I want to make compass with Arduino and QMC5883. Now, the magnetometer outputs me only X Y Z values, and I have to calculate the rest myself. So far, I've used this:
float azimuth = atan2(x, y) * 180.0/PI;
But it's pretty buggy, and vulnerable to tilting in any direction. Is there any better algorythm that - for example - phone manufactures use? I could use accelerometer for help, if it would be needed.
The BBC micro:bit's device abstraction layer (DAL) includes this code to do tilt adjustment based on angles derived from accelerometer data. From https://github.com/lancaster-university/microbit-dal/blob/master/source/drivers/MicroBitCompass.cpp
/**
* Calculates a tilt compensated bearing of the device, using the accelerometer.
*/
int MicroBitCompass::tiltCompensatedBearing()
{
// Precompute the tilt compensation parameters to improve readability.
float phi = accelerometer->getRollRadians();
float theta = accelerometer->getPitchRadians();
// Convert to floating point to reduce rounding errors
Sample3D cs = this->getSample(NORTH_EAST_DOWN);
float x = (float) cs.x;
float y = (float) cs.y;
float z = (float) cs.z;
// Precompute cos and sin of pitch and roll angles to make the calculation a little more efficient.
float sinPhi = sin(phi);
float cosPhi = cos(phi);
float sinTheta = sin(theta);
float cosTheta = cos(theta);
// Calculate the tilt compensated bearing, and convert to degrees.
float bearing = (360*atan2(x*cosTheta + y*sinTheta*sinPhi + z*sinTheta*cosPhi, z*sinPhi - y*cosPhi)) / (2*PI);
// Handle the 90 degree offset caused by the NORTH_EAST_DOWN based calculation.
bearing = 90 - bearing;
// Ensure the calculated bearing is in the 0..359 degree range.
if (bearing < 0)
bearing += 360.0f;
return (int) (bearing);
}
Related
I'm currently working on a game project and need to render a point in front of the current players vision, the game is written in a custom c++ engine. I have the current position (x,y,z) and the current rotation (pitch,yaw,roll). I need to extend the point forward along the known angle at a set distance.
edit:
What I Used As A Solution (Its slightly off but that's ok for me)
Vec3 LocalPos = {0,0,0};
Vec3 CurrentLocalAngle = {0,0,0};
float len = 0.1f;
float pitch = CurrentLocalAngle.x * (M_PI / 180);
float yaw = CurrentLocalAngle.y * (M_PI / 180);
float sp = sinf(pitch);
float cp = cosf(pitch);
float sy = sinf(yaw);
float cy = cosf(yaw);
Vec3 dir = { cp * cy, cp * sy, -sp };
LocalPos = { LocalPos.x + dir.x * len, LocalPos.y + dir.y * len,LocalPos.z + dir.z * len };
You can get the forward vector of the player from matrix column 3 if it is column based, then you multiply its normal by the distance you want then add the result to the player position you will get the point you need.
Convert the angle to a directional vector or just get the "forward vector" from the player if it's available in the engine you're using (it should be the same thing).
Directional vectors are normalized by nature (they have distance = 1), so you can just multiply them by the desired distance to get the desired offset. Multiply this vector by the distance you want the point to be relative to the reference point (the player's camera vector I presume), and then you just add one to the other to get the point in the world where this point belongs.
I need to calculate a distance vector from two GPS coordinates.
The purpose is to calculate the vector of one's change in position,
so the coordinates are not far from each other.
I would like to calculate the latitudinal and longitudinal distances in meters.
I found something here,
but this only gives the direction without distance.
Due to the fact that those coordinates are very close in my case,
I made the approximation that the center of the earth and those two points form a triangle.
Thus, I can use the Al Kashi theorem.
Here is the code:
// Common values
double b = EARTH_RADIUS + destination.altitude;
double c = EARTH_RADIUS + this.altitude;
double b2 = b*b;
double c2 = c*c;
double bc2 = 2*b*c;
// Longitudinal calculations
double alpha = destination.longitude - this.longitude;
// Conversion to radian
alpha = alpha * Math.PI / 180;
// Small-angle approximation
double cos = 1 - alpha*alpha/2; //Math.cos(alpha);
// Use the law of cosines / Al Kashi theorem
double x = Math.sqrt(b2 + c2 - bc2*cos);
// Repeat for latitudinal calculations
alpha = destination.latitude - this.latitude;
alpha = alpha * Math.PI / 180;
double cos = 1 - alpha*alpha/2; //Math.cos(alpha);
double y = Math.sqrt(b2 + c2 - bc2*cos);
// Obtain vertical difference, too
double z = destination.altitude - this.altitude;
return new Vector3D(x, y, z);
As you can see,
I have approximated the cosine because the angles are really small.
