I am learning XNA and in almost all of the educational kits found on http://creators.xna.com/en-US/. I always see a call to Normalize() on a vector. I understand that normalize basically converts the vector into unit length, so all it gives is direction.
Now my question is When to normalize and what exactly does it help me in. I am doing 2D programming so please explain in 2D concepts and not 3D.
EDIT: Here is code in the XNA kit, so why is the Normalize being called?
if (currentKeyboardState.IsKeyDown(Keys.Left)
|| currentGamePadState.DPad.Left == ButtonState.Pressed)
{
catMovement.X -= 1.0f;
}
if (currentKeyboardState.IsKeyDown(Keys.Right)
|| currentGamePadState.DPad.Right == ButtonState.Pressed)
{
catMovement.X += 1.0f;
}
if (currentKeyboardState.IsKeyDown(Keys.Up)
|| currentGamePadState.DPad.Up == ButtonState.Pressed)
{
catMovement.Y -= 1.0f;
}
if (currentKeyboardState.IsKeyDown(Keys.Down)
|| currentGamePadState.DPad.Down == ButtonState.Pressed)
{
catMovement.Y += 1.0f;
}
float smoothStop = 1;
if (catMovement != Vector2.Zero)
{
catMovement.Normalize();
}
catPosition += catMovement * 10 * smoothStop;
In your example, the keyboard presses give you movement in X or Y, or both. In the case of both X and Y, as when you press right and down at the same time, your movement is diagonal. But where movement just in X or Y alone gives you a vector of length 1, the diagonal vector is longer than one. That is, about 1.4 (the square root of 2).
Without normalizing the movement vector, then diagonal movement would be faster than just X or Y movement. With normalizing, the speed is the same in all 8 directions, which I guess is what the game calls for.
One common use case of vector normalization when you need to move something by a number of units in a direction. For example, if you have a game where an entity A moves towards an entity B at a speed of 5 units/second, you'll get the vector from A to B (which is B - A), you'll normalize it so you only keep the direction toward the entity B from A's viewpoint, and then you'll multiply it by 5 units/second. The resulting vector will be the velocity of A and you can then simply multiply it by the elapsed time to get the displacement by which you can move the object.
It depends on what you're using the vector for, but if you're only using the vectors to give a direction then a number of algorithms and formulas are just simpler if your vectors are of unit length.
For example, angles: the angle theta between two unit vectors u and v is given by the formula cos(theta) = u.v (where . is the dot product). For non-unit vectors, you have to compute and divide out the lengths: cos(theta) = (u.v) / (|u| |v|).
A slightly more complicated example: projection of one vector onto another. If v is a unit vector, then the orthogonal projection of u onto v is given by (u.v) v, while if v is a non-unit vector then the formula is (u.v / v.v) v.
In other words: normalize if you know that all you need is the direction and if you're not certain the vector is a unit vector already. It helps you because you're probably going to end up dividing your direction vectors by their length all the time, so you might as well just do it once when you create the vector and get it over with.
EDIT: I assume that the reason Normalize is being called in your example is so that the direction of velocity can be distinguished from the speed. In the final line of the code, 10 * smoothStop is the speed of the object, which is handy to know. And to recover velocity from speed, you need to multiply by a unit direction vector.
You normalize any vector by dividing each component by the magnitude of the vector, which is given by the square root of the sum of squares of components. Each component then has a magnitude that varies between [-1, 1], and the magnitude is equal to one. This is the definition of a unit vector.
Related
I have group of skeleton joints, for example 4 joints of leg. Each joint have (x, y, z) coordinates. So I want to find the combine speed or velocity of these group of joints because i believe I need to calculate velocity of each four joints of leg in order to find out the velocity of leg, please correct me If I'm wrong.
So what would be best possible way to calculate?
I'm using this formula
velocity= Sqrt((x[n]−x[n − 1] )2 + (y[n]−y[n − 1] )2 + (z[n]−z[n − 1] )2)
There is a simpler equation: Average Velocity = (Delta Distance)/(Delta Time)
If you want to know the average velocity at some point in the leg, calculate the location before it moved & the location on impact. Divide the magnitude of the vector of the difference between the 2 points by the amount of time it took to move between those points.
I am looking at this code and do not know what is this magnitude and normalizeddoing and how is this guy using them. In documentation there is only few thing but it doesn't explain much.
