10.1抛物線曲面之計算
利用MATLAB程式可以計算抛物線之運動參數，其程式內容如下，呼叫之名稱與參數分別為：function [y, yy, yyy]=parabol_cam(phi, phi_in, beta_range, direct, travel,rpm) 8-4.3.2 MATLAB計算程式 function [y, yy, yyy]=parabol_cam(phi,phi_in,beta_range,direct,travel,rpm) % code = 2 % [y, yy, yyy]=parabol_cam(phi,phi_in,beta_range,direct,travel,rpm) % phi=cam angle, degrees % phi_in=starting cam angle, degrees % beta_range=motion range, degrees % direct=motion type; +1 for upward, -1 for downward % travel=the follower travels(=1 for a unit of travel) % rpm=cam rotation speed, rpm(=0 for not effective) % Example: [y, yy, yyy]=parabol_cam(10,180,120,1,1,0) d2r=pi/180; th=phi*d2r; thinit=phi_in*d2r; beta=beta_range*d2r; speed=rpm*2*pi/60;if rpm==0, speed=1;end; theta=th-thinit; thmed=thinit+beta/2; thx=theta/beta; if direct==1, if th<thmed y=2*thx^2; yyy=4/beta/beta; yy=yyy*th; else y=1-2*(1-thx)^2; yy=4/beta*(1-thx); yyy=-4/beta/beta; end else if th<thmed y=1-2*thx^2; yyy=-4/beta/beta; yy=yyy*th; else y=2*(1-thx)^2; yy=-4/beta*(1-thx); yyy=4/beta/beta; end end y=y*travel; yy=yy*speed*travel; yyy=yyy*speed*speed*travel;
執行例：
1. 升程起始點120∘，β=60∘，h=0.8cm，角度為150∘時之位移、速度及加速度： >> [y, yy, yyy]=parabol_cam(150,120,60,1,0.8,0) y = 0.4 (cm) yy = 1.5279 (cm/s) yyy = -2.9181 (cm/s2) 2. 返程起始點210∘，β=150∘，h=0.8，角度為250∘時之位移、速度及加速度： >> [y, yy, yyy]=parabol_cam(250,210,150,-1,0.8,0) y = 0.68622(cm) yy = -0.3259(cm/s) yyy = -0.46689(cm/s2)
範例8.1
某凸輪開始時先在0-120∘區間滯留，其高度為零。升程落於120-180∘，並於180-210∘滯留，其總升程為8公分。設刻度區間為10∘，凸輪則以等角速度旋轉，且其升程與返程均採用等加速度運動。求其變化曲線。
[解]：
1. 設ψ凸輪之迴轉角度，ψ=120-180∘為升程，轉折點為150∘，起點120∘，令θ = ψ-120，β = 60及h = 8代入公式8.9，可得升程之第一段(θ=120-150∘)：
y(θ)=2h(θ/β)²=2(8)[(ψ-120)/60]²
2. 而升程之第二段(θ=150-180∘)則仍以ψ-120=θ，β=60及h=8代入公式8.10，得：
y(θ)=h[1-2(1-θ/β)²]=(8)[1-2(1-{ψ-120}/60]²
3. 返程之區間為210∘至360∘，轉折點為(210+360)/2=285∘，而β=360-210=150∘。設ψ-210=θ，代入返程一段的位移公式8.11，得第一區段(θ=210-285∘)：
y(θ)=h[1-2(θ/β)²]=(8)[1-2({ψ-210}/150]²
4. 返程第二段之區間為285至360度，其轉折點與β均與第3項同，代入返程二段的位移公式，可以得到：
y(θ)=2h[1-(θ/β)²]=2(8)[1-({ψ-210}/150]²
由上面所得之函數可以利用電腦程式運算。針對升程(Demo8.1)與返程( Demo8.2)計算結果如下：
% Demo8_1 % 計算升程對應點之資料:位移,速度及加速度 theta=120:10:180; for i=1:length(theta) [ss(i), vv(i), aa(i)]=parabol_cam(theta(i),120,60,1,8,0); end; %(theta:對應角度; ss:位移, vv:速度, aa:加速度) [theta' ss' vv' aa'] >> demo8_1 ans = 120.0000 0 0 29.1805 130.0000 0.4444 5.0930 29.1805 140.0000 1.7778 10.1859 29.1805 150.0000 4.0000 15.2789 -29.1805 160.0000 6.2222 10.1859 -29.1805 170.0000 7.5556 5.0930 -29.1805 180.0000 8.0000 -0.0000 -29.1805 %---------------------------------------- % Demo8_2 % 計算返程對應點之資料:位移,速度及加速度 theta=210:10:360; for i=1:length(theta) [ss(i), vv(i), aa(i)]=parabol_cam(theta(i),210,150,-1,8,0); end; %(theta:對應角度; ss:位移, vv:速度, aa:加速度) [theta' ss' vv' aa'] %------------------------------------------ >> demo8_2 ans = 210.0000 8.0000 0 -4.6689 220.0000 7.9289 -0.8149 -4.6689 230.0000 7.7156 -1.6297 -4.6689 240.0000 7.3600 -2.4446 -4.6689 250.0000 6.8622 -3.2595 -4.6689 260.0000 6.2222 -4.0744 -4.6689 270.0000 5.4400 -4.8892 -4.6689 280.0000 4.5156 -5.7041 -4.6689 290.0000 3.4844 -5.7041 4.6689 300.0000 2.5600 -4.8892 4.6689 310.0000 1.7778 -4.0744 4.6689 320.0000 1.1378 -3.2595 4.6689 330.0000 0.6400 -2.4446 4.6689 340.0000 0.2844 -1.6297 4.6689 350.0000 0.0711 -0.8149 4.6689 360.0000 0.0000 -0.0000 4.6689
若就從動件之位移，利用電腦之資料或依10度為區間實際計算，可以得出其運動圖如圖8.19。
圖 8.19 抛物線運動圖之例10.2 擺線曲線之函數
撰寫一套程式，仿抛物線之parabol_cam(phi, phi_in, beta_range, direct, travel,rpm)，作一產生擺線曲線之函數。並計算下面之情況：總升程為50mm，升程以擺線運動經歷180度，其後滯留90度，再以擺線返程。凸輪之最小半徑為25mm。輸出資料除位移、速度及加速度外，並計算對應點之曲率半徑ρ、壓力角Φ。
[解]
