I am currently working on a game using javascript and processing.js and I am having trouble trying to figure out how to move stuff diagonally. In this game, there is an object in the center that shoots other objects around it. Now I have no problem moving the bullet only vertically or only horizontally, however I am having difficulty implementing a diagonal motion for the bullet algorithm.
In terms of attempts, I tried putting on my math thinking cap and used the y=mx+b formula for motion along a straight line, but this is what my code ends up looking like:
ellipse(shuriken.xPos, shuriken.yPos, shuriken.width, shuriken.height); //this is what I want to move diagonally
if(abs(shuriken.slope) > 0.65) {
if(shuriken.targetY < shuriken.OrigYPos) {
shuriken.yPos -= 4;
} else {
shuriken.yPos += 4;
}
shuriken.xPos = (shuriken.yPos - shuriken.intercept)/shuriken.slope;
} else {
if(shuriken.targetX < shuriken.OrigXPos) {
shuriken.xPos -= 4;
} else {
shuriken.xPos += 4;
}
shuriken.yPos = shuriken.slope * shuriken.xPos + shuriken.intercept;
}
The above code is very bad and hacky as the speed varies with the slope of the line.
I tried implementing a trigonometry relationship but still in vain.
Any help/advice will be greatly appreciated!
Think of it this way: you want the shuriken to move s pixels. If the motion is horizontal, it should move s pixels horizontally; if vertical, s pixels vertically. However, if it's anything else, it will be a combination of pixels horizontally/vertically. What's the correct combination? Well, what shape do you get if you project s distance in any direction from a given point? That's right, a circle with radius s. Let's represent the direction in terms of an angle, a. So we have this picture:
How do we get the x and the y? If you notice, we have a triangle. If you recall your trigonometry, this is precisely what the sine, cosine, and tangent functions are for. I learned their definitions via the mnemonic SOHCAHTOA. That is: Sin (a) = Opposite/Hypotenuse, Cos(a) = Adjacent/Hypotenuse, Tan(a) = Opposite/Adjacent. In this case, opposite of angle a is y, and adjacent of angle a is x. Thus we have:
cos(a) = x / s
sin(a) = y / s
Solving for x and y:
x = s * cos(a)
y = s * sin(a)
So, given the angle a, and that you want to move your shuriken s pixels, you want to move it s * cos(a) horizontally and s * sin(a) vertically.
Just be sure you pass a in radians, not degrees, to javascript's Math.sin and Math.cos functions:
radians = degrees * pi / 180.0
This may be why your trigonometric solution didn't work as this has bitten me a bunch in the past.
If you know the angle and speed you are trying to move at, you can treat it as a polar coordinate, then convert to cartesian coordinates to get an x,y vector you would need to move the object by to go in that direction and speed.
If you don't know the angle, you could also come up with the vector by taking the difference in X and difference in Y (this I know you can do as you are able to calculate the slope between the 2 points). Then take the resulting vector and divide by the length of the vector to get a unit vector, which you can then scale to your speed to get a final vector in which you can move your object by.
(This is what probably what kennypu means by sticking with vectors?)
Related
I'm working through a tutorial to make the old arcade game Breakout - you have a paddle at the bottom of the screen and the goal is to deflect a moving ball into a series of blocks at the top of the screen.
The code to calculate the rebound effect is:
ball.dx = ball.speed * Math.sin(angle);
ball.dy = - ball.speed * Math.cos(angle);
The yellow circle represents the ball:
I understand sine and cosine as ratios of the hypotenuse; I just still can't seem to grasp how they are used to calculate the rebound angle here exactly. Can anyone explain how the resulting numbers, given an angle and a speed value, produce the directionality of the ball on rebound? I feel there's a simple conceptual piece of the puzzle I'm missing.
This is vector adding - the X and Y vector added give you the new speed value.