Code i am looking at is:
public float minDistance = 1.0f;
public float maxDistance = 4.0f;
public float smooth = 10.0f;
Vector3 dollyDir;
public Vector3 dollyDirAdjusted;
public float distance;
void Awake()
{
dollyDir = transform.localPosition.normalized;
//What has he got with this? Documentation says Returns this vector with
//a magnitude of 1 so if i get it it return's it's length as 1. So what is the point?
distance = transform.localPosition.magnitude;
//Have no idea what he got with this
}
void Update()
{
Vector3 desiredPos = transform.parent.TransformPoint(dollyDir * maxDistance);
//I know what TransfromPoint does but what is he getting with this dollyDir * maxDistance
//Some other code which i understand
}
And while i am here if someone could explain me Mathf.Clamp
Documentation for clamp is so unclear and how i got it is that i give top and bottom value 1 and 5 and if i set value = 3 it will return 3, but if i set value > 5 it will return 5 and if i set value < 1 it will return 1?
Thanks.
Vector3.magnitude returns a float, its a single dimensional value expressing vector length (so it looses directional information)
Vector3.normalized is a bit of an opposite operation - it returns vector direction, guaranteeing that the magnitude of the resulting vector is one, (it preserves the direction information but looses the magnitude information).
The two are often useful for when you want to seperate those two ways of looking at a vector, for example if you need to have an influcence between two bodies where the influence is inversly proportional to the distance between them, you can do
float mag=(targetpos-mypos).magnitude;
if (mag<maxRange)
{
Vector3 dir=(targetpos-mypos).normalized;
Vector3 newVector=dir*(maxRange-mag);
}
this is the most typical use case for me, I am not sure what the author of the original code meant by his as it does seem he could just use a vector difference without using the two extra calls
Mathf.Clamp returns value of value lies between min and max, returns min if its lower and max if is greater.
Another interesting feature of Vector3 is sqrMagnitude which returns a^2+b^2+c^c without computhing the square root. While it adds a bit to the complexity to the code (you need to compare it with squared distance), it saves a relatively expensive root computation; The slightly optimised but a tiny bit harder to read version would look like this
Vectior3 delta=targetpos-mypos;
if (delta.sqrMagnitude<maxRange*maxRange)
{
Vector3 dir=delta.normalized;
Vector3 newVector=dir*(maxRange-delta.magnitude);
}
A vector can be thought of as a direction and distance (although it is normally expressed as an amount in each axis; for example, +3 on x and -4 on y would have a magnitude of 5). A normalized vector has magnitude 1, so it is just the direction (+0.6,-0.8 for the +3,-4 example); the magnitude tells you the original distance part. You can then multiply a normalized vector (the direction) by any magnitude to express "go this far in that direction". In this case, they seem to be moving 4 units (maxDistance) in the same direction as the original vector.
A magnitude is the length of a vector from the origin. It is like using a tape measure from (0,0,0) to wherever your point is. A normalized vector is that same vector, but its length is exactly 1.
An important note is that the normalized vector is calculated is by dividing by its magnitude.
I am attempting to fuse skeletons from two separate Kinects.
From first Kinect I have skeleton1, from it I choose 4 points corresponding with 4 of it's joints. With those 4 points I can construct a plane on which they all are. The plane coefficients Ax + By + Cz = D are known. As I understand the planes Quaternion would be Q = (D; A; B; C)
From the second Kinect I have the same data, but in second Kinects coordinate system.
How can I rotate the plane from second Kinect so that it would have the same orientation as the plane from the first Kinect?
Unfortunately, quaternions don't work that way. They do not represent a plane, and cannot be created just by re-arranging the components of a plane equation (plane equation has two degrees of freedom, while quaternion is a non-parametric representation of rotation/orientation).
However, if you are looking for a rotation between two normal vectors (one for each plane), which will transform one plane to another, it is a relatively simple task:
if the two normal vectors are the same, there is no rotation needed and the result is identity,
if the two vectors are parallel (their dotproduct is zero) but not identical, the transformation between them is a rotation of 180 degrees around an arbitrary axis perpendicular to them,
in all other cases, you can compute the axis of rotation using a crossproduct (and normalization), and the angle using arc cosine of their dotproduct.
If you still need to create a quaternion out of them, you can do it with a bit of trigonometry.
