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.
Related
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)
Let's say we have a 100x100 coordinate system, like the one below. 0,0 is its left-top corner, 50,50 is its center point, 100,100 is its bottom right corner, etc.
Now we need to draw a line from the center outwards. We know the angle of the line, but need to calculate the coordinates of its end point. What do you think would be the best way to do it?
For example, if the angle of the line is 45 degrees, its end point coordinates would be roughly 75,15.
You need to use the trigonometric functions sin and cos.
Something like this:
theta = 45
// theta = pi * theta / 180 // convert to radians.
radius = 50
centerX = 50
centerY = 50
p.x = centerX + radius * cos(theta)
p.y = centerY - radius * sin(theta)
Keep in mind that most implementations assume that you're working with radians and have positive y pointing upwards.
Use the unit circle to calculate X and Y, but because your radius is 50, multiply by 50
http://en.wikipedia.org/wiki/Unit_circle
Add the offset (50,50) and bob's your uncle
X = 50 + (cos(45) * 50) ~ 85,36
Y = 50 - (sin(45) * 50) ~ 14,65
The above happens to be 45 degrees.
EDIT: just saw the Y axis is inverted
First you would want to calculate the X and Y coordinates as if the circle were the unit circle (radius 1). The X coordinate of a given angle is given by cos(angle), and the Y coordinate is given by sin(angle). Most implementations of sin and cos take their inputs in radians, so a conversion is necessary (1 degree = 0.0174532925 radians). Now, since your coordinate system is not in fact the unit circle, you need to multiply the resultant values by the radius of your circle. In this given instance, you would multiply by 50, since your circle extends 50 units in each direction. Finally, using a unit circle coorindate system assumes your circle is centered at the origin (0,0). To account for this, add (or subtract) the offset of your center from your calculated X and Y coordinates. In your scenario, the offset from (0,0) is 50 in the positive X direction, and 50 in the negative Y direction.
For example:
cos(45) = x ~= .707
sin(45) = y ~= .707
.707*50 = 35.35
35.35+50 = 85.35
abs(35.35-50) = 14.65
Thus the coordinates of the ending segment would be (85.35, 14.65).
Note, there is probably a built-in degrees-to-radians function in your language of choice, I provided the unit conversion for reference.
edit: oops, used degrees at first
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))
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