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How do you calculate the average of a set of angles?
If I have a set of bearings, ranging from 1-360, how can I find the average? Usually to find the average one would add them all up and divide by the number of items. The problem here is that doing that in the case of [1, 359], 2 bearings, would result in in 180, which in fact should be 360. Any ideas?
Represent the angles as vectors with Norm=1 and average the sum.
x1 = {cos(a),sin(a)}
x2 = {cos(b),sin(b)}
(x1+x2)/2 = {(cos(a)+cos(b))/2,(sin(a)+sin(b))/2}
which means the angle for the mean is
atan2((sin(a)+sin(b)) /(cos(a)+cos(b)))
Just beware of controlling the possible overflow when the denominator is close to zero.
It isn't clear from your question what you're trying to define the "average" to be... for directions on a circle there is no clear-cut obvious notion of average.
One interpretation is the value x that is the closest fit to the set of provided values, in the least-squares sense, where the distance between two bearings is defined as the smallest angle between them. Here is code to compute this average:
In[2]:= CircDist[a_, b_] := 180 - Mod[180 + a - b, 360]
In[6]:= Average[bearings_] :=
x /. NMinimize[
Sum[CircDist[x, bearings[[i]]]^2, {i, 1, Length[bearings]}],
x][[2]]
In[10]:= Average[{1, 359}]
Out[10]= -3.61294*10^-15
So what you want is the middle of two bearings - what happens if you have {90, 270}? Is the desired answer 0 or 180? This is something to consider.. also what's the middle of three bearings?
One thing you could do is:
Take the first two bearings in your set
Work out the difference between the two in either direction (i.e. [1, 359] would give 2 degrees in one direction, and 358 in the other)
If you want the desired angle to be the middle of the acutest of the two, take that as your difference and add to the anti-clockwise most of the pair (i.e. 359)
Use this as the new bearing and the next (i.e. 3rd in set) as the other bearing, and repeat, until all are 'middled'.
Off the top of my head, I don't think this is going to be fair, it'll probably bias it in one direction though (i.e. maybe in preference of the later values in your set).
Related
I need to sort a set of vectors in circular order. The simplest approach would be to use the angle between the vectors and a fixed axis. To get the angle, one would have to normalize the vectors which includes performing an expensive square root calculation.
As I want to avoid the costs and I don't need the particular angle - just some value that gives me the same order - I was wondering if there is a way to calculate a value for each vector that does not require the vector to be normalized and yields a similar value like the angle (i.e. if angle(x) > angle(y) then f(x) > f(y)).
The ratio of the y component to the x component should be enough to order the vectors without normalizing them. if the y:x ratio is larger, then the angle will be steeper. That'll work at least for the 1st quadrant (0 to 90 degrees), but the general idea should be enough to get you started.
I want to know how to measure the distance between two pixels in dicom . already done some google found pixel spacing (0028,0030) need to find the distance . could some one clearly explain ....
thanks
Assuming that you're trying to measure distances in the subject/animal/phantom/whatever, it all depends on whether you want to measure distances between different slices or just in the same slice.
Volumetric DICOM series typically have a slice spacing (0012,0088) in addition to the pixel spacing which you need to take into account. Note that there is also such a thing as slice thickness, which is distinct and should not be used for calculating distances, as there can be a gap or overlap between consecutive slices.
It is helpful to define a voxelspacing vector as follows (pseudocode):
voxelspacing.x = first element of PixelSpacing (0028,0030), i.e. before "\"
voxelspacing.y = second element of PixelSpacing (0028,0030), i.e. after "\"
voxelspacing.z = SliceSpacing (0018,0088) or 0 if 2D and/or not specified
Some brain-dead manufacturers and de-identification tools break the slice spacing tag in which case you'll have to calculate it from another source, such as difference in consecutive slice location, patient image position, etc, but that's another matter.
