I am using a WinSock connection to get the accelerometer info off and iPhone and into a Direct3D application. I have modified Apples GLGravity's sample code to get my helicopter moving in relation to gravity, however I need to "cap" the movement so the helicopter can't fly upside down! I have tried to limit the output of the accelerometer like so
if (y < -0.38f) {
y = -0.38f;
}
Except this doesn't seem to work!? The only thing I can think of is I need to modify the custom matrix, but I can't seem to get my head around what I need to be changing. The matrix is code is below.
_x = acceleration.x;
_y = acceleration.y;
_z = acceleration.z;
float length;
D3DXMATRIX matrix, t;
memset(matrix, '\0', sizeof(matrix));
D3DXMatrixIdentity(&matrix);
// Make sure acceleration value is big enough.
length = sqrtf(_x * _x + _y * _y + _z * _z);
if (length >= 0.1f && kInFlight == TRUE) { // We have a acceleration value good enough to work with.
matrix._44 = 1.0f; //
// First matrix column is a gravity vector.
matrix._11 = _x / length;
matrix._12 = _y / length;
matrix._13 = _z / length;
// Second matrix is arbitrary vector in the plane perpendicular to the gravity vector {Gx, Gy, Gz}.
// defined by the equation Gx * x + Gy * y + Gz * z = 0 in which we set x = 0 and y = 1.
matrix._21 = 0.0f;
matrix._22 = 1.0f;
matrix._23 = -_y / _z;
length = sqrtf(matrix._21 * matrix._21 + matrix._22 * matrix._22 + matrix._23 * matrix._23);
matrix._21 /= length;
matrix._22 /= length;
matrix._23 /= length;
// Set third matrix column as a cross product of the first two.
matrix._31 = matrix._12 * matrix._23 - matrix._13 * matrix._22;
matrix._32 = matrix._21 * matrix._13 - matrix._23 * matrix._11;
matrix._33 = matrix._11 * matrix._22 - matrix._12 * matrix._21;
}
If anyone can help it would be much appreciated!
I think double integration is probably over-complicating things. If I understand the problem correctly, the iPhone is giving you a vector of values from the accelerometers. Assuming the user isn't waving it around, that vector will be of roughly constant length, and pointing directly downwards with gravity.
There is one major problem with this, and that is that you can't tell when the user rotates the phone around the horizontal. Imagine you lie your phone on the table, with the bottom facing you as you're sitting in front of it; the gravity vector would be (0, -1, 0). Now rotate your phone around 90 degrees so the bottom is facing off to your left, but is still flat on the table. The gravity vector is still going to be (0, -1, 0). But you'd really want your helicopter to have turned with the phone. It's a basic limitation of the fact that the iPhone only has a 2D accelerometer, and it's extrapolating a 3D gravity vector from that.
So let's assume that you've told the user they're not allowed to rotate their phone like that, and they have to keep it with the bottom point to you. That's fine, you can still get a lot of control from that.
Next, you need to cap the input such that the helicopter never goes more than 90 degrees over on it's side. Imagine the vector that you're given as being a stick attached to your phone, and dangling with gravity. The vector you have is describing the direction of gravity, relative to the phone's flat surface. If it were (0, -1, 0) the stick is pointing directly downwards (-y). if it were (1, 0, 0), the stick is pointing to the right of the phone (+x), and implies that the phone has been twisted 90 degrees clockwise (looking away from you at the phone).
Assume in this metaphor that the stick has full rotational freedom. It can be pointing in any direction from the phone. So moving the stick around describes the surface of a sphere. But crucially, you only want the stick to be able to move around the lower half of that sphere. If the user twists the phone so that the stick would be in the upper half of the sphere, you want it to cap such that it's pointing somewhere around the equator of the sphere.
You can achieve this quite cleanly by using polar co-ordinates. 3D vectors and polar co-ordinates are interchangeable - you can convert to and from without losing any information.
Convert the vector you have (normalised of course) into a set of 3D polar co-ordinates (you should be able to find this logic on the web quite easily). This will give you an angle around the horizontal plane, and an angle for vertical plane (and a distance from the origin - for a normalised vector, this should be 1.0). If the vertical angle is positive, the vector is in the upper half of the sphere, negative it's in the lower half. Then, cap the vertical angle so that it is always zero or less (and so in the lower half of the sphere). Then you can take the horizontal and capped vertical angle, and convert it back into a vector.
