I am currently looking to implement an algorithm that will be able to compute the arc midpoint. From here on out, I will be referring to the diagram below. What is known are the start and end nodes (A and B respectively), the center (point C) and point P which is the intersection point of the line AB and CM (I am able to find this point without knowing point M because line AB is perpendicular to line CM and thus, the slope is -1/m). I also know the arc angle and the radius of the arc. I am looking to find point M.
I have been looking at different sources. Some suggest converting coordinates to polar, computing the mid point from the polar coordinates then reverting back to Cartesian. This involves sin and cos (and arctan) which I am a little reluctant to do since trig functions take computing time.
I have been looking to directly computing point M by treating the arc as a circle and having Line CP as a line that intersects the circle at Point M. I would then get two values and the value closest to point P would be the correct intersection point. However, this method, the algebra becomes long and complex. Then I would need to create special cases for when P = C and for when the line AB is horizontal and vertical. This method is ok but I am wondering if there are any better methods out there that can compute this point that are simpler?
Also, as a side note, I will be creating this algorithm in C++.
A circumference in polar form is expressed by
x = Cx + R cos(alpha)
y = Cy + R sin(alpha)
Where alpha is the angle from center C to point x,y. The goal now is how to get alpha without trigonometry.
The arc-midpoint M, the point S in the middle of the segment AB, and your already-calculated point P, all of them have the same alpha, they are on the same line from C.
Let's get vector vx,vy as C to S. Also calculate its length:
vx = Sx - Cx = (Ax + Bx)/2 - Cx
vy = Sy - Cy = (Ay + By)/2 - Cy
leV = sqrt(vx * vx + vy * vy)
I prefer S to P because we can avoid some issues like infinite CP slope or sign to apply to slope (towards M or its inverse).
By defintions of sin and cos we know that:
sin(alpha) = vy / leV
cos(alpha) = vx / leV
and finally we get
Mx = Cx + R * vx / leV
My = Cy + R * vy / leV
Note: To calculate Ryou need another sqrt function, which is not quick, but it's faster than sin or cos.
For better accuracy use the average of Ra= dist(AC) and Rb= dist(BC)
I would then get two values
This is algebraically unavoidable.
and the value closest to point P would be the correct intersection point.
Only if the arc covers less than 180°.
Then I would need to create special cases for when P = C
This is indeed the most tricky case. If A, B, C lie on a line, you don't know which arc is the arc, and won't be able to answer the question. Unless you have some additional information to start with, e.g. know that the arc goes from A to B in a counter-clockwise direction. In this case, you know the orientation of the triangle ABM and can use that to decide which solition to pick, instead of using the distance.
and for when the line AB is horizontal and vertical
Express a line as ax + by + c = 0 and you can treat all slopes the same. THese are homogeneous coordinates of the line, you can compute them e.g. using the cross product (a, b, c) = (Ax, Ay, 1) × (Bx, By, 1). But more detailed questions on how best to compute these lines or intersect it with the circle should probably go to the Math Stack Exchange.
if there are any better methods out there that can compute this point that are simpler?
Projective geometry and homogeneous coordinates can avoid a lot of nasty corner cases, like circles of infinite radius (also known as lines) or the intersection of parallel lines. But the problem of deciding between two solutions remains, so it probably doesn't make things as simple as you'd like them to be.
Related
In a geodetic coordinate system (wgs84), i have a pair of (latitude,longitude) say (45,50) and (60,20). Also i am said that a new pair of latitude,longitude lies along the line joining these two and at an offset of say 0.1 deg lat from (45,50) i.e. (45.1, x). How do i find this new point? What i tried was to apply the straight line equation
y = mx+c
m = (lat1 - lat2)/ long1-long2)
c = lat1 - m * long1
but that seemed to give wrong results.
Your problem is the calculation of m. You have turned it around!
The normal formula is:
a = (y1 - y2) / (x1 - x2)
so in your case it is:
m = (long2 -long1) / (lat1 - lat2)
so you'll get m = -2
And you also turned the calculation of c around.
Normal is:
b = y1 - a * x1
so you should do:
c = long1 - m * lat1
So you'll get c = 140.
The formula is:
long = -2 * lat + 140
Another way to think about it is given below. The result is the same, of cause.
The surface-line between two coordinates is not a straight line. It is a line drawn on the surface of a round object, i.e. earth. It will be a circle around the earth.
However all coordinates on that line will still go through a straight line.
That is because the coordinate represents the angles of a vector from center of earth to the point you are looking at. The two angles are compared to Equator (latitude) and compared to Greenwich (longitude).
So you need to setup a formula describing all coordinates for that line.
In your case the latitude goes from 45 to 60, i.e. increases by 15.