To easier understand how sin and cos work here, take the case of angel = 0 deg. The ball falls straight down, and should bounce back up:
ball.dx = ball.speed * Math.sin(0); // 0
ball.dy = - ball.speed * Math.cos(0); // 1
So there's no movement left or right, speed is the same but the vertical direction is reversed because of the minus sign.
Using sin and cos here takes care of having a constant speed, as well, as these always sum up to 1.
Hope that's a bit more clarifying than confusing, but I did some similar code tasks that got easily solved with basic vector operations.
Good day, I am trying to create a simple 2D solar system model in javascript, but am having some trouble understanding how to go about calculating where planets will be for the next frame, aswell as a few other bits which I'll go into detail with soon.
After watching this very nice video and a whole bunch of his others, I made a quick MS paint image to try and simplify my situation.
With the second scene, you can see that the new position is calulated using the velocity, gravitational pull, and the angle between these two 'directions'?
I cannot get my head around how to figure this all out.
Below is a JS fiddle of my code. You'll notice I'm trying my best to use real NASA given data to keep it accurate.
You'll want to look specifically at lines 138 which is where all the calculations for its next move are made.
https://jsfiddle.net/c8eru7mk/9/
attraction: function(p2) {
// Distance to other body
var dx = p2.position.x - this.position.x;
var dy = p2.position.y - this.position.y;
var d = Math.sqrt(dx ** 2 + dy ** 2); // Possibly correct
// Force of attracrtion
this.f = G * (this.mass * p2.mass) / (d ** 2); // Possibly Correct
// Direction of force, If you read it hard enough you should be able to hear my screams of pain
// Not sure if this is correct, most likely not.
var theta = Math.atan2(dy, dx);
var fx = Math.cos(theta) * this.f;
var fy = Math.sin(theta) * this.f;
this.velocity.x += fx / this.mass;
this.velocity.y += fy / this.mass;
this.position.x += this.velocity.x;
this.position.y += this.velocity.y;
}
The problems I'm currently facing are
If I am to use NASA values, the distance between planets is so big, they won't fit on the screen, and I can't simply scale the distances down by multiplying them by 0.0002 or whatever, as that'll mess with the gravitational constant, and the simulation will be completely off.
I have no idea how to caluclate the next position and my brain has imploded several times this past week trying to attempt it several times.
I have no idea on how to check if my configuration data of planets is wrong, or if the simulation is wrong, so I'm pretty much just guessing.
This is also my first time actually coding anything more complex than a button in javascript too, so feedback on code layout and whatnot is welcome!
Many thanks
Using NASA values is not a problem when using separate coordinates for drawing. Using an appropriate linear transfomration from real coordinates to screen coordinatees for displaying does not influence the physical values and computations.
For simulating the motion of a planet with iterative updates one can assume that the gravitational force and the velocity are constant for a small portion of time dt. This factor dt is missing in your conversions from accelration to velocity and from velocity to distance. Choosing an appropriate value for dt may need some experiments. If the value is too big the approximation will be too far off from reality. If the value is too small you may not see any movement or rounding errors may influence the result.
For the beginning let us assume that the sun is always at (0,0). Also for a start let us ignore the forces between the planets. Then here are the necessary formulas for a first not too bad approximation:
scalar acceleration of a planet at position (x,y) by the gravitational force of the sun (with mass M): a = G*M/(d*d) where d=sqrt(x*x+y*y). Note that this is indepent of the planet's mass.
acceleration vector: ax = -a*x/d, ay = -a*y/d (the vector (-x,-y) is pointing towards the sun and must be brought the length a)
change of the planet's velocity (vx,vy): vx += ax*dt, vy += ay*dt
change of the planet's position: x += vx*dt, y += vy*dt
Specifically, I'm working in canvas with javascript.
Basically, I have objects which have boundaries that I want to avoid, but still surround with a bezier curve. However, I'm not even sure where to begin to write an algorithm that would move control points to avoid colliding.
The problem is in the image below, even if you're not familiar with music notation, the problem should still be fairly clear. The points of the curve are the red dots
Also, I have access to the bounding boxes of each note, which includes the stem.
