Look at the gray circles along the lines, they need to be placed equally to represent the hours, here is my code for that, where did I go wrong?
for (int i=0; i<12; i++)
{
c.drawCircle(140*(float)Math.cos((double) (i*30))+ width/2,
140*(float) Math.sin((double) (i*30)) + height/2, 1, p);
}
The angle must be provided to Math.cos in radians (between 0 and 2*Math.PI).
So, instead of
Math.cos((double) (i*30))
Use
Math.cos(i*Math.PI/6.0)
Note that I also removed the useless cast to double : multiplying an int and a double already produces a double.
Math.sin and Math.cos operate on angles in radians and not degrees. To correct, multiply by pi and divide by 180.
Related
I'm trying to calculate the lateral surface of a frustum cone using the code below.
Input: 2 Nodes containing x, y, z values (absolut position) and radius (radius at position) all in double
What I am doing so far:
1. calculate length of frustum cone
2. calculate lateral surface of frustum cone
My Problem:
Hypothetically substraction of 2 floating point numbers with similar magnitude is problematic due to precision loss. (I didn't run into the issue)
My Question:
What can I do to improve the final result?
Possibilities I found / thought of:
- using BigDecimal (what I don't want to because of longer runtime)
- replacing (r1-r2)*(r1-r2) with r1^2 - 2*r1*r2 + r2^2
- implement a check on how close two double values are and if very close assume their difference to be 0. Would that even improve the accuracy of my final result? Doesn't the result of the substraction have a smaller absolute error than the one with 0 assumed?
public static double calculateLateralSurface(Node node1, Node node2) {
double x, y, z, length, r1, r2, lateralSurface;
// calculate length of frustum cone
x = node1.getPosX() - node2.getPosX();
y = node1.getPosY() - node2.getPosY();
z = node1.getPosZ() - node2.getPosZ();
length = Math.sqrt(x * x + y * y + z * z);
r1 = node1.getRadius();
r2 = node2.getRadius();
// calculate lateral surface of frustum cone
lateralSurface = (r1+r2)*Math.PI*Math.sqrt((r1-r2)*(r1-r2)+length*length);
return lateralSurface;
}
I hope someone can help me :)
double has more than enough accuracy for any practical and even not-so-practical use. So, unless your cone describes the field of view of an optical telescope from Earth all the way to Alpha Centauri, you should not have any precision problems.
I can point out to you that you are calculating length by taking a square root only to square it again later before using it, so eliminating that unnecessary pair of calculations might somehow improve things, but I doubt that this is your problem.
So, if your results do not look correct, then maybe a bug is to blame and not the precision of double.
In any case, why don't you provide a self-contained, compilable example, and actual input values, and actual output values, and your expected values, and we can look deeper into it.
I am currently working on a Processing (as in the language) sketch, which is driven by Turtle logic (see https://en.wikipedia.org/wiki/Turtle_graphics). This means that I draw a line from the current coordinate to a supplied coordinate. This supplied coordinate will then become the new current coordinate. I want to approximate a circle and have written a simple piece of code using trigonometrics. The code looks as follow:
void drawCircle(int radius){
// The circle center is radius amount to the left of the current xpos
int steps = 16;
double step = TWO_PI /steps;
for(double theta = step; theta <= TWO_PI; theta += step){
float deltaX = cos((float)theta) - cos((float)(theta - step));
float deltaY = sin((float)theta) - sin((float)(theta - step));
moveXY(deltaX*radius, deltaY*radius);
}
}
The program logic is simple. It will use the variable theta to loop through all the radians in a circle. The amount of steps will indicate how large each theta chunk is. It will then calculate the x,y values for the specific point in the circle governed by theta. It will then deduct the x,y values of the previous cycle (hence the theta-step) to get the amount it will have to move from this position to attain the desired x,y position. It will finally supply those delta values to a moveXY function, which draws a line from the current point to the supplied values and makes them the new current position.
The program seems to work quite well when using a limited amount of steps. However, when the step count is increased, the circles become more and more like a Fibonacci spiral. My guess is that this is due to imprecision with the float number and the sine and cosine calculations, and that this adds up with each iteration.
Have I interpreted something wrong? I am looking to port this code to Javascript eventually, so I am looking for a solution in the design. Using BigDecimal might not work, especially since it does not contain its own cosine and sine functions. I have included a few images to detail the problem. Any help is much appreciated!
Step count 16:
Step count 32:
Step count 64:
Step count 128:
Float and sine/cosine should be sufficiently precise. The question is: How precise is your position on the plane? If this position is measured in pixels, then each of your floating point values is rounded to integer after each step. The loss of precision then adds up.
At each iteration round the loop, you are calculating the delta without regard of what the current coordinate is. So effectively, you are "dead-reckoning", which is always going to be inaccurate, since errors at each step build up.
