVM. The article you linked doesn't, in my opinion, explain this very well and calls this term describing "the model's complexity", but perhaps this is not the best way of explaining it. Minimizing this term maximizes the margin (while still representing the data well), which is the predominant goal of using SVM's doing regression.
Warning, Math Heavy Explanation: The reason this is the case is that when maximizing the margin, you want to find the "farthest" non-outlier points right on the margin and minimize its distance. Let this farthest point be x_n. We want to find its Euclidean distance d from the plane f(w, x) = 0, which I will rewrite as w^T * x + b = 0 (where w^T is just the transpose of the weights matrix so that we can multiply the two). To find the distance, let us first normalize the plane such that |w^T * x_n + b| = epsilon, which we can do WLOG as w is still able to form all possible planes of the form w^T * x + b= 0. Then, let's note that w is perpendicular to the plane. This is obvious if you have dealt a lot with planes (particularly in vector calculus), but can be proven by choosing two points on the plane x_1 and x_2, then noticing that w^T * x_1 + b = 0, and w^T * x_2 + b = 0. Subtracting the two equations we get w^T(x_1 - x_2) = 0. Since x_1 - x_2 is just any vector strictly on the plane, and its dot product with w is 0, then we know that w is perpendicular to the plane. Finally, to actually calculate the distance between x_n and the plane, we take the vector formed by x_n' and some point on the plane x' (The vectors would then be x_n - x', and projecting it onto the vector w. Doing this, we get d = |w * (x_n - x') / |w||, which we can rewrite as d = (1 / |w|) * | w^T * x_n - w^T x'|, and then add and subtract b to the inside to get d = (1 / |w|) * | w^T * x_n + b - w^T * x' - b|. Notice that w^T * x_n + b is epsilon (from our normalization above), and that w^T * x' + b is 0, as this is just a point on our plane. Thus, d = epsilon / |w|. Notice that maximizing this distance subject to our constraint of finding the x_n and having |w^T * x_n + b| = epsilon is a difficult optimization problem. What we can do is restructure this optimization problem as minimizing 1/2 * w^T * w subject to the first two constraints in the picture you attached, that is, |y_i - f(x_i, w)| <= epsilon. You may think that I have forgotten the slack variables, and this is true, but when just focusing on this term and ignoring the second term, we ignore the slack variables for now, I will bring them back later. The reason these two optimizations are equivalent is not obvious, but the underlying reason lies in discrimination boundaries, which you are free to read more about (it's a lot more math that frankly I don't think this answer needs more of). Then, note that minimizing 1/2 * w^T * w is the same as minimizing 1/2 * |w|^2, which is the desired result we were hoping for. End of the Heavy Math
Now, notice that we want to make the margin big, but not so big that includes noisy outliers like the one in the picture you provided.
Thus, we introduce a second term. To motivate the margin down to a reasonable size the slack variables are introduced, (I will call them p and p* because I don't want to type out "psi" every time). These slack variables will ignore everything in the margin, i.e. those are the points that do not harm the objective and the ones that are "correct" in terms of their regression status. However, the points outside the margin are outliers, they do not reflect well on the regression, so we penalize them simply for existing. The slack error function that is given there is relatively easy to understand, it just adds up the slack error of every point (p_i + p*_i) for i = 1,...,N, and then multiplies by a modulating constant C which determines the relative importance of the two terms. A low value of C means that we are okay with having outliers, so the margin will be thinned and more outliers will be produced. A high value of C indicates that we care a lot about not having slack, so the margin will be made bigger to accommodate these outliers at the expense of representing the overall data less well.
A few things to note about p and p*. First, note that they are both always >= 0. The constraint in your picture shows this, but it also intuitively makes sense as slack should always add to the error, so it is positive. Second, notice that if p > 0, then p* = 0 and vice versa as an outlier can only be on one side of the margin. Last, all points inside the margin will have p and p* be 0, since they are fine where they are and thus do not contribute to the loss.
Notice that with the introduction of the slack variables, if you have any outliers then you won't want the condition from the first term, that is, |w^T * x_n + b| = epsilon as the x_n would be this outlier, and your whole model would be screwed up. What we allow for, then, is to change the constraint to be |w^T * x_n + b| = epsilon + (p + p*). When translated to the new optimization's constraint, we get the full constraint from the picture you attached, that is, |y_i - f(x_i, w)| <= epsilon + p + p*. (I combined the two equations into one here, but you could rewrite them as the picture is and that would be the same thing).
Hopefully after covering all this up, the motivation for the objective function and the corresponding slack variables makes sense to you.
If I understand the question correctly, you also want code to calculate this objective/loss function, which I think isn't too bad. I have not tested this (yet), but I think this should be what you want.