function [y, yy, yyy, rho, alpha]=cycloidal_cam(r0,theta,theta_in,beta_range,direct,travel,rpm) % code = 2 % [y, yy, yyy,rho,phi]=cycloidal_cam(r0,theta,theta_in,beta_range,direct,travel,rpm) % r0:min. radius of the pitch surface. % theta=cam angle, degrees % theta_in=starting cam angle, degrees % beta_range=motion range, degrees % direct=motion type; +1 for upward, -1 for downward % travel=the follower travels(=1 for a unit of travel) % rpm=cam rotation speed, rpm(=0 for not effective) % rho: radius of curvature % alpha: pressure angle, in degrees. % Example: [y, yy, yyy,rho,alpha]=cycloidal_cam(25,10,180,120,50,1,0) d2r=pi/180; th=theta*d2r; thinit=theta_in*d2r; beta=beta_range*d2r; speed=rpm*2*pi/60;if rpm==0, speed=1;end; theta=th-thinit; thmed=thinit+beta/2; thx=theta/beta;pi2=pi*2; if direct==1, y=thx-(1/pi2)*sin(pi2*thx); yy=(1/beta)*(1-cos(pi2*thx)); yyy=(pi2/beta/beta)*sin(pi2*thx); else y=1-thx+(1/pi2)*sin(pi2*thx); yy=-(1/beta)*(1-cos(pi2*thx)); yyy=-(pi2/beta/beta)*sin(pi2*thx); end y=y*travel; yy=yy*speed*travel; yyy=yyy*speed*speed*travel; r=r0+y; rho=(r.*r+yy.*yy).^1.5./(r.*r+2*yy.*yy-r.*yyy); alpha=atan2(yy,r)/d2r;
執行例：
>> [y, yy, yyy,rho,alpha]=cycloidal_cam(25,[0:30:180],0,180,1,50,0) y = 0 1.4417 9.7751 25 40.225 48.558 50 yy = 0 7.9577 23.873 31.831 23.873 7.9577 0 yyy = 0 27.566 27.566 3.8982e-015 -27.566 -27.566 -7.7963e-015 rho = 25 217.26 53.972 46.005 46.589 53.537 75 alpha = 0 16.749 34.47 32.482 20.103 6.1744 0
10.3 簡諧曲線之函數
8.5 撰寫一套程式，仿抛物線之parabol_cam(phi, phi_in, beta_range, direct, travel,rpm)，作一產生簡諧曲線之函數。並計算下面之情況：總升程為50mm，升程以擺線運動經歷90度，其後滯留90度，再以擺線返程。凸輪之最小半徑為25mm。輸出資料除位移、速度及加速度外，並計算對應點之曲率半徑ρ、壓力角Φ。
[解]
function [y, yy, yyy, rho, alpha]=harmonic_cam(r0,theta,theta_in,beta_range,direct,travel,rpm) % code = 2 % [y, yy, yyy,rho,phi]=harmonic_cam(r0,theta,theta_in,beta_range,direct,travel,rpm) % r0:min. radius of the pitch surface. % theta=cam angle, degrees % theta_in=starting cam angle, degrees % beta_range=motion range, degrees % direct=motion type; +1 for upward, -1 for downward % travel=the follower travels(=1 for a unit of travel) % rpm=cam rotation speed, rpm(=0 for not effective) % rho: radius of curvature % alpha: pressure angle, in degrees. % Example: [y, yy, yyy,rho,alpha]=harmonic_cam(25,10,180,120,50,1,0) d2r=pi/180; th=theta*d2r; thinit=theta_in*d2r; beta=beta_range*d2r; speed=rpm*2*pi/60;if rpm==0, speed=1;end; theta=th-thinit; thmed=thinit+beta/2; thx=theta/beta; if direct==1, y=0.5*(1-cos(pi*thx)); yy=(pi/2/beta)*sin(pi*thx); yyy=(pi/beta)^2/2*cos(pi*thx); else y=0.5*(1+cos(pi*thx)); yy=-(pi/2/beta)*sin(pi*thx); yyy=-(pi/beta)^2/2*cos(pi*thx); end y=y*travel; yy=yy*speed*travel; yyy=yyy*speed*speed*travel; r=r0+y; rho=(r.*r+yy.*yy).^1.5./(r.*r+2*yy.*yy-r.*yyy); alpha=atan2(yy,r)/d2r;
執行例：
>> [y, yy, yyy,rho,alpha]=harmonic_cam(25,10,180,120,50,1,0) y = 0.37059 yy = -0.72444 yyy = 0.29117 rho = 25.654 alpha = -1.6356
範例8.11
某凸輪開始在0-90∘區間在零高度滯留，然後提升後在150至180∘區間滯留，時高度為10公分，其餘則為返程。設刻度區間為10∘繪出其高度圖。凸輪則以等角速度旋轉，其升程與返程均採用等加速度運動。
[解]
1. 設ψ凸輪之迴轉角度，ψ=90-150∘為升程，轉折點為120∘，起點90∘，令θ = ψ-90，β = 60及h = 10代入公式8.9，可得升程之第一段(θ =90-120)：
y(θ)=2h(θ/β)²=2(10)[(ψ-90)/60]²
2. 而升程之第二段(θ =120-150)，則以ψ-90=θ，β=60及h=10代入公式8.10，得：
y(θ)=h[1-2(1-θ/β)²]=(10)[1-2{1-(ψ-90)/60}²]
3. 返程之區間為180∘至360∘，轉折點為(180+360)/2=270∘，β=360-180=180∘。設ψ-180=θ，代入返程一段的位移公式8.11，第一區段(θ =180-270)得：
y(θ)=h[1-2(1-θ/β)²]=(10)[1-2{1-(ψ-180)/180}²]
4. 返程第二段之區間θ =270-360度，代入返程一段的位移公式，可以得到：
y(θ)=2h(1-θ/β)²=2(10)[1-(ψ-180)/180]²
利用MATLAB程式可撰寫如下：
function [theta,y]=P8_11(theta,limit0,s)
% ex P8.11
% Inputs: theta: angles in row matrix
% limit0: mark angles [(dwell) t0 (rise) t1 (dwell) t2(return)t3]
% s:stroke
% Example: [theta,y]=P8_11([0:10:360],[90 150 180 360],30)
d2r=pi/180;
h=s;lm=limit0*d2r;
lm=[lm(1) (lm(1)+lm(2))/2 lm(2) lm(3) (lm(3)+lm(4))/2 lm(4)]/d2r;
LM=lm*d2r;
th=theta*d2r;
y=ones(size(th));
y(th<LM(1))=0;
y(th>=LM(3))=h;
y(th>=LM(1)&th<LM(2))=2*h*((th(th>=LM(1)&th<LM(2))-LM(1))/(LM(3)-LM(1))).^2;
y(th>=LM(2)&th<LM(3))=h*(1-2*(1-(th(th>=LM(2)&th<LM(3))-LM(1))/(LM(3)-LM(1))).^2);
y(th>=LM(4)&th<LM(5))=h*(1-2*((th(th>=LM(4)&th<LM(5))-LM(4))/(LM(6)-LM(4))).^2);
y(th>=LM(5)&th<=LM(6))=2*h*(1-(th(th>=LM(5)&th<=LM(6))-LM(4))/(LM(6)-LM(4))).^2;
plot(th/d2r,y);
xlabel('Theta in degrees');
ylabel('Follower displacement');
grid on;
執行結果：
>> [theta,y]=P8_11([0:10:360],[90 150 180 360],30) theta = 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 270 280 290 300 310 320 330 340 350 360 y = 0 0 0 0 0 0 0 0 0 0 1.6667 6.6667 15 23.333 28.333 30 30 30 30 29.815 29.259 28.333 27.037 25.37 23.333 20.926 18.148 15 11.852 9.0741 6.6667 4.6296 2.963 1.6667 0.74074 0.18519 0
