哈工大机械原理大作业任务书连杆机构参考模板
哈尔滨工业大学机械原理大作业-连杆机构运动分析
%打印图像
figure(1);
plot(fi1,sF);
title('位移变化曲线');
figure(2);
plot(fi1,vF);
title('速度变化曲线');
figure(3);
plot(fi1,aF);
title('加速度变化曲线');
六、计算结果
图8:推杆位移变化曲线
图9:推杆速度变化曲线
xD=400;yD=500;vDx=0;vDy=0;aDx=0;aDy=0;
xK=0;yK=600;vKx=0;vKy=0;aKx=0;aKy=0;
l1=150;l2=600;l3=500;lBE=480;l4=600;
fi5=pi;
n1=50;
w1=2*pi*n1/60;
fi1=linspace(0,2*pi,1000);
aEy(i)=aBy(i)-w2(i)^2*lBE*sin(fi2(i))+a2(i)*lBE*cos(fi2(i));
%求F点的运动参数
A1(i)=(yE(i)-yK)*cos(fi5)-(xE(i)-xK)*sin(fi5);
fi4(i)=fi5-asin(A1(i)/l4);
xF(i)=xE(i)+l4*cos(fi4(i));
(2)速度和加速度分析
B点的速度
B点的加速度
2、由“RRRⅡ级杆组”,已知B点和D点的运动参数,可求得构件2、构件3的运动参数
图5
D点的坐标方程
D点的速度
D点的加速度
(1)位置方程
构件2、构件3的长度
先求出 和
其中
则可求得
哈工大机械设计大作业平面连杆机构设计说明书
机械设计基础大作业计算说明书题目:平面连杆机构设计学院:材料学院班号:学号:姓名:日期: 2014年9月30日哈尔滨工业大学机械设计基础大作业任务书题目:平面连杆机构设计设计原始数据及要求:()目录1设计题目2设计原始数据3设计计算说明书3.1计算极位夹角3.2设计制图3.3验算最小传动角4参考文献1 设计题目平面连杆机构的图解法设计2 设计原始数据设计一曲柄摇杆机构。
已知摇杆长度,摆角,摇杆的行程速比系数,要求摇杆靠近曲柄回转中心一侧的极限位置与机架间的夹角为,试用图解法设其余三杆的长度,并检验(测量或计算)机构的最小传动角。
()3 设计计算说明书3.1 计算极位夹角极位夹角代入数值3.2 设计制图3.2.1 在图纸上取一点作为点,从点垂直向上引出一条长为的线段,终点为;3.2.2 从点在左侧引出一条与夹角为的射线;3.2.3 以点为圆心,以为半径画圆,与射线交于点;3.2.4 分别从、两点向下引两条射线,射线与夹角为,两射线交于点,点即为曲柄的回转中心;3.2.5 以点为圆心以为半径画圆;3.2.6 过点向左侧引出一条射线,射线与夹角,与圆交于点;3.2.7 连接,并量取其长度,以为半径画圆,直线,与圆的交点分别为,;3.2.8 在图中量取,,3.3 验算最小传动角3.3.1 在处根据余弦定理3.3.2 在处根据余弦定理所以最小传动角4 参考文献[1]宋宝玉,王瑜,张锋主编.机械设计基础.哈尔滨:哈尔滨工业大学出版,2010.[2]王瑜主编.机械设计基础大作业指导书.哈尔滨:哈尔滨工业大学出版社,2014.。
哈工大机械原理大作业连杆
Harbin Institute of Technology机械原理大作业一课程名称: 机械原理 设计题目: 连杆机构运动分析 院 系: 机电工程学院 班 级: 设 计 者: 学 号: 指导教师: 设计时间:1.运动分析题目(11)在图所示的六杆机构中,已知:AB l =150mm, AC l =550mm, BD l =80mm, DE l =500mm,曲柄以等角速度1w =10rad/s 沿逆时针方向回转,求构件3的角速度、角加速度和构件5的位移、速度、加速度。
2.机构的结构分析2.1建立以点A 为原点的固定平面直角坐标系A-x, y,如下图: 2.2机构结构分析该机构由Ⅰ级杆组RR (原动件1)、Ⅱ级杆组RPR (杆2及滑块3)和Ⅱ级杆组RRP (杆4及滑块5)组成。
3.建立组成机构的各基本杆组的运动分析数学模型3.1原动件1(Ⅰ级杆组RR )由图所示,原动件杆1的转角a=0-360°,角速度1w =10rad/s ,角加速度1a =0,运动副A 的位置坐标A x =A y =0,速度(A ,A),加速度(A ,A ),原动件1的长度AB l =150mm 。
求出运动副B 的位置坐标(B x , B y )、速度(B ,B )和加速度(B ,B )。
3.2杆2、滑块3杆组(RPR Ⅱ级杆组)已出运动副B 的位置(B x , B y )、速度(B ,B )和加速度(B ,B ),已知运动副C 的位置坐标C x =0, C y =550mm,速度,加速度,杆长AC l =550mm 。