This new vector, if plugged into the matrix code you already have, will give you the correct orientation, limited to the range of motion you need. It will also be stable if the user turns their phone slightly beyond the 90 degree mark - this logic will keep your directional vector as close to the user's current orientation as possible, without going beyond the limit you set.
Try normalizing the acceleration vector first. (edit: after you check the length) (edit edit: I guess I need to learn how to read... how do I delete my answer?)
So if I understand this correctly, the iPhone is feeding you accelerometer data, saying how hard you're moving the iPhone in 3 axes.
I'm not familiar with that apple sample, so I don't know what its doing. However, it sounds like you're mapping acceleration directly to orientation, but I think what you want to do is doubly integrate the acceleration in order to obtain a position and look at changes in position in order to orient the helicopter. Basically, this is more of a physics problem than a Direct3D problem.
It looks like you are using the acceleration vector from the phone to define one axis of a orthogonal frame of reference. And I suppose +Y is points towards the ground so you are concerned about the case when the vector points towards the sky.
Consider the case when the iphone reports {0, -6.0, 0}. You will change this vector to {0, -.38, 0}. But they both normalize to {0, -1.0, 0}. So, the effect of clamping y at -.38 is influenced by magnitude of the other two components of the vector.
What you really want is to limit the angle of the vector to the XZ plane when Y is negative.
Say you want to limit it to be no more than 30 degrees to the XZ plane when Y is negative. First normalize the vector then:
const float limitAngle = 30.f * PI/180.f; // angle in radians
const float sinLimitAngle = sinf(limitAngle);
const float XZLimitLength = sqrtf(1-sinLimitAngle*sinLimitAngle);
if (_y < -sinLimitAngle)
{
_y = -sinLimitAngle;
float XZlengthScale = XZLimitLength / sqrtf(_x*_x + _z*_z);
_x *= XZlengthScale;
_z *= XZlengthScale;
}
Related
So I am writing a game in C++, currently I am working on a 'Compass', but I am having some problems with the vector math..
Here is a little image I created to possibly help explain my question better
Ok, so as you can see the 2D position of A begins at (4, 4), but then I want to move A along the 45 degree angle until the 2D position reaches (16, 16), so basically there is a 12 distance between where A starts and where it ends. And my qustion is how would I calculate this?
the simplest way in 2D is to take angle 'ang', and distance 'd', and your starting point 'x' and 'y':
x1 = x + cos(ang) * distance;
y1 = y + sin(ang) * distance;
In 2D the rotation for any object can be just stored as a single value, ang.
using cos for x and sin for y is the "standard" way that almost everyone does it. cos(ang) and sin(ang) trace a circle out as ang increases. ang = 0 points right along the x-axis here, and as angle increases it spins counter-clockwise (i.e at 90 degrees it's pointing straight up). If you swap the cos and sin terms for x and y, you get ang = 0 pointing up along the y axis and clockwise rotation with increasing ang (since it's a mirror image), which could in fact be more convenient for making game, since y-axis is often the "forward" direction and you might like that increasing ang spins to the right.
x1 = x + sin(ang) * distance;
y1 = y + cos(ang) * distance;
Later you can get into vectors and matricies that do the same thing but in a more flexible manner, but cos/sin are fine to get started with in a 2D game. In a 3D game, using cos and sin for rotations starts to break down in certain circumstances, and you start really benefiting from learning the matrix-based approaches.
The distance between (4,4) and (16,16) isn't actually 12. Using pythagorean theorem, the distance is actually sqrt(12^2 + 12^2) which is 16.97. To get points along the line you want to use sine and cosine. E.g. If you want to calculate the point halfway along the line the x coordinate would be cos(45)(16.97/2) and the y would be sin(45)(16.97/2). This will work with other angles besides 45 degrees.
I am programming a fairly simple 3D minigolf game using OpenGl. I've run across a problem of slowing a minigolf ball that is rolling on a surface. The ball is described by velocity and position vectors.
When it moves on a flat surface, it must be slowed, but only in XZ plane (Y axis points up). The Y component must not be slowed, as only the ground applies friction - that of air versus ball is negligible, thus when the ball bounces, Y component comes into play. In order to change it, I add a gravity vector.
