Moving a star marker along a hexagon trajectory? - matlab

Question

I want to move a star marker along hexagon trajectory similar to "Circle trajectory" that I have added at the end of my question. Thanks.
This is the source code to that I have written yet for creating concentric hegzagons but I don't know how to move a star marker which traverses the concentric hexagons, I had written a similar simulation code for circle trajectory but I couldn't do it for hexagon.
%clc; % Clear the command window.
%close all; % Close all figures (except those of imtool.)
%clear; % Erase all existing variables. Or clearvars if you want.
workspace; % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 20;
angles = linspace(0, 360, 7);
radii = [20, 35, 50,70];
% First create and draw the hexagons.
numberOfHexagons = 4;
% Define x and y arrays. Each row is one hexagon.
% Columns are the vertices.
x1=radii(1) * cosd(angles)+50;
y1 = radii(1) * sind(angles)+50;
x2=radii(2) * cosd(angles)+50;
y2 = radii(2) * sind(angles)+50;
x3=radii(3) * cosd(angles)+50;
y3 = radii(3) * sind(angles)+50;
x4=radii(4) * cosd(angles)+50;
y4 = radii(4) * sind(angles)+50;
plot(x1 , y1, 'b');
hold on
plot(x2, y2, 'b');
hold on
plot(x3, y3, 'b');
hold on
plot(x4, y4, 'b');
hold on
% Connecting Line:
plot([70 100], [50 50],'color','b')
axis([0 100 0 100])
hold on
Circle trajectory:
% Initialization steps.
format long g;
format compact;
fontSize = 20;
r1 = 50;
r2 = 35;
r3= 20;
xc = 50;
yc = 50;
% Since arclength = radius * (angle in radians),
% (angle in radians) = arclength / radius = 5 / radius.
deltaAngle1 = 5 / r1;
deltaAngle2 = 5 / r2;
deltaAngle3 = 5 / r3;
theta1 = 0 : deltaAngle1 : (2 * pi);
theta2 = 0 : deltaAngle2 : (2 * pi);
theta3 = 0 : deltaAngle3 : (2 * pi);
x1 = r1*cos(theta1) + xc;
y1 = r1*sin(theta1) + yc;
x2 = r2*cos(theta2) + xc;
y2 = r2*sin(theta2) + yc;
x3 = r3*cos(theta3) + xc;
y3 = r3*sin(theta3) + yc;
plot(x1,y1,'color',[1 0.5 0])
hold on
plot(x2,y2,'color',[1 0.5 0])
hold on
plot(x3,y3,'color',[1 0.5 0])
hold on
% Connecting Line:
plot([70 100], [50 50],'color',[1 0.5 0])
% Set up figure properties:
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0, 0, 1, 1]);
drawnow;
axis square;
for i = 1 : length(theta1)
plot(x1(i),y1(i),'r*')
pause(0.1)
end
for i = 1 : length(theta2)
plot(x2(i),y2(i),'r*')
pause(0.1)
end
for i = 1 : length(theta3)
plot(x3(i),y3(i),'r*')
pause(0.1)
end

Answer 1

I would generalize your problem with parametric function for the trajectory. In it use rotation kernel which you want here few examples in C++/VCL/GDI (sorry I am not Matlab friendly but the equations should be the same in Matlab too) for circle,square and hexagon rotation kernels:
void getpnt_circle(double &x,double &y,double &pi2,double r,double t) // (x,y) = circle(r,t) t=<0,1>
{
pi2=2.0*M_PI; // circumference(r=1) 6.283185307179586476925286766559
t*=pi2;
x=r*cos(t);
y=r*sin(t);
}
//---------------------------------------------------------------------------
void getpnt_square(double &x,double &y,double &pi2,double r,double t) // (x,y) = square(r,t) t=<0,1>
{
pi2=8.0; // circumference(r=1)
// kernel
const int n=4; // sides
const double x0[n]={+1.0,+1.0,-1.0,-1.0}; // side start point
const double y0[n]={-1.0,+1.0,+1.0,-1.0};
const double dx[n]={ 0.0,-2.0, 0.0,+2.0}; // side tangent
const double dy[n]={+2.0, 0.0,-2.0, 0.0};
int ix;
t-=floor(t); // t = <0, 1.0)
t*=n; // t = <0,n)
ix=floor(t); // side of square
t-=ix; // distance from side start
x=r*(x0[ix]+t*dx[ix]);
y=r*(y0[ix]+t*dy[ix]);
}
//---------------------------------------------------------------------------
void getpnt_hexagon(double &x,double &y,double &pi2,double r,double t) // (x,y) = square(r,t) t=<0,1>
{
pi2=6.0; // circumference(r=1)
// kernel
const int n=6; // sides
const double c60=cos(60.0*M_PI/180.0);
const double s60=sin(60.0*M_PI/180.0);
const double x0[n]={+1.0,+c60,-c60,-1.0,-c60,+c60}; // side start point
const double y0[n]={ 0.0,+s60,+s60, 0.0,-s60,-s60};
const double dx[n]={-c60,-1.0,-c60,+c60,+1.0,+c60}; // side tangent
const double dy[n]={+s60, 0.0,-s60,-s60, 0.0,+s60};
int ix;
t-=floor(t); // t = <0, 1.0)
t*=n; // t = <0,n)
ix=floor(t); // side of square
t-=ix; // distance from side start
x=r*(x0[ix]+t*dx[ix]);
y=r*(y0[ix]+t*dy[ix]);
}
//---------------------------------------------------------------------------
void TMain::draw()
{
if (!_redraw) return;
// clear buffer
bmp->Canvas->Brush->Color=clBlack;
bmp->Canvas->FillRect(TRect(0,0,xs,ys));
int e;
double r,t,x,y,c,dr=15.0,dl=15.0;
int xx,yy,rr=3;
bmp->Canvas->MoveTo(xs2,ys2);
bmp->Canvas->Pen->Color=clAqua;
bmp->Canvas->Brush->Color=clBlue;
for (r=dr,t=0.0;;)
{
// get point from selected kernel
// getpnt_circle (x,y,c,r,t);
// getpnt_square (x,y,c,r,t);
getpnt_hexagon(x,y,c,r,t);
// render it
xx=xs2+x;
yy=ys2+y;
bmp->Canvas->LineTo(xx,yy);
bmp->Canvas->Ellipse(xx-rr,yy-rr,xx+rr,yy+rr);
// update position
r+=dr*dr/(r*c);
t+=dl/(r*c); t-=floor(t);
if (r>=xs2) break;
if (r>=ys2) break;
}
// render backbuffer
Main->Canvas->Draw(0,0,bmp);
_redraw=false;
}
//---------------------------------------------------------------------------
You can ignore the VCL/GDI rendering stuff.
xs,ys is full and xs2,ys2 is half resolution of window to scale the plot properly...
dl is distance between markers [pixels]
dr is distance between spiral screws [pixels]
the spiral is sampled with r,t step depending on the actual circumference (that is what the pi2 or c is for). The getpnt_xxxxx function will return x,y coordinate of your shape from parameter t=<0,1> and actual radius r. It also returns the actual value of circumference/r ratio called pi2
Here preview of the 3 kernels used for spiral ...