求出构件2的转角b,角速度2w 和角加速度2a . 3.3构件二上点D 的运动已知运动副B 的位置(B x , B y )、速度(B ,B )、加速度(B ,B ),已经求出构件2的转角b ,角速度2w 和角加速度2a ,杆BD 的长度BD l =80mm 。
根据Ⅰ级杆组RR 的运动分析数学模型求出点D 的位置坐标(D x ,D y )、速度(D ,D )和加速度(D ,D )。
哈工大-机械原理大作业-连杆机构运动分析
机械原理大作业(一)作业名称:连杆机构运动分析设计题目: 20院系:英才学院班级: XXXXXXX设计者:邵广斌学号: XXXXXXXXXX指导教师:林琳设计时间: 2013年05月19日哈尔滨工业大学机械设计1.运动分析题目如图所示机构,已知机构各构件的尺寸为150AB mm =,97β=︒,400BC mm =,300CD mm =,320AD mm =,100BE mm =,230EF mm =,400FG mm =,构件1的角速度为110/rad s ω=,试求构件2上点F 的轨迹及构件5上点G 的位移、速度和加速度,并对计算结果进行分析。
2. 机构分析该机构由原动件AB (Ⅰ级杆组)、BCD (RRR Ⅱ级杆组)和FG (RRP Ⅱ级杆组)组成。
3. 建立坐标系如图3,建立以定点A 为原点的平面直角坐标系A-xy 。
图1 运动机构结构图4. 运动分析数学模型4.1 原动件AB原动件AB 的转角: 10~2ψπ= 原动件AB 的角速度:110/rad s ω=原动件AB 的角加速度: 10α= 运动副A 的位置坐标: 0A x = 0A y =运动副A 的速度: 0xA v = 0yA v = 运动副A 的加速度: 0xA a = 0yA a =原动件AB 长度:150AB l mm =运动副B 的位置坐标: 1B A AB x x l cos ψ=+1B A AB y x l sin ψ=+运动副B 的速度: 11 xB xA AB v v l sin ωψ=-11 yB yA AB v v l cos ωψ=+运动副B 的加速度: 2 1111 xBxA AB AB a a l cos l sin ωψαψ=--21111yB yA AB AB a a l sin l cos ωψαψ=-+4.2 RRR Ⅱ级杆组BCD运动副D 的位置坐标: 320D x mm = 0D y = 运动副D 的速度: 0xD v = 0yD v = 运动副D 的加速度: 0xD a = 0yD a = 杆BC 长度: 400BC l mm = 杆CD 长度:300CD l mm =BC 相对于x 轴转角:200ψ=其中02BC D B A l x x =-() 0 2 BC D B B l y y =-()2220B B C C l C l D l D =+- 222())(BDD B D B l x x y y =-+- CD 相对于x 轴转角: 3C DC Dy y arctanx x ψ-=-求导可得BC 角速度2ω、角加速度2α以及CD 角速度3ω、角加速度3α。
哈工大机械原理大作业连杆机构9
机械原理大作业大作业一:连杆机构运动分析学生姓名:学号:指导教师:丁刚完成时间:机电工程学院机械设计系制二〇一八年四月连杆机构运动分析1题目(9)图1 设计题目在图1所示的机构中,已知l AB=60mm,l BC=180mm,l DE=200mm,l CD=120mm,l EF=300mm,h=80mm,h1=85mm,h2=225mm,构件1以等角速度ω1=100rad/s 转动。
求在一个运动循环中,滑块5的位移、速度和加速度曲线。
2分析结构1、杆1为RR主动件,绕A以ω1 转动,自由度1.2、4杆和滑块5为RRP II级杆组.,自由度0.3、2,3杆组成II级杆组RRR,自由度0.总共自由度为F=5*3-2*7=1 .由上述的杆组类型,确认出所需运动分析数学模型:同一构件上的点、RRP、RRR。
3.杆组法对平面连杆机构进行运动分析3.1对主动件杆1 RR I级构件的分析主动杆1转角:φ= [0°,360°) δ=0°,则φ’=ω1=100 rad/s角加速度φ’’=0.已知h2=225mm, h=80mm, l AB=60mm 所以A(225mm,80mm)A点速度(0,0),加速度(0,0)B点位置(x A+l AB*cos(φ), Y A+l AB*sin(φ))B点速度(-l AB*sin(φ), l AB*cos(φ)),加速度(-l AB*cos(φ), -l AB*sin(φ))3.2RRRII 级杆组分析(模型参考教材P37-38)图3 如图所示两个构件组成II 级杆组。
