the drawing algorithm which I currently use:
a_max = Pi*2 (float)(num_segments - 1.0f)/(float)num_segments;
for (unsigned int i = 0; i<=num_segments;i++)
{
const float a = (float)i / (float)num_segments * a_max;
SetPixel(centre.x + cos(a) *radius, centre.y +sin(a) *radius);
}
Works fine, but it starts drawing at (centre.x+radius, centre.y). I would like to have it to start at the top , because I want to draw a compass and zero degree is at the top, not on the right, so that I don't have to make a hacky solution.
Try rotating 90 degrees to the left before you start drawing, this should solve it for you.
A compass not only starts at "north" instead of "east" but also goes clockwise instead of counter-clockwise.
For this case, just swap sin(a) and cos(a):
x = centre.x + sin(a) * radius
y = centre.y + cos(a) * radius
I'm not sure what to search for or how to ask the question as I can't draw. Please bear with me.
If I have a rectangle with circular end caps. I want to remove some of the edges of the rectangle so there is a smooth path all round. Kinda like of you were to stretch the ends, the middle gets thinner.
I was trying to work out the chord of a larger, outer circle until I got stuck trying to work out where the circles should touch.
I can see some relationships for trigonometry, but my brain just won't go the extra mile.
Can anyone please help point me in the right direction.
Thanks.
Here is the answer:
// Small value for CSG
Delta = 0.01;
2Delta = 2 * Delta;
$fa=1; $fs=$fa;
module roudedArm(xl=50, yt=10, zh=5, in=2, bh=0.8) {
EndRadius = yt/2; // size of outer ends
EndSpacing = xl-yt; // distance between end point radii
ArmConcavity = in; // how much in does it go in on each side
ArmThickness = zh; // height in z
// Negative curve to narrow the Arm (calculated by pythagoras)
ArmCurveRadius = (pow((EndSpacing / 2), 2) - 2 * EndRadius * ArmConcavity + pow(ArmConcavity, 2)) / (2 * ArmConcavity);
// The orthogonal distance between the middle of the Arm the point it touches the round pillar sections
ArmSectionLength = (EndSpacing / 2) * ArmCurveRadius / (ArmCurveRadius + EndRadius);
// end points
lbxcylinder(r=EndRadius, h=ArmThickness);
translate([EndSpacing, 0, 0]) lbxcylinder(r=EndRadius, h=ArmThickness);
// inner curve
difference()
{
translate([EndSpacing / 2 - ArmSectionLength, -EndRadius -ArmThickness, 0])
translate([ArmSectionLength, (EndRadius + ArmThickness),0])
lbxcube([ArmSectionLength * 2, 2 * (EndRadius + ArmThickness), ArmThickness], bh=bh);
// Cut out Arm curve
translate([EndSpacing / 2, ArmCurveRadius + EndRadius - ArmConcavity, -Delta])
lbxcylinder(r = ArmCurveRadius, h = ArmThickness + 2Delta, bh=-bh);
translate([EndSpacing / 2, -(ArmCurveRadius + EndRadius - ArmConcavity), -Delta])
lbxcylinder(r = ArmCurveRadius, h = ArmThickness + 2Delta, bh=-bh);
}
}
module lbxcube(size, bh=0.8) {
// don't support bevelling in demo
translate([-size[0]/2, -size[1]/2, 0]) cube(size);
}
module lbxcylinder(r, h, bh=0.8) {
// don't support bevelling in demo
cylinder(r=r, h=h);
}
roudedArm(xl=50, yt=10, zh=5, in=2, bh=0.8);
Thanks to Rupert and his Curvy Door Handle on Thingiverse.
Given a line made up of several points, how do I make the line smoother/ curvier/ softer through adding intermediate points -- while keeping the original points completely intact and unmoved?
To illustrate, I want to go from the above to the below in this illustration:
Note how in the above picture, if we start at the bottom there will be a sharper right turn. In the bottom image however, this sharp right turn is made a bit "softer" by adding an intermediate point which is positioned in the middle of the two points, and using averages of the angles of the other lines. (Differently put, imagine the lines a race car would drive, as it couldn't abruptly change direction.) Note how, however, none of the original points was "touched", I just added more points.