Since you know that you want a circle, an alternative approach would be at each iteration, to first determine the actual point on the circle you want to get to, and then calculate the delta to get there - so something like the following (but I must admit I haven't tested it !):
void drawCircle(int radius){
// The circle center is radius amount to the left of the current xpos
int steps = 16;
double step = TWO_PI /steps;
float previousX = 0;
float previousY = radius;
for(double theta = step; theta <= TWO_PI; theta += step){
float thisX = radius * sin((float)theta);
float thisY = radius * cos((float)theta);
moveXY(thisX - previousX, thisY - previousY);
previousX = thisX;
previousY = thisY;
}
}
I have a character in my game that must rotate smoothly to get to a desired angle. Consider angle as the current angle and touchAngle as the desired angle which is always between 0 to 360. I want to add +1/-1 to current angle in every game update to get to the desired touchAngle. The problem is first it must chose direction and it must be between 0 to 360. this is my pseudo code:
int touchAngle;
float angle;
public void update()
{
if ((int)angle != touchAngle) angle += ???
}
Since you have values that are always normalized in the interval [0 360] this should not be too hard.
You just need to distinguish two different cases:
angle < touchAngle
angle > touchAngle
in the first case we want to rotate counterclockwise so the update has to be angle =+ 1 (assuming that you want to turn of 1 every update cycle).
In the second case we want to turn clockwise so the update should be angle -= 1.
The problem is that this is not always the shortest way to rotate. For instance if:
angle == 359
touchAngle == 1
we don't want to make all the way 358, 357, 356...instead we want to rotate counterclockwise for just 2 units: 360, 1.
This can be achieved comparing the distance between the angles abs(angle - touchAngle).
If this value is bigger than 180 it means we are going the wrong way, so we have to do the way around so
if(angle < touchAngle) {
if(abs(angle - touchAngle)<180)
angle += 1;
else angle -= 1;
}
else {
if(abs(angle - touchAngle)<180)
angle -= 1;
else angle += 1;
}
of course all of this until ((int)angale != touchAngle).
I might have made mistakes with the cases but this is the principle.
Generally you want to bring in time to the equation, so that you can smoothly change the angle over time. Most setups have a way to get a time it took to render the previous frame and the typical way to do this is to say..
int touchAngle;
float angle;
float deltaTime; //Time it took to render last frame, typically in miliseconds
float amountToTurnPerSecond;
public void update()
{
if((int)angle != touchAngle) angle += (deltaTime * amountToTurnPerSecond);
}
This will make it so that each second, your angle is changed by amountToTurnPerSecond, but changed slowly over each frame the correct amount of change so that it is smooth. Something to note about this is that you wont evenly end up at touchAngle most of the time, so checking to see if you go over and instead setting to touchAngle would be a good idea.
Edit to follow up on comment:
I think the easiest way to attain the correct direction for turn is actually not to use angles at all. You need to get the relative direction from your touch to your character in a 2d space. Typically you take the touch from screen space to world space, then do the calculations there (at least this is what I've done in the past). Start out by getting your touch into world space, then use the vector cross product to determine direction. This looks kind of like the following...
character position = cx, cy
target position = tx, ty
current facing direction of character = rx, ry
First we take the distance between the character and the target position:
dx = tx - cx
dy = ty - cy
This not only gives us how far it is from us, but essentially tells us that if we were at 0, 0, which quadrant in 2d space would the target be?
Next we do a cross product:
cross_product = dx * ry - dy * rx
If this is positive you go one way, if it's negative you go the other. The reason this works out is that if the distance is for instance (-5, 2) then we know that if we are facing directly north, the point is to our left 5 and forward 2. So we turn left.
I'm trying to use this method to rotate a point:
AffineTransform at = AffineTransform.getRotateInstance(Math.toRadians(90), 1, 2);
double[] pt = {0, 1};
at.transform(pt, 0, pt, 0, 1);
int x = (int)pt[0];
int y = (int)pt[1];
The expected result is x=0 and y=3 because we are rotating 90* clockwise, however the output is currently x=2 and y=1;
Why am I not getting the expected result?
The program is doing nothing wrong. By specifying Math.toRadians(90) as your theta, you are instructing it to rotate counter-clockwise, not clockwise. That is how rotation is defined in trigonometry.
Using -Math.toRadians(90) would get you the correct result.
Also, as a side note, casting the point values to int isn't the best idea for checking your code. Double-precision trigonometry is prone to very minor errors, and so a tiny bump in the wrong direction would get you the wrong integers. (I.e., 2.99999999995 would be cast to 2)
(int) Math.round(value) is much safer
Actually, you're rotating anti-clockwise. The Javadoc says:
Rotating by a positive angle theta rotates points on the positive X axis toward the positive Y axis
See the Javadoc
I have a 2D convex polygon in 3D space and a function to measure the area of the polygon.
public double area() {
if (vertices.size() >= 3) {
double area = 0;
Vector3 origin = vertices.get(0);
Vector3 prev = vertices.get(1).clone();
prev.sub(origin);
for (int i = 2; i < vertices.size(); i++) {
Vector3 current = vertices.get(i).clone();
current.sub(origin);
Vector3 cross = prev.cross(current);
area += cross.magnitude();
prev = current;
}
area /= 2;
return area;
} else {
return 0;
}
}
To test that this method works at all orientations of the polygon I had my program rotate it a little bit each iteration and calculate the area. Like so...
