I have two sequences of length n and m. Each is a sequence of points of the form (x,y) and represent curves in an image. I need to find how different (or similar) these sequences are given that fact that
one sequence is likely longer than the other (i.e., one can be half or a quarter as long as the other, but if they trace approximately the same curve, they are the same)
these sequences could be in opposite directions (i.e., sequence 1 goes from left to right, while sequence 2 goes from right to left)
I looked into some difference estimates like Levenshtein as well as edit-distances in structural similarity matching for protein folding, but none of them seem to do the trick. I could write my own brute-force method but I want to know if there is a better way.
Thanks.
Do you mean that you are trying to match curves that have been translated in x,y coordinates? One technique from image processing is to use chain codes [I'm looking for a decent reference, but all I can find right now is this] to encode each sequence and then compare those chain codes. You could take the sum of the differences (modulo 8) and if the result is 0, the curves are identical. Since the sequences are of different lengths and don't necessarily start at the same relative location, you would have to shift one sequence and do this again and again, but you only have to create the chain codes once. The only way to detect if one of the sequences is reversed is to try both the forward and reverse of one of the sequences. If the curves aren't exactly alike, the sum will be greater than zero but it is not straightforward to tell how different the curves are simply from the sum.
This method will not be rotationally invariant. If you need a method that is rotationally invariant, you should look at Boundary-Centered Polar Encoding. I can't find a free reference for that, but if you need me to describe it, let me know.
A method along these lines might work:
For both sequences:
Fit a curve through the sequence. Make sure that you have a continuous one-to-one function from [0,1] to points on this curve. That is, for each (real) number between 0 and 1, this function returns a point on the curve belonging to it. By tracing the function for all numbers from 0 to 1, you get the entire curve.
One way to fit a curve would be to draw a straight line between each pair of consecutive points (it is not a nice curve, because it has sharp bends, but it might be fine for your purpose). In that case, the function can be obtained by calculating the total length of all the line segments (Pythagoras). The point on the curve corresponding to a number Y (between 0 and 1) corresponds to the point on the curve that has a distance Y * (total length of all line segments) from the first point on the sequence, measured by traveling over the line segments (!!).
Now, after we have obtained such a function F(double) for the first sequence, and G(double) for the second sequence, we can calculate the similarity as follows:
double epsilon = 0.01;
double curveDistanceSquared = 0.0;
for(double d=0.0;d<1.0;d=d+epsilon)
{
Point pointOnCurve1 = F(d);
Point pointOnCurve2 = G(d);
//alternatively, use G(1.0-d) to check whether the second sequence is reversed
double distanceOfPoints = pointOnCurve1.EuclideanDistance(pointOnCurve2);
curveDistanceSquared = curveDistanceSquared + distanceOfPoints * distanceOfPoints;
}
similarity = 1.0/ curveDistanceSquared;
Possible improvements:
-Find an improved way to fit the curves. Note that you still need the function that traces the curve for the above method to work.
-When calculating the distance, consider reparametrizing the function G in such a way that the distance is minimized. (This means you have an increasing function R, such that R(0) = 0 and R(1)=1,
but which is otherwise general. When calculating the distance you use
Point pointOnCurve1 = F(d);
Point pointOnCurve2 = G(R(d));
Subsequently, you try to choose R in such a way that the distance is minimized. (to see what happens, note that G(R(d)) also traces the curve)).
Why not do some sort of curve fitting procedure (least-squares whether it be ordinary or non-linear) and see if the coefficients on the shape parameters are the same. If you run it as a panel-data sort of model, there are explicit statistical tests whether sets of parameters are significantly different from one another. That would solve the problem of the the same curve but sampled at different resolutions.
