The public keys ad the result of scalar multiplication in libsodium is always the U - coordinate of the point of curve25519 (RFC 7748). Is there a method in libsodium that helps to get the V coordinate from the U co-ordinate. Or is there any other library to derive the V co-ordinate?
Edit : While the problem is still unsolved , fortunately, after reworking on my use case , I realised that I do not need to get V co-ordinate for the next primitive.
The curve equation is v^2 = u^3 + 486662u^2 + u.
So, finding v just means solving the equation.
v = sqrt(u^3 + 486662u^2 + u)
Related
I am trying to understand the fisheye model by Scaramuzza, which is implemented in Matlab, see https://de.mathworks.com/help/vision/ug/fisheye-calibration-basics.html#mw_8aca38cc-44de-4a26-a5bc-10fb312ae3c5
The backprojection (uv to xyz) seems fairly straightforward according to the following equation:
, where rho=sqrt(u^2 +v^2)
However, how does the projection (from xyz to uv) work?! In my understanding we get a rather complex set of equations. Unfortunately, I don't find any details on that....
Okay, I believe I understand it now fully after analyzing the functions of the (windows) calibration toolbox by Scaramuzza, see https://sites.google.com/site/scarabotix/ocamcalib-toolbox/ocamcalib-toolbox-download-page
Method 1 found in file "world2cam.m"
For the projection, use the same equation above. In the projection case, the equation has three known (x,y,z) and three unknown variables (u,v and lambda). We first substitute lambda with rho by realizing that
u = x/lambda
v = y/lambda
rho=sqrt(u^2+v^2) = 1/lambda * sqrt(x^2+y^2) --> lambda = sqrt(x^2+y^2) / rho
After that, we have the unknown variables (u,v and rho)
u = x/lambda = x / sqrt(x^2+y^2) * rho
v = y/lambda = y / sqrt(x^2+y^2) * rho
z / lambda = z /sqrt(x^2+y^2) * rho = a0 + a2*rho^2 + a3*rho^3 + a4*rho^4
As you can see, the last equation now has only one unknown, namely rho. Thus, we can solve it easily using e.g. the roots function in matlab. However, the result does not always exist nor is it necessarily unique. After solving the unknown variable rho, calculating uv is very simple using the equation above.
This procedure needs to be performed for each point (x,y,z) separately and is thus rather computationally expensive for an image.
Method 2 found in file "world2cam_fast.m"
The last equation has the form rho(x,y,z). However, if we define m = z / sqrt(x^2+y^2) = tan(90°-theta), it only depends on one variable, namely rho(m).
Instead of solving this equation rho(m) for every new m, the authors "plot" the function for several values of m and fit an 8th order polynomial to these points. Using this polynomial they can calculate an approximate value for rho(m) much quicker in the following.
This becomes clear, because "world2cam_fast.m" makes use of ocam_model.pol, which is calculated in "undistort.m". "undistort.m" in turn makes use of "findinvpoly.m".
The set up of my game relies on placing several sprites along a curve that may look like a bow , ellipse, or be a bit more complex (assume it would be bezier curve then). I want to be able to place my sprites in somewhat equal distances apart.
Can anyone share how this could be done ?
Using cocos2d 2.1
I dont know if there is an easier method to do this ,but we could arrange the sprite along any curve by using the mathematical equation for that curve.
For a parabolic curve (advantage: symmetrical,so easy to place equidistant points)
Find out a equation that satisfies your start and EndPoints and obtain the y values for equidistant x points between start and end using the function below.
example: y = -x^2 + 20x - 1 (general equation : y = ax^2 + bx + c)
static inline parabolicYValue(float x, float a,float b, float c){
return (powf(a*x,2) + b*x + c);
}
You could come up with a similar function for Bezier curves :
(Bezier cubic curve)
static inline CGFloat bezierYValue( float a, float b, float c, float d, float x )
{
return (powf(1-x,3) * a +
3*x*(powf(1-x,2))*b +
3*powf(x,2)*(1-x)*c +
powf(x,3)*d );
}
However, obtaining equidistant points on a bezier curve is a bit of a chore. On the other hand if you by equidistant you mean , distance only along the x axis then this should not be problem.
After some time searching, I have revised my question.
I have found numerous examples of ball to ball collisions, but the only ones that seem to work use Vector2d or Vector2D.
This is a problem, because I am only allowed to use the regular java library, so my main question is: How do I convert the examples (which I will post below) to use what I can use?
I have several variables, both balls have the same mass, the velocities are broken into different variables, x and y. Also I have access to their x and y pos.
This is the ONLY problem left in my application.
I am at a total loss on how to convert the below example.
