1、3550改进正弦70100120二、 计算流程框图三、 建立数学模型1. 从动件运动规律方程首先,由于设计凸轮轮廓与凸轮角速度无关,所以不妨设凸轮运动角速度为w = 1rad/s。(1)推程运动规律 (0 90 s= v= a=式中:h=65mm,0=/2(2)远休程运动规律 (90 190s = 65mmv = 0a = 0(3)回程运动规律 (190 240 (190 196.25(196.25 233.75(233.75回程运动中的速度和加速度为位移对时间t的倒数:(4)近休程运动规律 (240 360s = 02. 从动件位移、速度、加速度线图(1)位移线图(2)速度线图(3)加速度线
2、图(4)位移、速度、加速度线图MATLAB源程序% 已知条件h = 65; %mmphi_0 = 90./180*pi; %radalpha_up_al = 35./180*pi; %升程许用压力角phi_00 = 50./180*pi;alpha_down_al = 70./180*pi; %回程许用压力角phi_s = 100./180*pi;phi_ss = 120./180*pi;w = 1; % 绘制从动件位移、速度、加速度线图% 推程阶段t_up = 0 : 0.5 : 90;t_up1 = t_up./180*pi;syms t_up1 phi_up s_up v_up a_up
3、phi_up = w.*t_up1;s_up = h./2.*(1 - cos(pi.*phi_up./phi_0);v_up = diff(s_up,t_up1);a_up = diff(v_up,t_up1);s_up1 = double(subs(s_up,t_up./180*pi);v_up1 = double(subs(v_up,t_up./180*pi);a_up1 = double(subs(a_up,t_up./180*pi);% 远休程t_s = 90 : (90+100);t_s1 = t_up./180*pi;s_s(1:201) = h;v_s(1:201) = 0;a
4、_s(1:% 回程阶段1t_down1 = (90+100) : (90+100+50/8);t_down11 = t_down1./180*pi;syms t_down11 phi_down1 s_down1 v_down1 a_down1phi_down1 = w.*t_down11;s_down1 = h - h./(4+pi).*(pi.*(phi_down1 - phi_0 - phi_s)./phi_00 - .sin(4.*pi.*(phi_down1 - phi_0 - phi_s)./phi_00)./4);v_down1 = diff(s_down1,t_down11);a
5、_down1 = diff(v_down1,t_down11);s_down11 = double(subs(s_down1,t_down1./180*pi);v_down11 = double(subs(v_down1,t_down1./180*pi);a_down11 = double(subs(a_down1,t_down1./180*pi);% 回程阶段2t_down2 = (90+100+50/8) : (90+100+7*50/8);t_down22 = t_down2./180*pi;syms t_down22 phi_down2 s_down2 v_down2 a_down2
6、phi_down2 = w.*t_down22;s_down2 = h - h./(4+pi).*(2+pi.*(phi_down2 - phi_0 - phi_s)./phi_00 - 9.*sin(pi./3 + 4.*pi.*(phi_down2 - phi_0 - phi_s)./(3.*phi_00)./4);v_down2 = diff(s_down2,t_down22);a_down2 = diff(v_down2,t_down22);s_down22 = double(subs(s_down2,t_down2./180*pi);v_down22 = double(subs(v_
7、down2,t_down2./180*pi);a_down22 = double(subs(a_down2,t_down2./180*pi);% 回程阶段3t_down3 = (90+100+7*50/8) : (90+100+50);t_down33 = t_down3./180*pi;syms t_down33 phi_down3 s_down3 v_down3 a_down3phi_down3 = w.*t_down33;s_down3 = h - h./(4+pi).*(4+pi.*(phi_down3 - phi_0 - phi_s)./phi_00 - sin(4.*pi.*(ph
8、i_down3 - phi_0 - phi_s)./phi_00)./4);v_down3 = diff(s_down3,t_down33);a_down3 = diff(v_down3,t_down33);s_down33 = double(subs(s_down3,t_down3./180*pi);v_down33 = double(subs(v_down3,t_down3./180*pi);a_down33 = double(subs(a_down3,t_down3./180*pi);% 近休程t_ss = (90+100+50) : 360;s_ss(1:241) = 0;v_ss(1
9、:a_ss(1:% 绘图位移t = t_up t_s t_down1 t_down2 t_down3 t_ss;phi = w .* t ./ 180 .*pi;s = s_up1 s_s s_down11 s_down22 s_down33 s_ss;v = v_up1 v_s v_down11 v_down22 v_down33 v_ss;a = a_up1 a_s a_down11 a_down22 a_down33 a_ss;figure(Name,从动件位移-时间线图);plot(t,s,klinewidth,1.0);grid on;title(xlabel(转角phi / 度yl
10、abel(位移h/mm% 绘图速度从动件速度-时间线图plot(t,v,速度v/mm*s-1% 绘图加速度从动件加速度-时间线图plot(t,a,加速度a/mm*s-23. 绘制ds/d线图并确定基圆半径和偏距(1) 绘制ds/d线图及源程序MATLAB源程序:% 绘制ds/dphi-s线图,确定基圆半径和偏距ds_dphi = v ./ w;凸轮ds/dphi - s线图plot(ds_dphi,s,1.5);hold on;axis(-150 150 -70 70);ds/dphi / (mm*s-2)s/mm% 三条临界线x = linspace(-150,150,301);k_up =
11、 tan(pi/2 - alpha_up_al);y_up = k_up.*x - 66;plot(x,y_up,k_down = - tan(pi/2 - alpha_down_al);y_down = k_down.*x - 24.7;plot(x,y_down,x0 = linspace(0,150,151);k0 = - tan(alpha_up_al);y0 = k0.*x0;plot(x0,y0,-% 由图像选取凸轮基圆半径为r0 = sqrt(232 + 342) = 41 mm,偏距e = 23mmplot(23,-34,orr0 = 41;e = 23;plot(linspa
12、ce(0,23,10),linspace(0,-34,10),r,linspace(0,23,10),linspace(-34,-34,10),linspace(23,23,10),linspace(0,-34,10),(2) 确定基圆半径和偏距在凸轮机构的ds/d-s线图里再作斜直线Dt-dt与升程的ds/d-s曲线相切并使与纵坐标夹角为升程许用压力角,则Dt-dt线的右下方为选择凸轮轴心的许用区。作斜直线Dt-dt与回程的ds/d-s曲线相切,并使与纵坐标夹角为回程的许用压力角,则Dt线的左下方为选择凸轮轴心的许用区。考虑到升程开始瞬时机构压力角也不超过许用值,自B0点作限制线B0-d0与纵坐标夹角为升程,则这三条直线的围成的下方区域为为选取凸轮中心的许用区。由图可取基圆半径r0=41mm,偏距e=23mm,s0=34mm。4. 绘制凸轮理论轮廓压力角、曲率半径线图(1) 压力角、曲率半径数学模型压力角计算公式:曲率半径计算公式:其中:(2) MATLAB程序% 凸轮理论轮廓压力角和曲率半径线图s0 = 34;% 压力角t = t_up t_s t_down1 t_dow
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