Step 1: Canonicalize the orientation. For example, let's say that all curved start at the endpoint with lowest lexicographic order.
def inCanonicalOrientation(path):
return path if path[0]<path[-1] else reversed(path)
Step 2: You can either be roughly accurate, or very accurate. If you wish to be very accurate, calculate a spline, or fit both curves to a polynomial of appropriate degree, and compare coefficients. If you'd like just a rough estimate, do as follows:
def resample(path, numPoints)
pathLength = pathLength(path) #write this function
segments = generateSegments(path)
currentSegment = next(segments)
segmentsSoFar = [currentSegment]
for i in range(numPoints):
samplePosition = i/(numPoints-1)*pathLength
while samplePosition > pathLength(segmentsSoFar)+currentSegment.length:
currentSegment = next(segments)
segmentsSoFar.insert(currentSegment)
difference = samplePosition - pathLength(segmentsSoFar)
howFar = difference/currentSegment.length
yield Point((1-howFar)*currentSegment.start + (howFar)*currentSegment.end)
This can be modified from a linear resampling to something better.
def error(pathA, pathB):
pathA = inCanonicalOrientation(pathA)
pathB = inCanonicalOrientation(pathB)
higherResolution = max([len(pathA), len(pathB)])
resampledA = resample(pathA, higherResolution)
resampledB = resample(pathA, higherResolution)
error = sum(
abs(pointInA-pointInB)
for pointInA,pointInB in zip(pathA,pathB)
)
averageError = error / len(pathAorB)
normalizedError = error / Z(AorB)
return normalizedError
Where Z is something like the "diameter" of your path, perhaps the maximum Euclidean distance between any two points in a path.
I would use a curve-fitting procedure, but also throw in a constant term, i.e. 0 =B0 + B1*X + B2*Y + B3*X*Y + B4*X^2 etc. This would catch the translational variance and then you can do a statistical comparison of the estimated coefficients of the curves formed by the two sets of points as a way of classifying them. I'm assuming you'll have to do bi-variate interpolation if the data form arbitrary curves in the x-y plane.
Related
I have a similar line graph plotted using R plot function (plot(df))
I want to get distance of the whole line between two points in the graph (e.g., between x(1) and x(3)). How can I do this?
If your function is defined over a fine grid of points, you can compute the length of the line segment between each pair of points and add them. Pythagoras is your friend here:
To the extent that the points are not close enough together that the function is essentially linear between the points, it will tend to (generally only slightly) underestimate the arc length.
Note that if your x-values are stored in increasing order, these ẟx and ẟy values can be obtained directly by differencing (in R that's diff)
If you have a functional form for y as a function of x you can apply the integral for the arc length -- i.e. integrate
∫ √[1+(dy/dx)²] dx
between a and b. This is essentially just the calculation in 1 taken to the limit.
If both x and y are parametric functions of another variable (t, say) you can simplify the parametric form of the above integral (if we don't forget the Jacobian) to integrating
∫ √[(dx/dt)²+(dy/dt)²] dt
between a and b
(Note the direct parallel to 1.)
if you don't have a convenient-to-integrate functional form in 2. or 3. you can use numerical quadrature; this can be quite efficient (which can be handy when the derivative function is expensive to evaluate).
The only equation to calculate this that I can find involves t in the range [0, 1], but I have no idea how long it will take to travel the entire path, so I can't calculate (1 - t).
I know the speed at which I'm traveling, but it seems to be a heavy idea to calculate the total time beforehand (nor do I actually know how to do that calculation). What is an equation to figure out the position without knowing the total time?
Edit To clarify on the cubic bezier curve: I have four control points (P0 to P1), and to get a value on the curve with t, I need to use the four points as such:
B(t) = (1-t)^3P0 + 3t(1-t)^2P1 + 3t^2(1-t)P2 + t^3P3
I am not using a parametric equation to define the curve. The control points are what define the curve. What I need is an equation that does not require the use of knowing the range of t.
I think there is a misunderstanding here. The 't' in the cubic Bezier curve's definition does not refer to 'time'. It is parameter that the x, y or even z functions based on. Unlike the traditional way of representing y as a function of x, such as y=f(x), an alternative way of representing a curve is by the parametric form that represents x, y and z as functions of an additional parameter t, C(t)=(x(t), y(t), z(t)). Typically the t value will range from 0 to 1, but this is not a must. The common representation for a circle as x=cos(t) and y=sin(t) is an example of parametric representation. So, if you have the parametric representation of a curve, you can evaluate the position on the curve for any given t value. It has nothing to do with the time it takes to travel the entire path.
