I am using Javascripts built in canvas feature to draw a graph showing home loan payments, loan balance, and equity based on user input. I am not able to use any other form of graphing package, as the code is part of an assessment.
My graph is drawn by converting data to X and Y coordinates. When a loan price is input, some home loan payment equations calculate the total amount payed, which is divided by the canvas width to get a spacing variable. This spacing variable is used to convert dollar amounts into pixels on the canvas. A similar setup is used to get the years and months spacing pixels.
The problem I am having is that the Y axis on Javascript's canvas is inverted, with 0 being the top of the canvas and 280, my canvas height, being at the bottom. So far, I have been able to work around this, simply by swapping "+" and "-" operators, however, I am currently creating the code that draws the Loan Balance line on the graph, and the inversion is causing issues that I can't seem to solve. It may be something simple that I'm just not seeing, or it may be a more complex problem that needs to be solved, but either way, I can't figure it out.
X = 0; // same as before, iterators both set back to 0 for the new line.
iterator = 0;
c.beginPath // this next line is for loan balance, it starts at 300000 and goes down with each payment made, then back up with each bit of interest accrued.
// due to the fact that the y axis begins at the top, this means that the pixels for payments is added to the pixel count, and the interest accrued is taken away.
c.moveTo(0, loanLocation) // set starting point to x=0 y= loanLocation
while (X <= 510)// loan balance loop
{
X = X + 0.001; // iterates X by .001 each time, allowing an accurate subpixel resolution loop, see above for why this is needed.
iterator = iterator + 0.001;
if (iterator >= monthSpacing)
{
loanBalance = loanBalance - monthlyPayment + (monthlyInterest * loanBalance);
//alert(loanBalance);
//interestY =
//alert(interestY);
//alert(X + " " + monthSpacing);
loanY = loanY + paymentY - (loanY * monthlyInterest);
//alert(loanY);
//loanY = loanBalance * paySpacing;
c.lineTo(X, loanY);
iterator = 0;
}
}
c.strokeStyle = "black"
c.stroke(); // there is no fill for this line, so it is just left as a stroke.
This is the set of code which draws the line, above it are a few variables which are being used here:
var X = 0;
var iterator = 0;
var monthSpacing = yearSpacing / 12;
//alert(yearSpacing);
//alert(monthSpacing);
var monthlyInterest = interest/1200; // this gives the montly interest rate, the monthly interest pixel amount is below
//alert(monthlyInterest);//debugging, comment out.
var paymentY = monthlyPayment * paySpacing;
var interestY = monthlyInterest * paySpacing; // this is inaccurate, the interestY needs to be gotten by multiplying the remaining loan balance by the
//monthly interest each month.
//var interestY; // will be used further down, must be calculated monthly so cannot be set outside of the line drawing loops.
var totalY = 280;
var equityY = 280;
var loanBalance = loan;
var loanY = loanLocation;
When run I get a strange inversion of the desired outcome, I want the loan balance line to curve down towards zero, but instead, the curve is happening in the opposite direction, I have tried two different ways to get the coordinates, the loanBalance way, which involved working with dollar values and converting that to pixels, and the loanY way, which involved working with pixel values directly.
loanBalance provided a line which was the exact inverse of the desired line, it began at the loan value, and curved upwards in the exact opposite direction to what I want, I am confident that the math I'm using for the loanBalance method is accurate, I simply cannot think of a way to convert that dollar value into pixels due to the inverted nature of the Y axis.
loanY provides a line which is headed "down", but is curving downwards at an increasingly shortened rate, this leads me to believe that while the subtraction (addition due to the inversion) of monthly repayments is accurately being calculated, the addition (subtraction) of monthly interest is being calculated incorrectly. Multiplication cannot be simply replaced with division like addition and subtraction can, so converting this value to pixels is proving difficult. The line drawn by the loanY way is definitely being affected by the inversion, but is not a perfect inverse of the desired line, the math being used for that way is clearly very wrong.
Ideally, I'd like to find a way to use the loanY way, it is consistent with the rest of the program, and can be used when not working with such obvious values as dollars. If I have to though, I will use the loanBalance way.
If you aren't entirely certain what I'm asking, or what the code being used is, I can post the program in it's entirety if that would help. I've not done that yet as I don't want to clutter the question more than I already have.
You can change to a Cartesian coordinate system like this:
// get a reference to your canvas element (eg it might have id='myCanvas')
var canvas=document.getElementById('myCanvas');
// get the context for the canvas
var context=canvas.getContext('2d');
// vertically flip the canvas so its Y origin is at the bottom
context.setTransform(1,0,0,-1,0,canvas.height);
This makes y==0 at the bottom of the canvas and increases upward.
If you're using other transformations, then put this transformation before the others.
Related
I am trying to calculate with higher precision numbers in JavaScript to be able to zoom in more on the Mandlebrot set.
