Interception of two objects in 3D space - vector

I am trying to workout the interception of two objects and try to code it into an application
A cannon is in Position A and a plane is in position B
The plane is moving with vector b (without affected by gravity)
A cannonball is shot (affected by gravity) with unit vector a and magnitude of m
They intercept at position C at T seconds
Knowns: A, m, B, b
Unknowns: a, C, T
The only thing I can think of to solve it in terms of code is split the equations into X, Y and Z component and substitute T as a value and increment it.
It will be nice if someone can kindly tell how to find one of the unknowns
Thanks

The answer is in here:
https://blog.forrestthewoods.com/solving-ballistic-trajectories-b0165523348c
from meowgoesthedog

Related

Having 3 vectors, how to check that a straight line can be drawn through them

I get the positions of 5 enemies in the game in vectors. Depending on the distance I choose, the number of enemies can vary from 0 to 5. I need to know their vectors each time to check whether it is possible to draw a straight line through a certain number of heroes (vectors).
After that, my hero will have to use his ability called wall. It consists of 2 start and end vectors. Thus, check whether my hero can put a wall on the enemies in the line to catch them
Let's say there are 3 enemy heroes whose positions I can get. I need to find out if I can pass through them directly, in order to use the ability on them.
Here's what using the ability looks like in the game
Here is getting the vector of one of the heroes
The ability itself can be twisted at a certain point. But anyway, it is necessary that the wall would touch several heroes
Wherever I move the mouse, I can put it in the desired position. But unfortunately it takes a lot of time, so I would like to automate
The coordinates of the wall itself, or rather its two edges, I can also get, but only after the ability has been used
If one prefers geometry to linear algebra...
Then One can compute the dot product of (unit-vector1. Unit-Vector2). That is equal to the SIN of the angle between them.
So if unit vector is the shooter position to target1, unit vector2 is the shooter to target2, etc... then when DOTPRODUCT(Vector1,vector2) = 1 and DOTPRODUCT(Vector1,vector3) = 1, then the three points are in syzygy.
And repeat from shooter to as many targets as you have to determine whether some or all of the points are in syzygy.
From your statement that there is a start and an endpoint I take that you select two enemys and want to trap anything in between.
So you're actually not looking for a straight line that can be drawn through your enemy positions but if they are withn a rectangle. It would be very unlikely and for more points nearly impossible that they are all collinear anyway.
So it becomes quite trivial. You draw a line through start and end enemy. Then you check the remaining enemies distance to that line vs the width of your AoE. Maye you want to also handle some body width in that calculation.
https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line
You can describe all points belonging to line (x0, y0) + (dx, dy)t = (x1, y1). Chose any two points and t as 1 and you will get (dx, dy) for line connecting two dots. Now you will need to find distance between this line and (x2, y2). it is distance between (x2, y2) and (xd, yd), where on one hand (xd, yd) = (x0, y0) + t1(dx, dy) and on other hand (xd, yd) = (x2, y2) + t2*(-dy, dx). Solving this two equations you will find t1, t2, (xd, yd) and distance between (x2, y2) and (xd, yd), which is distance between (x2, y2) and line, connecting (x0, y0) and (x1, y1).
Knowing this, you select dots with min_x and max_x and calculate ditance between line, connecting said dots and rest of the dots. If distance is lesser then some threshold of your choice, then you can assume that you can have line passing through all dots.
Any line in the plane can be described by an equation a*x + b*y + c = 0 with (a, b) ≠ (0, 0). Note that if you have an equation of this form, then multiplying each coefficient a, b, c with the same number yields an equation describing the same line. That's the reason (a, b, c) is called a homogeneous coordinate vector for that line.
How do you find a, b, c? One simple approach would be treating this as three linear equations in three unknowns. You plug in the x and y coordinates for all your three points, and get tree equations for a thorough c. However, there is a catch. Since the right hand side of each equation is zero, a = b = c = 0 is always a solution. In those cases where there is only one solution, that will be it. So in order for there to be a line, you need more than one solution. The mathematical tool to determine whether a set of equations had more than one solution is the determinant. It is zero if the system has no single unique solution.
Long story short: three points are collinear (on a line) if
⎛x1 y1 1⎞
det ⎜x2 y2 1⎟ = 0
⎝x3 y3 1⎠
The homogeneous coordinate vector describing the line world correspond to the kernel of that matrix.
Of course, if your input coordinates are floating point numbers, exact zero is unlikely. Presumably that wall does allow for some error in some way, and you'd need to tell us about that in order to get an answer that models this aspect correctly. In the mean time, know that the absolute value of the determinant above is proportional to the area of the triangle created by these three points. So if your were to pick a constant threshold value, the farther your enemies are apart along the direction of the wall, the less they could deviate from the straight line without violating that threshold.

