I am trying to draw a railway in the way it is implemented in railroad tycoon 3. I have understood some of the patterns by which this is done, but I cannot find a formula for their application. I am trying to build an arc using only two points and some initial direction. For simplicity, let's assume that I can only plot a path in the range from 0 to 90 degrees. The arc should always start in the direction of "Initial direction". I need a method that would implement this. I tried to depict the idea in the drawings. Any pseudocode would do I guess.
general idea
borderline case at alpha = 0
borderline case at alpha = 90 degrees
Center C lies at middle perpendicular to SE segment, so we should find vectors
se = (E.x - S.x, E.y - S.y)
perp = (-se.y, se.x) = (S.y - E.y, E.x - S.x) //MC direction
and middle point
M = ((E.x + S.x)/2, (E.y - S.y)/2)
and express C coordinates using parametric equation
C = M + perp*t
Note that initial arc direction d is perpendicular to radius CS, so their dot product is zero:
(C - S).dot.d = 0
(M.x + perp.x * t - S.x)*d.x + (M.y + perp.y * t - S.y)*d.y = 0
This is linear equation for unknown parameter t, after solving it we know center C and can build the arc.
Python code. Perhaps examples are too simple to reveal bugs, but the first with half-circle gives R=L/2, center at M, and the second one shows correct C position and R.
from math import hypot, atan2
def arcfromptsdir(sx, sy, ex, ey, dx, dy):
sex = ex - sx
sey = ey - sy
perpx = -sey
perpy = sex
mx = (ex + sx) / 2
my = (ey + sy) / 2
#equation
#(sex/2 + perpx * t )*dx + (sey/2 + perp.y * t )*dy = 0
p = perpx * dx + perpy * dy
if p:
t = -0.5*(sex*dx + sey*dx) / p
else:
return None
#centerx, centery, radius
cx = mx + perpx * t
cy = my + perpy * t
radius = hypot(cx-sx, cy-sy)
arcangle = atan2((sx-cx) * (ey-cy) - (sy-cy) * (ex-cx),
(sx-cx) * (ex-cx) + (sy-cy) * (ey-cy))
return (cx, cy, radius, arcangle)
print(arcfromptsdir(0, 0, 2, 0, 0, 1))
print(arcfromptsdir(0, 0, 2, 0, 0.7071, 0.7071))
(1.0, 0.0, 1.0, -3.141592653589793)
(1.0, -1.0, 1.4142135623730951, -1.5707963267948966)
Related
I have direction vector (0.000000, 0.707107, 0.707107) like this, i want to convert this vector to an angle between X, Y and Z direction planes and Direction vectors (0,45 deg,45 deg). Please provide the mathematical equations or VBA functions to get an angle.
To get angle between vector D = (dx, dy, dz) and coordinate planes, you can use scalar product of vector and its projection onto corresponding plane.
For example, to get projection on OYZ plane, you can just make x-component zero.
P(0yz) = (0, dy, dz)
S = D.dot.P = 0 + dy * dy + dz * dz
Fi(D, 0yz) = ArcCosine(S / (length(P) * length(D)) =
ArcCosine((dy*dy + dz*dz) / Sqrt((dx*dx + dy*dy + dz*dz)*(dy*dy + dz*dz)))=
ArcCosine(Sqrt((dy*dy + dz*dz) /(dx*dx + dy*dy + dz*dz))))=
ArcCosine(length(P) / length(D))
You can build similar formulas for OXY and OXZ planes
I have a bounding box of:
Left -122.27671
Bottom 37.80445
Right -122.26673
Top 37.81449
It could also be converted into NE Lat/Long and SW Lat/Long
Within that bounding box, I'd like to find the X,Y position of a specific Lat/Long. This would be using the Mercator projection.
I've seen answers that find the X,Y of a position on a world map using Mercator, but not within a specific lat/lon.
Any help appreciated!