Thanks!! For what it's worth, I'm implementing this using JavaScript and Canvas.
with each edge (e1 & e2) adjacent to each 'middle' edge (me) do
let X = 0.5 x length(me)
find 2 cubic bezier control points by extending the adjacent edges by X (see algorithm below)
get midpoint of cubic bezier (by applying formula below)
insert new 'midpoint' between me's two coordinates.
FloatPoint ExtendLine(const FloatPoint A, const FloatPoint B, single distance)
{
FloatPoint newB;
float lenAB = sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
newB.X = B.x - (B.x - A.x) / lenAB * distance;
newB.Y = B.Y - (B.Y - A.Y) / lenAB * distance;
return newB;
}
Edit: Formula for Bezier Curve midpoint: p(0.5) = 0.125(p0) + 0.375(p1) + 0.375(p2) + 0.125(p3)
The following code found elsewhere here does the job for me, in the specific context of JavaScript-Canvas which I'm using -- but please see Angus' answer for a more general approach:
var max = points.length;
context.beginPath();
var i = 0;
context.moveTo(points[i].x, points[i].y);
for (i = 1; i < max - 2; i++) {
var xc = (points[i].x + points[i + 1].x) * .5;
var yc = (points[i].y + points[i + 1].y) * .5;
context.quadraticCurveTo(points[i].x, points[i].y, xc, yc);
}
context.quadraticCurveTo(points[max - 2].x, points[max - 2].y, points[max - 1].x,points[max - 1].y);
context.closePath();
context.stroke();
Project a cubic bezier p1,p2,p3,p4 onto the line p1,p4. When p2 or p3 does not project onto the line segment between p1 and p4, the curve will bulge out from the anchor points. Is there a way to calculate the T value where the tangent of the curve is perpendicular to the anchor line?
This could also be stated as finding the T values where the projected curve is farthest from the center of the line segment p1,p4. When p2 and p3 project onto the line segment, then the solutions are 0 and 1 respectively. Is there an equation for solving the more interesting case?
The T value seems to depend only on the distance of the mapped control points from the anchor line segment.
I can determine the value by refining guesses, but I am hoping there is a better way.
Edit:
Starting with p1,..,p4 in 2d with values x1,y1, ..., x4,y4 I use the following code based on the answer from Philippe:
dx = x4 - x1;
dy = y4 - y1;
d2 = dx*dx + dy*dy;
p1 = ( (x2-x1)*dx + (y2-y1)*dy ) / d2;
p2 = ( (x3-x1)*dx + (y3-y1)*dy ) / d2;
tr = sqrt( p1*p1 - p1*p2 - p1 + p2*p2 );
t1 = ( 2*p1 - p2 - tr ) / ( 3*p1 - 3*p2 + 1 );
t2 = ( 2*p1 - p2 + tr ) / ( 3*p1 - 3*p2 + 1 );
In the sample I looked at, t2 had to be subtracted from 1.0 before it was correct.
Let's assume you got a 1D cubic Bézier curve with P0 = 0 and P3 = 1 then the curve is:
P(t) = b0,3(t)*0 + b1,3(t)*P1 + b2,3(t)*P2 + b3,3(t)*1
Where bi,3(t) are the Bernstein polynomials of degree 3. Then we're looking for the value of t where this P(t) is minimal and maximal, so we derive:
P'(t) = b1,3'(t)*P1 + b2,3'(t)*P2 + b3,3'(t)
= (3 - 12t + 9t^2)*P1 + (6t - 9t^2)*P2 + 3t^2
= 0
This has a closed-form but nontrivial solution. According to WolframAlpha, when 3P1 - 3P2 +1 != 0 it's:
t = [2*P1 - P2 +/- sqrt(P1^2-P1*P2-P1+P2^2)] / (3*P1 - 3*P2 + 1)
Otherwise it's:
t = 3P1 / (6P1 - 2)
For a general n-dimensional cubic Bézier P0*, P1*, P2*, P3* compute:
P1 = proj(P1*, P03*) / |P3* - P0*|
P2 = proj(P2*, P03*) / |P3* - P0*|
Where proj(P, P03*) is the signed distance from P0* to the point P projected on the line passing through P0* and P3*.