(after a certain amount of zooming the results get "pixelated", because of the low precision)
I have looked at this question, so I tried using a library such as BigNumber but it was unusably slow.
I have been trying to figure this out for a while and I think the only way is to use a slow library.
Is there a faster library?
Is there any other way to calculate with higher precision numbers?
Is there any other way to be able to zoom in more on the Mandlebrot set?
Probably unneceseary to add this code, but this is the function I use to check if a point is in the Mandlebrot set.
function mandelbrot(x, y, it) {
var z = [0, 0]
var c1 = [x, y]
for (var i = 0; i < it; i++) {
z = [z[0]*z[0] - z[1]*z[1] + c1[0], 2*z[0]*z[1] + c1[1]]
if (Math.abs(z[0]) > 2, Math.abs(z[1]) > 2) {
break
}
}
return i
}
The key is not so much the raw numeric precision of JavaScript numbers (though that of course has its effects), but the way the basic Mandelbrot "escape" test works, specifically the threshold iteration counts. To compute whether a point in the complex plane is in or out of the set, you iterate on the formula (which I don't exactly remember and don't feel like looking up) for the point over and over again until the point obviously diverges (the formula "escapes" from the origin of the complex plane by a lot) or doesn't before the iteration threshold is reached.
The iteration threshold when rendering a view of the set that covers most of it around the origin of the complex plane (about 2 units in all directions from the origin) can be as low as 500 to get a pretty good rendering of the whole set at a reasonable magnification on a modern computer. As you zoom in, however, the iteration threshold needs to increase in inverse proportion to the size of the "window" onto the complex plane. If it doesn't, then the "escape" test doesn't work with sufficient accuracy to delineate fine details at higher magnifications.
The formula I used in my JavaScript implementation is
maxIterations = 400 * Math.log(1/dz0)
where dz0 is (arbitrarily) the width of the window onto the plane. As one zooms into a view of the set (well, the "edge" of the set, where things are interesting), dz0 gets pretty small so the iteration threshold gets up into the thousands.
The iteration count, of course, for points that do "escape" (that is, points that are not part of the Mandelbrot set) can be used as a sort of "distance" measurement. A point that escapes within a few iterations is clearly not "close to" the set, while a point that escapes only after 2000 iterations is much closer. That distance quality can be used in various ways in visualizations, either to provide a color value (common) or possibly a z-axis value if the set is being rendered as a 3D view (with the set as a sort of "mesa" in three dimensions and the borders being a vertical "cliff" off the sides).
Given an SVG such as this fish bowl, I'm trying to calculate the volume of the area defined in pink as a percentage of the area between the "fill level" and "empty level".
I can't do a a simple percentage from top to bottom, as the fish bowl is irregularly shaped, and this will throw off the calculation by at least a few percentage points. I need to do this for many fish bowls of different shapes, so an algorithm is needed to determine the volume of each bowl.
Is there any way I can do this with javascript on an SVG element, and if so, is there any way I can go about figuring this out within element areas as a percentage?
Update: Uploaded sample SVG to jsfiddle
First you need to parse the SVG path to lines. Since they all don't cross
the Y axis, this reduces to finding the area under the curve caused by the fish bowl,
also known as the integral.
Let {x_0, x_1, ..., x_n} be the absolute value of the X coordinates of the line segments.
The function representing the graph of the fishbowl is the piecewise function:
f(x) =
{ (x - x_0)/(x_1 - x_0) if x_0 <= x < x_1
{ (x - x_1)/(x_2 - x_1) if x_1 <= x < x_2
{ ...
{ (x - x_(n-1))/(x_n - x_(n-1)) if x_(n-1) <= x < x_(n)
Then the volume of the fishbowl equals the integral of πf(x)2 (the solid of revolution formed by that function).
Let e be the empty level, v the fill level, and w the water level.
Then the ratio of the filled portion of the fishbowl is:
(∫ew πf(x)2 dx) / (∫ev πf(x)2 dx)
If instead your fishbowl is generated by the graph of a function, use that function as f(x) and then calculate the integral given above.
An integral can be approximated using numerical integration techniques such as Simpson's rule or
a Newton-Cotes method.
A I needed a solution to this that isn't prohibitive in terms of computational cost and I wasn't in the mood to write optimized code, I ended up rendering it against transparent background, converted to raster and then counted pixels. I'm sure someone with experience in graphics and geometry can come up with cleaner solutions, but I my optimized code in a high level language is unlikely to run faster than that of someone that's dedicated their lives to this and write in assembly.
Depending on the complexity of the geometry of your fish bowl you might need to up the rendering resolution of course.