You have the given curve and you have your speed. To calculate what you're asking for you need to divide the total distance by the speed you traveled given that time. That will give you the parametric (t) you need. So if the total curve has a distance of 72.2 units and your speed is 1 unit then your t is 1/72.2.
Your only missing bit is calculating the length of a given curve. This is typically done by subdividing it into line segments small enough that you don't care, and then adding up the total distance of those line segments. You could likely combine those two steps as well if you were so inclined. If you have your given speed, just iteration like 1000th of the curve add the line segment between the start and point 1000th of the way through the curve, and subtract that from how far you need to travel (given that you have speed and time, you have distance you need to travel), and keep that up until you've gone as far as you need to go.
The range for t is between 0 and 1.
x = (1-t)*(1-t)*(1-t)*p0x + 3*(1-t)*(1-t)*t*p1x + 3*(1-t)*t*t*p2x + t*t*t*p3x;
y = (1-t)*(1-t)*(1-t)*p0y + 3*(1-t)*(1-t)*t*p1y + 3*(1-t)*t*t*p2y + t*t*t*p3y;
Given the points of a line and a quadratic bezier curve, how do you calculate their nearest point?
There exist a scientific paper regarding this question from INRIA: Computing the minimum distance between two Bézier curves (PDF here)
I once wrote a tool to do a similar task. Bezier splines are typically parametric cubic polynomials. To compute the square of the distance between a cubic segment and a line, this is just the square of the distance between two polynomial functions, itself just another polynomial function! Note that I said the square of the distance, not the square root.
Essentially, for any point on a cubic segment, one could compute the square of the distance from that point to the line. This will be a 6th order polynomial. Can we minimize that square of the distance? Yes. The minimum must occur where the derivative of that polynomial is zero. So differentiate, getting a 5th order polynomial. Use your favorite root finding tool that generates all of the roots numerically. Jenkins & Traub, whatever. Choose the correct solution from that set of roots, excluding any solutions that are complex, and only picking a solution if it lies inside the cubic segment in question. Make sure you exclude the points that correspond to local maxima of the distance.
All of this can be efficiently done, and no iterative optimizer besides a polynomial root finder need be used, thus one does not require the use of optimization tools that require starting values, finding only a solution near that starting value.
For example, in the 3-d figure I show a curve generated by a set of points in 3-d (in red), then I took another set of points that lay in a circle outside, I computed the closest point on the inner curve from each, drawing a line down to that curve. These points of minimum distance were generated by the scheme outlined above.
I just wanna give you a few hints, in for the case Q.B.Curve // segment :
to get a fast enough computation, i think you should first think about using a kind of 'bounding box' for your algorithm.
Say P0 is first point of the Q. B. Curve, P2 the second point, P1 the control point, and P3P4 the segment then :
Compute distance from P0, P1, P2 to P3P4
if P0 OR P2 is nearest point --> this is the nearest point of the curve from P3P4. end :=).
if P1 is nearest point, and Pi (i=0 or 1) the second nearest point, the distance beetween PiPC and P3P4 is an estimate of the distance you seek that might be precise enough, depending on your needs.
if you need to be more acurate : compute P1', which is the point on the Q.B.curve the nearest from P1 : you find it applying the BQC formula with t=0.5. --> distance from PiP1' to P3P4 is an even more accurate estimate -but more costly-.
Note that if the line defined by P1P1' intersects P3P4, P1' is the closest point of QBC from P3P4.
if P1P1' does not intersect P3P4, then you're out of luck, you must go the hard way...
Now if (and when) you need precision :
think about using a divide and conquer algorithm on the parameter of the curve :
which is nearest from P3P4 ?? P0P1' or P1'P2 ??? if it is P0P1' --> t is beetween 0 and 0.5 so compute Pm for t=0.25.
Now which is nearest from P3P4?? P0Pm or PmP1' ?? if it is PmP1' --> compute Pm2 for t=0.25+0.125=0.375 then which is nearest ? PmPm2 or Pm2P1' ??? etc
you will come to accurate solution in no time, like 6 iteration and your precision on t is 0.004 !! you might stop the search when distance beetween two points becomes below a given value. (and not difference beetwen two parameters, since for a little change in parameter, points might be far away)
in fact the principle of this algorithm is to approximate the curve with segments more and more precisely each time.