已知了B 的位置(x B ,y B )= (x A +l AB *cos(φ), Y A +l AB *sin(φ)),速度(x ’B ,y ’B ) 和加速度(x ’’B ,y ’’B ), 已知运动副D (0,0), 还可知,x ’D =y ’D =0, x ’’D =y ’’D =0. l BC =180 mm, l CD = 120mm所以,x c =x D +l CD *cos(φi)= x B +l BC *cos(φj) y c =x D +l CD *sin(φi)= x B +l BC *sin(φi) 对于φ的求解: A 0=2*l CD (x B -x D ) B 0=2*l BC (y B -y D ) C 0=l CD 2+ l BD 2- l BC 2为了保证机构的装配正常:l BD ≤l CD + l BC AND l BD ≥Abs (l CD - l BC )可求3杆的转角φi=2*arctan((B 0±sqrt (A 02 + B 02- C 02))/(A 02+ C 02)),角速度w3=φi ’和角加速度α3= φi ’’3.3 同一构件上的点(模型参考书P35-36)Φiφjφi已知D(0,0),速度(0,0),加速度(0,0),3杆转角φi 角速度φi’角加速度φi’’,Φi和它的导数在3.2都有体现LDE= 200mm可求出E的坐标,速度,加速度.x E =x c+lCE*cos(φi)y E =x C+lCE*sin(φi)同样地,速度、加速度通过求导即可得出算式,可以编出程序。
哈工大机械原理大作业——连杆——8号
Harbin Institute of Technology机械原理大作业设计说明书课程名称:机械原理设计题目:连杆机构设计院系:机电工程学院班级:设计者:学号:指导教师:设计时间:哈尔滨工业大学运动分析题目——第二十七题如图所示机构,已知机构各构件的尺寸为AB=280mm ,BC=350mm ,CD=320mm, AD=160mm, BE=175mm, EF=220mm, xc=25mm, yg=80mm, 构件一的角速度为w1=10rad/s, 试求构件2上点F 的轨迹及构件5的角位移,角速度和角加速度,一、建立坐标系以A 点为原点建立如图所示坐标系x-y ,如上图所示二、机构结构分析该机构可认为由一个I 级杆组RR (杆AB )、II 级杆组RRR (杆2、3)、II 级杆组RPR (杆5及滑块4)组成。
如下图所示。
I 级杆组RRXII 级杆组RRRII 级杆组RPR三、确定已知参数和设计流程 一)AB (I 级杆组RR ) 运动副A 的位置坐标为 AB=280mmB 的位置坐标二)BCD 杆(II 级杆组RRR ) 运动副D 的位置坐标 BC=350mm, CD=320mm.0,0,0,0,0,0x ======yA xA yA xA A A a a v v y 加速度速度ϕϕϕϕϕϕsin 280,cos 280,cos 280,sin 280,sin 280,cos 280-=-==-===yB xB yB xB B B a a v v y x 加速度速度.0,0,0,0,0,0======yD xD yD xD D D a a v v y x 加速度速度由余弦定理,可求得)(2cos *222πϕ+-+=AB AD AB AD BD 由正弦定理得BD AB rc ADB )2sin(sina ϕπ+=∠由余弦定理得DB DC BC BD CD CDB ⨯⨯-+=∠2arccos222 )-sin(x BDC ADB CD ∠-∠=π )cos(y CDB ADB CD AD ∠-∠-+=π由此可以求出运动副C 的位置坐标(X,Y ),速度(vx ,vy )和加速度(ax,ay),杆BC 与x 轴的夹角,杆BC 的角速度,杆BC 的角加速度,杆CD 与x 轴的夹角,角速度,角加速度。
机械原理大作业(一)--连杆(34题)
机械原理大作业(一)作业名称:连杆机构运动分析设计题目:(34)题院系:船舶学院班级: 1213101设计者:学号:哈尔滨工业大学机械设计一、运动分析题目如图所示机构,已知机构各构件的尺寸为BF=200mm,EF=1.25BF,DE=1.13BF,EH=0.85BF,HF=0.65BF,CH=0.81BF,GC=1.56BF,BD=0.58BF,BG=1.85BF,GD=1.6BF,构件1的角速度为W1=10rad/s,试求构件2上点H的轨迹及构件5的角位移,角速度,角加速度,并对计算结果进行分析。
