Hi everyone I couldn't find any solution about this so I wanted to ask this on the forum.
While referencing the image below, I have a curve AB and I want to mirror the direction vector dB along the curve's surface. I only want the mirrored direction. So I need to use that curve as an axis to mirror that direction.
See image
2D Curve
Assuming that you have the normalized tangent t = normalize(a-b), use the following formula, where · is the dot product:
dA = dB - 2*t*(dB·t)
A bit of explanation: t*(dB·t) is the tangential component. Remove it once will give the normal component, a removing it twice will reverse the tangantial component, thus resulting in the reflected vector.
3D Surface
Assuming that you have the normal n of the surface, use the following formula:
dA = -(dB - n*(dB·n)) + n*(dB·n)
which can be simplified to dA = -dB + 2*n*(dB·n).
A bit of explanation: n*(dB·n) is the normal component. We first remove it from dB since we want to preserve it. dB - n*(dB·n) is thus the tangential component. Then, we negate the tangential component, and we add the normal component back to obtain the reflected vector.
Related
I'm working on a PyMEL script that allows the user to duplicate a selected object multiple times, using a CV curve and its points coordinates to transform & rotate each copy to a certain point in space.
In order to achieve this, Im using the adjacent 2 points of each CV (control vertex) to determine the rotation for the object.
I have managed to retrieve the coordinates of the curve's CVs
#Add all points of the curve to the cvDict dictionary
int=0
cvDict={}
while int<selSize:
pointName='point%s' % int
coords= pointPosition ('%s.cv[%s]' % (obj,int), w=1)
#Setup the key for the current point
cvDict[pointName]={}
#add coords to x,y,z subkeys to dict
cvDict[pointName]['x']= coords[0]
cvDict[pointName]['y']= coords[1]
cvDict[pointName]['z']= coords[2]
int += 1
Now the problem I'm having is figuring out how to get the angle for each CV.
I stumbled upon the angleBetween() function:
http://download.autodesk.com/us/maya/2010help/CommandsPython/angleBetween.html
In theory, this should be my solution, since I could find the "middle vector" (not sure if that's the mathematical term) of each of the curve's CVs (using the adjacent CVs' coordinates to find a fourth point) and use the above mentioned function to determine how much I'd have to rotate the object using a reference vector, for example on the z axis.
At least theoretically - the issue is that the function only takes 1 set of coords for each vector and I have absolutely no Idea how to convert my point coords to that format (since I always have at least 2 sets of coordinates, one for each point).
Thanks.
If you wanna go the long way and not grab the world transforms of the curve, definitely make use of pymel's datatypes module. It has everything that python's native math module does and a few others that are Maya specific. Also the math you would require to do this based on CVs can be found here.
Hope that puts you in the right direction.
If you're going to skip the math, maybe you should just create a locator, path-animate it along the curve, and then sample the result. That would allow you to get completely continuous orientations along the curve. The midpoint-constraint method you've outlined above is limited to 1 valid sample per curve segment -- if you wanted 1/4 of the way or 3/4 of the way between two cv's your orientation would be off. Plus you don't have to reinvent all of the manu different options for deciding on the secondary axis of rotation, reading curves with funky parameterization, and so forth.
I'm trying to take an array of 3D points and a plane and divide the points up into 2 arrays based on which side of the plane they are on. Before I get to heavily into debugging I wanted to post what I'm planning on doing to make sure my understanding of how to do this will work.
Basically I have the plane with 3 points and I use (pseudo code):
var v1 = new vector(plane.b.x-plane.a.x, plane.b.y-plane.a.y, plane.b.z-plane.a.z);
var v2 = new vector(plane.c.x-plane.a.x, plane.c.y-plane.a.y, plane.c.z-plane.a.z);
I take the cross product of these two vectors to get the normal vector.
Then I loop through my array of points and turn them into vectors and calculate the dot product against the normal.
Then i use the dot product to determine the side that the point is on.
Does this sound like it would work?
Let a*x+b*y+c*z+d=0 be the equation determining your plane.
Substitute the [x,y,z] coordinates of a point into the left hand side of the equation (I mean the a*x+b*y+c*z+d) and look at the sign of the result.
The points having the same sign are on the same side of the plane.
Honestly, I did not examine the details of what you wrote. I guess you agree that what I propose is simpler.