圖P8.11 運算結果10.4 DWELL函數簡介
上述各函數作為從動件運動件運動包括其上升下降之位移、速度、加速度等均可以利用程式表示。為求統一，可歸納到dwell.m這個函數之中。其呼叫法為：
function [y,yy,yyy]=dwell(ctheta,range,pattern)
其中參數定義如下：
ctheta = 需要計算之凸輪角度，單位為度數。可以使用矩陣輸入之型式。
pattern = 運動的型式，二元素之列矩陣，其代碼如下：
1:等速運動uniform 2:抛物線parabolic 3:簡諧simple harmonic
4:擺線cycloidal 5:多項式polynomial motion
e.g. [4 3]:升程為擺線運動cycloidal；返程為簡諧運動harmonic motion
range =升程及返程之範圍，三元素列矩陣 e.g.[90 180 240]升程始於90 度，止於180度；
返程始於240 度，止於360。
輸出: y：位移（其最大值為１，故得到對應值後應與從動件之衝程相乘，才是真正的位移，其第一及第二導數相同）；yy 對於凸輪角ctheta之第一導數； yyy 對於凸輪角ctheta之第二導數。
程式內容：
function [y,yy,yyy]=dwell(ctheta,range,pattern) % % This function determines the follower displacement and derivatives % for a full rotation cam. The routine is set up for the displacement % schedule in Examples 6.7 and 6.8 % The input values are: %ctheta = individual cam angles (deg)--can be a matrix to be processed %pattern = denote the types of motion used(a 2 element-row matrix). types % are: % 1:uniform 2:parabolic 3:simple harmonic 4: cycloidal % 5:polynomial motion % e.g. [4 3]:cycloidal motion for rise and harmonic motion for % return %range =the degrees the specific motion starts(array of 3 elements), e.g. % [90 180 240] starts to rise at 90 deg. and ends at 180 deg. % starts to return at 240 deg. and ends at 360 deg. % Output: y is for displacement, yy is the derivative of the displacement with % respect to theta, and yyy the second derivative with respect to theta. % Example dwell(60,[90 180 240],[4 3]); % Author:DSFon, BIME, NTU Revise Date:May 18, 2007 d2r=pi/180; theta=ctheta*d2r;range=range*d2r; dim=length(ctheta); y=zeros(size(ctheta));yy=y;yyy=y; for i=1:dim if theta(i)>=range(3) %for the last motion(downward) mode=pattern(2);betax=2*pi-range(3); switch mode, case 1, [y(i),yy(i),yyy(i)]=uniform(theta(i), range(3),betax,-1); case 2, [y(i),yy(i),yyy(i)]=parabolicm(theta(i), range(3),betax,-1); case 3, [y(i),yy(i),yyy(i)]=harmonicm(theta(i), range(3),betax,-1); case 4, [y(i),yy(i),yyy(i)]=cycloidm(theta(i), range(3),betax,-1); case 5, [y(i),yy(i),yyy(i)]=polynorm(theta(i), range(3),betax,-1); end; elseif theta(i)>=range(2) % dewell on the top y(i)=1; elseif theta(i)>=range(1) % for the 1st motion(upward) mode=pattern(1);betax=range(2)-range(1); switch mode, case 1, [y(i), yy(i), yyy(i)]=uniform(theta(i), range(1),betax,+1); case 2, [y(i), yy(i), yyy(i)]=parabolicm(theta(i), range(1),betax,+1); case 3, [y(i), yy(i), yyy(i)]=harmonicm(theta(i), range(1),betax,+1); case 4, [y(i), yy(i), yyy(i)]=cycloidm(theta(i), range(1),betax,+1); case 5, [y(i), yy(i), yyy(i)]=polynorm(theta(i), range(1),betax,+1); end end end %********************************************* function [t1, t2, t3]=uniform(th, thinit,beta,direct) % code = 1 for uniform motion %th=cam angle, radians %beta=motion range, radians %thinit=starting cam angle, radians %beta=motion range, radians %direct=motion type; +1 for upward, -1 for downward theta=th-thinit; t1=theta/beta; if direct==-1, t1=1-t1; end; t2=direct*1/beta; t3=0; %********************************************* function [t1, t2, t3]=parabolicm(th,thinit,beta,direct) % code = 2 for parabolic motion %th=cam angle, radians %beta=motion range, radians %thinit=starting cam angle, radians %beta=motion range, radians %direct=motion type; +1 for upward, -1 for downward theta=th-thinit; thmed=thinit+beta/2;thx=theta/beta; if direct==1, if th<thmed t1=2*thx^2; t3=4/beta/beta; t2=t3*th; else t1=1-2*(1-thx)^2; t2=4/beta*(1-thx); t3=-4/beta/beta; end else if th<thmed t1=1-2*thx^2; t3=-4/beta/beta; t2=t3*th; else t1=2*(1-thx)^2; t2=-4/beta*(1-th/beta); t3=4/beta/beta; end end %********************************************* function [t1, t2, t3]=harmonicm(th,thinit,beta,direct) % code = 3 for harmonic motion %th=cam angle, radians %beta=motion range, radians %thinit=starting cam angle, radians %beta=motion range, radians %direct=motion type; +1 for upward, -1 for downward theta=th-thinit; t1=0.5*(1-cos(pi*theta/beta)); if