Restricting a point to an arc on a circle's circumference

This question is a bit specific. Assume for simplicity that I have a unit circle, i.e. a circle centered on the origin (0,0) with radius 1. On this circle I have three points: A, B, and C. Assume that B and C are fixed, and that A can be moved along the circle's circumference.
Question: What conditions can I check efficiently to ensure that A is not moved outside the arc between B and C originally containing it?
As a point of clarification: I do not just wish to detect whether the point is on this arc or not (i.e., get a boolean), but to have a quantifiable measure (for example, of distance) so I can prevent it from being moved outside. You can imagine my setup like this, restricting point A to the red arc:
Here are some ideas I have pondered so far, but none were successful:
Limit the angle alpha of A = [cos(alpha),sin(alpha)] to the origin between the angles beta and gamma of B=[cos(beta),sin(beta)] and C=[cos(gamma),sin(gamma)], where alpha, beta, and gamma are angles in radians. Unfortunately, this naive case only works in scenario (a) above. If the arc to which A is restricted crosses the (in my case, western) discontinuity from +pi/-pi, a naive upper and lower bound won't do.
Calculate the length of the arcs between A-to-B and A-to-C, then ensure that as A is moved this sum does not change. Unfortunately, I have only found this solution to calculate the arcs between two points, and it always calculates the shorter arc. In scenario (c), however, the correct arc A-to-B is the larger one.
Given any two points on a circle, in your case B and C, there are of course two possible arcs. We use A to select between the two options. You say you want to prevent a point, say D, from moving outside this arc. I'm interpreting this to mean that we want a function which, if D lies on the arc defined by BC and A, returns D, otherwise it returns B or C, depending on which is nearer to D.
We can define the following function to implement this scheme:
def contstrainPoint(A, B, C, D)
M = midpoint(B, C)
If dot(MA, MD) >= 0
return D
else if(dot(MB, MD) >= 0
return B
else
return C
Where M is the midpoint of the chord BC and dot is the dot product function.
Take the cross product (A-B)x(A-C). This will have one non-zero component, normal to the plane. That component will vary smoothly, positive when A is on one arc, negative when it is on the other, and zero when it crosses B or C.

Calculate 3D Euler angle between two 3D vectors?

Say I have 2 3D vectors, each describing a direction in 3D space (or a rotation, but I'm not sure if that terminology is correct). How would I calculate the difference between the two vectors as an Euler angle? That is, if I applied the angle to the first vector, it would rotate to equal the other? I understand how Euler angles have issues and are implementation-dependant, but I don't know what the implications of this are on a question such as mine.
To clarify a little, when I say "3D vectors", I'm picturing the "translation" gizmo you get in most 3D modelling packages or in Unity (which is what I'm using).
EDIT: Actually I just reviewed the "vectors" that I'm using, and what I said is not quite correct. I actually have 6 vectors, 3 for each rotation. Each vector is a position in 3D space offset from the centre of rotation. This probably makes an already-difficult question near-impossible, right?
ADDITIONAL INFORMATION: Ok, so I've worked out what I actually want to ask (because this question is really badly done), and it applies more to Unity, so I've asked a more Unity-specific question over on Unity Answers.
I'm don't normally understand mathematical formulae that are posted online, so C++-styled pseudo-code would be by far the most helpful for me.
Any help would be much appreciated, and if my question lacks certain information, please just ask for more :)
If you are going to do 3d you need to understand linear algebra and matrix notation.[1] Affine 4x4 matrices are the basis of space transformations in all 3d applications I've ever seen, (Euler angles just give alternate means to describe that matrix). Even unity uses matrices, although to be able to efficiently do the Euler-Lagrange particle motion equation they prefer to have the decomposed form. A matrix encodes a entire space with 4 vectors. This is conceptually easy in this case (not the only use for matrices), the matrix encodes the directions x y z and offset vector w.
The reason matrix notation is useful is: It becomes possible to manipulate the things like normal math symbols. If you remember from school solving x from:
a * x = b
Divide both sides by a and you get
a/a * x = b /a ->
x = b / a
Now if you have 2 spaces with 3 vectors each you essentially have 2 fully formed spaces at origin. Assuming the vectors span a 3D space (in other words dont point all in one plane, its even better if they are orthogonal to each other in which case you can just use transformation functions directly). That means you have 3 spaces. So your problem is given you know 2 spaces. You need to know the space transform form space A -> space B (its customary to give matrices big letters to denote they are more complex). This is mathematically:
A * X = B
Where * is a matrix multiplications and A, X and B are transformation matrices. So then divide by A, but alas there's no matrix division, fortunately there is inverse and division is multiplication by inverse so that's what we do instead. Also one other note rotations are not commutative so we need to specify on which side we multiply so because A is before X we multiply on the left hand side with inverse of A. So we get:
A^-1 * A * X = A^-1 * B
Where ^-1 denotes matrix inverse. This simplifies to :
I * X = A^-1 * B ->
X = A^-1 * B
X is the rotation space. In unity code this looks like:
X = A.inverse * B
Where X, A and B are Matrix4x4 elements The language you use may have other conventions I'm using the java script reference here. You can covert this matrix to a quaternion and from there to Euler angles an example of this can be found here.
How to form A and B from the vectors? Just put the vector for starting space to A's columns 0-2 and destination spaces correspondingly to B columns[2].
[1] Yes its compulsory, its much simpler than it may seem at first. While you can live quite far without them they aren't any harder to use than saying rotate about x axis fro so and so. Also learn quats.
[2] I should check this, but unity seems to use column matrices so it should be right
PS: By the way if you have noisy data and more then 3 vectors per instance then you can use least squares to average the matrix t a 3 by 3 sub matrix.