For the curve / curve case i would first 'box' them also to avoid useless computation, so first use segment/segment computation, then (maybe) segment/curve computation, and only if needed curve/curve computation.
For curve/curve, divide and conquer works also, more difficult to explain but you might figure it out. :=)
hope you can find your good balance for speed/accuracy with this :=)
Edit : Think i found for the general case a nice solution :-)
You should iterate on the (inner) bounding triangles of each B.Q.C.
So we have Triangle T1, points A, B, C having 't' parameter tA, tB, tC.
and Triangle T2, points D, E, F, having t parameter tD, tE, tF.
Initially we have tA=0 tB=0.5 tC= 1.0 and same for T2 tD=0, tE=0.5, tF=1.0
The idea is to call a procedure recursivly that will split T1 and/or T2 into smaller rectangles until we are ok with the precision reached.
The first step is to compute distance from T1 from T2, keeping track of with segments were the nearest on each triangle. First 'trick': if on T1 the segment is AC, then stop recursivity on T1, the nearest point on Curve 1 is either A or C. if on T2 the nearest segment is DF, then stop recursivity on T2, the nearest point on Curve2 is either D or F. If we stopped recursivity for both -> return distance = min (AD, AF, CD, CF). then if we have recursivity on T1, and segment AB is nearest, new T1 becomes : A'=A B= point of Curve one with tB=(tA+tC)/2 = 0.25, C=old B. same goes for T2 : apply recursivityif needed and call same algorithm on new T1 and new T2. Stop algorithm when distance found beetween T1 and T2 minus distance found beetween previous T1 and T2 is below a threshold.
the function might look like ComputeDistance(curveParam1, A, C, shouldSplitCurve1, curveParam2, D, F, shouldSplitCurve2, previousDistance) where points store also their t parameters.
note that distance (curve, segment) is just a particular case of this algorithm, and that you should implement distance (triangle, triangle) and distance (segment, triangle) to have it worked. Have fun.
1.Simple bad method - by iteration go by point from first curve and go by point from second curve and get minimum
2.Determine math function of distance between curves and calc limit of this function like:
|Fcur1(t)-Fcur2(t)| ->0
Fs is vector.
I think we can calculate the derivative of this for determine extremums and get nearest and farest points
I think about this some time later, and post full response.
Formulate your problem in terms of standard analysis: You have got a quantity to minimize (distance), so you formulate an equation for this quantity and find the points where the first derivatives are zero. Parameterize with a single parameter by using the curve's parameter p, which is between 0 for the first point and 1 for the last point.
In the line case, the equation is fairly simple: Get the x/y coordinates from the spline's equation and compute the distance to the given line via vector equations (scalar product with the line's normal).
In the curve's case, the analytical solution could get pretty complicated. You might want to use a numerical minimization technique such as Nelder-Mead or, since you have a 1D continuous problem, simple bisection.
In the case of a Bézier curve and a line
There are three candidates for the closest point to the line:
The place on the Bézier curve segment that is parallel to the line (if such a place exists),
One end of the curve segment,
The other end of the curve segment.
Test all three; the shortest distance wins.
In the case of two Bézier curves
Depends if you want the exact analytical result, or if an optimised numerical result is good enough.
Analytical result
Given two Bézier curves A(t) and B(s), you can derive equations for their local orientation A'(t) and B'(s). The point pairs for which A'(t) = B'(s) are candidates, i.e. the (t, s) for which the curves are locally parallel. I haven't checked, but I assume that A'(t) - B'(s) = 0 can be solved analytically. If your curves are anything like those you show in your example, there should be either only one solution or no solution to that equation, but there could be two (or infinitely many in the case where the curves identical but translated -- in which case you can ignore this because the winner will always be one of the curve segment endpoints).
In an approach similar to the curve-line case outline above, test each of these point pairs, plus the curve segment endpoints. The shortest distance wins.
