Develop Reference

Moving an object along a path ontouch

I have a stroke circle path and a little fill circle path and i want to move the little circle along the stroke circle path with touch, the little circle should not be outside the stroke circle. How can i do this ? Please help me

Your touch might be slighty outside of your circle path. To provide object position at the circle, you can make the next:
find position of touch against circle center
dx = touch.x - center.x
dy = touch.y - center.y
find distance from the center
dist = sqrt(dx*dx +dy*dy)
make new position at the same disraction form center but at circle radius distance
newx = center.x + radius * dx / dist
newy = center.y + radius * dy / dist

Multiple colors for single path Android

Good day.I am creating a siri like wave for android and i encounter an big issue.I need the wave to be in 4 colors.Lets assume i only have one single line which is drawing on the screen accordingly to the voice decibels.Anyway i am able to do it but no way i am able to give 4 different colors for same path.Assume it is 1 single path which moves from screen start to screen end,i need that line to have 4 different colors,mainly i had to divide the path into 4 parts and draw the color for each parts,but neither google,nor any other source give me anything (not even found anything similar to what i want).
Meanwhile i am posting the code where actually i am drawing the lines.
for (int l = 0; l < mWaveCount; ++l) {
float midH = height / 2.0f;
float midW = width / 2.0f;
float maxAmplitude = midH / 2f - 4.0f;
float progress = 1.0f - l * 1.0f / mWaveCount;
float normalAmplitude = (1.5f * progress - 0.5f) * mAmplitude;
float multiplier = (float) Math.min(1.0, (progress / 3.0f * 2.0f) + (1.0f / 3.0f));
if (l != 0) {
mSecondaryPaint.setAlpha((int) (multiplier * 255));
}
mPath.reset();
for (int x = 0; x < width + mDensity; x += mDensity) {
float scaling = 1f - (float) Math.pow(1 / midW * (x - midW), 2);
float y = scaling * maxAmplitude * normalAmplitude * (float) Math.sin(
180 * x * mFrequency / (width * Math.PI) + mPhase) + midH;
// canvas.drawPoint(x, y, l == 0 ? mPrimaryPaint : mSecondaryPaint);
//
// canvas.drawLine(x, y, x, 2*midH - y, mSecondaryPaint);
if (x == 0) {
mPath.moveTo(x, y);
} else {
mPath.lineTo(x, y);
// final float x2 = (x + mLastX) / 2;
// final float y2 = (y + mLastY) / 2;
// mPath.quadTo(x2, y2, x, y);
}
mLastX = x;
mLastY = y;
}
if (l == 0) {
canvas.drawPath(mPath, mPrimaryPaint);
} else {
canvas.drawPath(mPath, mSecondaryPaint);
}
}
I tried to change color on if (l == 0) {
canvas.drawPath(mPath, mPrimaryPaint);
} but if i change it here,no result at all,either the line is separate and not moving at all,but it should,either the color is not applied,propably because i am doing it in loop as i had to and everytime the last color is picked to draw.Anyway can you help me out?Even an small reference is gold for me because really there is nothing at all in the internet.

Anyway even though Matt Horst answer fully correct,i found the simplest and easiest solution...i never thought it would be so easy.Anyway if in world there is someone who need to make an path divided into multiple colors,here is what you can do
int[] rainbow = getRainbowColors();
Shader shader = new LinearGradient(0, 0, 0, width, rainbow,
null, Shader.TileMode.REPEAT);
Matrix matrix = new Matrix();
matrix.setRotate(90);
shader.setLocalMatrix(matrix);
mPrimaryPaint.setShader(shader);
Where getRainbowColors() is an array of colors you wish your line to have and width is the length of the path so the Shader knows how to draw the colors in right way to fit the length of path.Anyway .....easy isnt it?and pretty purged me a lot to get into this simple point.Nowhere in internet you could find only if you are looking for something completelly different,than you might come across this.