Face f = poly.getFaces().get(0);
for (int i = 0; i < f.size(); i++) {
Vector3 v = f.getVertex(i);
v.rotate(0.1f, 0.2f, 0.3f);
}
if (blah % 1000 == 0)
System.out.println(blah + ":\t" + f.area());
My method seems correct when testing with a 20x20 square. However the rotate method (a method in the Vector3 class) seems to introduce some error into the position of each vertex in the polygon, which affects the area calculation. Here is the Vector3.rotate() method
public void rotate(double xAngle, double yAngle, double zAngle) {
double oldY = y;
double oldZ = z;
y = oldY * Math.cos(xAngle) - oldZ * Math.sin(xAngle);
z = oldY * Math.sin(xAngle) + oldZ * Math.cos(xAngle);
oldZ = z;
double oldX = x;
z = oldZ * Math.cos(yAngle) - oldX * Math.sin(yAngle);
x = oldZ * Math.sin(yAngle) + oldX * Math.cos(yAngle);
oldX = x;
oldY = y;
x = oldX * Math.cos(zAngle) - oldY * Math.sin(zAngle);
y = oldX * Math.sin(zAngle) + oldY * Math.cos(zAngle);
}
Here is the output for my program in the format "iteration: area":
0: 400.0
1000: 399.9999999999981
2000: 399.99999999999744
3000: 399.9999999999959
4000: 399.9999999999924
5000: 399.9999999999912
6000: 399.99999999999187
7000: 399.9999999999892
8000: 399.9999999999868
9000: 399.99999999998664
10000: 399.99999999998386
11000: 399.99999999998283
12000: 399.99999999998215
13000: 399.9999999999805
14000: 399.99999999998016
15000: 399.99999999997897
16000: 399.9999999999782
17000: 399.99999999997715
18000: 399.99999999997726
19000: 399.9999999999769
20000: 399.99999999997584
Since this is intended to eventually be for a physics engine I would like to know how I can minimise the cumulative error since the Vector3.rotate() method will be used on a very regular basis.
Thanks!
A couple of odd notes:
The error is proportional to the amount rotated. ie. bigger rotation per iteration -> bigger error per iteration.
There is more error when passing doubles to the rotate function than when passing it floats.
You'll always have some cumulative error with repeated floating point trig operations — that's just how they work. To deal with it, you basically have two options:
Just ignore it. Note that, in your example, after 20,000 iterations(!) the area is still accurate down to 13 decimal places. That's not bad, considering that doubles can only store about 16 decimal places to begin with.
Indeed, plotting your graph, the area of your square seems to be going down more or less linearly:
This makes sense, assuming that the effective determinant of your approximate rotation matrix is about 1 − 3.417825 × 10-18, which is well within normal double precision floating point error range of one. If that's the case, the area of your square would continue a very slow exponential decay towards zero, such that you'd need about two billion (2 × 109) 7.3 × 1014 iterations to get the area down to 399. Assuming 100 iterations per second, that's about seven and a half months 230 thousand years.
Edit: When I first calculated how long it would take for the area to reach 399, it seems I made a mistake and somehow managed to overestimate the decay rate by a factor of about 400,000(!). I've corrected the mistake above.
If you still feel you don't want any cumulative error, the answer is simple: don't iterate floating point rotations. Instead, have your object store its current orientation in a member variable, and use that information to always rotate the object from its original orientation to its current one.
This is simple in 2D, since you just have to store an angle. In 3D, I'd suggest storing either a quaternion or a matrix, and occasionally rescaling it so that its norm / determinant stays approximately one (and, if you're using a matrix to represent the orientation of a rigid body, that it remains approximately orthogonal).
Of course, this approach won't eliminate cumulative error in the orientation of the object, but the rescaling does ensure that the volume, area and/or shape of the object won't be affected.
You say there is cumulative error but I don't believe there is (note how your output desn't always go down) and the rest of the error is just due to rounding and loss of precision in a float.
I did work on a 2d physics engine in university (in java) and found double to be more precise (of course it is see oracles datatype sizes
In short you will never get rid of this behaviour you just have to accept the limitations of precision
EDIT:
Now I look at your .area function there is possibly some cumulative due to
+= cross.magnitude
but I have to say that whole function looks a bit odd. Why does it need to know the previous vertices to calculate the current area?