机构结构分析:二、机构的结构分析及基本杆组划分机构各构件都在同一平面内运动,活动构件数n=5,=7,=0则机构的自由度为:F=3×n-2×-1×=3×5-2×7-0=12.基本杆组划分(1)去除虚约束和局部自由度本机构中无虚约束或局部自由度。
(2)拆杆组。
从远离原动件(即杆1)进行拆分,就可以得到由杆4,5 组成的RRRⅡ级杆组GCH,2,3 组成的RRRⅡ级杆组EDF,最后剩下Ⅰ级机构杆1。
(3)确定机构的级别 由(2)知,机构为Ⅱ级机构三、各基本杆组的运动分析数学模型1)Ⅰ级杆组BF (原动件)在Ⅰ级杆组BF 中,即已知构件上B 点的运动参数,求同一构件上F 点(回转副)的运动参数。
调用Ⅰ级机构子程序即可求解 ①位置分析 由图可得F 点的矢量方程F B BF r r l =+x,y 轴上的投影坐标方程为cos sin F B BF BF F B BF BF x x l y y l ϕϕ=+⋅⎫⎬=+⋅⎭(1)②速度和加速度分析 将式(1)对时间t 求导即可得出速度方程:cos sin BF BF BF BF FF B BF FF B BF dx x x l dtdy y x l dt ϕϕϕϕ⋅⋅⋅⋅⋅⋅⎫⎪==-⎪⎬⎪==+⎪⎭ (2)2222cos sin sin cos BF BF BF BF BF BF BF FF B BF BF BF F F B BFBF d x x x l l dtd y y y l l dt ϕϕϕϕϕϕϕϕ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⎫⎪==--⎪⎬⎪==--⎪⎭(3)其中因为设B 为原点:B x =0;B y =0 ;B x ⋅=0 ;B y ⋅=0 ;B x ⋅⋅=0 ;B y ⋅⋅=0由上(1)(2)(3)方程可求出F 点的位移,速度,加速度2)RRR Ⅱ级杆组DEF 分析,求出F 点的角位移,角速度,角加速度上面1)中已求得F 点的位移,速度,加速度。
哈工大机械原理大作业23
机械原理大作业院系:船舶与海洋工程学院专业:机械设计制造及自动化班级:1413104学号:141310423姓名:田笑哈尔滨工业大学(威海)作业1 连杆机构运动分析1.运动分析题目2.对机构进行结构分析该机构由一级杆组RR(原动件1)、二级杆组RRR(杆2、杆3;杆4、杆5),二级干组RRP(杆6,滑块7)组成。
如图所示:一级杆组:二级杆组RRR(杆2、杆3):- 1 -二级杆组RRR(杆4、杆5):二级杆组RRP(杆6、滑块7):3.建立坐标系建立以A点为坐标原点的坐标系- 2 -- 3 -4.建立组成各基本杆组的运动分析数学模型(1)一级杆组运动分析: 如一级杆组图所示,设AB 杆与X 轴夹角为p,AB 逆时针转动。
位置分析:x _B=x_A+l.*cos(p);y_B=y_A+l.*sin(p); 速度分析:v_Bx=l.*sin(p).*-1.*w; v_By=l.*cos(p)*w;加速度分析:a_Bx=w^2*l.*cos(p).*-1-e*l.*cos(p); a_By=w^2*l.*sin(p).*-1+e*l.*sin(p); (2)二级杆组(RRR 型)运动分析:如二级杆组图(杆2、杆3)所示,设CB 与X 轴夹角为p_2。
角位移分析:d=sqrt((x_D-x_B).^2+(y_D-y_B).^2);b=atan((y_D-y_B)./(x_D-x_B));r=acos((d.^2+l_2^2-l_3^2)./(2*l_2.*d));x_C=x_B-l_2*cos(p_2); y_C=y_B-l_2*sin(p_2); p_2=b+M.*r;p_3=atan((y_C-y_D)./(x_C-x_D));角速度分析:x _D)-y_B)(x _C -(y_C -x _B)-y_D)(x _C -(y_C y_D)-v_By)(y_C -(v_Dy x _D)-v_Bx )(x _C -(v_Dx w_2+=x _D)-y_B)(x _C -(y_C -x _B)-y_D)(x _C -(y_C y_B)-v_By)(y_C -(v_Dy x _B)-v_Bx )(x _C -(v_Dx w_3+=角加速度分析:E=a_Dx-a_Bx+w_2^2*(x_C-x_B)-w_3^2*(x_C-x_D); F=a_Dy-a_By+w_2^2*(y_C-y_B)-w_3^2*(y_C-y_D); y_B)-x _D)(y_C -(x _C -y_D)-x _B)(y_C -(x _C y_D)-F(y_C x _D)-E(x _C e_2 +=y_B)-x _D)(y_C -(x _C -y_D)-x _B)(y_C -(x _C y_B)-F(y_C x _B)-E(x _C e_3 +=5.