Moving on, you now have the distance in millimeters between voxels for each dimension. You can then calculate the real-world euclidean distance given voxel coordinates in pointA and pointB:
delta = (pointA - pointB) * voxelspacing
distance = sqrt(delta.x^2 + delta.y^2 + delta.z^2);
Where all the operators are element-wise. It is critical to individually multiply the voxel coordinates with their respective spacings before computing distance, because voxels are typically not isotropic.
You need to know the dot pitch of the monitor. For example a jumbotron has huge pixels (guessing), so the distance is larger than it would be for a typical desktop monitor. Ask the manufacturer of the monitor for this information. After that use pythogorean theorum. sqrt(a^2 + b^2) = c c being the total distance and a/b are x and y distances. to find a and be you would find the coordinates of one pixel and subtract from the other. a = (x1-x2) b = (
I am new to this forum and not a native english speaker, so please be nice! :)
Here is the challenge I face at the moment:
I want to calculate the (approximate) relative coordinates of yet unknown points in a 3D euclidean space based on a set of given distances between 2 points.
In my first approach I want to ignore possible multiple solutions, just taking the first one by random.
e.g.:
given set of distances: (I think its creating a pyramid with a right-angled triangle as a base)
P1-P2-Distance
1-2-30
2-3-40
1-3-50
1-4-60
2-4-60
3-4-60
Step1:
Now, how do I calculate the relative coordinates for those points?
I figured that the first point goes to 0,0,0 so the second one is 30,0,0.
After that the third points can be calculated by finding the crossing of the 2 circles from points 1 and 2 with their distances to point 3 (50 and 40 respectively). How do I do that mathematically? (though I took these simple numbers for an easy representation of the situation in my mind). Besides I do not know how to get to the answer in a correct mathematical way the third point is at 30,40,0 (or 30,0,40 but i will ignore that).
But getting the fourth point is not as easy as that. I thought I have to use 3 spheres in calculate the crossing to get the point, but how do I do that?
Step2:
After I figured out how to calculate this "simple" example I want to use more unknown points... For each point there is minimum 1 given distance to another point to "link" it to the others. If the coords can not be calculated because of its degrees of freedom I want to ignore all possibilities except one I choose randomly, but with respect to the known distances.
Step3:
Now the final stage should be this: Each measured distance is a bit incorrect due to real life situation. So if there are more then 1 distances for a given pair of points the distances are averaged. But due to the imprecise distances there can be a difficulty when determining the exact (relative) location of a point. So I want to average the different possible locations to the "optimal" one.
Can you help me going through my challenge step by step?
You need to use trigonometry - specifically, the 'cosine rule'. This will give you the angles of the triangle, which lets you solve the 3rd and 4th points.
The rules states that
c^2 = a^2 + b^2 - 2abCosC
where a, b and c are the lengths of the sides, and C is the angle opposite side c.
In your case, we want the angle between 1-2 and 1-3 - the angle between the two lines crossing at (0,0,0). It's going to be 90 degrees because you have the 3-4-5 triangle, but let's prove:
50^2 = 30^2 + 40^2 - 2*30*40*CosC
CosC = 0
C = 90 degrees
This is the angle between the lines (0,0,0)-(30,0,0) and (0,0,0)- point 3; extend along that line the length of side 1-3 (which is 50) and you'll get your second point (0,50,0).
Finding your 4th point is slightly trickier. The most straightforward algorithm that I can think of is to firstly find the (x,y) component of the point, and from there the z component is straightforward using Pythagoras'.
Consider that there is a point on the (x,y,0) plane which sits directly 'below' your point 4 - call this point 5. You can now create 3 right-angled triangles 1-5-4, 2-5-4, and 3-5-4.
You know the lengths of 1-4, 2-4 and 3-4. Because these are right triangles, the ratio 1-4 : 2-4 : 3-4 is equal to 1-5 : 2-5 : 3-5. Find the point 5 using trigonometric methods - the 'sine rule' will give you the angles between 1-2 & 1-4, 2-1 and 2-4 etc.