direct==-1, t1=1-t1;end; t2=direct*(0.5*pi/beta)*sin(pi*theta/beta); t3=direct*0.5*(pi/beta)^2*cos(pi*theta/beta); %********************************************* function [t1, t2, t3]=cycloidm(th,thinit,beta,direct) % % code = 4 for cycloidal motion %th=cam angle, radians %thinit=starting cam angle, radians %beta=motion range, radians %direct=motion type; +1 for upward, -1 for downward theta=th-thinit; t1=theta/beta-(0.5/pi)*sin(2*pi*theta/beta); if direct==-1,t1=1-t1;end; t2=direct*(1-cos(2*pi*theta/beta)); t3=direct*2*pi/beta/beta*sin(2*pi*theta/beta); %********************************************* function [t1, t2, t3]=polynorm(th,thinit,beta,direct) % % code = 5 for polynormial motion %th=cam angle, radians %thinit=starting cam angle, radians %beta=motion range, radians %direct=motion type; +1 for upward, -1 for downward theta=th-thinit; thx=theta/beta; t1=thx*thx*thx*(10+thx*(-15+thx*6)); if direct==-1, t1=1-t1;end; t2=direct*(30/beta)*thx*thx*(1+thx*(-2+thx)); t3=direct*(60/beta/beta)*thx*(1+thx*(-3 +2*thx));
本程式將配合其他凸輪繪製時呼叫使用。
Dwell執行範例8.10
下面plot_dwell.m為一執行Dwell函數之範例。其指令如下：
function plot_dwell(ctheta,s,pattern,range)
相關參數請參照後節之程式介紹。
function plot_dwell(ctheta,s,pattern,range)
%ctheta = cam angle (deg)--can be a matrix
%pattern = denote the type of motion used(a 3 element-row matrix)
% 1:uniform 2:parabolic 3:simple harmonic 4: cycloidal
% 5:polynomial motion
% example [4 3]
%range =the degrees the specific motion starts
% Output: y is for displacement, yy is the derivative of the displacement with
% respect to theta, and yyy the second derivative with respect % to theta.
% Example plot_dwell(0:10:360,2,[4 3],[90 180 240]);
figure(1);clf;
[y,yy,yyy]=dwell(ctheta,range,pattern)
h1=plot(ctheta,y*s,'b-',ctheta,yy*s,'k-',ctheta,yyy*s,'r-')
legend('Displacement','Velocity','Acceleration',3)
xlabel('Elapsed Angle, degrees')
grid
執行例
>>plot_dwell(0:10:360,2,[4 3],[90 180 240]);
範例8.10
就梢從動件之凸輪通常可利用課本圖8.26及公式8.26之轉軸進行模擬各角度之位置，假設一從動件之梢長為Ｌ，其偏置量為e，凸輪之基圓半徑為r0，衝程為s。可以利用dwell函數求各角度對應衝程，然後依公式8.26進行轉軸。設此程式名稱為pincam，其呼叫方式如下：
function [x,y]=pincam(cth,r0,s,e,L,range,pattern,cw)
其中，各參數定義如下：
輸入參數： cth:凸輪角度，度數 r0:凸輪基圓半徑 e:偏置量 s:從動件衝程 L:從動件長度 cw:凸輪轉動方向（反時鐘為正，順時鐘為負） pattern = 運動的型式，二元素之列矩陣，其代碼如下： 1:等速運動uniform 2:抛物線parabolic 3:簡諧simple harmonic 4:擺線cycloidal 5:多項式polynomial motion e.g. [4 3]:升程為擺線運動cycloidal；返程為簡諧運動harmonic motion range =升程及返程之範圍，三元素列矩陣 e.g.[90 180 240]升程始於90 度， 止於180度；返程始於240 度，止於360。
程式內容
function [x,y]=pincam(cth,r0,s,e,L,range,pattern,cw) %Find the pin type cam with an offsect e %Inputs: % cth:angle of cam, degrees % r0:radius of base circle % e:offset % s:stroke % L:length of pin % cw:rotation direction of cam(-counterclockwise,+clockwise %pattern = denote the type of motion used(a 3 element-row matrix) % 1:uniform 2:parabolic 3:simple harmonic 4: cycloidal % 5:polynomial motion % example [4 3] %range =the degrees the specific motion starts, eg.[90 180 240] % Example: [x y]=pincam([10 60],5,2,1,10,[90 180 240],[4 3],-1) figure(1); clf; th=cth*pi/180; s0=sqrt(r0*r0-e*e); for i=1:length(cth) t=th(i)*cw; A=[cos(t) -sin(t);sin(t) cos(t)]; [ym,yy,yyy]=dwell(cth(i),range,pattern); x0=s0+ym*s; Sx=[0 x0 x0+L;e e e]; X=A\Sx; x(i)=X(1,2);y(i)=X(2,2); line(X(1,1:2),X(2,1:2)); line(X(1,2:3),X(2,2:3),'linewidth',3,'color','red') end hold on; plot([0 x],[0 y],'ro',x,y,'k-') axis equal
執行例一：偏置量為零時
>> [x y]=pincam([0:10:360],6,5,0,10,[90 180 240],[4 3],-1)
偏置量為零時執行例二：偏置量不為零時
>> [x y]=pincam([0:10:360],6,5,4,10,[90 180 240],[4 3],-1);
偏置量不為零時範例8.12
試就問題8.11所得從動子之抛物線運動之位移量，繪製一凸輪，設從動子屬刀緣接觸型，其基圓半徑為35mm，順時鐘方向旋轉。
[解]請參考圖8.6
繪圖程式撰寫如下：