I think adding the altitude to the earth's radius doesn't give a better approximation,
but since I have it…
I tested it compared to Google Maps for a distance of 38 meters and I got a result of 37.877.
My result might be more accurate! ^^
I'm using Wikipedia's spherical coordinate system article to create a sphere made out of particles in Three.js. Based on this article, I created a small Polarizer class that takes in polar coordinates with setPolar(rho, theta, phi) and it returns its corresponding x, y, z
Here's the setPolar() function:
// Rho: radius
// theta θ: polar angle on Y axis
// phi φ: azimuthal angle on Z axis
Polarizer.prototype.setPolar = function(rho, theta, phi){
// Limit values to zero
this.rho = Math.max(0, rho);
this.theta = Math.max(0, theta);
this.phi = Math.max(0, phi);
// Calculate x,y,z
this.x = this.rho * Math.sin(this.theta) * Math.sin(this.phi);
this.y = this.rho * Math.cos(this.theta);
this.z = this.rho * Math.sin(this.theta) * Math.cos(this.phi);
return this;
}
I'm using it to position my particles as follows:
var tempPolarizer = new Polarizer();
for(var i = 0; i < geometry.vertices.length; i++){
tempPolarizer.setPolar(
50, // Radius of 50
Math.random() * Math.PI, // Theta ranges from 0 - PI
Math.random() * 2 * Math.PI // Phi ranges from 0 - 2PI
);
// Set new vertex positions
geometry.vertices[i].set(
tempPolarizer.x,
tempPolarizer.y,
tempPolarizer.z
);
}
It works wonderfully, except that I'm getting high particle densities, or "pinching" at the poles:
I'm stumped as to how to avoid this from happening. I thought of passing a weighted random number to the latitude, but I'm hoping to animate the particles without the longitude also slowing down and bunching up at the poles.
Is there a different formula to generate a sphere where the poles don't get as much weight? Should I be using quaternions instead?
For random uniform sampling
use random point in unit cube , handle it as vector and set its length to radius of your sphere. For example something like this in C++:
x = 2.0*Random()-1.0;
y = 2.0*Random()-1.0;
z = 2.0*Random()-1.0;
m=r/sqrt(x*x+y*y+z*z);
x*=m;
y*=m;
z*=m;
where Random return number in <0.0,1.0>. For more info see:
Procedural generation of stars with skybox
For uniform non-random sampling
see related QAs:
Sphere triangulation by mesh subdivision
Make a sphere with equidistant vertices
In order to avoid high density at the poles, I had to lower the likelihood of theta (latitude) landing close to 0 and PI. My input of
Math.random() * Math.PI, for theta gives an equal likelihood to all values (orange).
Math.acos((Math.random() * 2) - 1) perfectly weights the output to make 0 and PI less likely along the sphere's surface (yellow)
Now I can't even tell where the poles are!
I need to calculate the angles to through the ball in that direction for a given speed and the point where it should land after thrown.
The horizontal angle is easy(We know both start and step points).How to calculate the vertical angle of projection.There is gravy applying on object.
Time of travel will be usual as bowling time(time between ball release and occurring step) as per video.
Is there a way directly in unity3d?
Watch this video for 8 seconds for clear understating of this question.
According to the Wikipedia page Trajectory of a projectile, the "Angle of reach" (The angle you want to know) is calculated as follows:
θ = 1/2 * arcsin(gd/v²)
In this formula, g is the gravitational constant 9.81, d is the distance you want the projectile to travel, and v is the velocity at which the object is thrown.
Code to calculate this could look something like this:
float ThrowAngle(Vector3 destination, float velocity)
{
const float g = 9.81f;
float distance = Vector3.Distance(transform.position, destination);
//assuming you want degrees, otherwise just drop the Rad2Deg.
return Mathf.Rad2Deg * (0.5f * Asin((g*distance)/Mathf.Pow(velocity, 2f)));
}
This will give you the angle assuming no air resistance etc. exist in your game.