# Function for calculating the error/loss for a SVM. I assume that:
# - 'x' is 2d array representing the vectors of the data points
# - 'y' is an array representing the values each vector actually gives
# - 'weights' is an array of weights that we tune for the regression
# - 'epsilon' is a scalar representing the breadth of our margin.
def optimization_objective(x, y, weights, epsilon):
# Calculates first term of objective (note that norm^2 = dot product)
margin_term = np.dot(weight, weight) / 2
# Now calculate second term of objective. First get the sum of slacks.
slack_sum = 0
for i in range(len(x)): # For each observation
# First find the absolute distance between expected and observed.
diff = abs(hypothesis(x[i]) - y[i])
# Now subtract epsilon
diff -= epsilon
# If diff is still more than 0, then it is an 'outlier' and will have slack.
slack = max(0, diff)
# Add it to the slack sum
slack_sum += slack
# Now we have the slack_sum, so then multiply by C (I picked this as 1 aribtrarily)
C = 1
slack_term = C * slack_sum
# Now, simply return the sum of the two terms, and we are done.
return margin_term + slack_term
I got this function working on my computer with small data, and you may have to change it a little to work with your data if, for example, the arrays are structured differently, but the idea is there. Also, I am not the most proficient with python, so this may not be the most efficient implementation, but my intent was to make it understandable.
Now, note that this just calculates the error/loss (whatever you want to call it). To actually minimize it requires going into Lagrangians and intense quadratic programming which is a much more daunting task. There are libraries available for doing this but if you want to do this library free as you are doing with this, I wish you good luck because doing that is not a walk in the park.
Finally, I would like to note that most of this information I got from notes I took in my ML class I took last year, and the professor (Dr. Abu-Mostafa) was a great help to have me learn the material. The lectures for this class are online (by the same prof), and the pertinent ones for this topic are here and here (although in my very biased opinion you should watch all the lectures, they were a great help). Leave a comment/question if you need anything cleared up or if you think I made a mistake somewhere. If you still don't understand, I can try to edit my answer to make more sense. Hope this helps!
I am using Point Cloud Library. I know there is a function to find lines using RANSAC method, but I want to do opposite of that. I have a point cloud, I have an equation of line, now, I would like to find all the points on or near(within given threshold) the line.
Is there any function/s I can use to achieve my goal?
I would really appreciate any kind of help.
I have attempted to use PCL a few times for Kinect processing but it hasn't worked out too well for me. So I attempted to create my own algorithms to do what I want, and for the application, they work much faster than the PCL ones :)
The project I am working on is on GitHub and you can find some code that may help in the bool ConvexHull::addPoint(double newX, double newY, double newZ) found here.
This utilises a 3D plane equation generated using RANSAC and then compares each point to it, calculating the distance between the point and the plane, just like Oscee said.
Here's the juicy bit of the code which I think may help you:
// Find the distance from point to plane.
// http://mathworld.wolfram.com/Point-PlaneDistance.html
dist = newX * plane.a;
dist += newY * plane.b;
dist += newZ * plane.c;
dist += plane.d;
dist /= sqrt(pow(plane.a, 2) + pow(plane.b, 2) + pow(plane.c, 2));
dist = (dist >= 0) ? dist : -dist; // Absolute distance.
if (dist > tolerance) {
return false; // Return false as point is outside of tolerance.
}
With this function I pass in every point from the 640*480 Kinect image that has a depth value greater than 0.
And for me, this works quite fast :)
I hope this helps.
I don't think you need any special function to do that - simply go through all your points, calculate the point-line distance and accept the ones within your threshold and reject/delete the ones outside.
I have a point that moves (in one dimension), and I need it to move smoothly. So I think that it's velocity has to be a continuous function and I need to control the acceleration and then calculate it's velocity and position.
The algorithm doesn't seem something obvious to me, but I guess this must be a common problem, I just can't find the solution.
Notes:
The final destination of the object may change while it's moving and the movement needs to be smooth anyway.
I guess that a naive implementation would produce bouncing, and I need to avoid that.
This is a perfect candidate for using a "critically damped spring".
Conceptually you attach the point to the target point with a spring, or piece of elastic. The spring is damped so that you get no 'bouncing'. You can control how fast the system reacts by changing a constant called the "SpringConstant". This is essentially how strong the piece of elastic is.
Basically you apply two forces to the position, then integrate this over time. The first force is that applied by the spring, Fs = SpringConstant * DistanceToTarget. The second is the damping force, Fd = -CurrentVelocity * 2 * sqrt( SpringConstant ).
The CurrentVelocity forms part of the state of the system, and can be initialised to zero.
In each step, you multiply the sum of these two forces by the time step. This gives you the change of the value of the CurrentVelocity. Multiply this by the time step again and it will give you the displacement.
We add this to the actual position of the point.