计算编程及输出结果Matlab 编程作图(具体程序见附录):(1)E 点的轨迹- 4 -0.10.150.20.250.30.350.40.450.50.550.6-0.0500.050.10.150.20.250.3x/mmy /m mE 点的轨迹(2)构件5的角位移、角速度和角加速度角位移:角速度:- 5 - 01234567-15-10-5510θ/radw /r a d /s构件5的角速度角加速度:1234567-100-50050100150200250300θ/radε/r a d /s 2构件5的角加速度附录:%一级杆组运动分析子程序%x_B=x_A+l.*cos(p);y_B=y_A+l.*sin(p);v_Bx=l.*sin(p).*-1.*w;v_By=l.*cos(p)*w;a_Bx=w^2*l.*cos(p).*-1-e*l.*cos(p);a_By=w^2*l.*sin(p).*-1+e*l.*sin(p);%二级杆组运动分析子程序%d=sqrt((x_D-x_B).^2+(y_D-y_B).^2);b=atan((y_D-y_B)./(x_D-x_B));f=(d.^2+l_2^2-l_3^2)./(2*l_2.*d);r=acos((d.^2+l_2^2-l_3^2)./(2*l_2.*d));p_2=b+M.*r;x_C=x_B-l_2.*cos(p_2);y_C=y_B-l_2.*sin(p_2);w_3=((0-v_Bx).*(x_C-x_B)+(0-v_By).*(y_C-y_B))./((y_C-y_D).* (x_C-x_B)-(y_C-y_B).*(x_C-x_D));%角速度%w_2=((0-v_Bx).*(x_C-x_D)+(0-v_By).*(y_C-y_D))./((y_C-y_D).* (x_C-x_B)-(y_C-y_B).*(x_C-x_D));E=0-a_Bx+w_2.^2.*(x_C-x_B)-w_3.^2.*(x_C-x_D);F=0-a_By+w_2.^2.*(y_C-y_B)-w_3.^2.*(y_C-y_D);e_2=(E.*(x_C-x_D)+F.*(y_C-y_D))./(x_C-x_B).*(y_C-y_D)-(x_C-x_D).*(y_C-y_B);e_3=(E.*(x_C-x_B)+F.*(y_C-y_B))./(x_C-x_B).*(y_C-y_D)-(x_C-x_D).*(y_C-y_B);%角加速度%p_3=atan((y_C-y_D)./(x_C-x_D));%角位移%- 6 -- 7 -大作业2 凸轮机构设计1.设计题目如图2-1所示直动从动件盘形凸轮机构,其原始参数见表2-1,据此参数设计该凸轮机构。
哈工大机械原理大作业连杆机构运动分析完美满分版哈尔滨工业大学
连杆机构运动分析说明书院(系)机电工程学院专业机械设计制造及其自动化姓名李乾学号1130810904班号1308109指导教师唐德威、赵永强日期2015年6月20日哈尔滨工业大学机电工程学院2015年6月一、题目如图1所示机构,已知机构各构件的尺寸为l AB=200mm,l BD=700mm,l AC=400mm,l AE=800mm,构件1的角速度为ω1=10rad/s,试求构件2上点D的轨迹及构件5的角位移、角速度和角加速度,并对计算结果进行分析。
(题中构件尺寸满足l BD-l AB<l AE<l BD+l AB)。
图 1 机构运动简图二、建立数学模型分析1.建立坐标系建立以点A为原点的平面直角坐标系A-x,y,如图2所示图 2 建立坐标系2.对机构进行结构分析该机构由Ⅰ级机构AB、两个RPRⅡ级基本杆组BCD、ED组成。
杆组拆分结果如图3、图4、图5所示。
图 3 Ⅰ级杆组AB图 4 RPRⅡ级基本杆组BCD图 5 RPRⅡ级基本组DE3.确定已知参数和求解流程(1)原动件AB(I级杆组)已知原动件1的转角φ=0~360°运动副A的运动参数x A=0y A=0原动件AB的长度l AB = 200mm代入I级杆组子程序,得到运动副B的位置坐标(x B,y B)根据《机械原理》第三版书中第36页的公式推导可知:A,B两点坐标在x轴,y轴上投影,得方程x B = x A+l AB*cosφy B = y A+l AB*sinφ(2)BCD(RPR II级杆组)已知运动副B的位置坐标(x B,y B)运动副C的坐标位置:x C=l AC=400mmy C=0代入RPR II级杆组子程序,求出构件2上D点的位置坐标(x D,y D)根据《机械原理》第三版书中第339页的公式推导可知:当杆件处于图所示位置,即x B>x D并且y B≥y D时,l j杆角位移:φj=arctan B0s+A0C0 A0s−B0C0式中:A0=x