I have multiple estimates for a transformation matrix, from mapping two point clouds to each other via ICP (Iterative Closest Point).
How can I generate the average transformation matrix for all these matrices?
Each matrix consists of a rigid translation and a rotation only, no scale or skew.
Ideally I would also like to calculate a weighted average, but an unweighted one is fine for now.
Averaging the translation vectors is of course trivial, but the rotations are problematic. One approach I found is averaging the individual base vectors for the rotations, but I am not sure that will result in a new orthonormal base, and the approach seems a little ad-hoc.
Splitting the transformation in translation and rotation is a good start. Averaging the translation is trivial.
Averaging the rotation is not that easy. Most approaches will use quaternions. So you need to transform the rotation matrix to a quaternion.
The easiest way to approximate the average is a linear blending, followed by renormalization of the quaternion:
q* = w1 * q1 + w2 * q2 + ... + w2 * qn
normalize q*
However, this is only an approximation. The reason for that is that the combination of two rotations is not performed by adding the quaternions, but by multiplying them. If we convert quaternions to a logarithmic space, we can use a simple linear blend (because multiplication will become additions). Then transform the quaternion back to the original space. This is the idea of the Spherical Average (Buss 2001). If you're lucky, you find a library that supports log and exp of quaternions:
start with q* as above
do until convergence
for each input quaternion i (index)
diff = q[i] * inverse(q*)
u[i] = log(diff, base q*)
//Now perform the linear blend
adapt := zero quaternion
weights := 0
for each input quaternion i
adapt += weight[i] * u[i]
weights += weight[i]
adapt *= 1/weights
adaptInOriginalSpace = q* ^ adapt (^ is the power operator)
q* = adaptInOriginalSpace * q*
You can define a threshold for adaptInOriginalSpace. If it is a very very small rotation, you can break the loop. This algorithm is proven to preserve geodesic distances on a sphere.
http://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation and http://en.wikipedia.org/wiki/Rotation_matrix#Quaternion will give you some elegant mathematics and a way to turn a rotation matrix into an angle of rotation round an axis of rotation. There will be two possible representations of each rotation, with different signs for both angle of rotation and axis of rotation.
You could convert everything and normalize them to have +ve angles of rotation, then work out the average angle of rotation and the average axis of rotation, renormalising this into a unit vector.
OTOH if your intention is to work out the most accurate possible estimate of the transformation, you need to write down some measure of the goodness of fit of any candidate transformation - a sum of squared errors is often mathematically convenient - and then solve an optimization problem to work out which transformation minimizes the sum of squared errors. This is at least easier to justify than taking an average of individually error-prone estimates, and may well be more accurate.
If you have an existing lerp method, then there is a trivial solution:
count = 1
average_transform = Matrix.Identity(4)
for new_transform in list_of_matrices:
factor = 1/count
average_transform = lerp(average_transform, new_transform, factor)
count += 1
This is only useful because lots more mathermatics packages have the ability to lerp matrices than to average lots of them.
Because I haven't come across this method elsewhere, here's an informal proof:
If there is one matrix, use just that matrix (factor will equal 1 for first matrix)
If there are two matrices, we need 50% of the second one (second factor is 50% so we lerp to half way between the existing first one and the new one)
If there are three matrices we need 33% of each, or 66% of the average of the first two and 33% of the third. The lerp factor of 0.3333 makes this happen.
And so on.
I haven't tested extensively with matrices, but I've used this successfully as a rolling average for other datatypes.
The singular value decomposition (SVD) can be used here.
Take the SVD of the sum of the rotation matricies, and then the average rotation matrix is simply given by Ravg = UV'.
"sdfgeoff" I can't comment in your answer because I'm new here, but you are the most correct, I think. Beutifull and elegant solution, by the way. Would be perfect if you use Spherical Linear Interpolation (SLERP) with quaternions, instead of Linear Interpolation (LERP) because quaternions that map rotations (quaternions with norm 1) define a sphere in 4D, and interpolating between then is in fact interpolate between two point in a sphere surface.