Following the 'put points into the plane's equation and check the sign' approach given previously. The equation can be easily obtained using SymPy. I used it to find location of points (saved as numpy arrays) in a list of points.
from sympy import Point3D, Plane
plane=Plane(Point3D(point1), Point3D(point2), Point3D(point3))
for point in pointList:
if plane.equation(x=point[0], y=point[1],z=point[2]) > 0:
print "point is on side A"
else:
print "point is on side B"
I haven't tested its speed compared to other methods mentioned above but is definitely the easiest method.
Your approach sounds good. However, when you say "and turn them into vectors", it might not be good (depending on the meaning of your sentence).
You should "turn your points into vector" by computing the difference in terms of coordinates between the current point and one of the points in the plane (for example, one of the 3 points defining the plane). As you wrote it, it sounds like you might have misunderstood that ; but apart from that, it's ok!
take into account the normal vector of the plane
example: for the point A=[-243.815437431962, -41.7407630281635, 10.0]
equation= -2663.1860000000006*Z +21305.488000000005=0
RESULt POSITIVE
but if equation= 2663.1860000000006*Z -21305.488000000005=0
RESULT NEGATIVE
I have a complicated problem and it involves an understanding of Maths I'm not confident with.
Some slight context may help. I'm building a 3D train simulator for children and it will run in the browser using WebGL. I'm trying to create a network of points to place the track assets (see image) and provide reference for the train to move along.
To help explain my problem I have created a visual representation as I am a designer who can script and not really a programmer or a mathematician:
Basically, I have 3 shapes (Figs. A, B & C) and although they have width, can be represented as a straight line for A and curves (B & C). Curves B & C are derived (bend modified) from A so are all the same length (l) which is 112. The curves (B & C) each have a radius (r) of 285.5 and the (a) angle they were bent at was 22.5°.
Each shape (A, B & C) has a registration point (start point) illustrated by the centre of the green boxes attached to each of them.
What I am trying to do is create a network of "track" starting at 0, 0 (using standard Cartesian coordinates).
My problem is where to place the next element after a curve. If it were straight track then there is no problem as I can use the length as a constant offset along the y axis but that would be boring so I need to add curves.
Fig. D. demonstrates an example of a possible track layout but please understand that I am not looking for a static answer (based on where everything is positioned in the image), I need a formula that can be applied no matter how I configure the track.
Using Fig. D. I tried to work out where to place the second curved element after the first one. I used the formula for plotting a point of the circumference of a circle given its centre coordinates and radius (Fig. E.).
I had point 1 as that was simply a case of setting the length (y position) of the straight line. I could easily work out the centre of the circle because that's just the offset y position, the offset of the radius (r) (x position) and the angle (a) which is always 22.5° (which, incidentally, was converted to Radians as per formula requirements).
After passing the values through the formula I didn't get the correct result because the formula assumed I was working anti-clockwise starting at 3 o'clock so I had to deduct 180 from (a) and convert that to Radians to get the expected result.
That did work and if I wanted to create a 180° track curve I could use the same centre point and simply deducted 22.5° from the angle each time. Great. But I want a more dynamic track layout like in Figs. D & E.
So, how would I go about working point 5 in Fig. E. because that represents the centre point for that curve segment? I simply have no idea.
Also, as a bonus question, is this the correct way to be doing this or am I over-complicating things?
This problem is the only issue stopping me from building my game and, as you can appreciate, it is a bit of a biggie so I thank anyone for their contribution in advance.
As you build up the track, the position of the next piece of track to be placed needs to be relative to location and direction of the current end of the track.
I would store an (x,y) position and an angle a to indicate the current point (with x,y starting at 0, and a starting at pi/2 radians, which corresponds to straight up in the "anticlockwise from 3-o'clock" system).
Then construct
fx = cos(a);
fy = sin(a);
lx = -sin(a);
ly = cos(a);
which correspond to the x and y components of 'forward' and 'left' vectors relative to the direction we are currently facing. If we wanted to move our position one unit forward, we would increment (x,y) by (fx, fy).
In your case, the rule for placing a straight section of track is then:
x=x+112*fx
y=y+112*fy
The rule for placing a curve is slightly more complex. For a curve turning right, we need to move forward 112*sin(22.5°), then side-step right 112*(1-cos(22.5°), then turn clockwise by 22.5°. In code,
x=x+285.206*sin(22.5*pi/180)*fx // Move forward
y=y+285.206*sin(22.5*pi/180)*fy
x=x+285.206*(1-cos(22.5*pi/180))*(-lx) // Side-step right
y=y+285.206*(1-cos(22.5*pi/180))*(-ly)
a=a-22.5*pi/180 // Turn to face new direction
Turning left is just like turning right, but with a negative angle.
