So the formula is basically:
xd = x2-x1
yd = y2-y1
Distance = sqrt(xd * xd + yd * yd)
But surely the formula has to be different depending on whether something is above, below, left, or right of the other object?
Like, if I have a sprite in the middle of the screen, and an enemy somewhere below, would that require changing the "x2-x1" (Let's just say the player sprite is x1, enemy is x2) the other way around if the enemy was above instead?
Distance in the sense you describe above will always be a positive value. The sum of the square of real numbers will always be positive, and the square root of a positive number will also always be positive. So, it doesn't matter whether you define xd = x2-x1 or xd = x1-x2. They only differ by their sign and so both have the same absolute value which means they both square to the same value.
So, there aren't really any special cases here. The formulation of the distance measure accommodates all of the concerns you raise.
Math.Sqrt(Math.Pow (a.X-b.X, 2) + Math.Pow (a.Y-b.Y, 2));
Try this. It should work!
yes, you are very right. In my case, I have to calculate distance between two points in 2D. I put x1 for swarm,x2 for intruder along X-Axis and y1 for intruder and y2 for swarm along Y-Axis.
d=sqrt((swarm(de,1) - (intruderX)).^2 + (swarm(de,2)-intruderY).^2);
[Distance is not calculated accurately, I want when intruder comes inside the circle of any swarm particle, it must be detected][1], some times, intruder comes inside the circle but not be detected. This is my problem. Anyone, who solve my problem, will be very grateful to them.
for de = 1:Ndrones
d = sqrt((swarm(de,1) - (intruderX)).^2 + (swarm(de,2)-intruderY).^2);
if(d<=rad) % intruder has been detected
x = intruderX;
y = intruderY;
title('Intruder Detected');
text(x,y+5,sprintf('Intruder'));
text(500,900,sprintf('Iterations: %.2f',iter));
plot(swarm(:,1),swarm(:,2));
for i=1:Ndrones
swarm(:, 9) = 100; %restart the minimum calculation
end
return;
end
end % end of de loop
[1]: http://i.stack.imgur.com/SBP27.png
Related
I've been trying to get the correct normals for a sphere I'm messing with using a vertex shader. The algorithm can be boiled down simply to
vert.xyz += max(0, sin(time + 0.004*vert.x))*10*normal.xyz
This causes a wave to roll across the sphere.
In order to make my normals correct, I need to transform them as well. I can take the tangent vector at a given x,y,z, get a perpendicular vector (0, -vert.z, vert.y), and then cross the tangent with the perp vector.
I've been having some issue with the math though, and it's become a personal vendetta at this point. I've solved for the derivative hundreds of times but I keep getting it incorrect. How can I get the tangent?
Breaking down the above line, I can make a math function
f(x,y,z) = max(0, sin(time + 0.004*x))*10*Norm(x,y,z) + (x,y,z)
where Norm(..) is Normalize((x,y,z) - CenterOfSphere)
After applying f(x,y,z), unchanged normals
What is the correct f '(x,y,z)?
I've accounted for the weirdness caused by the max in f(...), so that's not the issue.
Edit: The most successful algorithm I have right now is as follows:
Tangent vector.x = 0.004*10*cos(0.004*vert.x + time)*norm.x + 10*sin(0.004*vert.x + time) + 1
Tangent vector.y = 10*sin(0.004*vert.x + time) + 1
Tangent vector.z = 10*sin(0.004*vert.x + time) + 1
2nd Tangent vector.x = 0
2nd Tangent vector.y = -norm.z
2nd Tangent vector.z = norm.y
Normalize both, and perform Cross(Tangent2, Tangent1). Normalize again, and done (it should be Cross(Tangent1, Tangent2), but this seems to have better results... more hints of an issue in my math!).
This yields this
Get tangent/normal by derivate of function can sometimes fail if your surface points are nonlinearly distributed and or some math singularity is present or if you make a math mistake (which is the case in 99.99%). Anyway you can always use the geometric approach:
1. you can get the tangents easy by
U(x,y,z)=f(x+d,y,z)-f(x,y,z);
V(x,y,z)=f(x,y+d,z)-f(x,y,z);
where d is some small enough step
and f(x,y,z) is you current surface point computation
not sure why you use 3 input variables I would use just 2
but therefore if the shifted point is the same as unshifted
use this instead =f(x,y,z+d)-f(x,y,z);
at the end do not forget to normalize U,V size to unit vector
2. next step
if bullet 1 leads to correct normals
then you can simply solve the U,V algebraically
so rewrite U(x,y,z)=f(x+d,y,z)-f(x,y,z); to full equation
by substituting f(x,y,z) with the surface point equation
and simplify
[notes]
sometimes well selected d can simplify normalization to multipliyng by a constant
you should add normals visualization for example like this:
to actually see what is really happening (for debug purposses)
I'm learning Unity3d + some basic maths I've forgotten by messing around.