I'm looking for a way of decreasing speed proportionally in two axes. I tried decreasing exponentially/linearly both X and Z components, but that results in a false behavior - when moving along only one of these axes, ball slows at a lower rate than when moving in direction, say, 45 degrees, wherein both axes contribute to the velocity.
You need to find the magnitude of the 2D velocity vector in the XZ plane as a scalar quantity, then apply your friction function to that and then convert back to a 2D vector.
In C-like pseudo code:
// the current velocity in each direction
float velocityX, velocityY, velocityZ;
// compute magnitude using Pythagorean Theorem:
float magnitude = sqrt((velocityX * velocityX) + (velocityZ * velocityZ));
if(magnitude > 0.0) // is it moving?
{
// apply friction to magnitude using whatever function you want
// e.g. newMagnitude = magnitude * 0.95f;
float newMagnitude = applyFriction(magnitude);
float scaleFactor = newMagnitude / magnitude;
// compute the new velocity in each direction:
velocityX *= scaleFactor;
velocityZ *= scaleFactor;
}
Introducing:
I'm developing a little Tower defense game in opengl, currently I'm just despairing of a little problem....
I want the projectiles from the tower to aim with the head facing the unit. So my problem is more a mathmatical one but it belongs to opengl :)
I had the following idea; I could use a dot product to get an angle rotating around the x axis to get the head depending on the distance just straight down or flat to the ground and after that an additional angle to rotate around the y axis that the head of the arrow is everytime adjusted to the unit it's aiming on.
My code for the angle of rotation around the X axis (i called it m_fYNeigung because the height(Y) of the head changes by rotating around the x axis) looks like this:
plocalTowerArray[(sizeMapIndexY * 12) + sizeMapIndexX].Projektils[byteProjectilIndex].m_fYNeigung =
RADIANS_TO_DEGREES (acos ((float)
(
(faTowerPosition[0]) * (plocalTowerArray[(sizeMapIndexY * 12) + sizeMapIndexX].Projektils[byteProjectilIndex].m_faProDirectionVector[0]) +
(faTowerPosition[1] - 1) * (plocalTowerArray[(sizeMapIndexY * 12) + sizeMapIndexX].Projektils[byteProjectilIndex].m_faProDirectionVector[1]) +
(faTowerPosition[2]) * (plocalTowerArray[(sizeMapIndexY * 12) + sizeMapIndexX].Projektils[byteProjectilIndex].m_faProDirectionVector[2])
)
/
(
fabs (faTowerPosition[0]) * fabs (plocalTowerArray[(sizeMapIndexY * 12) + sizeMapIndexX].Projektils[byteProjectilIndex].m_faProDirectionVector[0]) +
fabs (faTowerPosition[1] - 1) * fabs (plocalTowerArray[(sizeMapIndexY * 12) + sizeMapIndexX].Projektils[byteProjectilIndex].m_faProDirectionVector[1]) +
fabs (faTowerPosition[2]) * fabs (plocalTowerArray[(sizeMapIndexY * 12) + sizeMapIndexX].Projektils[byteProjectilIndex].m_faProDirectionVector[2])
)
));
where faTowerPosition is the first vector, which is pointing down from the top of the tower (the arrow also starts at faTowerPosition[X/Y/Z]) the second vector for the dot product is m_faProDirectionVector which is a normalized direction vector describing the route of the arrow from the tower to the unit.
The Opengl Drawing part looks just as simple as this:
for (sizeJ = 0; sizeJ < localTowerArray[sizeI].m_byteProjectilAmount; sizeJ++)
{
if (localTowerArray[sizeI].Projektils[sizeJ].m_bOnFlight == true)
{
glPushMatrix();
glTranslatef (localTowerArray[sizeI].Projektils[sizeJ].m_faProPosition[0], localTowerArray[sizeI].Projektils[sizeJ].m_faProPosition[1], localTowerArray[sizeI].Projektils[sizeJ].m_faProPosition[2]);
//glRotatef (360.0f - localTowerArray[sizeI].Projektils[sizeJ].m_fXNeigung, 0, 1, 0);
glRotatef (localTowerArray[sizeI].Projektils[sizeJ].m_fYNeigung, 1, 0, 0);
DrawWaveFrontObject (m_pArrowProjektilObject);
glPopMatrix();
}
}
Just ignore the calculations I'm doing to the angle, I just did it to experiment with the acting of the arrows, i just noticed that it appears as would the arrow act different depending on the (i gotta say: the buildable map is scaled by x: -3.4 to 3.4 and z from 4 to -4) cords the tower was builded on -x/z,-z/x,z/x,-z/-x all these cases i guess are different and at least depending on the unit is running left or right side of the tower, the acting is also different.... so what i forgot to remind by using the dot product in this way?