UPDATE
Put this together from another question I saw. Can anyone validate if this seems legit?
map_width = 1240
map_height = 1279
map_lon_left = -122.296916
map_lon_right = -122.243380
map_lon_delta = map_lon_right - map_lon_left
map_lat_bottom = 37.782368
map_lat_bottom_degree = map_lat_bottom * Math::PI / 180
def convert_geo_to_pixel(lat, long)
x = (long - map_lon_left) * (map_width / map_lon_delta)
lat = lat * Math::PI / 180
world_map_width = ((map_width / map_lon_delta) * 360) / (2 * Math::PI)
map_offset_y = (world_map_width / 2 * Math.log((1 + Math.sin(map_lat_bottom_degree)) / (1 - Math.sin(map_lat_bottom_degree))))
y = map_height - ((world_map_width / 2 * Math.log((1 + Math.sin(lat)) / (1 - Math.sin(lat)))) - map_offset_y)
return [x, y]
end
Found a better solution that I've test and validated. Posting this for anyone else who might find it useful. It's written in Ruby but easy to convert to any other language
#north = to_radians(37.81449)
#south = to_radians(37.80445)
#east = to_radians(-122.26673)
#west = to_radians(-122.27671)
# Coordinates above are a subsection of Oakland, CA
#map_width = map_width
#map_height = map_height
def location_to_pixel(lat:, lon:)
lat = to_radians(lat)
lon = to_radians(lon)
ymin = mercator_y(#south)
ymax = mercator_y(#north)
x_factor = #map_width/(#east - #west)
y_factor = #map_height/(ymax - ymin)
y = mercator_y(lat);
x = (lon - #west) * x_factor
y = (ymax - y) * y_factor
[x, y]
end
def to_radians(deg)
deg * Math::PI/180
end
def mercator_y(lat)
Math.log(
Math.tan(lat/2 + Math::PI/4)
)
end
Let's s is shift of map in world space, bottom latitude in radians B, top latitude T. (I assume y=0 is bottom)
C * Sin(B) = 0 + s
C * Sin(T) = map_height + s
=>
C = map_height / (Sin(T) - Sin(B))
s = C * Sin(B)
y = C * Sin(Lat) - s =
C * Sin(Lat) - C * Sin(B) =
C * (Sin(Lat) - Sin(B)) =
map_height * (Sin(Lat) - Sin(B) / (Sin(T) - Sin(B))
// note - resembles linear interpolation is sine space
I want to calculate a point on a given line that is perpendicular from a given point.
I have a line segment AB and have a point C outside line segment. I want to calculate a point D on AB such that CD is perpendicular to AB.
I have to find point D.
It quite similar to this, but I want to consider to Z coordinate also as it does not show up correctly in 3D space.
Proof:
Point D is on a line CD perpendicular to AB, and of course D belongs to AB.
Write down the Dot product of the two vectors CD.AB = 0, and express the fact D belongs to AB as D=A+t(B-A).
We end up with 3 equations:
Dx=Ax+t(Bx-Ax)
Dy=Ay+t(By-Ay)
(Dx-Cx)(Bx-Ax)+(Dy-Cy)(By-Ay)=0
Subtitute the first two equations in the third one gives:
(Ax+t(Bx-Ax)-Cx)(Bx-Ax)+(Ay+t(By-Ay)-Cy)(By-Ay)=0
Distributing to solve for t gives:
(Ax-Cx)(Bx-Ax)+t(Bx-Ax)(Bx-Ax)+(Ay-Cy)(By-Ay)+t(By-Ay)(By-Ay)=0
which gives:
t= -[(Ax-Cx)(Bx-Ax)+(Ay-Cy)(By-Ay)]/[(Bx-Ax)^2+(By-Ay)^2]
getting rid of the negative signs:
t=[(Cx-Ax)(Bx-Ax)+(Cy-Ay)(By-Ay)]/[(Bx-Ax)^2+(By-Ay)^2]
Once you have t, you can figure out the coordinates for D from the first two equations.
Dx=Ax+t(Bx-Ax)
Dy=Ay+t(By-Ay)
function getSpPoint(A,B,C){
var x1=A.x, y1=A.y, x2=B.x, y2=B.y, x3=C.x, y3=C.y;
var px = x2-x1, py = y2-y1, dAB = px*px + py*py;
var u = ((x3 - x1) * px + (y3 - y1) * py) / dAB;
var x = x1 + u * px, y = y1 + u * py;
return {x:x, y:y}; //this is D
}
There is a simple closed form solution for this (requiring no loops or approximations) using the vector dot product.