It seems to me like you could set up one paint for each section, each with a different color. Then set up one path for each section too. Then as you draw across the screen, wherever the changeover point is between sections, start drawing with the new path. And make sure first to use moveTo() on the new path so it starts off where the old one left off.

For my solution, I tried changing the color of the linePaint in the onDraw Call. But it was drawing a single color.
So i used two different paints for two different colors and draw path on the canvas.
And it worked. Hope it helps someone out there.

How to draw watchface 'ticks' on a square watch?

I currently have this snippet generating the ticks around the outside of and android wear watchface
float innerMainTickRadius = mCenterX - 35;
for(int tickIndex = 0; tickIndex < 12; tickIndex++) {
float tickRot = (float) (tickIndex * Math.PI * 2 / 12);
float innerX = (float) Math.sin(tickRot) * innerMainTickRadius;
float innerY = (float) -Math.cos(tickRot) * innerMainTickRadius;
float outerX = (float) Math.sin(tickRot) * mCenterX;
float outerY = (float) -Math.cos(tickRot) * mCenterX;
canvas.drawLine(mCenterX + innerX, mCenterY + innerY, mCenterX + outerX, mCenterY + outerY, mTickPaint);
}
Which generates the ticks well on a round watchface but on a square it turns out like this:
but I'd like them to not be circular, but instead fit the shape a bit more suitably, e.g:
Is there a standard way to do this? I'm guessing I can't use trig again...

Of course you use geometry and trig. For example any line you put on the clock face you want to point to the center so one part will be the given (x,y) and the other will be arctan2(cy-y,cx-x) giving you the angle from the point you have towards the center (cx,cy) then simply draw the line in the direction of the center of a given length r, by drawing the line from x,y to cos(angle) * r, sin(angle) * r.
However, given your sample image you might want to draw the line from x,y to x+r,y then rotate the canvas by angle so that you can draw those numbers tweaked like that. Be sure to do canvas.save() before tweaking the canvas' matrix and canvas.restore() after the tweak.
This leaves the math of whatever shape you want to draw your ticks from and the positions thereto. You can do this within a Path. So define the path for a rounded rectangle and then use the PathMeasure class to get the getPosTan() and then ignore the tangent and just use the position it gives you to find your position around a rounded rectangle. That or simply calculate those positions as the positions through either a line segment or a bezier section depending on the decided shape.
For example:
static final int TICKS = 12;
static final float TICKLENGTH = 20;
In the draw routine,
float left = cx - 50;
float top = cy - 50;
float right = cx + 50;
float bottom = cy + 50;
float ry = 20;
float rx = 20;
float width = right-left;
float height = bottom-top;
Path path = new Path();
path.moveTo(right, top + ry);
path.rQuadTo(0, -ry, -rx, -ry);
path.rLineTo(-(width - (2 * rx)), 0);
path.rQuadTo(-rx, 0, -rx, ry);
path.rLineTo(0, (height - (2 * ry)));
path.rQuadTo(0, ry, rx, ry);
path.rLineTo((width - (2 * rx)), 0);
path.rQuadTo(rx, 0, rx, -ry);
path.rLineTo(0, -(height - (2 * ry)));
path.close();
PathMeasure pathMeasure = new PathMeasure();
pathMeasure.setPath(path,true);
float length = pathMeasure.getLength();
float[] pos = new float[2];
float r = TICKLENGTH;
for (int i = 0; i < TICKS; i++) {
pathMeasure.getPosTan(i * (length/TICKS),pos,null);
double angle = Math.atan2(cy - pos[1], cx - pos[0]); //yes, y then x.
double cos = Math.cos(angle);
double sin = Math.sin(angle);
canvas.drawLine(pos[0], pos[1], (float)(pos[0] + cos * r), (float)(pos[1] + sin * r), paint);
}
Admittedly it looks like:
So it would take a lot more work to get it looking like your image. But, it's totally doable. The path measure trick thing will work for any shape. I avoided using path.addRoundRect because of the Lollipop+ restriction. You can see my answer to that question here. And the other answers which are plenty fine to how to draw a rounded rectangle-esque shape. You can, if you would like to write an envelope function simply scale your current picture to the envelope of the rectangle according to the factor t, as it goes around the clock.

