機動學第十三次作業

1.
開始時，假設要設計一組複式齒輪使其轉速比為125。其考慮的因素及過程如下：
i 決定組合數：複式齒列可用串聯多個組合之方式達到整體轉速比。設計總是由最簡單的結構開始，由於每組之轉速比以維持在10以內為佳，超過此值時則需考慮增加齒輪組數。而轉速比125很明顯可以看出開平方的結果比10大，而剛好125又是5的三次方，所以直接開立方。

ii 設驅動之小齒輪數最小為12齒(N2>=12)*，則依序可以得到對應大齒輪之齒數如下：

N3=5x12=60
N3=5x13=65
N3=5x14=70
N3=5x15=75

與網頁上例子不同的地方在於，網頁上的複式齒輪的齒數並非整數，所以要選最接近整數的，而作業中的齒輪剛好都為整數
故不用再加以討論

可選用60:12
65:13
70:14
75:15等

依此類推，選出60:12;60:12;60:12三組(或改用其他數字)


2.

我覺得讓我覺得做的作好的一次作業就是第五次作業吧!
因為當時為了這個作業把整個假日，兩天的時間都花在
這個作業上，雖然做出來的手並不是非常的漂亮，但是
其在實際功能的說明對於沒有學過機動學的人也是相當
清楚，我想寫出來程式的那一瞬間應該是那幾天最快樂
的事情吧！

機動學第十二次作業

B94611029 林軍耀
1.

5/31日曾全程來上課。

2.1~2一組標準全齒輪齒輪之徑節為8（亦可使用自設值），
齒數分別為30T與48T，其工作壓力角為20度（可為
14.5或25度，自選）。

設其徑節為8，
齒數為30T和48T
工作壓力角20度
使用程式function [c_ratio,c_length,ad,pc,pb,d2,d3,ag]=contact_ratio(pd,n2,n3, phi)

其中pd=8
n2=30
n3=48
phi=20
可得

接觸比c_ratio=1.7005
接觸長度c_length =0.6275
齒冠ad =0.1250
周節pc =0.3927
基周節pb = 0.3690
齒輪一節圓直徑d2 = 3.7500
齒輪二節圓直徑d3 = 6
齒輪一之接近角、遠退角及作用角ag = 10.4850 9.9211 20.4061
齒輪二之接近角、遠退角及作用角 6.5532 6.2007 12.7538

其中

齒輪一的節徑為30/8=3.75
齒輪二的節徑為48/8=6

齒輪一的基圓直徑為3.75*cos(20)=3.523
齒輪二的基圓直徑為 6*cos(20)=5.638

單位皆為吋


2.3此組齒輪是否會產生干涉現象？試列式證明之。

(N2²+2N2 x N3)sin²>= 4 + 4N3
滿足此式就不會產生干涉

利用講義中公式可以知道

(20^2+2*20*48)(sin(20))^2 >= 4*(1+48)
271.1 > 196

故可知，不會產生干涉現象

2.4

使用draw_gear和move2_gear
利用move2_gear(8,30,48,20,5)
產生動畫


附上draw_gear和move2_gear程式

function move2_gear(Dpitch,nn1,nn2,phi,omega1)
% move2_gear(Dpitch,nn1,nn2,phi,omega1)
% To draw a whole gear
% Inputs:
% Inputs:
% Dpitch:dimetral pitch
% nn1,nn2: no. of teeth for both gears
% phi:pressure angle, degrees
% omega1: angular velocity of gear 1
% Example move2_gear(10,15,20,20,10)
clf;
d2r=pi/180;delt=0.01;
[coord1,r1,rb1]=one_tooth(Dpitch,nn1,phi,360,0,0);
[coord2,r2,rb2]=one_tooth(Dpitch,nn2,phi,360,0,0);
st=180/nn2;if nn1+nn2>2*fix((nn1+nn2)/2),st=0;end
coord2=rotate2D(coord2,180+st,0,0);
xc1=coord1(:,1);yc1=coord1(:,2);
xc2=coord2(:,1);yc2=coord2(:,2);
height=max(r1,r2)*1.2;
ar=min(abs(r1),abs(r2));
coord=bushing(ar/5,0,0); % Get the coordinates of 1st bushing
xb1=coord(:,1)-r1;yb1=coord(:,2);
xb2=coord(:,1)+r2;yb2=coord(:,2);
coord=bushing(-r1,-r1,0);%Get the 1st pitch circle
xp1=coord(:,1);yp1=coord(:,2);
coord=bushing(-r2,r2,0);% Get the 2nd pitch circle
xp2=coord(:,1);yp2=coord(:,2);
plot(xb1,yb1,'r-');hold on;
plot(xb2,yb2,'k-');
plot(xp1,yp1,'r:');
plot(xp2,yp2,'k:');
plot([-r1,r2]',[0,0]','r:');
xx1=min([r1,r2])/2;phir=(90-phi)*d2r;
plot([0,0]',[-xx1*2,xx1*2]','b:');
plot([-xx1 xx1]',[-xx1*tan(phir), xx1*tan(phir)]','b:');