With my experience from point cloud registration, I wuold like to say that this will not work. ICP don't return random rotations in the likehood of the correct rotation. You need to use a beter algorith to register you point clouds (Global Registration algorithms, like FPFH, 4PCS, K4PCS, BSC, FGR, etc). Or a better initial guess for the transformation. ICP will only give you totally wrong rotations (when stuck in local minima) or almost perfect rotations, when initialized with good initial transformations.
Conclusion: averaging it will not work.
I would suggest taking a look at "Average" of multiple quaternions? for a more elaborate discussion on how to compute the average of rotations.
Related
I was transforming an object with a matrix A using Object3D.applyMatrix4 and I found that at one point it didn't preserve an eigen vector's direction.
So I tried animating interpolation between Identity Matrix I and A and I found this:
How could the transformation be not continuous?
Linear interpolation of rotation matrices isn't mathematically sound. The vectors composing a rotation matrix need to be unit length.. or at least stay a consistent length.
Imagine a clock with a hand at 12, and a hand at 6.
If you Linearly interpolate the point at the tip of the 12 oclock hand, to the tip of the 6oclock hand, the point travels in a straight line from top of the clock to the bottom.
To interpolate the rotation represented by a 4x4 matrix, you can convert the rotations of the matrices, to quaternions, and .slerp (spherical linear interpolate) between those quaternions, then convert back to a matrix.
And then linearly interpolate the object.position. (although again.. this assumes linear motion between keyframes).
Now in the case that the rotation is small, you can get away with linearly interpolating the matrix, but you will need to orthonormalize it at each step, to reshape the mesh into one that has consistent length vectors that are orthogonal to each other. That isn't that hard.. you use a combination of dot products, multiplies and adds of the vectors forming the matrix rows (or columns, i forget) to orthonormalize the matrix. But its more of a pain, and less accurate than just using quaternions and .slerp.
#manthrax 's answer pointed out the fundamental problem of interpolating a matrix linearly which I wasn't aware of at the time and he was right about that. But the real problem was that Object3D.applyMatrix4 wasn't the right function for explicitly defining local matrix. I tried setting Object3D.matrix property directly and it worked. And the linear interpolation (although I shouldn't do that) became continuous.
Given three angular velocities vx, vy, vz about the x, y and z axes, measured in radians per second, as derived from an IMU's rate gyro, how do I produce an equivalent quaternion for the entire rotation between one sample and the next, i.e. the integral of rotation over time dt between the current sample and the previous sample?
The primary issue is that these three angular velocities are measured independently of each other, and yet rotations are not commutative. This means the order in which the angular velocities are applied during the integration would affect the computed quaternion, just as converting Euler angles to a quaternion produces a different quaternion depending on the order in which the Euler rotations are applied (e.g. x, then y, then z, vs. some other order).
I think the right thing to do is to split the timestep dt into a number of shorter time period samples, e.g. say N=10, then divide each velocity by that number, giving vx' = vx/N, vy' = vy/N, vz' = vz/N, and then applying the rotations N times in round robin fashion, in largest to smallest order, calculating the actual rotation over the interval dt/N in each case, and accumulating this into the final rotation quaternion.
I see a lot of references to quaternion derivatives when related questions are asked though, and I wonder if it might be possible to convert the angular velocities (which are derivatives of Euler angles) directly to a quaternion derivative (again though probably suffering from axis ordering sensitivity), then somehow integrate the quaternion derivative to convert back to a quaternion spanning time dt.
Seems like there should be a "right" way to do this, since every IMU that uses a rate gyro has to solve this problem. Any insights into this would be greatly appreciated!
I found the answer in this excellent post by Ashwin Narayan.
Update (1): the rowan library implements the necessary quaternion exponentiation in Python.
Update (2): User harold pointed to this answer, which shows the same quaternion exponentiation in C++ code, which is more legible than the NumPy code in rowan.