The 'sine rule' states that (in a right triangle)
a / SinA = b / SinB = c / SinC
So for triangle 1-2-4, although you don't know lengths 1-4 and 2-4, you do know the ratio 1-4 : 2-4. Similarly you know the ratios 2-4 : 3-4 and 1-4 : 3-4 in the other triangles.
I'll leave you to solve point 4. Once you have this point, you can easily solve the z component of 4 using pythagoras' - you'll have the sides 1-4, 1-5 and the length 4-5 will be the z component.
I'll initially assume you know the distances between all pairs of points.
As you say, you can choose one point (A) as the origin, orient a second point (B) along the x-axis, and place a third point (C) along the xy-plane. You can solve for the coordinates of C as follows:
given: distances ab, ac, bc
assume
A = (0,0)
B = (ab,0)
C = (x,y) <- solve for x and y, where:
ac^2 = (A-C)^2 = (0-x)^2 + (0-y)^2 = x^2 + y^2
bc^2 = (B-C)^2 = (ab-x)^2 + (0-y)^2 = ab^2 - 2*ab*x + x^2 + y^2
-> bc^2 - ac^2 = ab^2 - 2*ab*x
-> x = (ab^2 + ac^2 - bc^2)/2*ab
-> y = +/- sqrt(ac^2 - x^2)
For this to work accurately, you will want to avoid cases where the points {A,B,C} are in a straight line, or close to it.
Solving for additional points in 3-space is similar -- you can expand the Pythagorean formula for the distance, cancel the quadratic elements, and solve the resulting linear system. However, this does not directly help you with your steps 2 and 3...
Unfortunately, I don't know a well-behaved exact solution for steps 2 and 3, either. Your overall problem will generally be both over-constrained (due to conflicting noisy distances) and under-constrained (due to missing distances).
You could try an iterative solver: start with a random placement of all your points, compare the current distances with the given ones, and use that to adjust your points in such a way as to improve the match. This is an optimization technique, so I would look up books on numerical optimization.
If you know the distance between the nodes (fixed part of system) and the distance to the tag (mobile) you can use trilateration to find the x,y postion.
I have done this using the Nanotron radio modules which have a ranging capability.
I want to calculate the angle between two vectors a and b. Lets assume these are at the origin. This can be done with
theta = arccos(a . b / |a| * |b|)
However arccos gives you the angle in [0, pi], i.e. it will never give you an angle greater than 180 degrees, which is what I want. So how do you find out when the vectors have gone past the 180 degree mark? In 2D I would simply let the sign of the y-component on one of the vectors determine what quadrant the vector is in. But what is the easiest way to do it in 3D?
EDIT: I wanted to keep the question general but here we go. I'm programming this in c and the code I use to get the angle is theta = acos(dot(a, b)/mag(a)*mag(b)) so how would you programmatically determine the orientation?
This works in 2D because you have a plane defined in which you define the rotation.
If you want to do this in 3D, there is no such implicit 2D plane. You could transform your 3D coordinates to a 2D plane going through all three points, and do your calculation inside this plane.
But, there are of course two possible orientations for the plane, and that will affect which angles will be > 180 or smaller.
I came up with the following solution that takes advantage of the direction change of the cross product of the two vectors:
Make a vector n = a X b and normalize it. This vector is normal to the plane spanned by a and b.
Whenever a new angle is calculated compare it with the old normal. In the comparison, treat the old and the current normals as points and compute the distance between them. If this distance is 2 the normal (i.e. the cross product a X b has flipped).
You might want to have a threshold for the distance as the distance after a flip might be shorter than 2, depending on how the vectors a and b are oriented and how often you update the angle.
One solution that you could use:
What you effectively need to do is create a plane that one of the vectors is coplanar to.
Getting the cross product of both vectors will create a plane, then is you get the normal of this plane, you can get the angle between this and the vector you need to get the signed angle for, and you can use the angle to determine the sign.