cir1=line('xdata',[],'ydata',[],'erasemode','xor','linewidth',1,'color','r');
cir2=line('xdata',[],'ydata',[],'erasemode','xor','linewidth',1,'color','k');
line1=line('xdata',[],'ydata',[],'erasemode','xor','linewidth',2,'color','r');
line2=line('xdata',[],'ydata',[],'erasemode','xor','linewidth',2,'color','k');
lx1=[0 -r1]';ly1=[0,0]';
lx2=[r2,0]';ly2=[0,0]';
axis([-2.5*r1 2.5*r2 -height height]);
axis equal;
title('Press Ctl-C to stop');
theta1=180;theta2=180;s1=omega1*delt/d2r;
while 1,
z1=rotate2D([xc1,yc1],theta1,-r1,0);
z2=rotate2D([xc2,yc2],theta2,r2,0);
L1=rotate2D([lx1,ly1],theta1,-r1,0);
L2=rotate2D([lx2,ly2],theta2,r2,0);
set(cir1,'xdata',z1(:,1),'ydata',z1(:,2)); % For 1st circle moving
set(cir2,'xdata',z2(:,1),'ydata',z2(:,2)); % For 2nd circle moving
set(line1,'xdata',L1(:,1),'ydata',L1(:,2)); % For 1st line
set(line2,'xdata',L2(:,1),'ydata',L2(:,2)); % For 2nd line
drawnow;
pause(1/s1); %Stop for a while so we can see the graph
theta1=theta1+s1;
theta2=theta2-s1*r1/r2;
if theta1>360, theta1=theta1-360;end; %Reverse the direction at bondary line
if theta2>360,theta2=theta2-360;end;
end




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [coords,rp,rb]=one_tooth(Dp,N,phi,range,x0,y0)
% [coords]=draw_gear(Dp,N,phi,range,x0,y0)
% To draw a whole gear
% Inputs:
% Dp: Diametrial pitch
% N: no of teeth in a gear
% phi: pressure angle, degrees
% range: the section range to be drawn
% Example one_tooth(10,10,20,360)
nn=10;
d2r=pi/180;
phir=phi*d2r;
rp=N/Dp/2;
pc=pi/Dp;
ra=rp+1/Dp;
rb=rp*cos(phir);
rd=rp-1.25/Dp;
thpb=pc/rp;% angle respect to one pitch
tp=pc/2;
rr=linspace(rb,ra,nn)';
invphi=tan(phir)-phir;
for i=1:nn
beta=acos(rp/rr(i)*cos(phir));
tt(i)=(tp/rp/2+invphi-tan(beta)+beta);
end
coord1=[rr.*cos(tt') rr.*sin(tt')];
coord3=reverse(coord1);
th1=linspace(thpb/2,thpb/2-tt(nn),nn)';
coord0=[cos(th1) sin(th1)]*rd;
th2=linspace(tt(nn),-tt(nn),nn)';
coord2=[cos(th2) sin(th2)]*ra;
coord4=reverse(coord0);
coord=[coord0;coord1;coord2;coord3;coord4];
theta=thpb/d2r;
coords=[];i=0;
while i coord1=rotate2D(coord,-i,x0,y0);
coords=[coords;coord1];
i=i+theta;
end

function [B]=reverse(A)
nn=length(A); B=A;
for i=1:nn,B(i,:)=A(nn-i+1,:);end
B=[B(:,1) -B(:,2)];

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [coords] = bushing(rr,x0,y0)
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;
%%%%%%%%%%%%%%%%%%%
function [coords]=rotate2D(coord,theta,x0,y0)
th=theta*pi/180; c=cos(th);
s=sin(th);fact=[c s;-s c];
coords=coord*fact;
coords(:,1)=coords(:,1)+x0; coords(:,2)=coords(:,2)+y0;
function [coords]=draw_gear(Dp,N,phi,range,x0,y0)
% [coords]=draw_gear(Dp,N,phi,range,x0,y0)
%To draw a whole gear % Inputs: % Dp: Diametrical pitch
% N: no of teeth in a gear % phi: pressure angle, degrees
% range: the section range to be drawn % x0,y0: the location of the gear center
% Example [coords]=draw_gear(10,15,20,360,0,0)
[coord,theta,rp,rb]=tooth(Dp,N,phi); coords=[];
i=0; while i coord1=rotate2D(coord,-i,x0,y0);
coords=[coords;coord1];
i=i+theta;
end
plot(coords(:,1),coords(:,2));hold on;
[coord]=bushing(rp/8,x0,y0);
plot(coord(:,1),coord(:,2),'b-');
[coord]=bushing(-rp,x0,y0);
plot(coord(:,1),coord(:,2),'r:');
[coord]=bushing(-rb,x0,y0);
plot(coord(:,1),coord(:,2),'b:');
axis equal;




