For a new project I like to calculate the moon phases. So far I haven't seen any code that does that. I don't want to rely on online-services for this.
I have tried some functions, but they are not 100% reliable. Functions I have tried:
NSInteger r = iYear % 100;
r %= 19;
if (r>9){ r -= 19;}
r = ((r * 11) % 30) + iMonth + iDay;
if (iMonth<3){r += 2;}
r -= ((iYear<2000) ? 4 : 8.3);
r = floor(r+0.5);
other one:
float n = floor(12.37 * (iYear -1900 + ((1.0 * iMonth - 0.5)/12.0)));
float RAD = 3.14159265/180.0;
float t = n / 1236.85;
float t2 = t * t;
float as = 359.2242 + 29.105356 * n;
float am = 306.0253 + 385.816918 * n + 0.010730 * t2;
float xtra = 0.75933 + 1.53058868 * n + ((1.178e-4) - (1.55e-7) * t) * t2;
xtra = xtra + (0.1734 - 3.93e-4 * t) * sin(RAD * as) - 0.4068 * sin(RAD * am);
float i = (xtra > 0.0 ? floor(xtra) : ceil(xtra - 1.0));
float j1 = [self julday:iYear iMonth:iMonth iDay:iDay];
float jd = (2415020 + 28 * n) + i;
jd = fmodf((j1-jd + 30), 30);
and last one
NSInteger thisJD = [self julday:iYear iMonth:iMonth iDay:iDay];
float degToRad = 3.14159265 / 180;
float K0, T, T2, T3, J0, F0, M0, M1, B1, oldJ = 0.0;
K0 = floor((iYear-1900)*12.3685);
T = (iYear-1899.5) / 100;
T2 = T*T; T3 = T*T*T;
J0 = 2415020 + 29*K0;
F0 = 0.0001178*T2 - 0.000000155*T3 + (0.75933 + 0.53058868*K0) - (0.000837*T + 0.000335*T2);
M0 = 360*[self getFrac:((K0*0.08084821133)) + 359.2242 - 0.0000333*T2 - 0.00000347*T3];
M1 = 360*[self getFrac:((K0*0.07171366128)) + 306.0253 + 0.0107306*T2 + 0.00001236*T3];
B1 = 360*[self getFrac:((K0*0.08519585128)) + 21.2964 - (0.0016528*T2) - (0.00000239*T3)];
NSInteger phase = 0;
NSInteger jday = 0;
while (jday < thisJD) {
float F = F0 + 1.530588*phase;
float M5 = (M0 + phase*29.10535608)*degToRad;
float M6 = (M1 + phase*385.81691806)*degToRad;
float B6 = (B1 + phase*390.67050646)*degToRad;
F -= 0.4068*sin(M6) + (0.1734 - 0.000393*T)*sin(M5);
F += 0.0161*sin(2*M6) + 0.0104*sin(2*B6);
F -= 0.0074*sin(M5 - M6) - 0.0051*sin(M5 + M6);
F += 0.0021*sin(2*M5) + 0.0010*sin(2*B6-M6);
F += 0.5 / 1440;
oldJ=jday;
jday = J0 + 28*phase + floor(F);
phase++;
}
float jd = fmodf((thisJD-oldJ), 30);
All are working more and less, but none is really giving the correct dates of full moon for 2017 and 2018.
Does anyone have a function that will calculate the moon phases correctly - also based on time zone?
EDIT:
I only want the function for the Moonphases. SwiftAA offers a lot more and only produces not needed overhead in the app.
Related
Im trying to understand what this function does. It was given by my teacher and I just cant understands, whats logic behind the formulas finding x, and y coordinates. From my math class I know I my formulas for finding interception but its confusing translated in code. So I have some problems how they defined the formulas for a,b,c and for finding the coordinates x and y.
void Intersection::getIntersectionPoints(const Arc& arc, const Line& line) {
double a, b, c, mu, det;
std::pair<double, double> xPoints;
std::pair<double, double> yPoints;
std::pair<double, double> zPoints;
//(m2+1)x2+2(mc−mq−p)x+(q2−r2+p2−2cq+c2)=0.