If the angle is greater than 90 degrees, then it is below the created plane; less than 90 degrees, and it is above.
Depending on cost of calculations, the dot product can be used at this stage instead of the angle.
Just make sure that you always calculate the normals by the same order of vectors.
This is useable more easily if you're using the XYZ axes, and that's what you're comparing against, since you already have the vectors needed for the plane.
There are possbly more efficient solutions, but this is one I came up with.
Edit: clarification of created vectors
a X b = p. This is perpendicular to both a and b.
Then, do either:
a X p or b X p to create another vector that is the normal to the plane created by the 2 vectors. Choice of vector depends on which you're trying to find the angle for.
Strictly speaking, two 3D vectors always have two angles between them - one below or equal to 180, the other over or equal to 180. Arccos gives you one of them, you can get the other by subtracting from 360. Think of it that way: imagine two lines intersect. You have 4 angles there - 2 of one value, 2 of another. What's the angle between the lines? No single answer. Same here. Without some kind of extra criteria, you can not, in theory, tell which of the two angle values should be taken into account.
EDIT: So what you really need is an arbitrary example of fixing an orientation. Here's one: we look from the positive Z direction. If the plane between the two vectors contains the Z axis, we look from the positive Y direction. If the plane is YZ, we look from the positive X direction. I'll think how to express this in coordinate form, then edit again.
I am writing a physics engine/simulator which incorporates 3D space flight, planetary/stellar gravitation, ship thrust and relativistic effects. So far, it is going very well, however, one thing that I need help with is the math of the collision detection algorithm.
The iterative simulation of movement that I am using is basically as follows:
(Note: 3D Vectors are ALL CAPS.)
For each obj
obj.ACC = Sum(all acceleration influences)
obj.POS = obj.POS + (obj.VEL * dT) + (obj.ACC * dT^2)/2 (*EQ.2*)
obj.VEL = obj.VEL + (obj.ACC * dT)
Next
Where:
obj.ACC is the acceleration vector of the object
obj.POS is the position or location vector of the object
obj.VEL is the velocity vector of the object
obj.Radius is the radius (scalar) of the object
dT is the time delta or increment
What I basically need to do is to find some efficient formula that derives from (EQ.2) above for two objects (obj1, obj2) and tell if they ever collide, and if so, at what time. I need the exact time both so that I can determine if it is in this particular time increment (because acceleration will be different at different time increments) and also so that I can locate the exact position (which I know how to do, given the time)
For this engine, I am modelling all objects as spheres, all this formula/algorithm needs to do is to figure out at what points:
(obj1.POS - obj2.POS).Distance = (obj1.Radius + obj2.Radius)
where .Distance is a positive scalar value. (You can also square both sides if this is easier, to avoid the square root function implicit in the .Distance calculation).
(yes, I am aware of many, many other collision detection questions, however, their solutions all seem to be very particular to their engine and assumptions, and none appear to match my conditions: 3D, spheres, and acceleration applied within the simulation increments. Let me know if I am wrong.)
Some Clarifications:
1) It is not sufficient for me to check for Intersection of the two spheres before and after the time increment. In many cases their velocities and position changes will far exceed their radii.
2) RE: efficiency, I do not need help (at this point anyway) with respect to determine likely candidates for collisions, I think that I have that covered.
Another clarification, which seems to be coming up a lot:
3) My equation (EQ.2) of incremental movement is a quadratic equation that applies both Velocity and Acceleration:
obj.POS = obj.POS + (obj.VEL * dT) + (obj.ACC * dT^2)/2
In the physics engines that I have seen, (and certainly every game engine that I ever heard of) only linear equations of incremental movement that apply only Velocity:
obj.POS = obj.POS + (obj.VEL * dT)
This is why I cannot use the commonly published solutions for collision detection found on StackOverflow, on Wikipedia and all over the Web, such as finding the intersection/closest approach of two line segments. My simulation deals with variable accelerations that are fundamental to the results, so what I need is the intersection/closest approach of two parabolic segments.