Heres what I'm doing now..
As you can probably tell the sides of this shape form a parabola.
The distance they are out from the centre is the base radius + the height squared * by a constant (0.05 in this image)
The code generating this is very simple..
for (int changer = 1; changer > -2; changer-=2) {
Vector3 newPos = new Vector3(
transform.position.x
,transform.position.y + currentheight*changer
,transform.position.z - RadiusAtZero -(Mathf.Pow(currentheight,2)*CurveMultiplier)
);
var newFleck = Instantiate(Fleck, newPos, Quaternion.identity)as GameObject;
newFleck.transform.RotateAround(transform.position,Vector3.up,angle*changer);
FleckList.Add(newFleck );
}
Btw the for loop and 'changer' mirror everything so 'currentheight' is really just the distance from the centreline of the parabola.
Anyway I'd like to make the cubes (or flecks as I've called them) be angled so that they are tangentional to the parabola I have made.
I need to determine the angle of a tangent to the parabola at particular point.
I found this
to find the line tangent to y=x^2 -3 at (1, -2) we can simultaneously solve
y=x^2 -3 and y+2=m(x-1) and set the discriminant equal to zero
But I dont know how to implement this. Also I reckon my 'CurveMultiplier' constant makes my parabola equation different from that one.
Can someone write some code that determines the angle? (and also maybe explain it)
Update.
Here is fixed version using the derivative of the equation. (Also I have changed from boxes to tetrahedrons and few other superficial things)
The easiest solution is to use a derivative for the parabolic equation.
In your picture then I'll assume Y is vertical, X horizontal, and Z in/out of the screen. Then the parabola being rotated, based upon your description, is:
f(h) = 0.05*h^2 + R
(h is height, R is base radius). If you imagine a plane containing the Y axis, you can rotate the plane around the Y axis at any angle and the dual parabola looks the same.
The derivative of a parabolic equation of the form f(x) = C*h^2 + R is f'(x) = 2*C*h, which is the slope of the tangent at h. In this specific case, that would be:
f'(h) = 0.1*h
Since the cross-sectional plane has an angle relative to X and Z axes, then that tangent will also have the same angular component (you have a rotated parabola).
Depending upon the units given for the constants in f(h), particularly the 0.05 value, you may have to adjust this for the correct results.
Ok, I know this sounds really daft to be asking here, but it is programming related.
I'm working on a game, and I'm thinking of implementing a system that allows users to triangulate their 3D coordinates to locate something (eg for a task).
I also want to be able to let the user make the coordinates of the points they are using for triangulation have user-determined coordinates (so the location's coordinate is relative, probably by setting up a beacon or something).
I have a method in place for calculating the distance between the points, so essentially I can calculate the lengths of the sides of the triangle/pyramid as well as all but the coordinate I am after.
It has been a long time since I have done any trigonometry and I am rusty with the sin, cos and tan functions, I have a feeling they are required but have no clue how to implement them.
Can anyone give me a demonstration as to how I would go about doing this in a mathematical/programatical way?
extra info:
My function returns the exact distance between the two points, so say you set two points to 0,0,0 and 4,4,0 respectively, and those points are set to scale(the game world is divided into a very large 3d grid, with each 'block' area being represented by a 3d coordinate) then it would give back a value at around 5.6.
The key point about it varying is that the user can set the points, so say they set a point to read 0,0,0, the actual location could be something like 52, 85, 93. However, providing they then count the blocks and set their other points correctly (eg, set a point 4,4,0 at the real point 56, 89, 93) then the final result will return the relative position (eg the object they are trying to locate is at real point 152, 185, 93, it will return the relative value 100,100,0). I need to be able to calculate it knowing every point but the one it's trying to locate, as well as the distances between all points.
Also, please don't ask why I can't just calculate it by using the real coordinates, I'm hoping to show the equation up on screen as it calculates the result.7
Example:
Here is a diagram
Imagine these are points in my game on a flat plain.
I want to know the point f.
I know the values of points d and e, and the sides A,B and C.
Using only the data I know, I need to find out how to do this.
Answered Edit:
After many days of working on this, Sean Kenny has provided me with his time, patience and intellect, and thus I have now got a working implementation of a triangulation method.