Please forgive me for the naive question, I don't remember linear algebra at all.
To do that I use a matrix that is associated to an image to apply transformations,
The matrix of the image is a matrix, now I want to get the how much the matrix has been translated and scale.
It's OK when there are no rotations applied,
but rotation confuses things a lot.
Say your new matrix N = RTS, where R is a rotation, T is a translation, and S is a scaling. This means in order you scale, translate, then rotate. If you want to see the scaling and translation, left-multiply by R-inverse, which is the same as R's transpose. With respect to your original view, you will see a stretched and transformed matrix.
If instead N = TSR, you would have to right multiply by R inverse. Note: The two matrices N obtained by these operations will not in general be the same!
Alternatively you can change coordinate systems, but this is more involved as rotation and translation do not commute in general.
I am trying to implement Inverse Kinematics on a 2D arm(made up of three sticks with joints). I am able to rotate the lowest arm to the desired position. Now, I have some questions:
How can I make the upper arm move alongwith the third so the end point of the arm reaches the desired point. Do I need to use the rotation matrices for both and if yes can someone give me some example or an help and is there any other possibl;e way to do this without rotation matrices???
The lowest arm only moves in one direction. I tried google it, they are saying that cross product of two vectors give the direction for the arm but this is for 3D. I am using 2D and cross product of two 2D vectors give a scalar. So, how can I determine its direction???
Plz guys any help would be appreciated....
Thanks in advance
Vikram
I'll give it a shot, but since my Robotics are two decades in the past, take it with a grain of salt.
The way I learned it, every joint was described by its own rotation matrix, defined relative to its current position and orientation. The coordinate of the whole arm's endpoint was then calculated by combining the rotation matrices together.
This achieved exactly the effect you are looking for: you could move only one joint (change its orientation), and all the other joints followed automatically.
You won't have much chance in getting around matrices here - in fact, if you use homogeneous coordinates, all joint calculations (rotations as well as translations) can be modeled with matrix multiplications. The advantage is that the full arm position can then be described with a single matrix (plus the arm's origin).
With this transformation matrix, you can tackle the inverse kinematic problem: since the transformation matrix' elements will depend on the angles of the joints, you can treat the whole calculation 'endpoint = startpoint x transformation' as a system of equations, and with startpoint and endpoint known, you can solve this system to determine the unknown angles. The difficulty herein lies that the equation may not be solvable, or that there are multiple solutions.
I don't quite understand your second question, though - what are you looking for?
Instead of a rotation matrix, the rotation can be represented by its angle or by a complex number of the unit circle, but it's the same thing really. More importantly, you need a representation T of rigid body transformations, so that you can write stuff like t1 * t2 * t3 to compute the position and orientation of the third link.
Use atan2 to compute the angle between the vectors.
As the following Python example shows, those two things are enough to build a small IK solver.
from gameobjects.vector2 import Vector2 as V
from matrix33 import Matrix33 as T
from math import sin, cos, atan2, pi
import random
The gameobjects library does not have 2D transformations, so you have to write matrix33 yourself. Its interface is just like gameobjects.matrix44.
Define the forward kinematics function for the transformation from one joint to the next. We assume the joint rotates by angle and is followed by a fixed transformation joint:
def fk_joint(joint, angle): return T.rotation(angle) * joint
The transformation of the tool is tool == fk(joints, q) where joints are the fixed transformations and q are the joint angles:
def fk(joints, q):
prev = T.identity()
for i, joint in enumerate(joints):
prev = prev * fk_joint(joint, q[i])
return prev
If the base of the arm has an offset, replace the T.identity() transformation.