The angle is a function of the position now. I'm not immediately seeing the trick for getting a closed form in this case. But in the most general case, you could end up just storing the position of each tickmark, then you're just drawing the line that goes through that point and the center. so the angle at second i is just
theta(i)=arctan(y_pos(i) / x_pos(i))
assuming the center has coordinates (0,0). In this case, you only need to store the positions for 8 consecutive ticks because the face is periodic every 90 degrees and symmetric about the diagonals as well.

Drawing an arrow head in android

I am trying to draw an arrow to point to objects in am image. I have been able to write code to draw the line but cant seem to be able to find a way to draw the arrowhead.The code I wrote to draw a dragabble line is as follows.I need to draw an arrowhead on ACTION_UP event to the direction in which the line is pointing
if(event.getAction() ==MotionEvent.ACTION_DOWN) {
if (count==1){
x1 = event.getX();
y1 = event.getY();
System.out.println(count+"count of value a;skd");
Toast.makeText(getApplicationContext(), ""+(radius+count), Toast.LENGTH_LONG).show();
Log.i(TAG, "coordinate x1 : "+String.valueOf(x1)+" y1 : "+String.valueOf(y1));
}
}
else if(event.getAction() ==MotionEvent.ACTION_MOVE){
imageView.setImageBitmap(bmp2);
x2 = event.getX();
y2 = event.getY();
posX=(float)(x1+x2)/2;
posY=(float)(y1+y2)/2;
radius=(float) Math.sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2))/2;
onDraw();
Toast.makeText(getApplicationContext(), ""+radius, Toast.LENGTH_LONG).show();
}
Hi, for anyone still needing help .This is how I did it in the end
float h=(float) 30.0;
float phi = (float) Math.atan2(y2 - y1, x2 - x1);
float angle1 = (float) (phi - Math.PI / 6);
float angle2 = (float) (phi + Math.PI / 6);
float x3 = (float) (x2 - h * Math.cos(angle1));
float x4 = (float) (x2 - h * Math.cos(angle2));
float y3 = (float) (y2 - h * Math.sin(angle1));
float y4 = (float) (y2 - h * Math.sin(angle2));
c.drawLine(x1, y1,x2,y2 ,pnt);
c.drawLine(x2, y2,x3,y3 ,pnt);
c.drawLine(x2, y2,x4,y4 ,pnt);
I got help from the accepted answer and ios section in stackoverflow

How I would do this is to find the slope of the line, which is drawn between two points(start and end). The slope would be (dy/dx), and that would be a good start point for your arrow. Assuming you want the base of the arrowhead to be perpendicular to the line of the arrow, to find the slope of the base you would find the opposite reciprocal of the slope of the line. for example, lets say that your line has a slope of 2. The slope for the base of your triangle would be (-1/2), because you do (1/(oldslope)) and multiply by -1. I don't know android very well, but if I remember correctly, in Java, you would use a drawPolygon method, and you would have to specify 4 points(3 unique and 1 the same as the first to close it). Given the slope of the base of the tip, we can get our first two points and our final point. You should know before you start the dimensions of the arrowhead you wish to draw, so in this case b will be the length of your baseline. If you take ϴ=arctan(dy/dx), that will give you an angle between the x axis and your baseline. With that ϴ value, you can do ydif = b*sin(ϴ) to get the difference in y value between the two base corners of your arrow. Doing the same thing but with xdif = b*cos(ϴ) gives you the difference in the x value between the two base points. If the location of the final point of the line that the user drew is, say, (x1, y1), then the locations of the basepoints of the triangle would be (x1-(xdif/2), y1-(ydif/2)) and (x1+(xdif/2), y1+(ydif/2)). These two points, p1 and p2, are the first, second, and fourth points in the draw polygon method. To find the third point, we need to find the angle of the original line, by doing ϴ=arctan(dy/dx), this time using your original dy/dx. with that angle. Before we finish the actual calculation of the point, you first have to know how far from the end of your line the tip of the arrow should actually be, in my case, I will use the var h and h = 10. To get the cordinate, (x,y), assuming the cordinate for the line tip is (x1, y1)you would do (x1+hcosϴ, y1+hsinϴ). Use that for the third value in drawPolygon(), and you should be done. sorry if I kind of rushed at the end, I got kind of tired of typing, comment if you need help.