B-x DB0=y B-y DC0=l i+l ks=√A02+B02−C02而当x B<x D并且y B≥y D时,φj=arctan B0s+A0C0A0s−B0C0+180o 当x B<x D并且y B<y D时,φj=arctan B0s+A0C0A0s−B0C0+180o 当x B>x D并且y B<y D时,φj=arctan B0s+A0C0A0s−B0C0+360o图 6 RPR II级杆组分析内移动副C的位置:x C=x B-l i sinφjy C=y B-l i cosφj导杆上E点的位置:x E=x C+(l j-s)cosφjy E=y C+(l j-s)sinφj(3)DE(RPR II级杆组)已知运动副D的位置坐标(x D,y D),运动副E的坐标:x E=l AE=800mmy E=0代入RPR II级杆组子程序,求出构件5的转角φ5。
哈工大机械原理大作业_连杆35
Harbin Institute of Technology机械原理大作业一课程名称:机械原理设计题目:连杆机构设计院系:机电学院班级:1308102分析者:学号:指导教师:设计时间:2015年06月哈尔滨工业大学一、连杆机构运动分析题目以及坐标系的建立题目中的连杆机构可以简化为图示机构二、机构的结构分析,组成机构的基本杆组划分该机构由机架、一个原动件AB和两个II级杆组组成。
有题目分析易知B点的位置、速度和加速度是我们知道的由B点我们可以推导出c点的位移速度加速度从而得到D点的轨迹最终我们可以求得E点的各个参数。
五、计算编程利用vb软件进行编程,程序如下:Public q As SingleOption ExplicitPrivate Sub Command1_Click()Dim yd, xd, yc, xb, yb, a, b, t, q, i As Singleq = 0a1.ScaleWidth = 300a1.ScaleHeight = -300a1.ScaleLeft = -150a1.ScaleTop = 250a1.Line (-300, 0)-(300, 0)a1.Line (0, -300)-(0, 300)For i = -300 To 300If i Mod 50 = 0 Thena1.Line (-5, i)-(5, i)a1.Line (i, -5)-(i, 5)a1.CurrentX = 0a1.CurrentY = ia1.Print ia1.CurrentX = ia1.CurrentY = 0a1.Print iEnd IfNextFor t = 0 To 200q = q + t / 1000xb = 100 * Sin(q)yb = -100 * Cos(q)xd = 3 / 2 * xbyd = yb + Sqr(40000 - xb * xb) / 2 a1.Line (a, b)-(xd, yd)a = xdb = ydNextEnd SubPrivate Sub Command2_Click()Dim yd, xd, yc, xb, yb, a, b, t, q, i, ve, vc, ye, ye0, yc0, ve0, vc0, ae, ac, ae0, ac0, x1 As Singlea1.ScaleWidth = 10a1.ScaleHeight = -100a1.ScaleLeft = 150a1.ScaleTop = 50ve0 = 0vc0 = 0ye = 200x1 = 151For i = 1 To 15x1 = 3.14159 / 6 + x1a1.Line (x1, 1)-(x1, -1)a1.CurrentX = x1 - 3.14159 / 12a1.CurrentY = -2a1.Print i * 30NextFor t = 0 To 3000q = q + t / 10000xb = 100 * Sin(q)yb = -100 * Cos(q)xd = (3 / 2) * xbyd = yb + Sqr(40000 - xb * xb) / 2ye = yd + Sqr(40000 - xd * xd)yc = yb - Sqr(40000 - xb * xb)ve = (ye - ye0)vc = (yc - yc0)ae = ve - ve0ac = ae - ae0a1.Line (151, -100)-(151, 100)a1.Line (t - 50, 0)-(t, 0)If t Mod 5 = 0 Thena1.Line (150.9, t)-(151.1, t)a1.CurrentX = 151a1.CurrentY = ta1.Print