On the webpage AShelley referred to, the Closest Point of Approach method is developed for the case of two objects moving at constant velocity. However, I believe the same vector-calculus method can be used to derive a result in the case of two objects both moving with constant non-zero acceleration (quadratic time dependence).
In this case, the time derivative of the distance-squared function is 3rd order (cubic) instead of 1st order. Therefore there will be 3 solutions to the Time of Closest Approach, which is not surprising since the path of both objects is curved so multiple intersections are possible. For this application, you would probably want to use the earliest value of t which is within the interval defined by the current simulation step (if such a time exists).
I worked out the derivative equation which should give the times of closest approach:
0 = |D_ACC|^2 * t^3 + 3 * dot(D_ACC, D_VEL) * t^2 + 2 * [ |D_VEL|^2 + dot(D_POS, D_ACC) ] * t + 2 * dot(D_POS, D_VEL)
where:
D_ACC = ob1.ACC-obj2.ACC
D_VEL = ob1.VEL-obj2.VEL (before update)
D_POS = ob1.POS-obj2.POS (also before update)
and dot(A, B) = A.x*B.x + A.y*B.y + A.z*B.z
(Note that the square of the magnitude |A|^2 can be computed using dot(A, A))
To solve this for t, you'll probably need to use formulas like the ones found on Wikipedia.
Of course, this will only give you the moment of closest approach. You will need to test the distance at this moment (using something like Eq. 2). If it is greater than (obj1.Radius + obj2.Radius), it can be disregarded (i.e. no collision). However, if the distance is less, that means the spheres collide before this moment. You could then use an iterative search to test the distance at earlier times. It might also be possible to come up with another (even more complicated) derivation which takes the size into account, or possible to find some other analytic solution, without resorting to iterative solving.
Edit: because of the higher order, some of the solutions to the equation are actually moments of farthest separation. I believe in all cases either 1 of the 3 solutions or 2 of the 3 solutions will be a time of farthest separation. You can test analytically whether you're at a min or a max by evaluating the second derivative with respect to time (at the values of t which you found by setting the first derivative to zero):
D''(t) = 3 * |D_ACC|^2 * t^2 + 6 * dot(D_ACC, D_VEL) * t + 2 * [ |D_VEL|^2 + dot(D_POS, D_ACC) ]
If the second derivative evaluates to a positive number, then you know the distance is at a minimum, not a maximum, for the given time t.
Draw a line between the start location and end location of each sphere. If the resulting line segments intersect the spheres definitely collided at some point and some clever math can find at what time the collision occurred. Also make sure to check if the minimum distance between the segments (if they don't intersect) is ever less than 2*radius. This will also indicate a collision.
From there you can backstep your delta time to happen exactly at collision so you can correctly calculate the forces.
Have you considered using a physics library which already does this work? Many libraries use far more advanced and more stable (better integrators) systems for solving the systems of equations you're working with. Bullet Physics comes to mind.
op asked for time of collision. A slightly different approach will compute it exactly...
Remember that the position projection equation is:
NEW_POS=POS+VEL*t+(ACC*t^2)/2
If we replace POS with D_POS=POS_A-POS_B, VEL with D_VEL=VEL_A-VEL_B, and ACC=ACC_A-ACC_B for objects A and B we get:
$D_NEW_POS=D_POS+D_VEL*t+(D_ACC*t^2)/2
This is the formula for vectored distance between the objects. In order to get the squared scalar distance between them, we can take the square of this equation, which after expansion looks like:
distsq(t) = D_POS^2+2*dot(D_POS,D_VEL)*t + (dot(D_POS, D_ACC)+D_VEL^2)*t^2 + dot(D_VEL,D_ACC)*t^3 + D_ACC^2*t^4/4
In order to find the time where collision occurs, we can set the equation equal to the square of the sum of radii and solve for t:
0 = D_POS^2-(r_A+r_B)^2 + 2*dot(D_POS,D_VEL)*t + (dot(D_POS, D_ACC)+D_VEL^2)*t^2 + dot(D_VEL,D_ACC)*t^3 + D_ACC^2*t^4/4
Now, we can solve for the equation using the quartic formula.