function [rp,rb]=drawcam(r0,y,direct) % To draw a cam profile with base radius of r0 and rise in y % The program may work with P8_11 % Input: direct: +1 for clockwise; -1 for counterclockwise % % Example: [s,T]=drawcam(35,y) d2r=pi/180; nn=length(y); LM=max(y+r0); figure(2); line([-LM LM]',[0 0]');line([0 0]',[-LM LM]'); if direct==1, theta=linspace(0,360,nn)'*d2r; else theta=linspace(360,0,nn)'*d2r; end theta=theta+pi/2; rb=[r0*cos(theta) r0*sin(theta)]; rp=[(r0+y').*cos(theta) (r0+y').*sin(theta)]; line(rb(:,1),rb(:,2),'color','r') line(rp(:,1),rp(:,2)); axis equal; 執行結果： >> [s]=drawcam(35,y,1) s = 35 0 34.468 6.0777 32.889 11.971 30.311 17.5 26.812 22.498 22.498 26.812 17.5 30.311 11.971 32.889 6.0777 34.468 2.1431e-015 35 -6.3671 36.11 -14.251 39.154 -25 43.301 -37.496 44.686 -48.516 40.71 -56.292 32.5 -61.08 22.231 -64.013 11.287 -65 7.9602e-015 -63.83 -11.255 -60.384 -21.978 -54.848 -31.667 -47.523 -39.877 -38.805 -46.246 -29.167 -50.518 -19.128 -52.553 -9.2291 -52.341 -9.1849e-015 -50 8.1357 -46.14 15.074 -41.416 20.833 -36.084 25.473 -30.358 29.081 -24.402 31.754 -18.333 33.585 -12.224 34.651 -6.1098 35 -8.5725e-015
圖P8.12a 順時鐘方向旋轉
圖P8.12b 反時鐘方向旋轉範例8.13
試就問題8.11所得從動子之抛物線運動之位移量，繪製一凸輪，設從動子屬滾子型，其半徑為8mm，基圓半徑為40mm，順時鐘方向旋轉。
[解]請參考圖8.29
繪圖程式撰寫如下：
function [rp,rb]=draw_roller_cam(r0,r,y,direct) % To draw a cam profile with base radius of r0 and rise in y % The program may work with P8_11 % Input: direct: +1 for clockwise; -1 for counterclockwise % % Example: [rp,rb]=draw_roller_cam(40,6,y,1) d2r=pi/180; nn=length(y); LM=max(y+r0); figure(2); line([-LM LM]',[0 0]');line([0 0]',[-LM LM]'); if direct==1, theta=linspace(0,360,nn)'*d2r; else theta=linspace(360,0,nn)'*d2r; end theta=theta+pi/2; rb=[r0*cos(theta) r0*sin(theta)]; rp=[(r0+y').*cos(theta) (r0+y').*sin(theta)]; line(rb(:,1),rb(:,2),'color','r'); line(rp(:,1),rp(:,2)); th=linspace(0,2*pi,50)'; for i=1:nn, line(rp(i,1)+r*cos(th),rp(i,2)+r*sin(th)); end axis equal;
執行結果：
>> [rp,rb]=draw_roller_cam(40,8,y,1) rp = 2.4493e-015 40 -6.9459 39.392 -13.681 37.588 -20 34.641 -25.712 30.642 ------------------------------------------ 57.265 -10.097 55 -1.3471e-014 51.064 9.004 46.115 16.784 40.415 23.333 34.188 28.687 27.616 32.912 20.833 36.084 13.934 38.284 6.9781 39.575 1.2246e-014 40 rb = 2.4493e-015 40 -6.9459 39.392 -13.681 37.588 -20 34.641 -25.712 30.642 ------------------------------------ 25.712 30.642 20 34.641 13.681 37.588 6.9459 39.392 1.2246e-014 40
圖P8.13 滾子型之包絡線軌跡圖P8.1所得之結果作一切線切於各滾子內圈之弧即可得其工作曲線，而所採用的點數愈多，其內、外包絡線愈為顯明。
範例8.14
試就問題8.11所得從動子之抛物線運動之位移量，繪製一凸輪，設從動子屬平板型，基圓半徑為100mm，順時鐘方向旋轉。
[解]請參考圖8.32
繪圖程式撰寫如下：
function [rp,rb]=draw_plat_cam(nn,r0,s,d,range,pattern,cw) % The program work with dewell.m %Inputs: % nn:number of points in process % r0:radius of base circle % re:radius of the roller % a:distance between two axes of cam and roller % s:stroke % d:length of plate follower % cw:rotation direction of cam(-counterclockwise,+clockwise %pattern = denote the type of motion used(a 3 element-row matrix) % 1:uniform 2:parabolic 3:simple harmonic 4: cycloidal % 5:polynomial motion % example [4 3] %range =the degrees the specific motion starts, eg.[90 180 240] % Example: draw_plat_cam(36,5,2,2,[90 180 240],[4 3],-1) figure(1);clf;d2r=pi/180; th=linspace(0,360,nn); [y,ff]=dwell(th,range,pattern); theta=(th+90)*d2r;%The follower is in upright position if cw==-1,theta=pi-theta;end rb=[r0*cos(theta);r0*sin(theta)]; % %rp=[(r0+y*s).*cos(theta)+cw*ff*s.*sin(theta);... % (r0+y*s).*sin(theta)-cw*ff*s.*cos(theta)]; %line(rp(1,:),rp(2,:));%Draw working circle for i=1:nn, rr=r0+y(i); xx=rr*cos(theta(i));yy=rr*sin(theta(i)); line([0 xx], [0 yy]); dx=d*sin(theta(i));dy=d*cos(theta(i)); line([xx+dx xx-dx],[yy-dy yy+dy],'linewidth',2,'color','red'); end line(rb(1,:),rb(2,:),'color','r');%draw base circle axis equal;
執行結果：
>> draw_plat_cam(36,5,2,2,[90 180 240],[4 3],1)
圖P8.14 平板從動子之凸輪軌跡10.5 凸輪運動之進階分析
如何繪製凸輪之外緣曲線，是探討凸輪的主題。下面程式為針對板式從動件所產生之凸輪外緣曲線，並且依所需之運動型式進行設計。
平板從動件之凸輪
camwork程式為一平板式從動件對凸輪運動時之相對軌跡曲線之繪製，其法與前面之drawcam大致上相同。通常係讓凸輪固定，然後由從動件依設定之角度進行繪製。本程式之呼叫法如下：
function camwork(rb,rise,range,pattern,platdim,direction,ainc) 參數說明： pattern = 運動的型式，二元素之列矩陣，其代碼如下： 1:等速運動uniform 2:抛物線parabolic 3:簡諧simple harmonic 4:擺線cycloidal 5:多項式polynomial motion e.g. [4 3]:升程為擺線運動cycloidal；返程為簡諧運動harmonic motion range =升程及返程之範圍，三元素列矩陣 e.g.[90 180 240]升程始於90 度， 止於180度；返程始於240 度，止於360。 rb:基圓半徑 rise:從動件升程 platdim:從動件平板尺寸，其型式如 [Ll,Lr,t], 其中，L1,Lr分別為從動件中心線前與後之板面 長度，t:從動件之厚度 direction:凸輪轉動方向， +1 為逆時鐘；-1 為順時鐘。 ainc: 迴轉角度分量，度