I have read some of the duplicate answers about angle between two vectors, but I'm still stuck with my problem. I have two vectors and I want that the angle between them to always be 90 degrees. To achieve that I need to find the angle between them, so that I can subtract or add the correct amount of degrees so that the angle between them always is 90 degrees.
The picture illustrates a sprite and two vectors. How do I find the angle A between them two? I have tried to use this code to get the angle between two vectors, but I must have missed something out, because I don't get the correct results:
public float GetAngleBetween (Vector2 A, Vector2 B)
{
float DotProd = Vector2.Dot (A, B);
float Length = A.Length () * B.Length ();
return (float)MathHelper.ToDegrees ((float)Math.Acos (DotProd/Length));
}
Any input is welcome and thank you in advance for any answers.
The actual angle in radians is
Math.ACos(Vector2.Dot(a, b));
Make sure that a and b are normalized vectors or the results can get pretty weird.
I think you may be looking for the Vector2.Dot method which is used to calculate the product of two vectors, and can be used for angle calculations.
For example:
// the angle between the two vectors is less than 90 degrees.
Vector2.Dot(vector1.Normalize(), vector2.Normalize()) > 0
// the angle between the two vectors is more than 90 degrees.
Vector2.Dot(vector1.Normalize(), vector2.Normalize()) < 0
// the angle between the two vectors is 90 degrees; that is, the vectors are orthogonal.
Vector2.Dot(vector1.Normalize(), vector2.Normalize()) == 0
// the angle between the two vectors is 0 degrees; that is, the vectors point in the same direction and are parallel.
Vector2.Dot(vector1.Normalize(), vector2.Normalize()) == 1
// the angle between the two vectors is 180 degrees; that is, the vectors point in opposite directions and are parallel.
Vector2.Dot(vector1.Normalize(), vector2.Normalize()) == -1
Is this what you're looking for, or do you need the exact angle?
If I understand your question diagram and comments, the Dot product and Acos are not the only bits of info you need. You also need to account for when the sprite is not located at (0,0).
float angleInRadians = (float) Math.Acos(Vector2.Dot(Vector2.Normalize(vector1 - spritePosition), Vector2.Normalize(vector2 - spritePosition)));
int angleInDegrees = MathHelper.ToDegrees(angleInRadians);
I have to find the axis and angle of rotation of a camera with an UP and Direction vector(They both are perpendicular to each other). I have the initial and final positions of the UP and direction vectors of the camera that is rotated. I want to find the axis and angle of the rotation for the camera. I am using C# for my project. I am new to this 3D rotation. So pardon my questions if you find them silly.
From the direction (forward) vector f and up vector u you can get the side vector s by performing a vector cross product (s = f x u). All three vectors are now orthogonal. You should also make them orthonormal by normalizing each one of them. Taken together, these vectors form an orthonormal basis.
You now have two such basis: the one from your initial camera orientation and the one from your final camera orientation. Both basis can be represented as a rotation matrix. A rotation matrix is simply a 3x3 matrix where the 3 rows are respectively:
The forward vector
The up vector
The side vector
For example, the matrix:
[[1 0 0]
[0 1 0]
[0 0 1]]
could be your initial camera orientation at start-up with its forward vector, up vector and side vector pointing towards the positive x axis, y axis and z axis, respectively.
You can now convert these two basis (M1 and M2) to two unit quaternions (Q1 and Q2) using this algorithm which takes care about potential problems like divides by zero.
At this point, you have two unit quaternions representing your initial and final camera orientation. You must now find the quaternion qT that transforms Q1 into Q2, that is:
q2 = qT * q1
q2 * q1^-1 = qT * (q1 * q1^-1) = qT
=> qT = q2 * q1^-1
Knowing that the inverse of a unit quaternion is equal to its conjugate:
q1^-1 = q1* iif ||q1|| = 1
qT = q2 * q1^-1 = q2 * q1*
There is a single step left: extracting the axis and angle from quaternion qT:
angle = 2 * acos(qw)
x = qx / sqrt(1-qw*qw)
y = qy / sqrt(1-qw*qw)
z = qz / sqrt(1-qw*qw)
The angle is, of course, given in radian. Beware of the divide by zero when calculating x, y and z. This situation would happen when there is no rotation or a very small one, so you should test if angle > epsilon where you would choose epsilon to be quite small an angle (say 1/10 of a degree) and not calculate the vector if that is the case.