//a= m2;
//b= 2 * (mc - mq - p);
//c= q2−r2+p2−2cq+c2
a = pow((line.end().x - line.start().x), 2) + pow((line.end().y - line.start().y), 2) + pow((line.end().z - line.start().z), 2);
b = 2 * ((line.end().x - line.start().x)*(line.start().x - arc.center().x)
+ (line.end().y - line.start().y)*(line.start().y - arc.center().y)
+ (line.end().z - line.start().z)*(line.start().z - arc.center().z));
c = pow((arc.center().x), 2) + pow((arc.center().y), 2) +
pow((arc.center().z), 2) + pow((line.start().x), 2) +
pow((line.start().y), 2) + pow((line.start().z), 2) -
2 * (arc.center().x * line.start().x + arc.center().y * line.start().y +
arc.center().z * line.start().z) - pow((arc.radius()), 2);
det = pow(b, 2) - 4 * a * c;
/* Tangenta na kružnicu */
if (Math<double>::isEqual(det, 0.0, 0.00001)) {
if (!Math<double>::isEqual(a, 0.0, 0.00001))
mu = -b / (2 * a);
else
mu = 0.0;
// x = h + t * ( p − h )
xPoints.second = xPoints.first = line.start().x + mu * (line.end().x - line.start().x);
yPoints.second = yPoints.first = line.start().y + mu * (line.end().y - line.start().y);
zPoints.second = zPoints.first = line.start().z + mu * (line.end().z - line.start().z);
}
if (Math<double>::isGreater(det, 0.0, 0.00001)) {
// first intersection
mu = (-b - sqrt(pow(b, 2) - 4 * a * c)) / (2 * a);
xPoints.first = line.start().x + mu * (line.end().x - line.start().x);
yPoints.first = line.start().y + mu * (line.end().y - line.start().y);
zPoints.first = line.start().z + mu * (line.end().z - line.start().z);
// second intersection
mu = (-b + sqrt(pow(b, 2) - 4 * a * c)) / (2 * a);
xPoints.second = line.start().x + mu * (line.end().x - line.start().x);
yPoints.second = line.start().y + mu * (line.end().y - line.start().y);
zPoints.second = line.start().z + mu * (line.end().z - line.start().z);
}
Denoting the line's start point as A, end point as B, circle's center as C, circle's radius as r and the intersection point as P, then we can write P as
P=(1-t)*A + t*B = A+t*(B-A) (1)
Point P will also locate on the circle, therefore
|P-C|^2 = r^2 (2)
Plugging equation (1) into equation (2), you will get
|B-A|^2*t^2 + 2(B-A)\dot(A-C)*t +(|A-C|^2 - r^2) = 0 (3)
This is how you get the formula for a, b and c in the program you posted. After solving for t, you shall obtain the intersection point(s) from equation (1). Since equation (3) is quadratic, you might get 0, 1 or 2 values for t, which correspond to the geometric configurations where the line might not intersect the circle, be exactly tangent to the circle or pass thru the circle at two locations.
I am writing a simple monte carlo code for simulation of electron scattering. I ran the Kernel for 10 million electron and it runs fine, but when I increase the number of electrons to a higher number, say 50 million, the code just wouldn't finish and the computer freezes. I wanted to know if this is a hardware issue or if there is a possible bug in the code. I am running the code on a iMac with ATI Radeon HD 5870.