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [coords,theta,rp,rb]=tooth(Dp,N,phi)
% Example tooth(10,10,20)
nn=10;
d2r=pi/180;
phir=phi*d2r;
rp=N/Dp/2;
pc=pi/Dp;
ra=rp+1/Dp;
rb=rp*cos(phir);
rd=rp-1.25/Dp;
thpb=pc/rp;% angle respect to one pitch
tp=pc/2;
rr=linspace(rb,ra,nn)';
invphi=tan(phir)-phir;
for i=1:nn
beta=acos(rp/rr(i)*cos(phir));
tt(i)=(tp/rp/2+invphi-tan(beta)+beta);
end
coord1=[rr.*cos(tt') rr.*sin(tt')];
coord3=reverse(coord1);
th1=linspace(thpb/2,thpb/2-tt(nn),nn)';
coord0=[cos(th1) sin(th1)]*rd;
th2=linspace(tt(nn),-tt(nn),nn)';
coord2=[cos(th2) sin(th2)]*ra;
coord4=reverse(coord0);
coords=[coord0;coord1;coord2;coord3;coord4];
theta=thpb/d2r;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [B]=reverse(A)
nn=length(A); B=A;
for i=1:nn,B(i,:)=A(nn-i+1,:);end
B=[B(:,1) -B(:,2)];

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [coords] = bushing(rr,x0,y0)
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;

%%%%%%%%%%%%%%%%%%%%%%%%

function [coords]=rotate2D(coord,theta,x0,y0)
th=theta*pi/180;
c=cos(th);s=sin(th);fact=[c s;-s c];
coords=coord*fact;
coords(:,1)=coords(:,1)+x0;
coords(:,2)=coords(:,2)+y0;

機動學第十一次作業

b94611029林軍耀

1.本人本週(5/24)有來上課。

2.
使用程式
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;



因為等加速度的位移軌跡
就是拋物線
所以在pattern中輸入2


返程等速
plot_dwell(0:10:360,5,[2 1],[100 200 260]);


返程等加速度
plot_dwell(0:10:360,5,[2 2],[100 200 260]);


返程簡諧
plot_dwell(0:10:360,5,[2 3],[100 200 260]);


返程擺線
plot_dwell(0:10:360,5,[2 4],[100 200 260]);


返程多項式
plot_dwell(0:10:360,5,[2 5],[100 200 260]);



求出各種返程型態之圖形

其中
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

使用function [y,yy,yyy]=dwell(ctheta,range,pattern)

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        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));

3.

利用[x y]=pincam([0:10:360],15,s,0,10,[100 200 260],[2 1],-1)
以及[s]=drawcam(15,y,1)
作出圖形
與第四小題的動畫一同呈現

4.

利用上面的程式
[s]=drawcam(15,y,1)

再用
for i=1:1:360
x2=s(:,1);
y2=s(:,2);
x3=x2*cosd(i)-y2*sind(i);
y3=x2*sind(i)+y2*cosd(i);
axis([-30 30 -30 30]);
plot(x3,y3);
pause(0.01);
axis([-30 30 -30 30]);
axis equal;
grid on;
end;

作出動畫


附件1
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-')

附件2
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;
grid on;

機動學第十次作業

本週(5/17)有來上課。

10.2



當一桿以桿上一點M做等角速度迴轉運動時，端點之速度為端點至M點之距離乘以角速度w
若M點為桿外一點，V=仍然為端點至M點之距離*ω
a=端點至M之距離*ω*ω

假若M以V等速水平移動
θ為桿與水平面之夾角
圖請見網頁
則P點此時之速度為

V+r*ω*sinθ - r*ω*cosθ

加速度維持不變



假若M又具有加速度a)時
則P點之速度假設與上述相同，
則加速度為

V+r*ω^2*cosθ + r*ω^2*sinθ


10.3

function sldwork(R,L,e)
th1=slider_limit1(R,L,e);
th2=90;
angle=linspace(th1,th2,100);
d=slider_solved(angle,R,L,e,1);
x=R*cosd(angle);
y=R*sind(angle);

for i=1:100
hold on
plot([0,x(i),d(i)],[0,y(i),e],'linewidth',4);
plot([d(i)-2,d(i)+2,d(i)+2,d(i)-2,d(i)-2],[e-2,e-2,e+2,e+2,e-2]);
plot([x(i),0],[y(i),e-d(i)*(y(i)-e)/(x(i)-d(i))],'b.','linewidth',8)
plot([d(i),d(i)],[0,y(i)*d(i)/x(i)],'b.')
axis equal;
axis ([-10 50 -10 50]);
pause(0.03);
end

程式如上

動畫中出現的兩個點為瞬心
還有三個瞬心分別出現在旋轉結以及滑塊與桿之連結點上
第六個瞬心在無限遠的地方
為滑塊的瞬心

機動學第九次作業

本人本週(5/3)有來上課

b94611029 林軍耀


由於偏置量會影響滑桿之活動範圍
所以都曲柄如果要轉360度時
其連桿與曲桿及滑塊三者必須構成
某種三角形的關係

連桿長-曲桿長 大於 偏置量

由課本4-35頁的slider_limit程式可知道，
由於偏置量e大於連桿與曲桿之差，
所以會產生極限點
在theta = 6.9199度時，有右極限角
　
令R=29+10=39
　L=39+5=44
　e=10
參考下圖


以及利用move_sldpaths([10,R,L,10],0,0,3,10,10,0,1,0,4,100)以及
slider_draw(39,44,10)做出動畫