So naturally, collisions must be detected between the bounding boxes and the curves (some direction here would be good, but I've been browsing and see that there's a decent amount of info on this). But what happens after collisions have been detected? What would have to happen to calculate control points locations to make something that looked more like:
Bezier approach
Initially the question is a broad one - perhaps even to broad for SO as there are many different scenarios that needs to be taken into consideration to make a "one solution that fits them all". It's a whole project in its self. Therefor I will present a basis for a solution which you can build upon - it's not a complete solution (but close to one..). I added some suggestions for additions at the end.
The basic steps for this solutions are:
Group the notes into two groups, a left and a right part.
The control points are then based on the largest angle from the first (end) point and distance to any of the other notes in that group, and the last end point to any point in the second group.
The resulting angles from the two groups are then doubled (max 90°) and used as basis to calculate the control points (basically a point rotation). The distance can be further trimmed using a tension value.
The angle, doubling, distance, tension and padding offset will allow for fine-tuning to get the best over-all result. There might be special cases which need additional conditional checks but that is out of scope here to cover (it won't be a full key-ready solution but provide a good basis to work further upon).
A couple of snapshots from the process:
The main code in the example is split into two section, two loops that parses each half to find the maximum angle as well as the distance. This could be combined into a single loop and have a second iterator to go from right to middle in addition to the one going from left to middle, but for simplicity and better understand what goes on I split them into two loops (and introduced a bug in the second half - just be aware. I'll leave it as an exercise):
var dist1 = 0, // final distance and angles for the control points
dist2 = 0,
a1 = 0,
a2 = 0;
// get min angle from the half first points
for(i = 2; i < len * 0.5 - 2; i += 2) {
var dx = notes[i ] - notes[0], // diff between end point and
dy = notes[i+1] - notes[1], // current point.
dist = Math.sqrt(dx*dx + dy*dy), // get distance
a = Math.atan2(dy, dx); // get angle
if (a < a1) { // if less (neg) then update finals
a1 = a;
dist1 = dist;
}
}
if (a1 < -0.5 * Math.PI) a1 = -0.5 * Math.PI; // limit to 90 deg.
And the same with the second half but here we flip around the angles so they are easier to handle by comparing current point with end point instead of end point compared with current point. After the loop is done we flip it 180°:
// get min angle from the half last points
for(i = len * 0.5; i < len - 2; i += 2) {
var dx = notes[len-2] - notes[i],
dy = notes[len-1] - notes[i+1],
dist = Math.sqrt(dx*dx + dy*dy),
a = Math.atan2(dy, dx);
if (a > a2) {
a2 = a;
if (dist2 < dist) dist2 = dist; //bug here*
}
}
a2 -= Math.PI; // flip 180 deg.
if (a2 > -0.5 * Math.PI) a2 = -0.5 * Math.PI; // limit to 90 deg.
(the bug is that longest distance is used even if a shorter distance point has greater angle - I'll let it be for now as this is meant as an example. It can be fixed by reversing the iteration.).