int rand_r (unsigned int seed)
{
unsigned int next = seed;
int result;
next *= 1103515245;
next += 12345;
result = (unsigned int) (next / 65536) % 2048;
next *= 1103515245;
next += 12345;
result <<= 10;
result ^= (unsigned int) (next / 65536) % 1024;
next *= 1103515245;
next += 12345;
result <<= 10;
result ^= (unsigned int) (next / 65536) % 1024;
seed = next;
return result;
}
__kernel void MC(const float E, __global float* bse, const int count) {
int tx, ty;
tx = get_global_id(0);
ty = get_global_id(1);
float RAND_MAX = 2147483647.0f;
int rand_seed;
int seed = count*ty + tx;
float rand;
float PI;
PI = 3.14159f;
float z;
z = 28.0f;
float rho;
rho = 8.908f;
float A;
A = 58.69f;
int num;
num = 10000000/(count*count);
int counter, counter1, counter2;
counter = 0;
float4 c_new, r_new;
float E_new, alpha, de_ds, phi, psi, mfp,sig_eNA,step, dsq, dsqi, absc0z;
float J;
J = (9.76f*z + 58.5f*powr(z,-0.19f))*1E-3f;
float4 r0 = (float4)(0.0f, 0.0f, 0.0f, 0.0f);
float2 tilt = (float2)((70.0f/180.0f)*PI , 0.0f);
float4 c0 = (float4)(cos(tilt.y)*sin(tilt.x), sin(tilt.y)*sin(tilt.x), cos(tilt.x), 0.0f);
for (int i = 0; i < num; ++i){
rand_seed = rand_r(seed);
seed = rand_seed;
rand = rand_seed/RAND_MAX; //some random no. generator in gpu
r0 = (float4)(0.0f, 0.0f, 0.0f, 0.0f);
c0 = (float4)(cos(tilt.y)*sin(tilt.x), sin(tilt.y)*sin(tilt.x), cos(tilt.x), 0.0f);
E_new = E;
c_new = c0;
alpha = (3.4E-3f)*powr(z,0.67f)/E_new;
sig_eNA = (5.21f * 602.3f)*((z*z)/(E_new*E_new))*((4.0f*PI)/(alpha*(1+alpha)))*((E_new + 511.0f)*(E_new + 511.0f)/((E_new + 1024.0f)*(E_new + 1024.0f)));
mfp = A/(rho*sig_eNA);
step = -mfp * log(rand);
r_new = (float4)(r0.x + step*c_new.x, r0.y + step*c_new.y, r0.z + step*c_new.z, 0.0f);
r0 = r_new;
counter1 = 0;
counter2 = 0;
while (counter1 < 1000){
alpha = (3.4E-3f)*powr(z,0.67f)/E_new;
sig_eNA = (5.21f * 602.3f)*((z*z)/(E_new*E_new))*((4*PI)/(alpha*(1+alpha)))*((E_new + 511.0f)*(E_new + 511.0f)/((E_new + 1024.0f)*(E_new + 1024.0f)));
mfp = A/(rho*sig_eNA);
rand_seed = rand_r(seed);
seed = rand_seed;
rand = rand_seed/RAND_MAX; //some random no. generator in gpu
step = -mfp * log(rand);
de_ds = -78500.0f*(z/(A*E_new)) * log((1.66f*(E_new + 0.85f*J))/J);
rand_seed = rand_r(seed);
seed = rand_seed;
rand = rand_seed/RAND_MAX; //new random no.
phi = acos(1 - ((2*alpha*rand)/(1 + alpha - rand)));
rand_seed = rand_r(seed);
seed = rand_seed;
rand = rand_seed/RAND_MAX; //third random no.
psi = 2*PI*rand;
if ((c0.z >= 0.999f) || (c0.z <= -0.999f) ){
absc0z = abs(c0.z);
c_new = (float4)(sin(phi) * cos(psi), sin(phi) * sin(psi), (c0.z/absc0z)*cos(phi), 0.0f);
}
else {
dsq = sqrt(1-c0.z*c0.z);
dsqi = 1/dsq;
c_new = (float4)(sin(phi)*(c0.x*c0.z*cos(psi) - c0.y*sin(psi))*dsqi + c0.x*cos(phi), sin(phi) * (c0.y * c0.z * cos(psi) + c0.x * sin(psi)) * dsqi + c0.y * cos(phi), -sin(phi) * cos(psi) * dsq + c0.z * cos(phi), 0.0f);
}
r_new = (float4)(r0.x + step*c_new.x, r0.y + step*c_new.y, r0.z + step*c_new.z, 0.0f);
r0 = r_new;
c0 = c_new;
E_new += step*rho*de_ds;
if (r0.z <= 0 && counter2 == 0){
counter++ ;
counter2 = 1;
}
counter1++ ;
}
}
bse[count*ty + tx] = counter;
}
I am working on Maya-like camera implementation, and I've done track and dolly functions correctly but I just cannot implement tumble.
I am working in PhiloGL engine (WebGL base), so I would really appreciate some help with code in this engine.