本程式呼叫 dwell.m, bush.m, camfollow.m及 frame.m等函數，若需要板面傾斜某一角度，則需另呼叫trans4.m程式，此函數在第二章中有關轉動座標之函數中已經介紹過，可以引用。
function camwork(rb,rise,range,pattern,platdim,direction,ainc)
%Draw a plate cam in diferent positions
% Call dwell.m, bush.m, camfollow.m, frame.m
%Inputs:
% rb:radius of the cam base circle
% rise:rise of the follower
%range =the degrees the specific motion starts(array of 3 elements), e.g.
% [90 180 240] starts to rise at 90 deg. and ends at 180 deg.
% starts to return at 240 deg. and ends at 360 deg.
%pattern = denote the types of motion used(a 2 element-row matrix). types
% are:
% 1:uniform 2:parabolic 3:simple harmonic 4: cycloidal
% 5:polynomial motion
% e.g. [4 3]:cycloidal motion for rise and harmonic motion for
% return
%platdim:the plate dimension, in forms of [Ll,Lr,t], in which
% L1,Lr:length of plage in front of and behind the centerline,
% t:thickness of the axle of the plate follower.
% direction:cam direction, +1 for counterclockwise; -1 for clockwise
% ainc: angular increment, degrees
%Examples: camwork(5,2,[100 180 270],[3 3],[2 3 1],1,60)
platdim(find(platdim<=0.2))=0.2;
d2r=pi/180;
camangle=0:1:360;
th=camangle*d2r;
[s,ss,sss]=dwell(camangle,range,pattern);
s=s*rise;ss=ss*rise;
xcam=(rb+s).*cos(th)-ss.*sin(th);
ycam=(rb+s).*sin(th)+ss.*cos(th);
ycam=ycam*direction;
[coords]=bushing(-rb,0,0);
xb=coords(:,1);yb=coords(:,2);
[coords]=bushing(rb/7,0,0);
xbush=coords(:,1);ybush=coords(:,2);
figure(1);
clf;
axis equal;
thi=[0:ainc:360]';
nx=length(thi);
bush=line('xdata',[],'ydata',[],'linewidth',1.5,'color','r','erasemode','none');
base=line('xdata',[],'ydata',[],'linewidth',1,'color','b','erasemode','none');
cam=line('xdata',[],'ydata',[],'linewidth',2,'color','k','erasemode','none');
frame1=line('xdata',[],'ydata',[],'linewidth',1,'color','r','erasemode','none');
frame2=line('xdata',[],'ydata',[],'linewidth',1,'color','r','erasemode','none');
for k=1:nx-1,
follow(k)=line('xdata',[],'ydata',[],'linewidth',1,'color','k','erasemode','none');
end
line1=line('xdata',[],'ydata',[],'linewidth',1,'color','k','erasemode','none');
bead=line('xdata', [], 'ydata', [],'marker','o','markersize', 6, 'erasemode',...
'xor');
set(bush,'xdata',xbush,'ydata',ybush);
set(base,'xdata',xb,'ydata',yb);
set(cam,'xdata',xcam,'ydata',ycam);
%set(bush,'xdata',xbush,'ydata',ybush);
thr=thi*d2r;
tho=thi;if direction==-1,tho=360-thi;end;
[coordf]=camfollow(platdim(1),platdim(2),platdim(3),rise);
coords=coordf;
linest=[];
[frame0]=frame(platdim(3),rise,1,0);
set(frame1,'xdata',frame0(:,1)+rb,'ydata',frame0(:,2))
[frame0]=frame(platdim(3),rise,-1,0);
set(frame2,'xdata',frame0(:,1)+rb,'ydata',frame0(:,2))
for i=1:length(thr)-1
[s,ss,sss]=dwell(thi(i),range,pattern);
s=s*rise;ss=ss*rise;
xx=(rb+s)*cos(thr(i))-ss*sin(thr(i));
yy=((rb+s)*sin(thr(i))+ss*cos(thr(i)))*direction;
coords=coordf;
coords(:,1)=coordf(:,1)+rb+s;
xf=rb+s;yf=0;
[coords]=trans4(coords,tho(i),5);
[linef]=trans4([0 0;xf yf],tho(i),5);
text(linef(2,1),linef(2,2),num2str(tho(i)));
linest=[linest;0 0;linef];
set(bead,'xdata',xx,'ydata',yy);
set(follow(i),'xdata',coords(:,1),'ydata',coords(:,2));
end
set(line1,'xdata',linest(:,1),'ydata',linest(:,2));
axis equal;
function [coords] = bushing(rr,x0,y0)
% Determine the coordinates of the anchor seat
% rr: radius of the shaft
% x0,y0: center coordinates of the shaft
d2r=pi/180;
theta=[360:-10:0]*d2r;
r=abs(rr);
rx=r*cos(theta);ry=r*sin(theta);
if rr<0,
rx=rx+x0;
ry=ry+y0;
coords=[rx' ry'];
return;
end;
rx1=rx/2;rx2=-rx1;
ry1=ry/2;ry2=-ry1;
r4=r+r/4;r3=r/3;
bx=[ 0 0 0 -r -r -r4 -r4 r4 r4 -r r r];
by=[r3 -r3 0 0 -r -r -r4 -r4 -r -r -r 0];
coords(:,1)=[bx rx rx1 rx2]'+x0;
coords(:,2)=[by ry ry1 ry2]'+y0;
hold off;
function [coords]=camfollow(tl,tr,thickness,stroke,theta)
%Detemine the coordinates of the plate follower
%Inputs:
% tl:length of plate in front of center line
% tr:length of plate behind the cneter line
% thickness:thichness of the plate
% stroke:rise of the follower
% theta:the slant angle(degrees) of plate