I hope to place the different language equivalents of the code as I test them so that future coders may use this code and not have the same problem I have had.
I spent a bit of time working on a solution but I think the implementer, i.e you, should know what it's doing, so any errors encountered can be tackled later on. As such, I'll give my answer in the form of strong hints.
First off, we have a vector from d to e which we can work out: if we consider the coordinates as position vectors rather than absolute coordinates, how can we determine what the vector pointing from d to e is? Think about how you would determine the displacement you had moved if you only knew where you started and where you ended up? Displacement is a straight line, point A to B, no deviation, not: I had to walk around that house so I walked further. A straight line. If you started at the point (0,0) it would be easy.
Secondly, the cosine rule. Do you know what it is? If not, read up on it. How can we rearrange the form given in the link to find the angle d between vectors DE and DF? Remember you need the angle, not a function of the angle (cos is a function remember).
Next we can use a vector 'trick' called the scalar product. Notice there is a cos function in there. Now, you may be thinking, we've just found the angle, why are we doing it again?
Define DQ = [1,0]. DQ is a vector of length 1, a unit vector, along the x-axis. Which other vector do we know? Do we know of two position vectors?
Once we have two vectors (I hope you worked out the other one) we can use the scalar product to find the angle; again, just the angle, not a function of it.
Now, hopefully, we have 2 angles. Could we take one from the other to get yet another angle to our desired coordinate DF? The choice of using a unit vector earlier was not arbitrary.
The scalar product, after some cancelling, gives us this : cos(theta) = x / r
Where x is the x ordinate for F and r is the length of side A.
The end result being:
theta = arccos( xe / B ) - arccos( ( (A^2) + (B^2) - (C^2) ) / ( 2*A*B ) )
Where theta is the angle formed between a unit vector along the line y = 0 where the origin is at point d.
With this information we can find the x and y coordinates of point f relative to d. How?
Again, with the scalar product. The rest is fairly easy, so I'll give it to you.
x = r.cos(theta)
y = r.sin(theta)
From basic trigonometry.
I wouldn't advise trying to code this into one value.
Instead, try this:
//pseudo code
dx = 0
dy = 0 //initialise coordinates somehow
ex = ex
ey = ey
A = A
B = B
C = C
cosd = ex / B
cosfi = ((A^2) + (B^2) - (C^2)) / ( 2*A*B)
d = acos(cosd) //acos is a method in java.math
fi = acos(cosfi) //you will have to find an equivalent in your chosen language
//look for a method of inverse cos
theta = fi - d
x = A cos(theta)
y = A sin(theta)
Initialise all variables as those which can take decimals. e.g float or double in Java.
The green along the x-axis represents the x ordinate of f, and the purple the y ordinate.
The blue angle is the one we are trying to find because, hopefully you can see, we can then use simple trig to work out x and y, given that we know the length of the hypotenuse.
This yellow line up to 1 is the unit vector for which scalar products are taken, this runs along the x-axis.
We need to find the black and red angles so we can deduce the blue angle by simple subtraction.
Hope this helps. Extensions can be made to 3D, all the vector functions work basically the same for 3D.
If you have the displacements from an origin, regardless of whether this is another user defined coordinate or not, the coordinate for that 3D point are simply (x, y, z).
If you are defining these lengths from a point, which also has a coordinate to take into account, you can simply write (x, y, z) + (x1, y1, z1) = (x2, y2, z2) where x2, y2 and z2 are the displacements from the (0, 0, 0) origin.
If you wish to find the length of this vector, i.e if you defined the line from A to B to be the x axis, what would the x displacement be, you can use Pythagoras for 3D vectors, it works just the same as with 2D:
Length l = sqrt((x^2) + (y^2) + (z^2))
EDIT:
Say you have a user defined point A (x1, y1, z1) and you want to define this as the origin (0,0,0). You have another user chosen point B (x2, y2, z2) and you know the distance from A to B in the x, y and z plane. If you want to work out what this point is, in relation to the new origin, you can simply do
B relative to A = (x2, y2, z2) - (x1, y1, z1) = (x2-x1, y2-y1, z2-z1) = C
C is the vector A>B, a vector is a quantity which has a magnitude (the length of the lines) and a direction (the angle from A which points to B).
If you want to work out the position of B relative to the origin O, you can do the opposite:
B relative to O = (x2, y2, z2) + (x1, y1, z1) = (x1+x2, y1+y2, z1+z2) = D
D is the vector O>B.