Imagine your points as vectors where point A is at the origin (0,0) and all other points are referenced from it (you can easily transform your points to this reference frame by subtracting point A from every point).
In this reference frame point D is simply the vector projection of point C on the vector B which is expressed as:
// Per wikipedia this is more efficient than the standard (A . Bhat) * Bhat
Vector projection = Vector.DotProduct(A, B) / Vector.DotProduct(B, B) * B
The result vector can be transformed back to the original coordinate system by adding point A to it.
A point on line AB can be parametrized by:
M(x)=A+x*(B-A), for x real.
You want D=M(x) such that DC and AB are orthogonal:
dot(B-A,C-M(x))=0.
That is: dot(B-A,C-A-x*(B-A))=0, or dot(B-A,C-A)=x*dot(B-A,B-A), giving:
x=dot(B-A,C-A)/dot(B-A,B-A) which is defined unless A=B.
What you are trying to do is called vector projection
Here i have converted answered code from "cuixiping" to matlab code.
function Pr=getSpPoint(Line,Point)
% getSpPoint(): find Perpendicular on a line segment from a given point
x1=Line(1,1);
y1=Line(1,2);
x2=Line(2,1);
y2=Line(2,1);
x3=Point(1,1);
y3=Point(1,2);
px = x2-x1;
py = y2-y1;
dAB = px*px + py*py;
u = ((x3 - x1) * px + (y3 - y1) * py) / dAB;
x = x1 + u * px;
y = y1 + u * py;
Pr=[x,y];
end
I didn't see this answer offered, but Ron Warholic had a great suggestion with the Vector Projection. ACD is merely a right triangle.
Create the vector AC i.e (Cx - Ax, Cy - Ay)
Create the Vector AB i.e (Bx - Ax, By - Ay)
Dot product of AC and AB is equal to the cosine of the angle between the vectors. i.e cos(theta) = ACx*ABx + ACy*ABy.
Length of a vector is sqrt(x*x + y*y)
Length of AD = cos(theta)*length(AC)
Normalize AB i.e (ABx/length(AB), ABy/length(AB))
D = A + NAB*length(AD)
For anyone who might need this in C# I'll save you some time:
double Ax = ;
double Ay = ;
double Az = ;
double Bx = ;
double By = ;
double Bz = ;
double Cx = ;
double Cy = ;
double Cz = ;
double t = ((Cx - Ax) * (Bx - Ax) + (Cy - Ay) * (By - Ay)) / (Math.Pow(Bx - Ax, 2) + Math.Pow(By - Ay, 2));
double Dx = Ax + t*(Bx - Ax);
double Dy = Ay + t*(By - Ay);
Here is another python implementation without using a for loop. It works for any number of points and any number of line segments. Given p_array as a set of points, and x_array , y_array as continues line segments or a polyline.
This uses the equation Y = mX + n and considering that the m factor for a perpendicular line segment is -1/m.
import numpy as np
def ortoSegmentPoint(self, p_array, x_array, y_array):
"""
:param p_array: np.array([[ 718898.941 9677612.901 ], [ 718888.8227 9677718.305 ], [ 719033.0528 9677770.692 ]])
:param y_array: np.array([9677656.39934991 9677720.27550726 9677754.79])
:param x_array: np.array([718895.88881594 718938.61392781 718961.46])
:return: [POINT, LINE] indexes where point is orthogonal to line segment
"""
# PENDIENTE "m" de la recta, y = mx + n
m_array = np.divide(y_array[1:] - y_array[:-1], x_array[1:] - x_array[:-1])
# PENDIENTE INVERTIDA, 1/m
inv_m_array = np.divide(1, m_array)
# VALOR "n", y = mx + n
n_array = y_array[:-1] - x_array[:-1] * m_array
# VALOR "n_orto" PARA LA RECTA PERPENDICULAR
n_orto_array = np.array(p_array[:, 1]).reshape(len(p_array), 1) + inv_m_array * np.array(p_array[:, 0]).reshape(len(p_array), 1)
# PUNTOS DONDE SE INTERSECTAN DE FORMA PERPENDICULAR
x_intersec_array = np.divide(n_orto_array - n_array, m_array + inv_m_array)
y_intersec_array = m_array * x_intersec_array + n_array
# LISTAR COORDENADAS EN PARES
x_coord = np.array([x_array[:-1], x_array[1:]]).T
y_coord = np.array([y_array[:-1], y_array[1:]]).T
# FILAS: NUMERO DE PUNTOS, COLUMNAS: NUMERO DE TRAMOS
maskX = np.where(np.logical_and(x_intersec_array < np.max(x_coord, axis=1), x_intersec_array > np.min(x_coord, axis=1)), True, False)
maskY = np.where(np.logical_and(y_intersec_array < np.max(y_coord, axis=1), y_intersec_array > np.min(y_coord, axis=1)), True, False)
mask = maskY * maskX
return np.argwhere(mask == True)
As Ron Warholic and Nicolas Repiquet answered, this can be solved using vector projection. For completeness I'll add a python/numpy implementation of this here in case it saves anyone else some time:
import numpy as np
# Define some test data that you can solve for directly.
first_point = np.array([4, 4])
second_point = np.array([8, 4])
target_point = np.array([6, 6])
# Expected answer
expected_point = np.array([6, 4])
# Create vector for first point on line to perpendicular point.
point_vector = target_point - first_point
# Create vector for first point and second point on line.
line_vector = second_point - first_point
# Create the projection vector that will define the position of the resultant point with respect to the first point.
projection_vector = (np.dot(point_vector, line_vector) / np.dot(line_vector, line_vector)) * line_vector
# Alternative method proposed in another answer if for whatever reason you prefer to use this.
_projection_vector = (np.dot(point_vector, line_vector) / np.linalg.norm(line_vector)**2) * line_vector
# Add the projection vector to the first point
projected_point = first_point + projection_vector
# Test
(projected_point == expected_point).all()
Since you're not stating which language you're using, I'll give you a generic answer:
Just have a loop passing through all the points in your AB segment, "draw a segment" to C from them, get the distance from C to D and from A to D, and apply pithagoras theorem. If AD^2 + CD^2 = AC^2, then you've found your point.
Also, you can optimize your code by starting the loop by the shortest side (considering AD and BD sides), since you'll find that point earlier.
Here is a python implementation based on Corey Ogburn's answer from this thread.
It projects the point q onto the line segment defined by p1 and p2 resulting in the point r.
It will return null if r falls outside of line segment:
def is_point_on_line(p1, p2, q):
if (p1[0] == p2[0]) and (p1[1] == p2[1]):
p1[0] -= 0.00001
U = ((q[0] - p1[0]) * (p2[0] - p1[0])) + ((q[1] - p1[1]) * (p2[1] - p1[1]))
Udenom = math.pow(p2[0] - p1[0], 2) + math.pow(p2[1] - p1[1], 2)
U /= Udenom
r = [0, 0]
r[0] = p1[0] + (U * (p2[0] - p1[0]))
r[1] = p1[1] + (U * (p2[1] - p1[1]))
minx = min(p1[0], p2[0])
maxx = max(p1[0], p2[0])
miny = min(p1[1], p2[1])
maxy = max(p1[1], p2[1])
is_valid = (minx <= r[0] <= maxx) and (miny <= r[1] <= maxy)
if is_valid:
return r
else:
return None
I have a square bitmap of a circle and I want to compute the normals of all the pixels in that circle as if it were a sphere of radius 1:
The sphere/circle is centered in the bitmap.
What is the equation for this?
Don't know much about how people program 3D stuff, so I'll just give the pure math and hope it's useful.
Sphere of radius 1, centered on origin, is the set of points satisfying:
x2 + y2 + z2 = 1
We want the 3D coordinates of a point on the sphere where x and y are known. So, just solve for z:
z = ±sqrt(1 - x2 - y2).
Now, let us consider a unit vector pointing outward from the sphere. It's a unit sphere, so we can just use the vector from the origin to (x, y, z), which is, of course, <x, y, z>.
Now we want the equation of a plane tangent to the sphere at (x, y, z), but this will be using its own x, y, and z variables, so instead I'll make it tangent to the sphere at (x0, y0, z0). This is simply:
x0x + y0y + z0z = 1
Hope this helps.