Numerical result
Let's say the points on the two Bézier curves are defined as A(t) and B(s). You want to minimize the distance d( t, s) = |A(t) - B(s)|. It's a simple two-parameter optimization problem: find the s and t that minimize d( t, s) with the constraints 0 ≤ t ≤ 1 and 0 ≤ s ≤ 1.
Since d = SQRT( ( xA - xB)² + (yA - yB)²), you can also just minimize the function f( t, s) = [d( t, s)]² to save a square root calculation.
There are numerous ready-made methods for such optimization problems. Pick and choose.
Note that in both cases above, anything higher-order than quadratic Bézier curves can giver you more than one local minimum, so this is something to watch out for. From the examples you give, it looks like your curves have no inflexion points, so this concern may not apply in your case.
The point where there normals match is their nearest point. I mean u draw a line orthogonal to the line. .if that line is orthogonal to the curve as well then the point of intersection is the nearest point
I am trying to plot large amounts of points using some library. The points are ordered by time and their values can be considered unpredictable.
My problem at the moment is that the sheer number of points makes the library take too long to render. Many of the points are redundant (that is - they are "on" the same line as defined by a function y = ax + b). Is there a way to detect and remove redundant points in order to speed rendering ?
Thank you for your time.
The following is a variation on the Ramer-Douglas-Peucker algorithm for 1.5d graphs:
Compute the line equation between first and last point
Check all other points to find what is the most distant from the line
If the worst point is below the tolerance you want then output a single segment
Otherwise call recursively considering two sub-arrays, using the worst point as splitter
In python this could be
def simplify(pts, eps):
if len(pts) < 3:
return pts
x0, y0 = pts[0]
x1, y1 = pts[-1]
m = float(y1 - y0) / float(x1 - x0)
q = y0 - m*x0
worst_err = -1
worst_index = -1
for i in xrange(1, len(pts) - 1):
x, y = pts[i]
err = abs(m*x + q - y)
if err > worst_err:
worst_err = err
worst_index = i
if worst_err < eps:
return [(x0, y0), (x1, y1)]
else:
first = simplify(pts[:worst_index+1], eps)
second = simplify(pts[worst_index:], eps)
return first + second[1:]
print simplify([(0,0), (10,10), (20,20), (30,30), (50,0)], 0.1)
The output is [(0, 0), (30, 30), (50, 0)].
About python syntax for arrays that may be non obvious:
x[a:b] is the part of array from index a up to index b (excluded)
x[n:] is the array made using elements of x from index n to the end
x[:n] is the array made using first n elements of x
a+b when a and b are arrays means concatenation
x[-1] is the last element of an array
An example of the results of running this implementation on a graph with 100,000 points with increasing values of eps can be seen here.
I came across this question after I had this very idea. Skip redundant points on plots. I believe I came up with a far better and simpler solution and I'm happy to share as my first proposed solution on SO. I've coded it and it works well for me. It also takes into account the screen scale. There may be 100 points in value between those plot points, but if the user has a chart sized small, they won't see them.
So, iterating through your data/plot loop, before you draw/add your next data point, look at the next value ahead and calculate the change in screen scale (or value, but I think screen scale for the above-mentioned reason is better). Now do the same for the next value ahead (getting these values is just a matter of peeking ahead in your array/collection/list/etc adding the for next step increment (probably 1/2) to the current for value whilst in the loop). If the 2 values are the same (or perhaps very minor change, per your own preference), you can skip this one point in your chart by simply adding 'continue' in the loop, skipping adding the data point as the point lies exactly on the slope between the point before and after it.
Using this method, I reduce a chart from 963 points to 427 for example, with absolutely zero visual change.
I think you might need to perhaps read this a couple of times to understand, but it's far simpler than the other best solution mentioned here, much lighter weight, and has zero visual effect on your plot.
I would probably apply a "least squares" algorithm to obtain a line of best fit. You can then go through your points and downfilter consecutive points that lie close to the line. You only need to plot the outliers, and the points that take the curve back to the line of best fit.
Edit: You may not need to employ "least squares"; if your input is expected to hover around "y=ax+b" as you say, then that's already your line of best fit and you can just use that. :)
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.