To place the subsequent pieces, just run this procedure again, calculating fx,fy, lx and ly with the now-updated value of a, and then incrementing x and y depending on what type of track piece is next.
There is one other point that you might consider; in my experience, building tracks which form closed loops with these sort of pieces usually works if you stick to making 90° turns or rather symmetric layouts. However, it's quite easy to make tracks which don't quite join up, and it's not obvious to see how they should be modified to allow them to join. Something to bear in mind perhaps if your program allows children to design their own layouts.
Point 5 is equidistant from 3 as 2, but in the opposite direction.
Apologies if this is considered a repeat question, but the answers I've seen on here are too complex for my needs.
I simply need to find out if a line segment intersects a circle. I don't need to find the distance to the line from the circle center, I don't need to solve for the points of intersection.
The reason I need something simple is that I have to code this in SQL and am unable to call out to external libraries, and need to write this formula in a WHERE clause... basicaly it has to be done in a single statement that I can plug values in to.
Assuming 2 points A (Ax,Ay) and B (Bx,By) to describe the line segment, and a circle with center point C (Cx,Cy) and radius R, the formula I am currently using is:
( RR ( (Ax-Bx)(Ax-Bx) + (Ay-By)(Ay-By) ) )
-( ((Ax-Cx)(By-Cy))-((Bx-Cx)(Ay-Cy)) ) > 0
This formula is taken from link text, and is based on a 0,0 centered circle.
The reason I am posting is that I am getting weird results and I wondered if I did something stupid. :(
although this doesn't exactly answer your question: Do you really have to calculate this on the fly on a SQL-Select? This means that the DB-system has to calculate the formula for every single row in the table (or every single row for which the remaining where conditions hold, respectively) which might result in bad performance.
Instead, you might consider creating a separate boolean column and calculate its value in an on-insert/on-update trigger. There, in turn, you wouldn't even need to put the test in a single line formula. Using a separate column has another advantage: You can create an index on that column which allows you to get your set of intersecting/non-intersecting records very fast.
I am trying to animate an object, let's say its a car. I want it go from point
x1,y1,z1
to point x2,y2,z2 . It moves to those points, but it appears to be drifting rather than pointing in the direction of motion. So my question is: how can I solve this issue in my updateframe() event? Could you point me in the direction of some good resources?
Thanks.
First off how do you represent the road?
I recently done exactly this thing and I used Catmull-Rom splines for the road. To orient an object and make it follow the spline path you need to interpolate the current x,y,z position from a t that walks along the spline, then orient it along the Frenet Coordinates System or Frenet Frame for that particular position.
Basically for each point you need 3 vectors: the Tangent, the Normal, and the Binormal. The Tangent will be the actual direction you will like your object (car) to point at.
I choose Catmull-Rom because they are easy to deduct the tangents at any point - just make the (vector) difference between 2 other near points to the current one. (Say you are at t, pick t-epsilon and t+epsilon - with epsilon being a small enough constant).
For the other 2 vectors, you can use this iterative method - that is you start with a known set of vectors on one end, and you work a new set based on the previous one each updateframe() ).
You need to work out the initial orientation of the car, and the final orientation of the car at its destination, then interpolate between them to determine the orientation in between for the current timestep.
This article describes the mathematics behind doing the interpolation, as well as some other things to do with rotating objects that may be of use to you. gamasutra.com in general is an excellent resource for this sort of thing.
I think interpolating is giving the drift you are seeing.
You need to model the way steering works .. your update function should 1) move the car always in the direction of where it is pointing and 2) turn the car toward the current target .. one should not affect the other so that the turning will happen and complete more rapidly than the arriving.
In general terms, the direction the car is pointing is along its velocity vector, which is the first derivative of its position vector.
For example, if the car is going in a circle (of radius r) around the origin every n seconds then the x component of the car's position is given by:
x = r.sin(2πt/n)
and the x component of its velocity vector will be:
vx = dx/dt = r.(2π/n)cos(2πt/n)
Do this for all of the x, y and z components, normalize the resulting vector and you have your direction.
Always pointing the car toward the destination point is simple and cheap, but it won't work if the car is following a curved path. In which case you need to point the car along the tangent line at its current location (see other answers, above).
going from one position to another gives an object a velocity, a velocity is a vector, and normalising that vector will give you the direction vector of the motion that you can plug into a "look at" matrix, do the cross of the up with this vector to get the side and hey presto you have a full matrix for the direction control of the object in motion.