// get the mtd
Vector2d delta = (position.subtract(ball.position));
float d = delta.getLength();
// minimum translation distance to push balls apart after intersecting
Vector2d mtd = delta.multiply(((getRadius() + ball.getRadius())-d)/d);
// resolve intersection --
// inverse mass quantities
float im1 = 1 / getMass();
float im2 = 1 / ball.getMass();
// push-pull them apart based off their mass
position = position.add(mtd.multiply(im1 / (im1 + im2)));
ball.position = ball.position.subtract(mtd.multiply(im2 / (im1 + im2)));
// impact speed
Vector2d v = (this.velocity.subtract(ball.velocity));
float vn = v.dot(mtd.normalize());
// sphere intersecting but moving away from each other already
if (vn > 0.0f) return;
// collision impulse
float i = (-(1.0f + Constants.restitution) * vn) / (im1 + im2);
Vector2d impulse = mtd.multiply(i);
// change in momentum
this.velocity = this.velocity.add(impulse.multiply(im1));
ball.velocity = ball.velocity.subtract(impulse.multiply(im2));
Here is the URL for the question:
http://stackoverflow.com/questions/345838/ball-to-ball-collision-detection-and-handling
And I have taken a look at his source code.
Thank you for taking the time to read this issue.
SUCCESS!
I have found how to use Vector2d, and it works PERFECTLY!
Will edit later with answer!
I'm implementing my own 3d engine in c# based on a really basic 3d open-source engine in JavaScript called a3. I don't know If I have 100% understand you but It sounds like you can only find examples with Vector2d but you are not allowed to use that class?
I that is the case, as you can imagine javascript does not have native Vector2d types so someone had to implement. Don't be afraid of giving it a try, is just a few high school maths functions, you should be able to implement your own Vector2d class in just a few minutes
The following link contain implementations if vector2d, vector3d, vector4d, matrix3, and matrix4 in javascript: https://github.com/paullewis/a3/tree/master/src/js/core/math hope it helps :)
The best example I can give is located at:
http://www.mathopenref.com/arclength.html
In that Java applet, imagine C is the object to be rotated around and A is the camera. I wish to move the camera to point B, but I do not know how to work out B's co-ordinates. How do you do it? In my case, I know the positions of C and A, and the angle theta to rotate.
I know you can use:
x = Xcentre + radius * sin(theta)
y = Ycentre + radius * cos(theta)
but this fails to take into account the camera current position.
I can't help but feel there's some simple solution I'm missing.
Solved by using the equations listed and just reversing the calculation to derive theta. Then applied a check to ensure 360 degree rotations can be done (else only 180 degrees can).
Firstly, I have done much googling and checking other stackoverflow posts about this, but cannot get a working reply or a snippet of working code. Maths is not my strength.
I need to have a routine that takes a camera point (CX,CY,CZ) and rotate it about a lookat point (LX,LY,LZ) by three rotation angles (RX,RY,RZ). Using euler rotations leads to gimbal lock in some cases which I need to avoid. So I heard about using quaternions.
I found this to convert the rotations into a quaternion
http://www.euclideanspace.com/maths/geometry/rotations/conversions/eulerToQuaternion/index.htm
and this to convert from a quaternion back to euler XYZ rotations
http://www.euclideanspace.com/maths/geometry/rotations/conversions/quaternionToEuler/index.htm
They seem to work fine, but I need to know how to use the quaternion to rotate the CX,CY,CZ around LX,LY,LZ and then return the new CX,CY,CZ without issues of gimbal lock.
There is so much out there about this, that I am sure a good explanation and snippet of code will help not only me but many others in the future.
So please help if you can. Many thanks.
The short answer, if your quaternion is Q and the new camera point is C':
C' = Q*(C-L)*Q^-1 + L
where the points are augmented with Cw=0 and multiplication and inverse are according to quaternion rules.
Specifically, let D = C - L. Then we let F = Q*D:
Fw = Qw*0 - Qx*Dx - Qy*Dy - Qz*Dz
Fx = Qw*Dx + Qx*0 + Qy*Dz - Qz*Dy
Fy = Qw*Dy - Qx*Dz + Qy*0 + Qz*Dx
Fz = Qw*Dz + Qx*Dy - Qy*Dx + Qz*0
Finally, we get C' = F*Q^-1 + L:
Cw' = 0
Cx' = Fw*Qx - Fx*Qw + Fy*Qz - Fz*Qy + Lx
Cy' = Fw*Qy - Fx*Qz - Fy*Qw + Fz*Qx + Ly
Cz' = Fw*Qz + Fx*Qy - Fy*Qx - Fz*Qw + Lz
However, be aware that if you're creating the quaternion from an Euler representation, you'll still end up with gimbal lock. The gimbal lock is a property of the Euler representation and the quaternion will just represent the same transformation. To get rid of gimbal lock, you'll need to avoid the Euler representation altogether, unless I misunderstand how you're using it.