%Outputs:coords:coordinates of the plate
%coordinates for bock and outside form
t=thickness;s=stroke*1.5;tm=t/2;
coords=[0 0;0 tl; tm tl; t tm;t+s tm;t+s -tm;t -tm;tm -tr;0 -tr;0 0];
if nargin==5 % draw a slant follower
coords=trans4(coords,theta,5);
end
function [coords]=frame(thickness,stroke,mode,theta)
% determine the coordinate of follower seat
% Inputs:
% thickness:the thckness of the follower shaft
% stroke:rise of the follower
% theta:slant angle of the follower, degrees
% mode:+1 for one side, -1 for another side
t=thickness*1.05;s=stroke*1.2;
t0=thickness+stroke;tx=t/2;tm=t/3;
if mode==1
coords=[t0 -tx];
tt=t0+tm;
while 1
coords=[coords;tt -tx;tt-tm -tx-tm;tt -tx];
tt=tt+tm;
if tt-t0>s, break;end
end
end
if mode==-1
coords=[t0 tx];
tt=t0+tm;
while 1
coords=[coords;tt tx;tt-tm tx+tm;tt tx];
tt=tt+tm;
if tt-t0>s, break;end
end
end
if nargin==4,
coords=trans4(coords,theta,5);
end
執行範例：
基圓半徑5公分，升程2公分，分佈範圍為[100 180 270]度，升程與返程均使用簡諧運動型式，從動件為平板，左2公分，右3公分，板軸厚度為1公分，試將圓周以60均分繪出其從動件位置及對應之凸輪曲線。
>>camwork(5,2,[100 180 270],[3 3],[2 3 1],1,60)
10.6 滾輪從動件之凸輪
以滾子從動件運轉之凸輪可以利用camwork2.m進行繪製，其呼叫法如下：
function camwork2(rb,ro,rise,offset,range,pattern,direction,ainc) 參數說明： pattern = 運動的型式，二元素之列矩陣，其代碼如下： 1:等速運動uniform 2:抛物線parabolic 3:簡諧simple harmonic 4:擺線cycloidal 5:多項式polynomial motion e.g. [4 3]:升程為擺線運動cycloidal；返程為簡諧運動harmonic motion range =升程及返程之範圍，三元素列矩陣 e.g.[90 180 240]升程始於90 度， 止於180度；返程始於240 度，止於360。 rb:基圓半徑 ro:滾子從動件半徑 rise:從動件升程 offset:從動件之中心線偏置量 direction:凸輪轉動方向， +1 為逆時鐘；-1 為順時鐘。 ainc: 迴轉角度分量，度
camwork2程式呼叫 dwell.m, bush.m, camfollow2.m及 frame.m等函數。同camwork，若滾子軸需要板面傾斜某一角度，則需另呼叫trans4.m程式，此函數在第二章中有關轉動座標之函數中已經介紹過，可以引用。
function camwork2(rb,ro,rise,offset,range,pattern,direction,ainc)
%
%function camwork2(rb,ro,rise,offset,range,pattern,direction,ainc)
%find roller cam curve using envelop technique
% Call dwell.m, bush.m, camfollow.m, frame.m
%Inputs:
% rb:radius of the cam base circle
% ro:radius of the roller follower
% rise:rise of the follower
% offset:the offset of centerline of follower
% range =the degrees the specific motion starts(array of 3 elements), e.g.
% [90 180 240] starts to rise at 90 deg. and ends at 180 deg.
% starts to return at 240 deg. and ends at 360 deg.
% pattern= denote the types of motion used(a 2 element-row matrix). types
% are:
% 1:uniform 2:parabolic 3:simple harmonic 4: cycloidal
% 5:polynomial motion
% e.g. [4 3]:cycloidal motion for rise and harmonic motion for
% return
% direction:cam direction, +1 for counterclockwise; -1 for clockwise
% ainc: angular increment, degrees
%Examples: camwork2(5,2,1.5,1,[100 180 270],[3 3],1,60)
%Designed by D S Fon, BIME, NYU. Dated June 10, 2002
d2r=pi/180;
camangle=0:1:360;
th=camangle*d2r;
[s,ss,sss]=dwell(camangle,range,pattern);
s=s*rise;ss=ss*rise;
rho=rb+ro+s;
alpha=atan((rho./(offset*offset+rho.*rho)-offset).*ss);
phi=alpha-atan(offset./rho)/d2r;
offset0=abs(offset);
dd=sqrt((rb+ro)^2-offset*offset);%distance from center of roller to x-axis
ct=cos(th);st=sin(th);
UU=(dd+s).*st-(offset+ss).*ct;
VV=(dd+s).*ct+(offset+ss).*st;
xcam=offset*st+(dd+s).*ct-sign(ct)*ro./sqrt(1+(UU./VV).^2);
ycam=(xcam.*UU+(dd+s).*ss)./VV;
ycam=ycam*direction;
xf=offset*st+(dd+s).*ct;
yf=(-offset*ct+(dd+s).*st)*direction;
arg=ycam.^2+xcam.^2-(yf.^2+xf.^2);
ycam(arg>0)=[];
xcam(arg>0)=[];
[coords]=bushing(-rb,0,0);
xb=coords(:,1);yb=coords(:,2);
[coords]=bushing(-abs(offset),0,0);
xoff=coords(:,1);yoff=coords(:,2);
figure(1);
clf;
thi=[0:ainc:360]';
nx=length(thi);
base=line('xdata',[],'ydata',[],'linewidth',1,'color','b','erasemode','none');
offcircle=line('xdata',[],'ydata',[],'linewidth',1,'color','r','erasemode','none');
pitch=line('xdata',[],'ydata',[],'linewidth',1,'color','r','erasemode','none');
cam=line('xdata',[],'ydata',[],'linewidth',2,'color','k','erasemode','none');
frame1=line('xdata',[],'ydata',[],'linewidth',1,'color','r','erasemode','none');
frame2=line('xdata',[],'ydata',[],'linewidth',1,'color','r','erasemode','none');
for k=1:nx-1,
follow(k)=line('xdata',[],'ydata',[],'linewidth',1,'color','k',...