Edit 2:
//pseudo code
userx = x;
usery = y;
userz = z;
//move origin
for (every block i){
xi = xi-x;
yi = yi - y;
zi = zi -z;
}
I am having readings of Yaw, pitch, Roll, Rotation matrix, Quaternion and Acceleration. These reading are taken with frequency of 20 (per second). They are collected from the mobile device which is moving in 3D space from one position to other position.
I am setting reference plane by multiplying inverse matrix to the start postion. The rest of the readings are taken by considering the first one as a reference one. Now I want to convert these readings to 3D cartesian system.
How to convert it? Can anyone please help?
Ok basically yaw, pitch and roll are the euler angles, with them you got your rotation matrix already.
Quaternions are aquivalent to that, with them you can also calculate the rotation matrix you need.
If you have rotation matrices R_i for every moment i in your l=20secs interval. Than these rotations are relative the the one applied at R_(i-1) you can calculate their rotation relative to the first position. So A_i = R_1*...*R_i but after all you could also just safe the new direction of travel (safes calculations).
So asuming that the direction of travel is d_0 = (1,0,0) at first. You can calculate the next by d_i = R_i*d_(i-1) (always norm d_(i-1) because it might get smaller or bigger due to error). The first position is p and your start speed is v_0 = (0,0,0) and finally the acceleration is a_i. You need to calculate the vectorial speed v_i for every moment:
v_i = v_(i-1) + l*a_i*A_i*d_0 = v_(i-1) + l*a_i*d_i
Now you basically know where you are moving, and what kind of speed you use, so your position p_i at the moment i is given by:
`p_i = p_0 + l * ( v_1 + v_2 + ... + v_i)`
For the units:
a_i = [m/s^2]^3
v_i = [m/s]^3
p_i = [m]^3
Precision
Now some points to the precision of your position calculation (just if you want to know how good it will work). Suppose you have an error e>= ||R_i*v-w|| (where w is the correct vector). in the data you calculate the rotation matrices with. Your error is multipling itself so your error in the i moment is e_i <= e^i.
Then because you apply l and a_i to it, it becomes:
f_i <= l*a_i*e^i
But you are also adding up the error when you add up the speed, so now its g_i <= f_1+...+f_i. And yeah you also add up for the position (both sums over i):
h_i <= g_1+...+g_i = ΣΣ (l*a_i*e^i) = l* ΣΣ (a_i*e^i)
So this is basically the maximum difference from your position p_i to the correct position w (||p_i - w|| <= h_i).
This is still not taking in account that you don't get the correct acceleration from your device (I don't know how they normally do this), because correct would be:
a_i = ||∫a_i(t) dt|| (where a_i(t) is vectorial now)
And you would need to calculate the difference in direction (your rotation matrix) as:
Δd_i = (∫a_i(t) dt)/a_i (again a_i(t) is vectorial)
So apart from the errors you get from the error in your rotations from your device (and from floating point arithmetic), you have an error in your acceleration, I won't calculate that now but you would substitute a_i = a_i + b_i.
So I'm pretty sure it will be far off from the real position. You even have to take in account that you're speed might be non zero when it should be!
But that beeing said, I would really like to know the precision you get after implementing it, that's what always keept me from trying it.
Although the context of this question is about making a 2d/3d game, the problem i have boils down to some math.
Although its a 2.5D world, lets pretend its just 2d for this question.
// xa: x-accent, the x coordinate of the projection
// mapP: a coordinate on a map which need to be projected
// _Dist_ values are constants for the projection, choosing them correctly will result in i.e. an isometric projection
xa = mapP.x * xDistX + mapP.y * xDistY;
ya = mapP.x * yDistX + mapP.y * yDistY;
xDistX and yDistX determine the angle of the x-axis, and xDistY and yDistY determine the angle of the y-axis on the projection (and also the size of the grid, but lets assume this is 1-pixel for simplicity).
x-axis-angle = atan(yDistX/xDistX)
y-axis-angle = atan(yDistY/yDistY)
a "normal" coordinate system like this
--------------- x
|
|
|
|
|
y
has values like this:
xDistX = 1;
yDistX = 0;
xDistY = 0;
YDistY = 1;
So every step in x direction will result on the projection to 1 pixel to the right end 0 pixels down. Every step in the y direction of the projection will result in 0 steps to the right and 1 pixel down.
When choosing the correct xDistX, yDistX, xDistY, yDistY, you can project any trimetric or dimetric system (which is why i chose this).
So far so good, when this is drawn everything turns out okay. If "my system" and mindset are clear, lets move on to perspective.