I've looked at how Maya's camera actually work, but I cannot find out. Here is my code so-far
if(mode == "rot")
{
var angleX = diffx / 150;
var angleY = diffy / 150;
//var angleZ = sign * Math.sqrt((diffx * diffx)+(diffy * diffy)) / 150;
e.stop();
//axe Z
//camera.position.x = x * Math.cos(angleX) - y * Math.sin(angleX);
//camera.position.y = x * Math.sin(angleX) + y * Math.cos(angleX);
//axe X
//camera.position.y = y * Math.cos(angleY) - z * Math.sin(angleY);
//camera.position.z = y * Math.sin(angleY) + z * Math.cos(angleY);
//camera.update();
//axe Y
camera.position.z = z * Math.cos(angleX) - x * Math.sin(angleX);
camera.position.x = z * Math.sin(angleX) + x * Math.cos(angleX);
camera.update();
position.x = e.x;
position.y = e.y;
position.z = e.z;
}
This isn't working nor do I know what am I doing wrong.
Any clues?
I use this in inka3d (www.inka3d.com) but it does not depend on inka3d. The output is a 4x4 matrix. Can you make use of that?
// turntable like camera, y is up-vector
// tx, ty and tz are camera target position
// rx, ry and rz are camera rotation angles (rad)
// di is camera distance from target
// fr is an array where the resulting view matrix is written into (16 values, row major)
control.cameraY = function(tx, ty, tz, rx, ry, rz, di, fr)
{
var a = rx * 0.5;
var b = ry * 0.5;
var c = rz * 0.5;
var d = Math.cos(a);
var e = Math.sin(a);
var f = Math.cos(b);
var g = Math.sin(b);
var h = Math.cos(c);
var i = Math.sin(c);
var j = f * e * h + g * d * i;
var k = f * -e * i + g * d * h;
var l = f * d * i - g * e * h;
var m = f * d * h - g * -e * i;
var n = j * j;
var o = k * k;
var p = l * l;
var q = m * m;
var r = j * k;
var s = k * l;
var t = j * l;
var u = m * j;
var v = m * k;
var w = m * l;
var x = q + n - o - p;
var y = (r + w) * 2.0;
var z = (t - v) * 2.0;
var A = (r - w) * 2.0;
var B = q - n + o - p;
var C = (s + u) * 2.0;
var D = (t + v) * 2.0;
var E = (s - u) * 2.0;
var F = q - n - o + p;
var G = di;
var H = -(tx + D * G);
var I = -(ty + E * G);
var J = -(tz + F * G);
fr[0] = x;
fr[1] = A;
fr[2] = D;
fr[3] = 0.0;
fr[4] = y;
fr[5] = B;
fr[6] = E;
fr[7] = 0.0;
fr[8] = z;
fr[9] = C;
fr[10] = F;
fr[11] = 0.0;
fr[12] = x * H + y * I + z * J;
fr[13] = A * H + B * I + C * J;
fr[14] = D * H + E * I + F * J;
fr[15] = 1.0;
};
I'm trying to derive a CATransform3D that will map a quad with 4 corner points to another quad with 4 new corner points. I've spent a little bit of time researching this and it seems the steps involve converting the original Quad to a Square, and then converting that Square to the new Quad. My methods look like this (code borrowed from here):
- (CATransform3D)quadFromSquare_x0:(float)x0 y0:(float)y0 x1:(float)x1 y1:(float)y1 x2:(float)x2 y2:(float)y2 x3:(float)x3 y3:(float)y3 {
float dx1 = x1 - x2, dy1 = y1 - y2;
float dx2 = x3 - x2, dy2 = y3 - y2;
float sx = x0 - x1 + x2 - x3;
float sy = y0 - y1 + y2 - y3;
float g = (sx * dy2 - dx2 * sy) / (dx1 * dy2 - dx2 * dy1);
float h = (dx1 * sy - sx * dy1) / (dx1 * dy2 - dx2 * dy1);
float a = x1 - x0 + g * x1;
float b = x3 - x0 + h * x3;
float c = x0;
float d = y1 - y0 + g * y1;
float e = y3 - y0 + h * y3;
float f = y0;
CATransform3D mat;
mat.m11 = a;
mat.m12 = b;
mat.m13 = 0;
mat.m14 = c;
mat.m21 = d;
mat.m22 = e;
mat.m23 = 0;
mat.m24 = f;
mat.m31 = 0;
mat.m32 = 0;
mat.m33 = 1;
mat.m34 = 0;
mat.m41 = g;
mat.m42 = h;
mat.m43 = 0;
mat.m44 = 1;
return mat;
}