If your destination and your "throwing point" are not at the same height, you may want to set both to y=0 first, otherwise, errors may occur.
EDIT:
Considering that your launch point is higher up than the destination, this formula from the same page should work:
θ = arctan(v² (+/-) √(v^4-g(gx² + 2yv²))/gx)
Here, x is the range, or distance, and y is the altitude (relative to the launch point).
Code:
float ThrowAngle(Vector3 start, Vector3 destination, float v)
{
const float g = 9.81f;
float xzd = Mathf.Sqrt(Mathf.Pow(destination.x - start.x, 2) + Mathf.Pow(destination.z - start.z, 2));
float yd = destination.y - start.y;
//assuming you want degrees, otherwise just drop the Rad2Deg. Split into two lines for better readability.
float sqrt = (Mathf.Pow(v,4) - g * (g*Mathf.Pow(xzd,2) + 2*yd*Mathf.Pow(v,2))/g*xzd);
//you could also implement a solution which uses both values in some way, but I left that out for simplicity.
return Mathf.Atan(Mathf.Pow(v, 2) + sqrt);
}
I have two vectors in a game. One vector is the player, one vector is an object. I also have a vector that specifies the direction the player if facing. The direction vector has no z part. It is a point that has a magnitude of 1 placed somewhere around the origin.
I want to calculate the angle between the direction the soldier is currently facing and the object, so I can correctly pan some audio (stereo only).
The diagram below describes my problem. I want to calculate the angle between the two dashed lines. One dashed line connects the player and the object, and the other is a line representing the direction the player is facing from the point the player is at.
At the moment, I am doing this (assume player, object and direction are all vectors with 3 points, x, y and z):
Vector3d v1 = direction;
Vector3d v2 = object - player;
v1.normalise();
v2.normalise();
float angle = acos(dotProduct(v1, v2));
But it seems to give me incorrect results. Any advice?
Test of code:
Vector3d soldier = Vector3d(1.f, 1.f, 0.f);
Vector3d object = Vector3d(1.f, -1.f, 0.f);
Vector3d dir = Vector3d(1.f, 0.f, 0.f);
Vector3d v1 = dir;
Vector3d v2 = object - soldier;
long steps = 360;
for (long step = 0; step < steps; step++) {
float rad = (float)step * (M_PI / 180.f);
v1.x = cosf(rad);
v1.y = sinf(rad);
v1.normalise();
float dx = dotProduct(v2, v1);
float dy = dotProduct(v2, soldier);
float vangle = atan2(dx, dy);
}
You shoud always use atan2 when computing angular deltas, and then normalize.
The reason is that for example acos is a function with domain -1...1; even normalizing if the input absolute value (because of approximations) gets bigger than 1 the function will fail even if it's clear that in such a case you would have liked an angle of 0 or PI instead. Also acos cannot measure the full range -PI..PI and you'd need to use explicitly sign tests to find the correct quadrant.
Instead atan2 only singularity is at (0, 0) (where of course it doesn't make sense to compute an angle) and its codomain is the full circle -PI...PI.
Here is an example in C++
// Absolute angle 1
double a1 = atan2(object.y - player.y, object.x - player.x);
// Absolute angle 2
double a2 = atan2(direction.y, direction.x);
// Relative angle
double rel_angle = a1 - a2;
// Normalize to -PI .. +PI
rel_angle -= floor((rel_angle + PI)/(2*PI)) * (2*PI) - PI;
In the case of a general 3d orientation you need two orthogonal directions, e.g. the vector of where the nose is pointing to and the vector to where your right ear is.
In that case the formulas are just slightly more complex, but simpler if you have the dot product handy:
// I'm assuming that '*' is defined as the dot product
// between two vectors: x1*x2 + y1*y2 + z1*z2
double dx = (object - player) * nose_direction;
double dy = (object - player) * right_ear_direction;
double angle = atan2(dx, dy); // Already in -PI ... PI range
In 3D space, you also need to compute the axis:
Vector3d axis = normalise(crossProduct(normalise(v1), normalise(v2)));