Your longitude goes from 50 to 20, i.e. decreses by 30.
So your formula will be:
(lat(t), long(t)) = (45, 50) + (15*t, -30*t) for t in [0:1]
Now you can calculate the value of t that will hit (45.1, x) and afterwards you can calculate x.
The equations you use describe a straight line in an 2D cartesian coordinate system.
Longitude and latitude describe a point in a spherical coordinate system.
A spherical coordinate system is not cartesian.
A similar question was answered here.
I have a ground set up of various points, some of which are flat and others are at an angle, I'm trying to check if there is a collision between the angled points (non-axis aligned).
I have a vector array consisting of two floats at each point - This is each of the points of the ground.
Here's an image representation of what the ground looks like.
http://i.imgur.com/cgEMqUv.png?1?4597
At the moment I want to check collisions between points 1 and 2 and then go onto the others.
I shall use points 1 and 2 as an example.
g1x = 150; g2x = 980;
g2x = 500; g2y = 780;
The dxdy of this is dx = 350 and dy = -200
The normal x of this is dy and the normal y is -dx
nx = -200;
ny = -350;
normalized it is the length between points 1 and 2 which is 403.11
nx/normalized = -0.496
ny/normalized = -0.868
//get position of object - Don't know if its supposed to be velocity or not
float vix = object->getPosition().x;
float viy = object->getPosition().y;
//calculate dot product - unsure if vix/viy are supposed to be minused
float dot = ((-vix * nrmx) + (-viy * nrmy)) * nrmx; //= -131.692
Is this information correct to calculate the normal and dot product between the two points.
How can I check if there is a collision with this line and then reflect according to the normal.
Thanks :) any and all changes are welcome.
Say you have a particle at position x travelling at velocity v and a boundary defined by the line between a and b.
We can find how far along the boundary (as a fraction) the particle collides by projecting c-a onto b-a and dividing by the length ||b-a||. That is,
u = ((c-a).((b-a)/||b-a||))/||b-a|| == (c-a).(b-a) / ||b-a||2.
If u > 1 then the particle travels past the boundary on the b side, if u < 0 then the particle travels past the boundary on the a side. The point of collision would be
c = a + u b.
The time to collision could be found by solving
x + t v = a + s (b-a)
for t. The reflection matrix can be found here. But it will need to be rotated by 90 deg (or pi/2) so that you're reflecting orthogonal to the line, not across it.
In terms of multiple boundaries, calculate the time to collision for each of them, sort by that time (discarding negative times) and check for collisions through the list. Once you've found the one that you will collide with then you can move your particle to the point of collision, reflect it's velocity, change the delta t and redo the whole thing again (ignoring the one you just collided with) as you may collide with more than one boundary in a corner case (get it? It's a maths pun).
Linear algebra can be fun, and you can do so much more with it, getting to grips with linear algebra allows you to do some powerful things. Good luck!
This is quite complicated to explain, so I will do my best, sorry if there is anything I missed out, let me know and I will rectify it.
My question is, I have been tasked to draw this shape,
(source: learnersdictionary.com)
This is to be done using C++ to write code that will calculate the points on this shape.
Important details.
User Input - Centre Point (X, Y), number of points to be shown, Font Size (influences radius)
Output - List of co-ordinates on the shape.
The overall aim once I have the points is to put them into a graph on Excel and it will hopefully draw it for me, at the user inputted size!
I know that the maximum Radius is 165mm and the minimum is 35mm. I have decided that my base Font Size shall be 20. I then did some thinking and came up with the equation.
Radius = (Chosen Font Size/20)*130. This is just an estimation, I realise it probably not right, but I thought it could work at least as a template.
I then decided that I should create two different circles, with two different centre points, then link them together to create the shape. I thought that the INSIDE line will have to have a larger Radius and a centre point further along the X-Axis (Y staying constant), as then it could cut into the outside line.
So I defined 2nd Centre point as (X+4, Y). (Again, just estimation, thought it doesn't really matter how far apart they are).
I then decided Radius 2 = (Chosen Font Size/20)*165 (max radius)
So, I have my 2 Radii, and two centre points.
Now to calculate the points on the circles, I am really struggling. I decided the best way to do it would be to create an increment (here is template)
for(int i=0; i<=n; i++) //where 'n' is users chosen number of points
{
//Equation for X point
//Equation for Y Point
cout<<"("<<X<<","<<Y<<")"<<endl;
}
Now, for the life of me, I cannot figure out an equation to calculate the points. I have found equations that involve angles, but as I do not have any, I'm struggling.
I am, in essence, trying to calculate Point 'P' here, except all the way round the circle.
