So coming from a flash background I have an OK understanding of some simple 2D trig. In 2d with I circle, I know the math to place an item on the edge given an angle and a radius using.
x = cos(a) * r;
y = sin(a) * r;
Now if i have a point in 3d space, i know the radius of my sphere, i know the angle i want to position it around the z axis and the angle i want to position it around, say, the y axis. What is the math to find the x, y and z coordinates in my 3d space (assume that my origin is 0,0,0)? I would think i could borrow the Math from the circle trig but i can't seem to find a solution.
Your position in 3d is given by two angles (+ radius, which in your case is constant)
x = r * cos(s) * sin(t)
y = r * sin(s) * sin(t)
z = r * cos(t)
here, s is the angle around the z-axis, and t is the height angle, measured 'down' from the z-axis.
The picture below shows what the angles represent, s=theta in the range 0 to 2*PI in the xy-plane, and t=phi in the range 0 to PI.
The accepted answer did not seem to support negative x values (possibly I did something wrong), but just in case, using notation from ISO convention on coordinate systems defined in this Wikipedia entry, this system of equations should work:
import math
x = radius * sin(theta) * cos(phi)
y = radius * sin(theta) * sin(phi)
z = radius * cos(theta)
radius = math.sqrt(math.pow(x, 2) + math.pow(y, 2) + math.pow(z, 2))
phi = math.atan2(y, x)
theta = math.acos((z / radius))
Related
What is best way or an algorithm for generating a random 3d point [x,y,z] inside the volume of the circular cylinder if radius r and height h of the cylinder are given?
How about -- in Python pseudocode, letting R be the radius and H be the height:
s = random.uniform(0, 1)
theta = random.uniform(0, 2*pi)
z = random.uniform(0, H)
r = sqrt(s)*R
x = r * cos(theta)
y = r * sin(theta)
z = z # .. for symmetry :-)
The problem with simply taking x = r * cos(angle) and y = r * sin(angle) is that then when r is small, i.e. at the centre of the circle, a tiny change in r doesn't change the x and y positions very much. IOW, it leads to a nonuniform distribution in Cartesian coordinates, and the points get concentrated toward the centre of the circle. Taking the square root corrects this, at least if I've done my arithmetic correctly.
[Ah, it looks like the sqrt was right.]
(Note that I assumed without thinking about it that the cylinder is aligned with the z-axis and the cylinder centre is located at (0,0,H/2). It'd be less arbitrary to set (0,0,0) at the cylinder centre, in which case z should be chosen to be between -H/2 and H/2, not 0,H.)
Generate a random point inside the rectangular solid circumscribing the cylinder; if it's inside the cylinder (probability pi/4), keep it, otherwise discard it and try again.
Generate a random angle (optionally less than 2π), a random r less than the radius, and a random z less than the height.
x = r * cos(angle)
y = r * sin(angle)
The z axis is easy: -0.5 * h <= z <= 0.5 * h
The x and y are equal to a circle will be:
x^2 + y^2 <= r^2
Buth math is long ago for me :-)
I'm trying to use latitude and longitude coordinates to plot a map in Processing. Is there a way to set the zero coordinates of the X and Y axis to the center of the display window.
Or does anyone know how to convert spherical coordinates to cartesian?
Thanks
I'll assume you have spherical coordinates of r, radius; theta, horizontal angle around Z-axis starting at (1,0,0) and rotating toward (0,1,0); and phi, vertical angle from positive Z-axis toward negative Z-axis; that being how I remember it from back when. Remember that angles are in radians in most programming languages; 2*pi radians = 180 degrees.
x = r * cos(theta) * sin(phi)
y = r * sin(theta) * sin(phi)
z = r * cos(phi)
How can I calculate a point (X,Y) a specified distance away, on a rotated axis? I know what angle I'd like the point "moving" along (in degrees).
x = cos(a) * d
y = sin(a) * d
where a is the angle and d is the distance.
If the trigonometry functions takes radians intead of degrees, you have to convert the angle by dividing by 180/pi.
Convert to polar coordinates and then rotate the point through the angle you want:
x = r * cos( theta );
y = r * sin( theta );
Note: theta in radians ( deg = rad * 180 / pi )
More info on polar coordinates.
Do you mean the 3d formulas? They are easy as well. But we need to know what's your convention for specifying the axis.
how do i find out pixel value at certain degree on the circumference of a circle if I know the pixel co-ordinates of the center of the circle, radius of the circle ,and perpendicular angle.
Basically, I am trying to draw the hands of a clock at various times ( 1 o clock , 2 o clock etc )
Let h be the hour as a floating point number (h=2.25 would be 02:15, etc.) between 0 and 12. (cX,cY) are the coordinates of the center. hLength and mLength are the lengths of the hour and min hands.
// Hour hand
hAngle = 2.0*Pi*h/12.0; // 0..12 mapped to 0..2*Pi
hX = cX + hLength * sin(hAngle);
hY = cY - hLength * cos(hAngle);
// Min hand
mAngle = 2.0*Pi*h; // 0..1 mapped to 0..2*Pi, etc.
mX = cX + mLength * sin(mAngle);
mY = cY - mLength * cos(mAngle);
Where the centre of the circle is (X0, Y0), the radius is R and the angle with the x-axis is theta:
X1 = (R * cos theta) + X0
and
Y1 = (R * sin theta) + Y0
If (x1,y1) is a point on the circumference and (x,y) is the center, then x1 = x + r * cos(angle) and y1 = y + r * sin(angle)
if center is at x0, y0, and 0,0 iz at bottom-left corner, then 1 o'clock is at x0 + rsin(2π/3), y0+rcos(2π/3).
Draw lines from the center to coordinates computed with sin for the y coordinates and cos for the x coordinates (both multiplied by the length of the hand).
Wikipedia has more information on how sin and cos "work".
Convert angle in degrees to a point
How could I convert an angle (in degrees/radians) to a point (X,Y) a fixed distance away from a center-point.
Like a point rotating around a center-point.
Exactly the opposite of atan2 which computes the angle of the point y/x (in radians).
Note: I kept the original title because that's what people who do not understand will be searching by!
Let the fixed distance be D, then X = D * cos(A) and Y = D * sin(A), where A is the angle.
If center-point (Xcp, Ycp) isn't the origin you also need to add it's coordinates to (X,Y) i.e. X = Xcp + D * cos(A) and Y = Ycp + D * sin(A)
What PolyThinker said.
Also, if you need the distance from the origin, it's sqrt(x^2 + y^2).
t = angle
r = radius (fixed distance)
x = rcost
y = rsint