(OP):
you mean something like:
const int R = 31, SZ = power_of_two(R*2);
std::vector<vec4_t> p;
for(int y=0; y<SZ; y++) {
for(int x=0; x<SZ; x++) {
const float rx = (float)(x-R)/R, ry = (float)(y-R)/R;
if(rx*rx+ry*ry > 1) { // outside sphere
p.push_back(vec4_t(0,0,0,0));
} else {
vec3_t normal(rx,sqrt(1.-rx*rx-ry*ry),ry);
p.push_back(vec4_t(normal,1));
}
}
}
It does make a nice spherical shading-like shading if I treat the normals as colours and blit it; is it right?
(TZ)
Sorry, I'm not familiar with those aspects of C++. Haven't used the language very much, nor recently.
This formula is often used for "fake-envmapping" effect.
double x = 2.0 * pixel_x / bitmap_size - 1.0;
double y = 2.0 * pixel_y / bitmap_size - 1.0;
double r2 = x*x + y*y;
if (r2 < 1)
{
// Inside the circle
double z = sqrt(1 - r2);
.. here the normal is (x, y, z) ...
}
Obviously you're limited to assuming all the points are on one half of the sphere or similar, because of the missing dimension. Past that, it's pretty simple.
The middle of the circle has a normal facing precisely in or out, perpendicular to the plane the circle is drawn on.
Each point on the edge of the circle is facing away from the middle, and thus you can calculate the normal for that.
For any point between the middle and the edge, you use the distance from the middle, and some simple trig (which eludes me at the moment). A lerp is roughly accurate at some points, but not quite what you need, since it's a curve. Simple curve though, and you know the beginning and end values, so figuring them out should only take a simple equation.
I think I get what you're trying to do: generate a grid of depth data for an image. Sort of like ray-tracing a sphere.
In that case, you want a Ray-Sphere Intersection test:
http://www.siggraph.org/education/materials/HyperGraph/raytrace/rtinter1.htm
Your rays will be simple perpendicular rays, based off your U/V coordinates (times two, since your sphere has a diameter of 2). This will give you the front-facing points on the sphere.
From there, calculate normals as below (point - origin, the radius is already 1 unit).
Ripped off from the link above:
You have to combine two equations:
Ray: R(t) = R0 + t * Rd , t > 0 with R0 = [X0, Y0, Z0] and Rd = [Xd, Yd, Zd]
Sphere: S = the set of points[xs, ys, zs], where (xs - xc)2 + (ys - yc)2 + (zs - zc)2 = Sr2
To do this, calculate your ray (x * pixel / width, y * pixel / width, z: 1), then:
A = Xd^2 + Yd^2 + Zd^2
B = 2 * (Xd * (X0 - Xc) + Yd * (Y0 - Yc) + Zd * (Z0 - Zc))
C = (X0 - Xc)^2 + (Y0 - Yc)^2 + (Z0 - Zc)^2 - Sr^2
Plug into quadratic equation:
t0, t1 = (- B + (B^2 - 4*C)^1/2) / 2
Check discriminant (B^2 - 4*C), and if real root, the intersection is:
Ri = [xi, yi, zi] = [x0 + xd * ti , y0 + yd * ti, z0 + zd * ti]
And the surface normal is:
SN = [(xi - xc)/Sr, (yi - yc)/Sr, (zi - zc)/Sr]
Boiling it all down:
So, since we're talking unit values, and rays that point straight at Z (no x or y component), we can boil down these equations greatly:
Ray:
X0 = 2 * pixelX / width
Y0 = 2 * pixelY / height
Z0 = 0
Xd = 0
Yd = 0
Zd = 1
Sphere:
Xc = 1
Yc = 1
Zc = 1
Factors:
A = 1 (unit ray)
B
= 2 * (0 + 0 + (0 - 1))
= -2 (no x/y component)
C
= (X0 - 1) ^ 2 + (Y0 - 1) ^ 2 + (0 - 1) ^ 2 - 1
= (X0 - 1) ^ 2 + (Y0 - 1) ^ 2
Discriminant
= (-2) ^ 2 - 4 * 1 * C
= 4 - 4 * C
From here:
If discriminant < 0:
Z = ?, Normal = ?