(source: tutorvista.com)
Another point I am thinking may be a problem is imposing limits on the values calculated to only display the values that are on the shape.? Not sure how to chose limits exactly other than to make the outside line a full Half Circle so I have a maximum radius?
So. Does anyone have any hints/tips/links they can share with me on how to proceed exactly?
Thanks again, any problems with the question, sorry will do my best to rectify if you let me know.
Cheers
UPDATE;
R1 = (Font/20)*130;
R2 = (Font/20)*165;
for(X1=0; X1<=n; X1++)
{
Y1 = ((2*Y)+(pow(((4*((pow((X1-X), 2)))+(pow(R1, 2)))), 0.5)))/2;
Y2 = ((2*Y)-(pow(((4*((pow((X1-X), 2)))+(pow(R1, 2)))), 0.5)))/2;
cout<<"("<<X1<<","<<Y1<<")";
cout<<"("<<X1<<","<<Y2<<")";
}
Opinion?
As per Code-Guru's comments on the question, the inner circle looks more like a half circle than the outer. Use the equation in Code-Guru's answer to calculate the points for the inner circle. Then, have a look at this question for how to calculate the radius of a circle which intersects your circle, given the distance (which you can set arbitrarily) and the points of intersection (which you know, because it's a half circle). From this you can draw the outer arc for any given distance, and all you need to do is vary the distance until you produce a shape that you're happy with.
This question may help you to apply Code-Guru's equation.
The equation of a circle is
(x - h)^2 + (y - k)^2 = r^2
With a little bit of algebra, you can iterate x over the range from h to h+r incrementing by some appropriate delta and calculate the two corresponding values of y. This will draw a complete circle.
The next step is to find the x-coordinate for the intersection of the two circles (assuming that the moon shape is defined by two appropriate circles). Again, some algebra and a pencil and paper will help.
More details:
To draw a circle without using polar coordinates and trig, you can do something like this:
for x in h-r to h+r increment by delta
calculate both y coordinates
To calculate the y-coordinates, you need to solve the equation of a circle for y. The easiest way to do this is to transform it into a quadratic equation of the form A*y^2+B*y+C=0 and use the quadratic equation:
(x - h)^2 + (y - k)^2 = r^2
(x - h)^2 + (y - k)^2 - r^2 = 0
(y^2 - 2*k*y + k^2) + (x - h)^2 - r^2 = 0
y^2 - 2*k*y + (k^2 + (x - h)^2 - r^2) = 0
So we have
A = 1
B = -2*k
C = k^2 + (x - h)^2 - r^2
Now plug these into the quadratic equation and chug out the two y-values for each x value in the for loop. (Most likely, you will want to do the calculations in a separate function -- or functions.)
As you can see this is pretty messy. Doing this with trigonometry and angles will be much cleaner.
More thoughts:
Even though there are no angles in the user input described in the question, there is no intrinsic reason why you cannot use them during calculations (unless you have a specific requirement otherwise, say because your teacher told you not to). With that said, using polar coordinates makes this much easier. For a complete circle you can do something like this:
for theta = 0 to 2*PI increment by delta
x = r * cos(theta)
y = r * sin(theta)
To draw an arc, rather than a full circle, you simply change the limits for theta in the for loop. For example, the left-half of the circle goes from PI/2 to 3*PI/2.
I have a function in my program which rotates a point (x_p, y_p, z_p) around another point (x_m, y_m, z_m) by the angles w_nx and w_ny.
The new coordinates are stored in global variables x_n, y_n, and z_n. Rotation around the y-axis (so changing value of w_nx - so that the y - values are not harmed) is working correctly, but as soon as I do a rotation around the x- or z- axis (changing the value of w_ny) the coordinates aren't accurate any more. I commented on the line I think my fault is in, but I can't figure out what's wrong with that code.
void rotate(float x_m, float y_m, float z_m, float x_p, float y_p, float z_p, float w_nx ,float w_ny)
{
float z_b = z_p - z_m;
float x_b = x_p - x_m;
float y_b = y_p - y_m;
float length_ = sqrt((z_b*z_b)+(x_b*x_b)+(y_b*y_b));
float w_bx = asin(z_b/sqrt((x_b*x_b)+(z_b*z_b))) + w_nx;
float w_by = asin(x_b/sqrt((x_b*x_b)+(y_b*y_b))) + w_ny; //<- there must be that fault
x_n = cos(w_bx)*sin(w_by)*length_+x_m;
z_n = sin(w_bx)*sin(w_by)*length_+z_m;
y_n = cos(w_by)*length_+y_m;
}
What the code almost does:
compute difference vector
convert vector into spherical coordinates
add w_nx and wn_y to the inclination and azimuth angle (see link for terminology)
convert modified spherical coordinates back into Cartesian coordinates
There are two problems:
the conversion is not correct, the computation you do is for two inclination vectors (one along the x axis, the other along the y axis)
even if computation were correct, transformation in spherical coordinates is not the same as rotating around two axis
Therefore in this case using matrix and vector math will help:
b = p - m
b = RotationMatrixAroundX(wn_x) * b
b = RotationMatrixAroundY(wn_y) * b
n = m + b
basic rotation matrices.
