沿六角形軌跡移動星形標記？

Question

我想沿着六角形軌跡移動星形標記，類似於我在問題末尾添加的「圓形軌跡」。謝謝。

這是源代碼，我寫了又創造同心hegzagons，但我不知道如何移動星標記橫穿同心六邊形，我寫了一個類似的模擬代碼爲圓形軌道，但我無法做到六角形。

%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 

圓軌跡：

% 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

我會與軌跡參數功能概括您的問題。在用它旋轉的內核，你想在這裏舉幾個例子在C++/VCL/GDI（對不起，我不是Matlab的友好的，但公式應該是相同的Matlab的太）爲圓形，方形和六邊形旋轉的內核：

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; 
    } 
//--------------------------------------------------------------------------- 

可以忽略VCL/GDI渲染的東西。
xs,ys已滿，xs2,ys2是窗口的一半分辨率縮放正確的情節......
dl是標誌之間的距離[pixels]
dr是螺旋形的螺釘之間的距離[pixels]

螺旋與根據r,t步驟中取樣實際周長（這是pi2或c的用途）。 getpnt_xxxxx函數將從參數t=<0,1>和實際半徑r返回形狀的x,y座標。它還返回circumference/r比實際值稱爲pi2

這裏用於螺旋3個內核預覽...