I wanted to add some perspective to this grid so i added some extra's like this:
camera = new MapPoint(60, 60);
dx = mapP.x - camera.x; // delta x
dy = mapP.y - camera.y; // delta y
dist = Math.sqrt(dx * dx + dy * dy); // dist is the distance to the camera, Pythagoras etc.. all objects must be in front of the camera
fac = 1 - dist / 100; // this formula determines the amount of perspective
xa = fac * (mapP.x * xDistX + mapP.y * xDistY) ;
ya = fac * (mapP.x * yDistX + mapP.y * yDistY );
Now the real hard part... what if you got a (xa,ya) point on the projection and want to calculate the original point (x,y).
For the first case (without perspective) i did find the inverse function, but how can this be done for the formula with the perspective. May math skills are not quite up to the challenge to solve this.
( I vaguely remember from a long time ago mathematica could create inverse function for some special cases... could it solve this problem? Could someone maybe try?)
The function you've defined doesn't have an inverse. Just as an example, as user207422 already pointed out anything that's 100 units away from the camera will get mapped to (xa,ya)=(0,0), so the inverse isn't uniquely defined.
More importantly, that's not how you calculate perspective. Generally the perspective scaling factor is defined to be viewdist/zdist where zdist is the perpendicular distance from the camera to the object and viewdist is a constant which is the distance from the camera to the hypothetical screen onto which everything is being projected. (See the diagram here, but feel free to ignore everything else on that page.) The scaling factor you're using in your example doesn't have the same behaviour.
Here's a stab at trying to convert your code into a correct perspective calculation (note I'm not simplifying to 2D; perspective is about projecting three dimensions to two, trying to simplify the problem to 2D is kind of pointless):
camera = new MapPoint(60, 60, 10);
camera_z = camera.x*zDistX + camera.y*zDistY + camera.z*zDistz;
// viewdist is the distance from the viewer's eye to the screen in
// "world units". You'll have to fiddle with this, probably.
viewdist = 10.0;
xa = mapP.x*xDistX + mapP.y*xDistY + mapP.z*xDistZ;
ya = mapP.x*yDistX + mapP.y*yDistY + mapP.z*yDistZ;
za = mapP.x*zDistX + mapP.y*zDistY + mapP.z*zDistZ;
zdist = camera_z - za;
scaling_factor = viewdist / zdist;
xa *= scaling_factor;
ya *= scaling_factor;
You're only going to return xa and ya from this function; za is just for the perspective calculation. I'm assuming the the "za-direction" points out of the screen, so if the pre-projection x-axis points towards the viewer then zDistX should be positive and vice-versa, and similarly for zDistY. For a trimetric projection you would probably have xDistZ==0, yDistZ<0, and zDistZ==0. This would make the pre-projection z-axis point straight up post-projection.
Now the bad news: this function doesn't have an inverse either. Any point (xa,ya) is the image of an infinite number of points (x,y,z). But! If you assume that z=0, then you can solve for x and y, which is possibly good enough.
To do that you'll have to do some linear algebra. Compute camera_x and camera_y similar to camera_z. That's the post-transformation coordinates of the camera. The point on the screen has post-tranformation coordinates (xa,ya,camera_z-viewdist). Draw a line through those two points, and calculate where in intersects the plane spanned by the vectors (xDistX, yDistX, zDistX) and (xDistY, yDistY, zDistY). In other words, you need to solve the equations:
x*xDistX + y*xDistY == s*camera_x + (1-s)*xa
x*yDistX + y*yDistY == s*camera_y + (1-s)*ya
x*zDistX + y*zDistY == s*camera_z + (1-s)*(camera_z - viewdist)
It's not pretty, but it will work.
I think that with your post i can solve the problem. Still, to clarify some questions:
Solving the problem in 2d is useless indeed, but this was only done to make the problem easier to grasp (for me and for the readers here). My program actually give's a perfect 3d projection (i checked it with 3d images rendered with blender). I did left something out about the inverse function though. The inverse function is only for coordinates between 0..camera.x * 0.5 and 0.. camera.y*0.5. So in my example between 0 and 30. But even then i have doubt's about my function.
In my projection the z-axis is always straight up, so to calculate the height of an object i only used the vieuwingangle. But since you cant actually fly or jumpt into the sky everything has only a 2d point. This also means that when you try to solve the x and y, the z really is 0.
I know not every funcion has an inverse, and some functions do, but only for a particular domain. My basic thought in this all was... if i can draw a grid using a function... every point on that grid maps to exactly one map-point. I can read the x and y coordinate so if i just had the correct function i would be able to calculate the inverse.
But there is no better replacement then some good solid math, and im very glad you took the time to give a very helpfull responce :).