Try to use vector math. Decide in which order you rotate, first along x, then along y perhaps.
If you rotate along z-axis, [z' = z]
x' = x*cos a - y*sin a;
y' = x*sin a + y*cos a;
The same repeated for y-axis: [y'' = y']
x'' = x'*cos b - z' * sin b;
z'' = x'*sin b + z' * cos b;
Again rotating along x-axis: [x''' = x'']
y''' = y'' * cos c - z'' * sin c
z''' = y'' * sin c + z'' * cos c
And finally the question of rotating around some specific "point":
First, subtract the point from the coordinates, then apply the rotations and finally add the point back to the result.
The problem, as far as I see, is a close relative to "gimbal lock". The angle w_ny can't be measured relative to the fixed xyz -coordinate system, but to the coordinate system that is rotated by applying the angle w_nx.
As kakTuZ observed, your code converts point to spherical coordinates. There's nothing inherently wrong with that -- with longitude and latitude, one can reach all the places on Earth. And if one doesn't care about tilting the Earth's equatorial plane relative to its trajectory around the Sun, it's ok with me.
The result of not rotating the next reference axis along the first w_ny is that two points that are 1 km a part of each other at the equator, move closer to each other at the poles and at the latitude of 90 degrees, they touch. Even though the apparent purpose is to keep them 1 km apart where ever they are rotated.
if you want to transform coordinate systems rather than only points you need 3 angles. But you are right - for transforming points 2 angles are enough. For details ask Wikipedia ...
But when you work with opengl you really should use opengl functions like glRotatef. These functions will be calculated on the GPU - not on the CPU as your function. The doc is here.
Like many others have said, you should use glRotatef to rotate it for rendering. For collision handling, you can obtain its world-space position by multiplying its position vector by the OpenGL ModelView matrix on top of the stack at the point of its rendering. Obtain that matrix with glGetFloatv, and then multiply it with either your own vector-matrix multiplication function, or use one of the many ones you can obtain easily online.
But, that would be a pain! Instead, look into using the GL feedback buffer. This buffer will simply store the points where the primitive would have been drawn instead of actually drawing the primitive, and then you can access them from there.
This is a good starting point.
I have a point in 3D space and two angles, I want to calculate the resulting line from this information. I have found how to do this with 2D lines, but not 3D. How can this be calculated?
If it helps: I'm using C++ & OpenGL and have the location of the user's mouse click and the angle of the camera, I want to trace this line for intersections.
In trig terms two angles and a point are required to define a line in 3d space. Converting that to (x,y,z) is just polar coordinates to cartesian coordinates the equations are:
x = r sin(q) cos(f)
y = r sin(q) sin(f)
z = r cos(q)
Where r is the distance from the point P to the origin; the angle q (zenith) between the line OP and the positive polar axis (can be thought of as the z-axis); and the angle f (azimuth) between the initial ray and the projection of OP onto the equatorial plane(usually measured from the x-axis).
Edit:
Okay that was the first part of what you ask. The rest of it, the real question after the updates to the question, is much more complicated than just creating a line from 2 angles and a point in 3d space. This involves using a camera-to-world transformation matrix and was covered in other SO questions. For convenience here's one: How does one convert world coordinates to camera coordinates? The answers cover converting from world-to-camera and camera-to-world.
The line can be fathomed as a point in "time". The equation must be vectorized, or have a direction to make sense, so time is a natural way to think of it. So an equation of a line in 3 dimensions could really be three two dimensional equations of x,y,z related to time, such as:
x = ax*t + cx
y = ay*t + cy
z = az*t + cz
To find that set of equations, assuming the camera is at origin, (0,0,0), and your point is (x1,y1,z1) then
ax = x1 - 0
ay = y1 - 0
az = z1 - 0
cx = cy = cz = 0
so
x = x1*t
y = y1*t
z = z1*t
Note: this also assumes that the "speed" of the line or vector is such that it is at your point (x1,y1,z1) after 1 second.
So to draw that line just fill in the points as fine as you like for as long as required, such as every 1/1000 of a second for 10 seconds or something, might draw a "line", really a series of points that when seen from a distance appear as a line, over 10 seconds worth of distance, determined by the "speed" you choose.