Tangent circle(s) for two other circles?

There are two circles: a centered at point A, and circle b (center at B). What is the equation to calculate 2D position of all or none tangent circles possible. Main constraint is, that radius is the same for all the circles. As far as I know, there should be either no solution (figure 2), or 2 solutions (figure 1). How to find out if there are solutions, and also position of centers of those solutions (C and D).
Figure 1: 2 solutions should be possible here
Figure 2: No solutions!
Update (solution):
1) Calculate distance from A to B -> |AB|:
2) Checks whether a solution exist, it exist only if:
3) If it exist, calculate half-point between points A and B:
4) Create normalized perpendicular vector to line segment AB:
5) Calculate distance from this H point to C point -> |HC|:
6) Finally calculate point C along the (HC) starting at X at distance |HC|:
I suppose this question should migrate to a more math related site.
Try to imagine where these two tangent circles go when the circles a and b get further and further apart. They get closer to the line AB. Once the AB segment equals 4r these two tangent circles will overlap. From now on, once circles a and b get further apart, there's no tangent circles whatsoever.
If you want to calculate the position of these circles, just assume that the distance between the centers is always 2r:
You should get two, one or none solutions for xC and yC, which will be the centers of your tangent circles. I hope I haven't messed something up.
Solutions
Provided you do know there are solutions ( just check if d(A,B) <= 4r ), these are the coordinates of your two circles:
http://pastebin.com/LeW7Ws98
A little scary, eh? But it's working. There are the following variables:
x_A, y_A - the coordinates of the circle A,
x_B, y_B - the coordinates of the circle B,
r - the radius.
I've checked the solutions with the values from one of my comments below. I think that you can copy these solutions and inject them into your code straight away (provided there's a sqrt function) and get the results after declaring some variables.
These solutions are loosely derived from the Save's proposition but I couldn't comment below his answer - I've got less than 50 reputation points, duh ... ( thanks SO! You're the man! ). However I'm pretty sure they should be valid for my system anyways. Cheers
A solution exists iff d(A,B) = sqrt(2)*2*r
To find the center of the solution circles, that will let you draw the circonferences, you can intersect the circle with center (x_m,y_m), that is the medium point of the segment AB, of radius sqrt(2)*r, with the line perpendicular to AB and passing from (x_m,y_m)
This should give you all the needed information to check if a solution exixsts, and if it does, to draw it.

How to determine if a point approximates an extension to the line between two other points?

I have a three points (A,B,C) that denote objects moving in 2D space. For each node I know its position and its velocity vector. All three objects are moving in the same direction.
I would like to know whether a point C (x3, y3) approximates a "positive" extension to the line formed by points A(x1, y1) and B(x2, y2). That is, I would like to know whether point C is "ahead" of point B (i.e "A->B->C" and not "C->A->B").
I know that checking if points A, B, C are collinear will give me an indication of all three points are lying on the same line, however, i cannot figure out whether point C approximates a positive extension to the line.
Any suggestion would be highly appreciated.
You can calculate the scalar product of the difference vectors AB and BC. If that is positive, then C is what you call 'in front of B. It may be way off to the left or right, though.
The scalar product would be calculated as
(b1-a1)x(c1-b1) + (b2-a2)x(c2-b2).
when A=(a1, a2), B=(b1, b2), C= (c1,c2) - it is the cos of the angle between the two vectors times the lengths of the vectors, and cos is positive for angles less than 90 degree.

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