The OP is solving the IK problem for position by cyclic coordinate descent. The idea is to move the tool closer to the goal position by adjusting one joint variable at a time. Let q be the angle of a joint and prev be the transformation of the base of the joint. The joint should be rotated by the angle between the vectors to the tool and goal positions:
def ccd_step(q, prev, tool, goal):
a = tool.get_position() - prev.get_position()
b = goal - prev.get_position()
return q + atan2(b.get_y(), b.get_x()) - atan2(a.get_y(), a.get_x())
Traverse the joints and update the tool configuration for every change of a joint value:
def ccd_sweep(joints, tool, q, goal):
prev = T.identity()
for i, joint in enumerate(joints):
next = prev * fk_joint(joint, q[i])
q[i] = ccd_step(q[i], prev, tool, goal)
prev = prev * fk_joint(joint, q[i])
tool = prev * next.get_inverse() * tool
return prev
Note that fk() and ccd_sweep() are the same for 3D; you just have to rewrite fk_joint() and ccd_step().
Construct an arm with n identical links and run cnt iterations of the CCD sweep, starting from a random arm configuration q:
def ccd_demo(n, cnt):
q = [random.uniform(-pi, pi) for i in range(n)]
joints = [T.translation(0, 1)] * n
tool = fk(joints, q)
goal = V(0.9, 0.75) # Some arbitrary goal.
print "i Error"
for i in range(cnt):
tool = ccd_sweep(joints, tool, q, goal)
error = (tool.get_position() - goal).get_length()
print "%d %e" % (i, error)
We can try out the solver and compare the rate of convergence for different numbers of links:
>>> ccd_demo(3, 7)
i Error
0 1.671521e-03
1 8.849190e-05
2 4.704854e-06
3 2.500868e-07
4 1.329354e-08
5 7.066271e-10
6 3.756145e-11
>>> ccd_demo(20, 7)
i Error
0 1.504538e-01
1 1.189107e-04
2 8.508951e-08
3 6.089372e-11
4 4.485040e-14
5 2.601336e-15
6 2.504777e-15
In robotics we most often use DH parameters for the forward and reverse kinematics. Wikipedia has a nice introduction.
The DH (Denavit-Hartenberg) notation is part of the solution. It helps you collect a succinct set of values that describe the mechanics of your robot such as link length and joint type.
From there it becomes easier to calculate forward kinematics. The first think you have to understand is how to translate a coordinate frame from one place to another coordinate frame. For example, given your robot (or the DH table of it), what is the set of rotations and translations you have to apply to one coordinate frame (the world for example) to know the location of a point (or vector) in the robot's wrist coordinate frame.
As you may already know, homogeneous transform matrices are very useful for such transformations. They are 4x4 matrices that encapsulate rotation and translation. Another very useful property of those matrices is that if you have two coordinate frames linked and defined by some rotation and translation, if you multiply the two matrices together, then you just need to multiply your transformation target by the product of that multiplication.
So the DH table will help you build that matrix.
Inverse kinematics is a bit more complicated though and depends on your application. The complication arises from having multiple solutions for the same problem. The greater the number of DOF, the greater the number of solutions.
Think about your arm. Pinch something solid around you. You can move your arm to several locations in the space and still keep your pinching vector unchanged. Solving the inverse kinematics problem involves deciding which solution to choose as well.
Finding the angle between two vectors is not hard using the cosine rule. However, because I am programming for a platform with very limited resources, I would like to avoid calculations such as sqrt and arccos. Even simple divisions should be limited as much as possible.
Fortunately, I do not need the angle per se, but only need some value that is proportional to said angle.
So I am looking for some computationally cheap algorithm to calculate a quantity that is related to the angle between two vectors. So far, I haven't found something that fits the bill, nor have I been able to come up with something myself.
If you don't need the actual euclidean angle, but something that you can use as a base for angle comparisons, then changing to taxicab geometry may be a choice, because you can drop trigonometry and it's slowness while MAINTAINING THE ACCURACY (or at least with really minor loosing of accuracy, see below).
In main modern browser engines the speedup factor is between 1.44 - 15.2 and the accuracy is nearly the same as in atan2. Calculating diamond angle is average 5.01 times faster than atan2 and using inline code in Firefox 18 the speedup reaches factor 15.2. Speed comparison: http://jsperf.com/diamond-angle-vs-atan2/2.