In C++ code:
float CriticallyDampedSpring( float a_Target,
float a_Current,
float & a_Velocity,
float a_TimeStep )
{
float currentToTarget = a_Target - a_Current;
float springForce = currentToTarget * SPRING_CONSTANT;
float dampingForce = -a_Velocity * 2 * sqrt( SPRING_CONSTANT );
float force = springForce + dampingForce;
a_Velocity += force * a_TimeStep;
float displacement = a_Velocity * a_TimeStep;
return a_Current + displacement;
}
In systems I was working with a value of around 5 was a good point to start experimenting with the value of the spring constant. Set it too high will result in too fast a reaction, and too low the point will react too slowly.
Note, you might be best to make a class that keeps the velocity state rather than have to pass it into the function over and over.
I hope this is helpful, good luck :)
EDIT: In case it's useful for others, it's easy to apply this to 2 or 3 dimensions. In this case you can just apply the CriticallyDampedSpring independently once for each dimension. Depending on the motion you want you might find it better to work in polar coordinates (for 2D), or spherical coordinates (for 3D).
I'd do something like Alex Deem's answer for trajectory planning, but with limits on force and velocity:
In pseudocode:
xtarget: target position
vtarget: target velocity*
x: object position
v: object velocity
dt: timestep
F = Ki * (xtarget-x) + Kp * (vtarget-v);
F = clipMagnitude(F, Fmax);
v = v + F * dt;
v = clipMagnitude(v, vmax);
x = x + v * dt;
clipMagnitude(y, ymax):
r = magnitude(y) / ymax
if (r <= 1)
return y;
else
return y * (1/r);
where Ki and Kp are tuning constants, Fmax and vmax are maximum force and velocity. This should work for 1-D, 2-D, or 3-D situations (magnitude(y) = abs(y) in 1-D, otherwise use vector magnitude).
It's not quite clear exactly what you're after, but I'm going to assume the following:
There is some maximum acceleration;
You want the object to have stopped moving when it reaches the destination;
Unlike velocity, you do not require acceleration to be continuous.
Let A be the maximum acceleration (by which I mean the acceleration is always between -A and A).
The equation you want is
v_f^2 = v_i^2 + 2 a d
where v_f = 0 is the final velocity, v_i is the initial (current) velocity, and d is the distance to the destination (when you switch from acceleration A to acceleration -A -- that is, from speeding up to slowing down; here I'm assuming d is positive).
Solving:
d = v_i^2 / (2A)
is the distance. (The negatives cancel).
If the current distance remaining is greater than d, speed up as quickly as possible. Otherwise, begin slowing down.
Let's say you update the object's position every t_step seconds. Then:
new_position = old_position + old_velocity * t_step + (1/2)a(t_step)^2
new_velocity = old_velocity + a * t_step.
If the destination is between new_position and old_position (i.e., the object reached its destination in between updates), simply set new_position = destination.
You need an easing formula, which you would call at a set interval, passing in the time elapsed, start point, end point and duration you want the animation to be.
Doing time-based calculations will account for slow clients and other random hiccups. Since it calculates on time elapsed vs. the time in which it has to compkete, it will account for slow intervals between calls when returning how far along your point should be in the animation.
The jquery.easing plugin has a ton of easing functions you can look at:
http://gsgd.co.uk/sandbox/jquery/easing/
I've found it best to pass in 0 and 1 as my start and end point, since it will return a floating point between the two, you can easily apply it to the real value you are modifying using multiplication.
I am working on a geographic project in Java.
The input are coordinates : 24.4444 N etc
Output: a PLAIN map (not round) showing the point of the coordinates.
I don't know the algorithm to transform from coordinates to x,y on a JComponent, can somebody help me?
The map looks like this:
http://upload.wikimedia.org/wikipedia/commons/7/74/Mercator-projection.jpg
Thank you
Given your sparse example, the range of your inputs will be (90.0N - 90.0S) and (180W - 180E). It is easiest - and standard - if you convert South and West to negatives giving you latitudes of (90.0..-90.0) and longitudes of (180.0..-180.0).
Given the size of your canvas - let's say it is 140x120 pixels - you get:
x = (latitude * canvas_height / 180.0) + (canvas_height / 2)
y = (longitude * canvas_width / 360.0) + (canvas_width / 2)
or:
x = (longitude * 120.0 / 180.0) + (120/2)
y = (latitude * 140.0 / 360.0) + (140/2)
where I have ordered the operations to minimize rounding error. This assumes the canvas has point (0,0) in the upper-left or, if not, that you are Australian.
Added: you just threw in the bit about Mercator projections making my simple answer incorrect (but possibly still usable by you if you don't actually care about projection)
MSW provided a good example. Ultimately the algorithm depends on the map projection used. Here are a couple of good resources I used in the past.
The following link is a good reference to a number of different map projections with enough math formulas to choke a horse.
http://www.remotesensing.org/geotiff/proj_list/
Here is a decent reference for doing this in PhP. While not Java, it should be straight forward enough to apply the principles outlined.
http://www.web-max.ca/PHP/article_1.php