'erasemode','none');
linex(k)=line('xdata',[],'ydata',[],'linewidth',1,'color','k',...
'erasemode','none');
end
set(base,'xdata',xb,'ydata',yb);
set(cam,'xdata',xcam,'ydata',ycam);
set(offcircle,'xdata',xoff,'ydata',yoff);
set(pitch,'xdata',xf,'ydata',yf);
thr=thi*d2r;
tho=thi;if direction==-1,tho=360-thi;end;
[coordf]=camfollow2(ro,ro/2,rise);
coordf(:,2)=coordf(:,2);
coords=coordf;
maxline=rb+ro*2+rise;
offset0=-offset*direction;
[frame0]=frame(ro/2,rise,1,0);
set(frame1,'xdata',frame0(:,1)+rb+ro,'ydata',frame0(:,2)+offset0);
[frame0]=frame(ro/2,rise,-1,0);
set(frame2,'xdata',frame0(:,1)+rb+ro,'ydata',frame0(:,2)+offset0);
for i=1:length(thr)-1
[s,ss,sss]=dwell(thi(i),range,pattern);
s=s*rise;ss=ss*rise;
ct=cos(thr(i));st=sin(thr(i));
xf=offset*st+(dd+s)*ct;
yf=(-offset*ct+(dd+s)*st)*direction;
[coords]=trans4(coordf,tho(i),5);
coords=trans4(coords,[xf yf],1);
linef=[maxline 0;0 0;0 offset0];
[linef]=trans4(linef,tho(i),5);
linef=[linef;xf yf];
text(xf,yf,num2str(tho(i)));
set(linex(i),'xdata',linef(:,1),'ydata',linef(:,2));
set(follow(i),'xdata',coords(:,1),'ydata',coords(:,2));
end
axis equal;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [coords] = bushing(rr,x0,y0)
% Determine the coordinates of the anchor seat
% rr: radius of the shaft
% x0,y0: center coordinates of the shaft
d2r=pi/180;
theta=[360:-10:0]*d2r;
r=abs(rr);
rx=r*cos(theta);ry=r*sin(theta);
if rr<0,
rx=rx+x0;
ry=ry+y0;
coords=[rx' ry'];
return;
end;
rx1=rx/2;rx2=-rx1;
ry1=ry/2;ry2=-ry1;
r4=r+r/4;r3=r/3;
bx=[ 0 0 0 -r -r -r4 -r4 r4 r4 -r r r];
by=[r3 -r3 0 0 -r -r -r4 -r4 -r -r -r 0];
coords(:,1)=[bx rx rx1 rx2]'+x0;
coords(:,2)=[by ry ry1 ry2]'+y0;
hold off;
function [coords]=camfollow2(r,thickness,stroke,theta,npts)
%[coords]=camfollow2(r,thickness,stroke,theta,npts)
%This program gives coordinates for different type of cam followers
% Inputs:
% r : radius of roller if any
% thickness: thickness of rod
% stroke : the maximum stroke movement
% theta : angle that foller inclines
% npts : No. of points to draw the roller, default=30, npts=3 for point follower
% if thickness >2r, the orign will be transfered to the edge of circle or the point,
% otherwise, it is located at the center of the circle.
%Designed by D.S.Fon, BIME, NTU, dated June 6, 2002
t=thickness;tm=t/2;
s=stroke;
if stroke>0,s=stroke*1.5;end
th1=0;th2=2*pi;
loc=0;
if t<2*r,
th1=asin(abs(tm)/r);
th2=th2-th1;
else
t=2*r,th1=pi/2;th2=th2-th1;
loc=1;
end
nn=30;if nargin==5,nn=npts;end;
th=linspace(th1,th2,nn);
rx=r*cos(th);ry=r*sin(th);
coords=[rx' ry';s+r -tm;s+r tm;rx(1) ry(1)];
if loc,coords=trans4(coords,[r 0],1);end
if nargin==4 % draw a slant follower
coords=trans4(coords,theta,5);
end
function [coords]=frame(thickness,stroke,mode,theta)
% determine the coordinate of follower seat
% Inputs:
% thickness:the thckness of the follower shaft
% stroke:rise of the follower
% theta:slant angle of the follower, degrees
% mode:+1 for one side, -1 for another side
t=thickness*1.05;s=stroke*1.2;
t0=thickness+stroke;tx=t/2;tm=t/3;
if mode==1
coords=[t0 -tx];
tt=t0+tm;
while 1
coords=[coords;tt -tx;tt-tm -tx-tm;tt -tx];
tt=tt+tm;
if tt-t0>s, break;end
end
end
if mode==-1
coords=[t0 tx];
tt=t0+tm;
while 1
coords=[coords;tt tx;tt-tm tx+tm;tt tx];
tt=tt+tm;
if tt-t0>s, break;end
end
end
if nargin==4,
coords=trans4(coords,theta,5);
end
執行範例：
基圓半徑5公分之凸輪，其從動件為滾子，直徑為2公分。設此組凸輪之升程1.5公分，從動件之偏置量為1公分。分佈範圍為[100 180 270]度，升程與返程均使用簡諧運動型式，從動件為平板，左2公分，右3公分，板軸厚度為1公分，試將圓周以60均分繪出其從動件位置及對應之凸輪曲線。將上述資訊化為指令參數，執行結果如下：
>> camwork2(5,2,1.5,1,[100 180 270],[3 3],1,60)
其他凸輪動畫
Interactive Cam Mechanism Design
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