The code is very simple:
function DiamondAngle(y, x)
{
if (y >= 0)
return (x >= 0 ? y/(x+y) : 1-x/(-x+y));
else
return (x < 0 ? 2-y/(-x-y) : 3+x/(x-y));
}
The above code gives you angle between 0 and 4, while atan2 gives you angle between -PI and PI as the following table shows:
Note that diamond angle is always positive and in the range of 0-4, while atan2 gives also negative radians. So diamond angle is sort of more normalized. And another note is that atan2 gives a little more precise result, because the range length is 2*pi (ie 6.283185307179586), while in diamond angles it is 4. In practice this is not very significant, eg. rad 2.3000000000000001 and 2.3000000000000002 are both in diamond angles 1.4718731421442295, but if we lower the precision by dropping one zero, rad 2.300000000000001 and 2.300000000000002 gives both different diamond angle. This "precision loosing" in diamond angles is so small, that it has some significant influence only if the distances are huge. You can play with conversions in http://jsbin.com/bewodonase/1/edit?output (Old version: http://jsbin.com/idoyon/1):
The above code is enough for fast angle comparisons, but in many cases there is need to convert diamond angle to radians and vice verca. If you eg. have some tolerance as radian angles, and then you have a 100,000 times loop where this tolerance is compared to other angles, it's not wise to make comparisons using atan2. Instead, before looping, you change the radian tolerance to taxicab (diamond angles) tolerance and make in-loop comparisons using diamond tolerance and this way you don't have to use slow trigonometric functions in speed-critical parts of the code ( = in loops).
The code that makes this conversion is this:
function RadiansToDiamondAngle(rad)
{
var P = {"x": Math.cos(rad), "y": Math.sin(rad) };
return DiamondAngle(P.y, P.x);
}
As you notice there is cos and sin. As you know, they are slow, but you don't have to make the conversion in-loop, but before looping and the speedup is huge.
And if for some reason, you have to convert diamond angle to radians, eg. after looping and making angle comparisons to return eg. the minimum angle of comparisons or whatever as radians, the code is as follows:
function DiamondAngleToRadians(dia)
{
var P = DiamondAngleToPoint(dia);
return Math.atan2(P.y,P.x);
}
function DiamondAngleToPoint(dia)
{
return {"x": (dia < 2 ? 1-dia : dia-3),
"y": (dia < 3 ? ((dia > 1) ? 2-dia : dia) : dia-4)};
}
Here you use atan2, which is slow, but idea is to use this outside any loops. You cannot convert diamond angle to radians by simply multiplying by some factor, but instead finding a point in taxicab geometry of which diamond angle between that point and the positive X axis is the diamond angle in question and converting this point to radians using atan2.
This should be enough for fast angle comparisons.
Of course there is other atan2 speedup techniques (eg. CORDIC and lookup tables), but AFAIK they all loose accuracy and still may be slower than atan2.
BACKGROUND: I have tested several techniques: dot products, inner products, law of cosine, unit circles, lookup tables etc. but nothing was sufficient in case where both speed and accuracy are important. Finally I found a page in http://www.freesteel.co.uk/wpblog/2009/06/encoding-2d-angles-without-trigonometry/ which had the desired functions and principles.
I assumed first that also taxicab distances could be used for accurate and fast distance comparisons, because the bigger distance in euclidean is bigger also in taxicab. I realized that contrary to euclidean distances, the angle between start and end point has affect to the taxicab distance. Only lengths of vertical and horizontal vectors can be converted easily and fast between euclidean and taxicab, but in every other case you have to take the angle into account and then the process is too slow (?).
So as a conclusion I am thinking that in speed critical applications where is a loop or recursion of several comparisons of angles and/or distances, angles are faster to compare in taxicab space and distances in euclidean (squared, without using sqrt) space.
Have you tried a CORDIC algorithm? It's a general framework for solving polar ↔ rectangular problems with only add/subtract/bitshift + table, essentially doing rotation by angles of the form tan-1 (2-n). You can trade off accuracy with execution time by altering the number of iterations.