Else:
t = (2 + (discriminant) ^ 1 / 2) / 2
If t < 0 (hopefully never or always the case)
t = -t
Then:
Z: t
Nx: Xi - 1
Ny: Yi - 1
Nz: t - 1
Boiled farther still:
Intuitively it looks like C (X^2 + Y^2) and the square-root are the most prominent figures here. If I had a better recollection of my math (in particular, transformations on exponents of sums), then I'd bet I could derive this down to what Tom Zych gave you. Since I can't, I'll just leave it as above.
There's a line A-B and C at the center between A and B. It forms a circle as in the figure. If we assume A-B line as a diameter of the circle and then C is it's center. My problem is I have no idea how to draw another three lines (in blue) each 45 degree away from AC or AB. No, this is not a homework, it's part of my complex geometry in a rendering.
alt text http://www.freeimagehosting.net/uploads/befcd84d8c.png
There are a few ways to solve this problem, one of which is to find the angle of the line and then add 45 degrees to this a few times. Here's an example, it's in Python, but translating the math should be easy (and I've tried to write the Python in a simplistic way).
Here's the output for a few lines:
The main function is calc_points, the rest is just to give it A and B that intersect the circle, and make the plots.
from math import atan2, sin, cos, sqrt, pi
from matplotlib import pyplot
def calc_points(A, B, C):
dx = C[0]-A[0]
dy = C[1]-A[1]
line_angle = atan2(dy, dx)
radius = sqrt(dy*dy + dx*dx)
new_points = []
# now go around the circle and find the points
for i in range(3):
angle = line_angle + (i+1)*45*(pi/180) # new angle in radians
x = radius*cos(angle) + C[0]
y = radius*sin(angle) + C[1]
new_points.append([x, y])
return new_points
# test this with some reasonable values
pyplot.figure()
for i, a in enumerate((-20, 20, 190)):
radius = 5
C = [2, 2]
# find an A and B on the circle and plot them
angle = a*(pi/180)
A = [radius*cos(pi+angle)+C[0], radius*sin(pi+angle)+C[1]]
B = [radius*cos(angle)+C[0], radius*sin(angle)+C[1]]
pyplot.subplot(1,3,i+1)
pyplot.plot([A[0], C[0]], [A[1], C[1]], 'r')
pyplot.plot([B[0], C[0]], [B[1], C[1]], 'r')
# now run these through the calc_points function and the new lines
new_points = calc_points(A, B, C)
for np in new_points:
pyplot.plot([np[0], C[0]], [np[1], C[1]], 'b')
pyplot.xlim(-8, 8)
pyplot.ylim(-8, 8)
for x, X in (("A", A), ("B", B), ("C", C)):
pyplot.text(X[0], X[1], x)
pyplot.show()
If you want to find coordinates of blue lines, may be you will find helpful some information about tranformations (rotations):
http://en.wikipedia.org/wiki/Rotation_matrix
You need to rotate for example vector AC and then you can find coordinate of end point of blue line.
start with this and add a button with code:
private void btnCircleLined_Click(object sender, System.EventArgs e)
{
Graphics graph = Graphics.FromImage(DrawArea);
int x = 100, y = 100, diameter = 50;
myPen.Color = Color.Green;
myPen.Width = 10;
graph.DrawEllipse(myPen, x, y, diameter, diameter);
myPen.Color = Color.Red;
double radian = 45 * Math.PI / 180;
int xOffSet = (int)(Math.Cos(radian) * diameter / 2);
int yOffSet = (int)(Math.Sin(radian) * diameter / 2);
graph.DrawLine(myPen, x, y + yOffSet + myPen.Width + diameter / 2, x + xOffSet + myPen.Width + diameter / 2, y);
graph.DrawLine(myPen, x, y, x + xOffSet + myPen.Width + diameter / 2, y + yOffSet + myPen.Width + diameter / 2);
graph.Dispose();
this.Invalidate();
}
edit: could not see your picture so I misinterpeted your question, but this should get you started.
Translate A with C at the origin (i.e. A-C), rotate CW 45°, then translate back. Repeat three more times.
If I were doing this I'd use polar co-ordinates (apologies for including the link if you are already well aware what they are) as an easy way of figuring out the co-ordinates of the points on the circumference that you need. Then draw lines to there from the centre of the circle.