First at all, your code is very difficult to understand, so I'm guessing a lot to try to answer you. If I assume something wrong, my apologize for it.
I am assuming that you want to use the euler angle rotation to align correctly your projectiles. So, first you will do a X rotation and after that, a Y rotation.
To do a X rotation, your vectors, for the dot product, must be on an YZ plane and assuming that your projectile start at Z direction, your first vector is (0, 0, 1). The second vector, as you said, is a vector pointing to unit and could be expressed by (px, py, pz). You must project this vector to the plane YZ to get the second vector for your dot product, so this vector will be (0, py, pz)
Now, to calculate the dot product you apply the following formule
x1.x2+y1.y2+z1.z2 = |p1|.|p2|.cos a, where |p1| and |p2| is the module of vector (its length)
In this example, the first vector is unitary, but the second not. So |p2| = sqrt(py^2 +pz^2). Thereafter:
acos(a) = pz/sqrt(py^2 + pz^2)
This will give you the angle around X axis. Do the same calculation to achieve Y angle rotation
PS. After I wrote this answer, I noted that you use the function "fabs". I guess you want to find the module of you second vector, but fabs give you the absolute value of a escalar. To calculate a module of a vector (its length) you need to use the above formulae as cited.
I hope this is an acceptable question for this forum, as it's more of a maths question than a programming question.
I am developing a basic space shooter game using the cocos2d framework, where the user controls a spaceship and shoots asteroids/enemies, and the enemies also shoot back at the spaceship (original eh?).
The 'lasers' that are fired by the enemies are a tube like shape, and generally only travel form right to left on the x axis. This is easy to achieve, and the movement is created with this line of code:
currentEnemyProjectile.position = ccp(currentEnemyProjectile.position.x - ((screenSize.width/kLaserSpeed)* dtSixty), currentEnemyProjectile.position.y);
I also have one particular enemy, which resides in the middle of the screen and rotates to fire its projectiles at the spaceship, so if the space ship is at a different y position than the enemy then the projectile sprite will be rotated and then will move to the spaceship's position, so it will need to also move through the y axis.
I'm currently using code to achieve this:
float xDiff = ((screenSize.width/kLaserSpeed)* dtSixty);
float yDiff = (xDiff / 57) * currentEnemyProjectile.rotation;
currentEnemyProjectile.position = ccp(currentEnemyProjectile.position.x - xDiff, currentEnemyProjectile.position.y + yDiff);
I just worked this out as a temporary fix, I know it is not the optimal way of doing things. It sort of works, but as I am working out the y offset from the movement along the x axis, the projectiles will move faster if they have further to travel along the y axis. Ideally they should move slower along the x axis if they are moving further upwards or downwards, so that the rate of travel is constant for any trajectory. (I hope I've explained that well enough)
It's been roughly 10 years since I last had a maths lesson, so my geometry/trigonometry is rather hazy/non-existent, can anyone shed any light on how this can be done?
Also, it's worth noting that I would like to do this without the use of cocos2d actions. I would rather move my sprites each frame as I have been currently doing as I manipulating the flow of time as part of the game, so I need more control than the cocos2d actions can offer.
Thanks in advance for any help.
To find a position using a starting position, an angle, and a distance, the formula is as follows:
ccp( x + ( distance * sin( angle ) ), y + ( distance * cos( angle) ) )
Keep in mind that the angle for this formula will have to be in radians, not degrees. You can convert degrees into radians like this:
radians = degrees * PI / 180
Try this. I realized since you know the vector to the player, you don't need the trig. You only need to calculate v and h once so you may want to save those values and not calculate them on every frame.