If you managed to draw a line from the input event, you might additionally draw a triangle on its end indicating the direction.
On another project I drew a square everytime a magnetic point on a grid was touched (as you can see here) Sorry I can not provide you any sample code right now. But if that's a suitable approach for you, I might post it later.

Here is a good code, its not mine, It was a Java Graphics2D code that I converted to Canvas. All credit go to the original guy/lady who wrote it
private void drawArrowHead(Canvas canvas, Point tip, Point tail)
{
double dy = tip.y - tail.y;
double dx = tip.x - tail.x;
double theta = Math.atan2(dy, dx);
int tempX = tip.x ,tempY = tip.y;
//make arrow touch the circle
if(tip.x>tail.x && tip.y==tail.y)
{
tempX = (tip.x-10);
}
else if(tip.x<tail.x && tip.y==tail.y)
{
tempX = (tip.x+10);
}
else if(tip.y>tail.y && tip.x==tail.x)
{
tempY = (tip.y-10);
}
else if(tip.y<tail.y && tip.x==tail.x)
{
tempY = (tip.y+10);
}
else if(tip.x>tail.x || tip.x<tail.x)
{
int rCosTheta = (int) ((10)*Math.cos(theta)) ;
int xx = tip.x - rCosTheta;
int yy = (int) ((xx-tip.x)*(dy/dx) + tip.y);
tempX = xx;
tempY = yy;
}
double x, y, rho = theta + phi;
for(int j = 0; j < 2; j++)
{
x = tempX - arrowLength * Math.cos(rho);
y = tempY - arrowLength * Math.sin(rho);
canvas.drawLine(tempX,tempY,(int)x,(int)y,this.paint);
rho = theta - phi;
}
}
Just call this for both sides of your line and it will draw an arrow at each side!

Draw perpendicular line to given line on canvas android

I drawn line passing through points - (x1,y1) , (x2,y2)
Now i want to draw another line perpendicular to this line with same length.
Please guide me for this ..

Think of your line as a vector from (x1,y1) to (x2,y2). Then we get the x and y components of this vector according to
vX = x2-x1
vY = y2-y1
The vector of equal size to this but perpendicular to it in the plane has x and y components
vXP = -(y2-y1)
vYP = x2-x1
you can verify these 2 vectors are perpendicular by taking the scalar product of the 2 vectors which will be zero. Now you have your vector of equal length and perpendicular to your first vector, you simply need to decide the start point of your line. We will call that (a,b). Then using your start point, the end point of your line is given by
(a - (y2-y1), b + (x2-x1))
or if you want it to point in the reverse direction (still perpendicular) it will be
(a + (y2-y1), b - (x2-x1))

Well, it's simple maths :
int dx = x2 - x1;
int dy = y2 - y1;
int ox,oy; // Origin of new line
//...
drawLine( ox, oy, ox+dy, oy-dx) // This line will be perpendicular to original one
All you have to do is to choose the origin.
For example, if you want that the lines cuts at their center, let :
ox = x1 + (dx - dy) / 2;
oy = y1 + (dx + dy) / 2;