The relationship I found works good is the angle difference between the floor and the point times two:
var da1 = Math.abs(a1); // get angle diff
var da2 = a2 < 0 ? Math.PI + a2 : Math.abs(a2);
a1 -= da1*2; // double the diff
a2 += da2*2;
Now we can simply calculate the control points and use a tension value to fine tune the result:
var t = 0.8, // tension
cp1x = notes[0] + dist1 * t * Math.cos(a1),
cp1y = notes[1] + dist1 * t * Math.sin(a1),
cp2x = notes[len-2] + dist2 * t * Math.cos(a2),
cp2y = notes[len-1] + dist2 * t * Math.sin(a2);
And voila:
ctx.moveTo(notes[0], notes[1]);
ctx.bezierCurveTo(cp1x, cp1y, cp2x, cp2y, notes[len-2], notes[len-1]);
ctx.stroke();
Adding tapering effect
To create the curve more visually pleasing a tapering can be added simply by doing the following instead:
Instead of stroking the path after the first Bezier curve has been added adjust the control points with a slight angle offset. Then continue the path by adding another Bezier curve going from right to left, and finally fill it (fill() will close the path implicit):
// first path from left to right
ctx.beginPath();
ctx.moveTo(notes[0], notes[1]); // start point
ctx.bezierCurveTo(cp1x, cp1y, cp2x, cp2y, notes[len-2], notes[len-1]);
// taper going from right to left
var taper = 0.15; // angle offset
cp1x = notes[0] + dist1*t*Math.cos(a1-taper);
cp1y = notes[1] + dist1*t*Math.sin(a1-taper);
cp2x = notes[len-2] + dist2*t*Math.cos(a2+taper);
cp2y = notes[len-1] + dist2*t*Math.sin(a2+taper);
// note the order of the control points
ctx.bezierCurveTo(cp2x, cp2y, cp1x, cp1y, notes[0], notes[1]);
ctx.fill(); // close and fill
Final result (with pseudo notes - tension = 0.7, padding = 10)
FIDDLE
Suggested improvements:
If both groups' distances are large, or angles are steep, they could probably be used as a sum to reduce tension (distance) or increase it (angle).
A dominance/area factor could affect the distances. Dominance indicating where the most tallest parts are shifted at (does it lay more in the left or right side, and affects tension for each side accordingly). This could possibly/potentially be enough on its own but needs to be tested.
Taper angle offset should also have a relationship with the sum of distance. In some cases the lines crosses and does not look so good. Tapering could be replaced with a manual approach parsing Bezier points (manual implementation) and add a distance between the original points and the points for the returning path depending on array position.
Hope this helps!
Cardinal spline and filtering approach
If you're open to use a non-Bezier approach then the following can give an approximate curve above the note stems.
This solutions consists of 4 steps:
Collect top of notes/stems
Filter away "dips" in the path
Filter away points on same slope
Generate a cardinal spline curve
This is a prototype solution so I have not tested it against every possible combination there is. But it should give you a good starting point and basis to continue on.
The first step is easy, collect points representing the top of the note stem - for the demo I use the following point collection which slightly represents the image you have in the post. They are arranged in x, y order:
var notes = [60,40, 100,35, 140,30, 180,25, 220,45, 260,25, 300,25, 340,45];
which would be represented like this:
Then I created a simple multi-pass algorithm that filters away dips and points on the same slope. The steps in the algorithm are as follows:
While there is a anotherPass (true) it will continue, or until max number of passes set initially
The point is copied to another array as long as the skip flag isn't set
Then it will compare current point with next to see if it has a down-slope
If it does, it will compare the next point with the following and see if it has an up-slope
If it does it is considered a dip and the skip flag is set so next point (the current middle point) won't be copied
The next filter will compare slope between current and next point, and next point and the following.
If they are the same skip flag is set.
If it had to set a skip flag it will also set anotherPass flag.
If no points where filtered (or max passes is reached) the loop will end
The core function is as follows:
while(anotherPass && max) {
skip = anotherPass = false;
for(i = 0; i < notes.length - 2; i += 2) {
if (!skip) curve.push(notes[i], notes[i+1]);
skip = false;
// if this to next points goes downward
// AND the next and the following up we have a dip
if (notes[i+3] >= notes[i+1] && notes[i+5] <= notes[i+3]) {
skip = anotherPass = true;
}
// if slope from this to next point =
// slope from next and following skip
else if (notes[i+2] - notes[i] === notes[i+4] - notes[i+2] &&
notes[i+3] - notes[i+1] === notes[i+5] - notes[i+3]) {
skip = anotherPass = true;
}
}
curve.push(notes[notes.length-2], notes[notes.length-1]);
max--;
if (anotherPass && max) {
notes = curve;
curve = [];
}
}
The result of the first pass would be after offsetting all the points on the y-axis - notice that the dipping note is ignored:
After running through all necessary passes the final point array would be represented as this:
The only step left is to smoothen the curve. For this I have used my own implementation of a cardinal spline (licensed under MIT and can be found here) which takes an array with x,y points and smooths it adding interpolated points based on a tension value.