(I haven't checked this, so please confirm there is nothing wrong in my reasoning.)
Having searched the web, I see various people in various forums alluding to approximating a cubic curve with a quadratic one. But I can't find the formula.
What I want is this:
input: startX, startY, control1X, control1Y, control2X, control2Y, endX, endY
output: startX, startY, controlX, controlY, endX, endY
Actually, since the starting and ending points will be the same, all I really need is...
input: startX, startY, control1X, control1Y, control2X, control2Y, endX, endY
output: controlX, controlY
As mentioned, going from 4 control points to 3 is normally going to be an approximation. There's only one case where it will be exact - when the cubic bezier curve is actually a degree-elevated quadratic bezier curve.
You can use the degree elevation equations to come up with an approximation. It's simple, and the results are usually pretty good.
Let's call the control points of the cubic Q0..Q3 and the control points of the quadratic P0..P2. Then for degree elevation, the equations are:
Q0 = P0
Q1 = 1/3 P0 + 2/3 P1
Q2 = 2/3 P1 + 1/3 P2
Q3 = P2
In your case you have Q0..Q3 and you're solving for P0..P2. There are two ways to compute P1 from the equations above:
P1 = 3/2 Q1 - 1/2 Q0
P1 = 3/2 Q2 - 1/2 Q3
If this is a degree-elevated cubic, then both equations will give the same answer for P1. Since it's likely not, your best bet is to average them. So,
P1 = -1/4 Q0 + 3/4 Q1 + 3/4 Q2 - 1/4 Q3
To translate to your terms:
controlX = -0.25*startX + .75*control1X + .75*control2X -0.25*endX
Y is computed similarly - the dimensions are independent, so this works for 3d (or n-d).
This will be an approximation. If you need a better approximation, one way to get it is by subdividing the initial cubic using the deCastlejau algorithm, and then degree-reduce each segment. If you need better continuity, there are other approximation methods that are less quick and dirty.
The cubic can have loops and cusps, which quadratic cannot have. This means that there are not simple solutions nearly never. If cubic is already a quadratic, then the simple solution exists. Normally you have to divide cubic to parts that are quadratics. And you have to decide what are the critical points for subdividing.
http://fontforge.org/bezier.html#ps2ttf says:
"Other sources I have read on the net suggest checking the cubic spline for points of inflection (which quadratic splines cannot have) and forcing breaks there. To my eye this actually makes the result worse, it uses more points and the approximation does not look as close as it does when ignoring the points of inflection. So I ignore them."
This is true, the inflection points (second derivatives of cubic) are not enough. But if you take into account also local extremes (min, max) which are the first derivatives of cubic function, and force breaks on those all, then the sub curves are all quadratic and can be presented by quadratics.
I tested the below functions, they work as expected (find all critical points of cubic and divides the cubic to down-elevated cubics). When those sub curves are drawn, the curve is exactly the same as original cubic, but for some reason, when sub curves are drawn as quadratics, the result is nearly right, but not exactly.
So this answer is not for strict help for the problem, but those functions provide a starting point for cubic to quadratic conversion.
