Develop Reference

Rotate item in a circle shape using phone accelerometer/gravity sensor vector

I'm trying to make a small circle move in another bigger circle as a ball moving in circle relative to Android phone tilting. I'm doing this in Godot but I think the principle is the same in all game engines probably. I make a scene in Godot and add two sprites as the two circles as the following picture. I get the accelerometer 3D vector, use x and y values and calculate the angle in x and y plate (relative to y axis i.e Vector2(0, 1) ) and rotate the mentioned scene to that degree using an animation, using this code
func _process(delta: float) -> void:
var vec3 = Input.get_accelerometer()
accelVec = Vector2(-stepify(vec3.x, 0.1), -stepify(vec3.y, 0.1))
var angle = accelVec.angle_to(Vector2(0, 1))
rotateTween.interpolate_property(self, "rotation", rotation, angle, 0.2,
Tween.TRANS_LINEAR)
rotateTween.start()
return
But the problem lies in here that when the x value of accelerometer 3D vector changes from a positive to negative value i.e when the ball is at top of the circle and is going to go to the other half of the circle, it actually moves from bottom of the circle to the desired point rather than from top of the circle.
I hope I explained the problem well, though I doubt it.
I added the Gif that shows actual test on an android phone here Testing in Android.
Any idea how to solve this problem?
Thanks in advance.

This is because Tween is interpolating linear values. It doesn't know it's working with angles, or that they should wrap around.
So when you're at -179 degrees and you tween to 179--a difference of 2 degrees--Tween just sees -179 -> 179 and goes through the whole circle.
You probably don't need a Tween here at all because _process() happens every frame.

Calculate angle of touched point and rotate it on a fixed image or canvas or bitmap image in Android

Ohh.. damm Math !! once again stuck. It seems to be easy but i think its not that easy,
Problem Statement: I want to rotate the 3 fixed points which lies on a fixed circle.
1.when 1 point is selected remaining 2 points should be static mode and only selected point should move/rotate on circumference of circle.
2.And all 3 points are connected via 3 lines as shown in images..when we select a point and rotate it,connected lines also increase and decrease..
I already tried to solve this problem finding angle at each instant after touch.but its not quite working as per my need..
something like this

I hope the following explanation enable you to put the steps into your coding language.
Presumption is that the vertex to be moved has already selected and so the calculation of (xcnd,ycnd) as defined below is used to set the selected vertex of the triangle.
Let the constraining circle have centre at (cx,cy) and radius r.
Let the coordinates of where the screen is touched be (xtch,ytch)
Let the coordinates of where the screen is touched relative to the centre be (xrel,yrel)
then xrel = xtch - cx and yrel = ytch - cy
Let the coordinates of the point on the constraining circle when the screen is touched at (xtch,ytch) be (xcnd,ycnd).
xcndrel = xcnd - cx, and ycndrel = ycnd - cy give the coordinates on the constraining circle relative to its centre,
Note that
xrel and xcndrel will have the same signs (ie both positive or both negative)
and yrel and ycndrel will also have the same signs.
the function abs(x) = x if x>=0 and -x if x<0 should be available in whatever language you are using
the function sign(x) may or may not be available, sign(x) =1 if x>0 and -1 if x<0 and undefined for x=0
If not available then sign(x)=x/abs(x)
Check if xrel=0
if xrel=0 xcndrel=0, ycndrel=r*sign(yrel)
Otherwise work in first quadrant ie where x>0 and y>0 using abs(xrel) and abs(yrel)
find angle where screen is touched relative to centre of circle using
theta=arctan(abs(yrel)/abs(xrel))
find the coordinates (xcndrel, ycndrel) by using theta in the first quadrant and then placing in the correct quadrant using the signs of xrel and yrel
xcndrel = sign(xrel)*r*COS(theta)
ycndrel = sign(yrel)*r*SIN(theta)
Screen coordinates can now be found
xcnd = xcndrel +cx
ycnd = ycndrel + cy

provide circular motion for a custom drawing on touch: Android

I have a custom drawn rectangle which i want to move in a circular path based on touch events.
It follows the direction of the touch for clockwise or anticlockwise movement but basically move in circular motion, as if moving on the edge of the circle.
My current thought process is as follows:
Based on the users current and previous x,y i shall find the angle in degrees and then move this rectangle by the same angle by re-drawing in the new position, just making sure that it moves on the edge of a circle.
But this leads to some confusion on the following:
1. how do i decide whether angle movement is clockwise or anti-clockwise.
2. I am not being able to figure out the math for this properly.
Would this be the best approach or is there a better idea for doing this?
Also, if this is the best approach, could someone please tell me the formula for calculating the angle by which i should move it while taking care of the clocking and anticlockwise ?
could someone please help?
please let me know if any more details are required.
Thanks

Steps
Here are a few steps in order to move your rectangle along a circle's rim when the user taps and holds to the side of the circle:
1. Obtain direction desired.
2. Obtain angle from current x and y coordinates.
3. Add direction (+1 if counterclockwise, -1 if clockwise) to angle.
4. Calculate new x and y coordinates.
5. Update/display rectangle.
Details
1. In pseudocode, direction = sign(Rectangle1.x - UsersFingerPosition.x). Here sign is a function returning -1 if the number was negative, 0 if it is 0, and 1 if it is positive. Note that sign(0) will only result when the user is on the exact x and y of your rectangle's location. In that case, the rectangle would not move (which should be good). In Java, the sign function is Math.signum().
2. To obtain the current angle use the following java code:
double angle = Math.toDegrees(Math.atan2(Circle.y-Rectangle1.y, Rectangle1.x-Circle.x));
Note the order of Circle.y-Rectangle.y and Rectangle.x...Circle.x. This is a result of the coordinate (0, 0) being in the top left corner instead of the center of the screen.
3. Simple enough, just add direction to angle. If desired, do something like
angle += direction*2; //So it will move more quickly
4. To get the new x and y coordinates of your rectangle, use the trigonometric functions sine and cosine:
Rectangle1.x = Math.cos(Math.toRadians(angle))*Circle.radius + Circle.x - Rectangle1.width;
Rectangle1.y = Math.sin(Math.toRadians(angle))*Circle.radius + Circle.y - Rectangle1.height;
(where Circle.x and Circle.y are the coordinates of the center of your circle and Circle.radius is naturally it's radius).
5. This one you'll have to take care of (or have already) :)!
Hope this helps you!