- (CATransform3D)squareFromQuad_x0:(float)x0 y0:(float)y0 x1:(float)x1 y1:(float)y1 x2:(float)x2 y2:(float)y2 x3:(float)x3 y3:(float)y3 {
CATransform3D mat = [self quadFromSquare_x0:x0 y0:y0 x1:x1 y1:y1 x2:x2 y2:y2 x3:x3 y3:y3];
// invert through adjoint
float a = mat.m11, d = mat.m21, /* ignore */ g = mat.m41;
float b = mat.m12, e = mat.m22, /* 3rd col*/ h = mat.m42;
/* ignore 3rd row */
float c = mat.m14, f = mat.m24;
float A = e - f * h;
float B = c * h - b;
float C = b * f - c * e;
float D = f * g - d;
float E = a - c * g;
float F = c * d - a * f;
float G = d * h - e * g;
float H = b * g - a * h;
float I = a * e - b * d;
// Probably unnecessary since 'I' is also scaled by the determinant,
// and 'I' scales the homogeneous coordinate, which, in turn,
// scales the X,Y coordinates.
// Determinant = a * (e - f * h) + b * (f * g - d) + c * (d * h - e * g);
float idet = 1.0f / (a * A + b * D + c * G);
mat.m11 = A * idet; mat.m21 = D * idet; mat.m31 = 0; mat.m41 = G * idet;
mat.m12 = B * idet; mat.m22 = E * idet; mat.m32 = 0; mat.m42 = H * idet;
mat.m13 = 0 ; mat.m23 = 0 ; mat.m33 = 1; mat.m43 = 0 ;
mat.m14 = C * idet; mat.m24 = F * idet; mat.m34 = 0; mat.m44 = I * idet;
return mat;
}
After calculating both matrices, multiplying them together, and assigning to the view in question, I end up with a transformed view, but it is wildly incorrect. In fact, it seems to be sheared like a parallelogram no matter what I do. What am I missing?
UPDATE 2/1/12
It seems the reason I'm running into issues may be that I need to accommodate for FOV and focal length into the model view matrix (which is the only matrix I can alter directly in Quartz.) I'm not having any luck finding documentation online on how to calculate the proper matrix, though.
I was able to achieve this by porting and combining the quad warping and homography code from these two URLs:
http://forum.openframeworks.cc/index.php/topic,509.30.html
http://forum.openframeworks.cc/index.php?topic=3121.15
UPDATE: I've open sourced a small class that does this: https://github.com/dominikhofmann/DHWarpView
I have a space ship that I want to turn to a destination angle. Currently it works like 90% of the time, but sometimes, it 'jumps' to the destination angle rather than moving smoothly. Here is my code:
a = System.Math.Sin(.destStoppingAngle + System.Math.PI)
b = System.Math.Cos(.destStoppingAngle + System.Math.PI)
c = System.Math.Sin(.msngFacing)
d = System.Math.Cos(.msngFacing)
det = a * d - b * c
If det > 0 Then
.msngFacing = .msngFacing - .ROTATION_RATE * TV.TimeElapsed
If det < 0.1 Then
.msngFacing = .destStoppingAngle
.turning = False
End If
Else
.msngFacing = .msngFacing + .ROTATION_RATE * TV.TimeElapsed
If det > 0.1 Then
.msngFacing = .destStoppingAngle
.turning = False
End If
End If
I would do it like this. First you need a function to lerp an angle (C code, port it yourself):
float lerpangle(float from, float to, float frac) {
float a;
if ( to - from > 180 ) {
to -= 360;
}
if ( to - from < -180 ) {
to += 360;
}
a = from + frac * (to - from);
return a;
}
Then, when starting the rotation you have the duration and stoppingangle as your own parameters. Get the startingangle from your object and startingtime (in something decently precise, milliseconds) and save them. The rotation then goes like this:
current_rotation = lerpangle(startingangle, stoppingangle,
(time.now - startingtime) / duration)