To find both local extremes and inflection points, the following get_t_values_of_critical_points() should provide them. The
function compare_num(a,b) {
if (a < b) return -1;
if (a > b) return 1;
return 0;
}
function find_inflection_points(p1x,p1y,p2x,p2y,p3x,p3y,p4x,p4y)
{
var ax = -p1x + 3*p2x - 3*p3x + p4x;
var bx = 3*p1x - 6*p2x + 3*p3x;
var cx = -3*p1x + 3*p2x;
var ay = -p1y + 3*p2y - 3*p3y + p4y;
var by = 3*p1y - 6*p2y + 3*p3y;
var cy = -3*p1y + 3*p2y;
var a = 3*(ay*bx-ax*by);
var b = 3*(ay*cx-ax*cy);
var c = by*cx-bx*cy;
var r2 = b*b - 4*a*c;
var firstIfp = 0;
var secondIfp = 0;
if (r2>=0 && a!==0)
{
var r = Math.sqrt(r2);
firstIfp = (-b + r) / (2*a);
secondIfp = (-b - r) / (2*a);
if ((firstIfp>0 && firstIfp<1) && (secondIfp>0 && secondIfp<1))
{
if (firstIfp>secondIfp)
{
var tmp = firstIfp;
firstIfp = secondIfp;
secondIfp = tmp;
}
if (secondIfp-firstIfp >0.00001)
return [firstIfp, secondIfp];
else return [firstIfp];
}
else if (firstIfp>0 && firstIfp<1)
return [firstIfp];
else if (secondIfp>0 && secondIfp<1)
{
firstIfp = secondIfp;
return [firstIfp];
}
return [];
}
else return [];
}
function get_t_values_of_critical_points(p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y) {
var a = (c2x - 2 * c1x + p1x) - (p2x - 2 * c2x + c1x),
b = 2 * (c1x - p1x) - 2 * (c2x - c1x),
c = p1x - c1x,
t1 = (-b + Math.sqrt(b * b - 4 * a * c)) / 2 / a,
t2 = (-b - Math.sqrt(b * b - 4 * a * c)) / 2 / a,
tvalues=[];
Math.abs(t1) > "1e12" && (t1 = 0.5);
Math.abs(t2) > "1e12" && (t2 = 0.5);
if (t1 >= 0 && t1 <= 1 && tvalues.indexOf(t1)==-1) tvalues.push(t1)
if (t2 >= 0 && t2 <= 1 && tvalues.indexOf(t2)==-1) tvalues.push(t2);
a = (c2y - 2 * c1y + p1y) - (p2y - 2 * c2y + c1y);
b = 2 * (c1y - p1y) - 2 * (c2y - c1y);
c = p1y - c1y;
t1 = (-b + Math.sqrt(b * b - 4 * a * c)) / 2 / a;
t2 = (-b - Math.sqrt(b * b - 4 * a * c)) / 2 / a;
Math.abs(t1) > "1e12" && (t1 = 0.5);
Math.abs(t2) > "1e12" && (t2 = 0.5);
if (t1 >= 0 && t1 <= 1 && tvalues.indexOf(t1)==-1) tvalues.push(t1);
if (t2 >= 0 && t2 <= 1 && tvalues.indexOf(t2)==-1) tvalues.push(t2);
var inflectionpoints = find_inflection_points(p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y);
if (inflectionpoints[0]) tvalues.push(inflectionpoints[0]);
if (inflectionpoints[1]) tvalues.push(inflectionpoints[1]);
tvalues.sort(compare_num);
return tvalues;
};
And when you have those critical t values (which are from range 0-1), you can divide the cubic to parts:
function CPoint()
{
var arg = arguments;
if (arg.length==1)
{
this.X = arg[0].X;
this.Y = arg[0].Y;
}
else if (arg.length==2)
{
this.X = arg[0];
this.Y = arg[1];
}
}
function subdivide_cubic_to_cubics()
{
var arg = arguments;
if (arg.length!=9) return [];
var m_p1 = {X:arg[0], Y:arg[1]};
var m_p2 = {X:arg[2], Y:arg[3]};
var m_p3 = {X:arg[4], Y:arg[5]};
var m_p4 = {X:arg[6], Y:arg[7]};
var t = arg[8];
var p1p = new CPoint(m_p1.X + (m_p2.X - m_p1.X) * t,
m_p1.Y + (m_p2.Y - m_p1.Y) * t);
var p2p = new CPoint(m_p2.X + (m_p3.X - m_p2.X) * t,
m_p2.Y + (m_p3.Y - m_p2.Y) * t);
var p3p = new CPoint(m_p3.X + (m_p4.X - m_p3.X) * t,
m_p3.Y + (m_p4.Y - m_p3.Y) * t);
var p1d = new CPoint(p1p.X + (p2p.X - p1p.X) * t,
p1p.Y + (p2p.Y - p1p.Y) * t);
var p2d = new CPoint(p2p.X + (p3p.X - p2p.X) * t,
p2p.Y + (p3p.Y - p2p.Y) * t);
var p1t = new CPoint(p1d.X + (p2d.X - p1d.X) * t,
p1d.Y + (p2d.Y - p1d.Y) * t);
return [[m_p1.X, m_p1.Y, p1p.X, p1p.Y, p1d.X, p1d.Y, p1t.X, p1t.Y],
[p1t.X, p1t.Y, p2d.X, p2d.Y, p3p.X, p3p.Y, m_p4.X, m_p4.Y]];
}
subdivide_cubic_to_cubics() in above code divides an original cubic curve to two parts by the value t. Because get_t_values_of_critical_points() returns t values as an array sorted by t value, you can easily traverse all t values and get the corresponding sub curve. When you have those divided curves, you have to divide the 2nd sub curve by the next t value.