In your case, take one vector as a fixed reference, and copy the other to a temporary vector, which you rotate using the cordic angles towards the first vector (roughly bisection) until you reach a desired angular accuracy.
(edit: use sign of dot product to determine at each step whether to rotate forward or backward. Although if multiplies are cheap enough to allow using dot product, then don't bother with CORDIC, perhaps use a table of sin/cos pairs for rotation matrices of angles π/2n to solve the problem with bisection.)
(edit: I like Eric Bainville's suggestion in the comments: rotate both vectors towards zero and keep track of the angle difference.)
Back in the day of a few K of RAM and machines with limited mathematical capabilities I used lookup tables and linear interpolation. The basic idea is simple: create an array with as much resolution as you need (more elements reduce the error created by interpolation). Then interpolate between lookup values.
Here is an example in processing (original dead link).
You can do this with your other trig functions as well. On the 6502 processor this allowed for full 3D wire frame graphics to be computed with an order of magnitude speed increase.
Here on SO I still don't have the privilege to comment (though I have at math.se) so this is actually a reply to Timo's post on diamond angles.
The whole concept of diamond angles based on the L1 norm is most interesting and if it were merely a comparison of which vector has a greater/lesser w.r.t. the positive X axis it would be sufficient. However, the OP did mention angle between two generic vectors, and I presume the OP wants to compare it to some tolerance for finding smoothness/corner status or sth like that, but unfortunately, it seems that with only the formulae provided on jsperf.com or freesteel.co.uk (links above) it seems it is not possible to do this using diamond angles.
Observe the following output from my Asymptote implementation of the formulae:
Vectors : 50,20 and -40,40
Angle diff found by acos : 113.199
Diff of angles found by atan2 : 113.199
Diamond minus diamond : 1.21429
Convert that to degrees : 105.255
Rotate same vectors by 30 deg.
Vectors : 33.3013,42.3205 and -54.641,14.641
Angle diff found by acos : 113.199
Diff of angles found by atan2 : 113.199
Diamond minus diamond : 1.22904
Convert that to degrees : 106.546
Rotate same vectors by 30 deg.
Vectors : 7.67949,53.3013 and -54.641,-14.641
Angle diff found by acos : 113.199
Diff of angles found by atan2 : 113.199
Diamond minus diamond : 1.33726
Convert that to degrees : 116.971
So the point is you can't do diamond(alpha)-diamond(beta) and compare it to some tolerance unlike you can do with the output of atan2. If all you want to do is diamond(alpha)>diamond(beta) then I suppose diamond is fine.
The cross product is proportional to the angle between two vectors, and when the vectors are normalized and the angle is small the cross product is very close to the actual angle in radians due to the small angle approximation.
specifically:
I1Q2-I2Q1 is proportional to the angle between I1Q1 and I2Q2.
The solution would be trivial if the vectors were defined/stored using polar coordinates instead of cartesian coordinates (or, 'as well as' using cartesian coordinates).
dot product of two vectors (x1, y1) and (x2, y2) is
x1 * x2 + y1 * y2
and is equivilent to the product of the lengths of the two vectors times the cosine of the angle between them.
So if you normalize the two vectors first (divide the coordinates by the length)
Where length of V1 L1 = sqrt(x1^2 + y1^2),
and length of V2 L2 = sqrt(x2^2 + y2^2),
Then normalized vectors are
(x1/L1, y1/L1), and (x2/L2, y2/L2),
And dot product of normalized vectors (which is the same as the cosine of angle between the vectors) would be
(x1*x2 + y1*y2)
-----------------
(L1*L2)
of course this may be just as computationally difficult as calculating the cosine
if you need to compute the square root, then consider using the invsqrt hack.
acos((x1*x2 + y1*y2) * invsqrt((x1*x1+y1*y1)*(x2*x2+y2*y2)));
The dot product might work in your case. It's not proportional to the angle, but "related".