// Vector from enemy to player
CGPoint v = enemy.position - player.position;
// distance from enemy to player (sqrt of the dot product of v with v)
float h = sqrt(v.x*v.x+v.y*v.y);
// Desired distance to move along v
delta = ((screenSize.width/kLaserSpeed)* dtSixty);
// delta/h is the % moved along v, find the % moved in the x and y direction
xDiff = v.x * delta/h;
yDiff = v.y * delta/h;
I have a function in my program which rotates a point (x_p, y_p, z_p) around another point (x_m, y_m, z_m) by the angles w_nx and w_ny.
The new coordinates are stored in global variables x_n, y_n, and z_n. Rotation around the y-axis (so changing value of w_nx - so that the y - values are not harmed) is working correctly, but as soon as I do a rotation around the x- or z- axis (changing the value of w_ny) the coordinates aren't accurate any more. I commented on the line I think my fault is in, but I can't figure out what's wrong with that code.
void rotate(float x_m, float y_m, float z_m, float x_p, float y_p, float z_p, float w_nx ,float w_ny)
{
float z_b = z_p - z_m;
float x_b = x_p - x_m;
float y_b = y_p - y_m;
float length_ = sqrt((z_b*z_b)+(x_b*x_b)+(y_b*y_b));
float w_bx = asin(z_b/sqrt((x_b*x_b)+(z_b*z_b))) + w_nx;
float w_by = asin(x_b/sqrt((x_b*x_b)+(y_b*y_b))) + w_ny; //<- there must be that fault
x_n = cos(w_bx)*sin(w_by)*length_+x_m;
z_n = sin(w_bx)*sin(w_by)*length_+z_m;
y_n = cos(w_by)*length_+y_m;
}
What the code almost does:
compute difference vector
convert vector into spherical coordinates
add w_nx and wn_y to the inclination and azimuth angle (see link for terminology)
convert modified spherical coordinates back into Cartesian coordinates
There are two problems:
the conversion is not correct, the computation you do is for two inclination vectors (one along the x axis, the other along the y axis)
even if computation were correct, transformation in spherical coordinates is not the same as rotating around two axis
Therefore in this case using matrix and vector math will help:
b = p - m
b = RotationMatrixAroundX(wn_x) * b
b = RotationMatrixAroundY(wn_y) * b
n = m + b
basic rotation matrices.
Try to use vector math. Decide in which order you rotate, first along x, then along y perhaps.
If you rotate along z-axis, [z' = z]
x' = x*cos a - y*sin a;
y' = x*sin a + y*cos a;
The same repeated for y-axis: [y'' = y']
x'' = x'*cos b - z' * sin b;
z'' = x'*sin b + z' * cos b;
Again rotating along x-axis: [x''' = x'']
y''' = y'' * cos c - z'' * sin c
z''' = y'' * sin c + z'' * cos c
And finally the question of rotating around some specific "point":
First, subtract the point from the coordinates, then apply the rotations and finally add the point back to the result.
The problem, as far as I see, is a close relative to "gimbal lock". The angle w_ny can't be measured relative to the fixed xyz -coordinate system, but to the coordinate system that is rotated by applying the angle w_nx.
As kakTuZ observed, your code converts point to spherical coordinates. There's nothing inherently wrong with that -- with longitude and latitude, one can reach all the places on Earth. And if one doesn't care about tilting the Earth's equatorial plane relative to its trajectory around the Sun, it's ok with me.
The result of not rotating the next reference axis along the first w_ny is that two points that are 1 km a part of each other at the equator, move closer to each other at the poles and at the latitude of 90 degrees, they touch. Even though the apparent purpose is to keep them 1 km apart where ever they are rotated.
if you want to transform coordinate systems rather than only points you need 3 angles. But you are right - for transforming points 2 angles are enough. For details ask Wikipedia ...
But when you work with opengl you really should use opengl functions like glRotatef. These functions will be calculated on the GPU - not on the CPU as your function. The doc is here.
Like many others have said, you should use glRotatef to rotate it for rendering. For collision handling, you can obtain its world-space position by multiplying its position vector by the OpenGL ModelView matrix on top of the stack at the point of its rendering. Obtain that matrix with glGetFloatv, and then multiply it with either your own vector-matrix multiplication function, or use one of the many ones you can obtain easily online.
But, that would be a pain! Instead, look into using the GL feedback buffer. This buffer will simply store the points where the primitive would have been drawn instead of actually drawing the primitive, and then you can access them from there.
This is a good starting point.