When all splitting is proceeded, you have the control points of all sub curves. Now there are left only the cubic control point conversion to quadratic. Because all sub curves are now down-elevated cubics, the corresponding quadratic control points are easy to calculate. The first and last of quadratic control points are the same as cubic's (sub curve) first and last control point and the middle one is found in the point, where lines P1-P2 and P4-P3 crosses.
Conventions/terminology
Cubic defined by: P1/2 - anchor points, C1/C2 control points
|x| is the euclidean norm of x
mid-point approx of cubic: a quad that shares the same anchors with the cubic and has the control point at C = (3·C2 - P2 + 3·C1 - P1)/4
Algorithm
pick an absolute precision (prec)
Compute the Tdiv as the root of (cubic) equation sqrt(3)/18 · |P2 - 3·C2 + 3·C1 - P1|/2 · Tdiv ^ 3 = prec
if Tdiv < 0.5 divide the cubic at Tdiv. First segment [0..Tdiv] can be approximated with by a quadratic, with a defect less than prec, by the mid-point approximation. Repeat from step 2 with the second resulted segment (corresponding to 1-Tdiv)
0.5<=Tdiv<1 - simply divide the cubic in two. The two halves can be approximated by the mid-point approximation
Tdiv>=1 - the entire cubic can be approximated by the mid-point approximation
The "magic formula" at step 2 is demonstrated (with interactive examples) on this page.
Another derivation of tfinniga's answer:
First see Wikipedia Bezier curve
for the formulas for quadratic and cubic Bezier curves (also nice animations):
Q(t) = (1-t)^2 P0 + 2 (1-t) t Q + t^2 P3
P(t) + (1-t)^3 P0 + 3 (1-t)^2 t P1 + 3 (1-t) t^2 P2 + t^3 P3
Require these to match at the middle, t = 1/2:
(P0 + 2 Q + P3) / 4 = (P0 + 3 P1 + 3 P2 + P3) / 8
=> Q = P1 + P2 - (P0 + P1 + P2 + P3) / 4
(Q written like this has a geometric interpretation:
Pmid = middle of P0 P1 P2 P3
P12mid = midway between P1 and P2
draw a line from Pmid to P12mid, and that far again: you're at Q.
Hope this makes sense -- draw a couple of examples.)
In general, you'll have to use multiple quadratic curves - many cases of cubic curves can't be even vaguely approximated with a single quadratic curve.
There is a good article discussing the problem, and a number of ways to solve it, at http://www.timotheegroleau.com/Flash/articles/cubic_bezier_in_flash.htm (including interactive demonstrations).
I should note that Adrian's solution is great for single cubics, but when the cubics are segments of a smooth cubic spline, then using his midpoint approximation method causes slope continuity at the nodes of the segments to be lost. So the method described at http://fontforge.org/bezier.html#ps2ttf is much better if you are working with font glyphs or for any other reason you want to retain the smoothness of the curve (which is most probably the case).
Even though this is an old question, many people like me will see it in search results, so I'm posting this here.
I would probably draw a series of curves instead of trying to draw one curve using a different alg. Sort of like drawing two half circles to make up a whole circle.
Try looking for opensource Postcript font to Truetype font converters. I'm sure they have it. Postscript uses cubic bezier curves, whereas Truetype uses quadratic bezier curves. Good luck.