It won't generate a perfect curve but the result from this would be:
FIDDLE
There are ways to improve the visual result which I haven't addressed, but I will leave it to you to do that if you feel it's needed. Among those could be:
Find center of points and increase the offset depending on angle so it arcs more at top
The end points of the smoothed curve sometimes curls slightly - this can be fixed by adding an initial point right below the first point as well at the end. This will force the curve to have better looking start/end.
You could draw double curve to make a taper effect (thin beginning/end, thicker in the middle) by using the first point in this list on another array but with a very small offset at top of the arc, and then render it on top.
The algorithm was created ad-hook for this answer so it's obviously not properly tested. There could be special cases and combination throwing it off but I think it's a good start.
Known weaknesses:
It assumes the distance between each stem is the same for the slope detection. This needs to be replaced with a factor based comparison in case the distance varies within a group.
It compares the slope with exact values which may fail if floating point values are used. Compare with an epsilon/tolerance
I have a small little game I'm making in javascript and Raphael.js(which i'm fairly new to) and I'm making a turret essentially, just a circle that has a rectangle swivel around it. And that works fine and dandy!
Code for transform is :
this.self = this.self.animate({ transform : this.transform }, 250);
However, I need to find the coords of the rectangle after I animate it, but getBBox() keeps getting the same coords. Does anyone have any suggestions? A visual picture of the transform would be:
So I need the turret coords after the transformation. I need to find the front of the turret so I know where the bullet needs to come out of! Any advice will be appreciated!
By using the rotation number, will help you to find the coordinates. Lets say the rotation angel is q = 45 degrees.
This means that y changes by asin(q) and x changes by a - acos(q).
EDIT
Pay attention to all cases. In this particular case, both coordinates got decreased, but if you turn to southeast, then y increases and x decreases. Or if northwest: y and x decrease.
Transform is just a visual effect, it's not affects on coordinates.
You know width of turret and you know rotation angle.
Use sin & cos to calculate new coords.
X = Math.cos((i * Math.PI) / 180) * R + x;
Y = Math.sin((i * Math.PI) / 180) * R + y;
i - angle
R - width of turret
x and y - turret offset
So I'm working on a particle emitter with javascript and canvas.
And I want to be able to set what direction the particles are emitting based on an angle.
This can be done with this function:
y = Math.tan(45 * Math.PI/180);
Which returns 1 if the angle is 45. etc.
But I don't exacly know how I should implement this since pixels are calculated a little different. Think -1 as removing one pixel each step and 1 as adding one pixel.
If the angle is 45, Y is 1 and X is 1 which is correct.
But to get a pixel traveling at 315 degrees Y is -1 and X should be 1.
And at 225 degrees Y should be -1 (but is 1) and X should be -1.
How should the function look like if it should work like this?
Here is an image of how im thinking:
(The emitter is in the origin.)
Actually it's simple,
angle = (angle * Math.PI/180) % 360;
tangent = Math.tan(angle);
Since you do not know where is x;
section_x_positive = (angle<90||angle>270?1:-1);
section_y_positive = (angle>0&&angle<180?1:-1);
x = abs(tangent) * section_x_positive;
y = abs(tangent) * section_y_positive;
It sounds to me like your problem is that you're thinking about direction, which is a vector quantity, as if it were a scalar.
You need to remember that a 2D vector is represented as two components:
You can work in terms of unit vectors, so the magnitude r = 1.
So if you have a direction angle, which should be measured in radians, increasing in the counterclockwise direction, and starting at the x = 0 horizontal axis, you'll end up with two components of the unit vector that points in the direction you want.