Steps
Here are a few steps in order to move your rectangle along a circle's rim:
1. Obtain finger position/Check that it's still dragging the rectangle.
2. Obtain angle from current x and y coordinates.
3. Calculate new x and y coordinates.
4. Update/display rectangle.
Details
1. This one is probably specific to your code, however, make sure that when the user starts dragging the rectangle, you set a variable like rectangleDragging to true. Before you run the next steps (in the code), check that rectangleDragging == true. Set it to false once the user lets go.
2. To obtain the current angle use the following java code:
double angle = Math.toDegrees(Math.atan2(Circle.y-Finger.y, Finger.x-Circle.x));
Note the order of Circle.y-Finger.y and Finger.x...Circle.x. This is a result of the coordinate (0, 0) being in the top left corner instead of the center of the screen.
3. To get the new x and y coordinates of your rectangle, use the trigonometric functions sine and cosine:
Rectangle1.x = Math.cos(Math.toRadians(angle))*Circle.radius + Circle.x - Rectangle1.width;
Rectangle1.y = Math.sin(Math.toRadians(angle))*Circle.radius + Circle.y - Rectangle1.height;
(where Circle.x and Circle.y are the coordinates of the center of your circle and Circle.radius is naturally it's radius). Subtracting the width and height of the rectangle should center it on the circle's border instead of placing the left, upper corner on the circle.
4. This one you'll have to take care of (or have already) :)!
Hope this helps you!

finding specifically where an object is for collision detection purposes

I have a square that rotates to a random angle and then travels in a straight line in the direction it is pointing. It does this by using a variable as its x axis and then calling
Variable++
Each frame.
unfortunatley i cannot work out how to return the exact position of the square because the square can be travelling at any angle and therefore doesn't rigidly follow the world coordinte grid. This means that the x variable is not the shapes x coordinate.
How do i return the shapes exact coordinates and how do i do it in such a way that i can have two squares drawn from the same class behaving differently.

So you've got a measure of distance from where the object started along its internal sideways axis and a measure of the angle between that axis and the horizontal?
If so then the formula you want is simple trigonometry. Assuming the object started at (x, y) and has travelled 'distance' units along an axis at an angle of 'angle' with the horizontal then the current position (x', y') is:
x' = x + distance * cos(angle)
y' = y + distance * sin(angle)
If you have the origin in the lower left of the screen and axes arranged graph paper style with x increasing to the right and y increasing as you go upward, that assumes that the angle is measured anticlockwise and that the object is heading along positive x when angle is zero.
If you'll permit a hand waving explanation, the formula works because one definition of sine and cosine is that they're the (x, y) coordinates of the point on the outside of a unit circle at the angle specified. It also matches with the very first thing most people learn about trigonometry, that sine is 'opposite over hypotenuse', and cosine is 'adjacent over hypotenuse'. In this case your hypotenuse has length 'distance' and and you want to get the 'opposite' and 'adjacent' lengths of a right angled triangle that coincides with the axes.
Assuming Android follows J2SE in this area, the one thing to watch out for is that Math.sin and Math.cos take an angle in radians, whereas OpenGL's rotatef takes an argument in degrees. Math.toDegrees and Math.toRadians can do the conversion for you.

When you made the shape you should have already specified its X & Y coordinates. Im not too sure what you mean when you say you cant find the coordinates?
Also make sure you do fame independent movement; currently you are adding one to your variable on every loop of your program. This means if it runs a 60 Frame Per Second(FPS) it will move 60 units, but if it runs at 30FPS it will move at half the speed

Random direction without visually rotating shape opengl-es android

I want to be able to assign a shape a random direction to go in at a regular speed.
I tried assigning a random number to the x and y values of a translation but that caused the shape to sometimes move way too fast or just blip off the screen.
Is there a way to choose a random direction for a shape to move in without rotating the shape (at least as far as the user can tell)?
Also is there a way of re-genereating the random number when an event is called (ie: button click) which would allow a change in direction?
EDIT: Just checked. Using the rotate function with a random angle will shoot the square off in a random direction, but still is there a way to do this without altering the orientation of the shape?

Ok, so you want a shape to move in either the postive or negative x, y or z directions?
If you want this to be random then use the Random class. From this you can use Random.nextInt(1) which will either be 0 or 1, if the result is 0 then set the value to be -1.
The +1 or -1 is now used for your direction.
Now you know which way the object travels you can simply multiply this by the speed of your object
Here is some rough code
Random rand = new Random(); int direction = rand.nextInt(1);
if(direction == 0) direction = -1;
Shape theShape.position.x += direction * theShape.speed
This may not be how your interface is setup (such as theShape.position or theShape.speed) but the idea is the same
After reading this again I think you are trying to rotate your shape lol, rotation is not just adding to the x and y values. Do you use matrices in your application? If not your should use trig to figure out what the x and y values will be after a rotation of x degrees