The quartic formula will yield 4 roots, but we are only interested in real roots. If there is a double real root, then the two objects touch edges at exactly one point in time. If there are two real roots, then the objects continuously overlap between root 1 and root 2 (i.e. root 1 is the time when collision starts and root 2 is the time when collision stops). Four real roots means that the objects collide twice, continuously between root pairs 1,2 and 3,4.
In R, I used polyroot() to solve as follows:
# initial positions
POS_A=matrix(c(0,0),2,1)
POS_B=matrix(c(2,0),2,1)
# initial velocities
VEL_A=matrix(c(sqrt(2)/2,sqrt(2)/2),2,1)
VEL_B=matrix(c(-sqrt(2)/2,sqrt(2)/2),2,1)
# acceleration
ACC_A=matrix(c(sqrt(2)/2,sqrt(2)/2),2,1)
ACC_B=matrix(c(0,0),2,1)
# radii
r_A=.25
r_B=.25
# deltas
D_POS=POS_B-POS_A
D_VEL=VEL_B-VEL_A
D_ACC=ACC_B-ACC_A
# quartic coefficients
z=c(t(D_POS)%*%D_POS-r*r, 2*t(D_POS)%*%D_VEL, t(D_VEL)%*%D_VEL+t(D_POS)%*%D_ACC, t(D_ACC)%*%D_VEL, .25*(t(D_ACC)%*%D_ACC))
# get roots
roots=polyroot(z)
# In this case there are only two real roots...
root1=as.numeric(roots[1])
root2=as.numeric(roots[2])
# trajectory over time
pos=function(p,v,a,t){
T=t(matrix(t,length(t),2))
return(t(matrix(p,2,length(t))+matrix(v,2,length(t))*T+.5*matrix(a,2,length(t))*T*T))
}
# plot A in red and B in blue
t=seq(0,2,by=.1) # from 0 to 2 seconds.
a1=pos(POS_A,VEL_A,ACC_A,t)
a2=pos(POS_B,VEL_B,ACC_B,t)
plot(a1,type='o',col='red')
lines(a2,type='o',col='blue')
# points of a circle with center 'p' and radius 'r'
circle=function(p,r,s=36){
e=matrix(0,s+1,2)
for(i in 1:s){
e[i,1]=cos(2*pi*(1/s)*i)*r+p[1]
e[i,2]=sin(2*pi*(1/s)*i)*r+p[2]
}
e[s+1,]=e[1,]
return(e)
}
# plot circles with radius r_A and r_B at time of collision start in black
lines(circle(pos(POS_A,VEL_A,ACC_A,root1),r_A))
lines(circle(pos(POS_B,VEL_B,ACC_B,root1),r_B))
# plot circles with radius r_A and r_B at time of collision stop in gray
lines(circle(pos(POS_A,VEL_A,ACC_A,root2),r_A),col='gray')
lines(circle(pos(POS_B,VEL_B,ACC_B,root2),r_B),col='gray')
Object A follows the red trajectory from the lower left to the upper right. Object B follows the blue trajectory from the lower right to the upper left. The two objects collide continuously between time 0.9194381 and time 1.167549. The two black circles just touch, showing the beginning of overlap - and overlap continues in time until the objects reach the location of the gray circles.
Seems like you want the Closest Point of Approach (CPA). If it is less than the sum of the radiuses, you have a collision. There is example code in the link. You can calculate each frame with the current velocity, and check if the CPA time is less than your tick size. You could even cache the cpa time, and only update when acceleration was applied to either item.