[2021 addition]
This SO answer calculates the area of one <path> using the brute-force method described above: Scaling the filling portion of an SVG path
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
Im working on a little project, I have some particles i want to move towards target positions without exceeding a max velocity, first i tried capping the X and Y velocities seperately which caused the hypotenuse of the two to be able to go over the max speed, i then remembered my maths classes and attempted this:
var totalVel = Math.sqrt(Math.pow(curVelocity[0],2) + Math.pow(curVelocity[1],2));
if(totalVel > maxSpeed){
//sin(θ) = Opposite / Hypotenuse
var angle = Math.asin(curVelocity[1]/totalVel);
var newHyp = maxSpeed;
var newOp = Math.sin(angle)*newHyp;
var newAdj = Math.sqrt(Math.pow(newHyp,2) - Math.pow(newOp,2));
curVelocity[1] = newOp;
curVelocity[0] = newAdj;
}
(curVelocity is an array where index 0 is X and index 1 is Y)
This works well hald the time, the other time it curves away from the target its trying to reach.. matches it on the Y plane but heads in the whole wrong direction in the X plane. im guessing its something to do with using math.sin when perhaps it no longer applies in the direction its traveling but i wouldnt know where ot begin differentiating what to use, or if that idea is even correct.
A live example of what im talking about can be found at this location here, refreshing the page will change the starting location and target location, the black circle is the particle the green circle is the target location
My aim is to find a smooth best fit line between this two circuit curvy shapes.
Is there any algorithm betten than mine that can find a set of points (or a curve) between two lines like this example?
The algorithm I have so far takes the inner part and for each point finds the closest, however this doesnt work because (look at the first corner).
(Red is the inner part, green is the outer part, blue is the optimised dots I have found)
Here is my jsfiddle:
http://jsfiddle.net/STLuG/
This is the algorithm:
for (i = 0; i < coords[0].length; i++) {
var currentI = coords[0][i];
j = 0;
var currentJ = coords[0][j];
currentDist = dist(currentI,currentJ);
for (j=1; j < coords[1].length; j++) {
possibleJ = coords[1][j];
possibleDist = dist(currentI, possibleJ);
if (possibleDist < currentDist) {
currentJ = possibleJ;
currentDist = possibleDist;
} else {
}
}
b_context.fillRect(
(currentI.x+currentJ.x)/2+maxX,
(currentI.y+currentJ.y)/2+maxY,
1, 1);
}
Thanks
I would try least-squares-algorithm.
You have a number of points: y0 and x0 and y1 and x1 for the first and the second curve respectively.
You want to find a curve y(t) and x(t) which is smooth and in-between the two given curves.
So there is the distance between the first curve(x0(t), y0(t)) to your to be calculated curve(x(t), y(t)):
S0=SumOverAllT(x0(t)-x(t))^2 + (y0(t) - y(t))^2
The same for the second curve:
S1=SumOverAllT(x1(t)-x(t))^2 + (y1(t) - y(t))^2
The sum of both sums:
S=S0+S1
You will have a set of parameters which you want to determine.
E.g. if you use polynomials:
x(t)=ax+bx*t+cx*t^2+dx*t^3....
y(t)=ay+by*t+cy*t^2+dy*t^3....
You will then calculate
dS/dax, dS/dbx, dS/dcx, ....
for all parameters to be calculated
and set these derivatives to zero:
dS/dax==0
dS/dbx==0
....
This will give you a set of linear equations which can be attacked by the gauss algorithm or whatever method to solve a system of linear equations.
If you're using polynomials it might happen that the curve calculated oscillates strongly.
In this case I would suggest to try to minimize the integral of the square of the second derivative:
I=integral((d^2x/dt^2)^2 + (d^2y/dt^2)^2, dt)
you would calculate the differential of I vs. some additional parameters which you did not use for above system of equation -- adding a parameter rx and calculating dI/drx==0 -- thus you have one more parameter and one more equation.
Anybody with a PHD in mathematics please advice me on any stupidity I mentioned above.
Also search the internet for:
Curve fitting
Spline approximation
A better approach would be to use splines -- piecewise continuous polynomials, so that
the 0 derivative
the first derivative
the second derivative
is continuous.
Look up or buy Numerical recipes to find code which does exactly this.
For spline approximation you would have a set of polynomials:
x0(t)=a0x + b0x*(t - t0) + c0x*(t-t0)^2 + d0x*(t - t0)^3....
x1(t)=a1x + b1x*(t - t0) + c1x*(t-t0)^2 + d1x*(t - t0)^3....
Every polynomial would only be used to cover the matching t=t0..t1 between two given points.
You would then add equations to make certain that the value, first and second derivatives are identical for two neighboring polynomials.
And set the 2 derivative for the first and last polynomial to zero.
Potentially you could calculate two splines -- one for every of the two input curves you have:
x0(t)
y0(t)
x1(t)
y1(t)
And then you could derive the middle of the two splines:
x(t)=(x0(t) + (x1(t)-x0(t))/2
y(t)=(y0(t) + (y1(t)-y0(t))/2
make certain that the distance between any of the given curves and you resulting curve is never zero so that they don't cross each other
To make certain, that your calculated line does not cross one of the given lines, you could minimize (sum(sum(1/(x0-x)^2)) + sum(sum(1/(x1-x)^2)))