t * 55a1.CurrentX = 151a1.CurrentY = -ta1.Print -t * 55End Ifa1.CurrentX = 151a1.CurrentY = 0a1.Line (q - 0.2, vc0)-(q, vc)a1.Line (q - 0.2, ve0)-(q, ve)yc0 = ycye0 = yeve0 = vevc0 = vcae0 = aeac0 = acNextEnd SubPrivate Sub Command4_Click()Dim yd, xd, yc, xb, yb, a, b, t, q, i, ve, vc, ye, ye0, yc0, ve0, vc0, ae, ac, ae0, ac0, x1 As Singlea1.ScaleWidth = 10a1.ScaleHeight = -30a1.ScaleLeft = 150a1.ScaleTop = 15ve0 = 0vc0 = 0ye = 200x1 = 151For i = 1 To 15x1 = 3.14159 / 6 + x1a1.Line (x1, 0.5)-(x1, -0.5)a1.CurrentX = x1 - 3.14159 / 12a1.CurrentY = -2a1.Print i * 30Nexta1.Line (151, -100)-(151, 100)a1.Line (0, 0)-(300, 0)For t = 0 To 3000q = q + t / 10000xb = 100 * Sin(q)yb = -100 * Cos(q)xd = (3 / 2) * xbyd = yb + Sqr(40000 - xb * xb) / 2 ye = yd + Sqr(40000 - xd * xd)yc = yb - Sqr(40000 - xb * xb)ve = (ye - ye0)vc = (yc - yc0)ae = ve - ve0ac = ae - ae0a1.Line (q - 0.2, ae0)-(q, ae)a1.Line (q - 0.2, ac0)-(q, ac)If t Mod 2 = 0 Thena1.Line (150.9, t)-(151.1, t)a1.CurrentX = 151a1.CurrentY = ta1.Print t * 300a1.CurrentX = 151a1.CurrentY = -ta1.Print -t * 300End Ifyc0 = ycye0 = yeve0 = vevc0 = vcae0 = aeac0 = acNextEnd SubPrivate Sub Command5_Click()Dim yd, xd, yc, xb, yb, a, b, t, q, i, ve, vc, ye, ye0, yc0, ve0, vc0, ae, ac, ae0, ac0, x1 As Singlea1.ScaleWidth = 8.5a1.ScaleHeight = -500a1.ScaleLeft = 25a1.ScaleTop = 300ye = 200x1 = 26For t = 0 To 3000q = q + t / 10000xb = 100 * Sin(q)yb = -100 * Cos(q)xd = (3 / 2) * xbyd = yb + Sqr(40000 - xb * xb) / 2ye = yd + Sqr(40000 - xd * xd)yc = yb - Sqr(40000 - xb * xb)a1.Line (q - 1 / 10, yc0 + 300)-(q, yc + 300) a1.Line (q - 1 / 10, ye0 - 150)-(q, ye - 150) yc0 = ycye0 = yeIf t Mod 27 = 0 Thena1.Line (25.8, t)-(26.2, t)a1.CurrentX = 26a1.CurrentY = ta1.Print 16500 * (t / 270 - 1)End IfNexta1.Line (0, 270)-(50, 270)a1.Line (26, 2270)-(26, -270)For i = 1 To 15x1 = 3.14159 / 6 + x1a1.Line (x1, 275)-(x1, 265)a1.CurrentX = x1 - 3.14159 / 12a1.CurrentY = 270a1.Print i * 30NextEnd SubPrivate Sub Command3_Click()a1.ClsTimer1.Enabled = FalseEnd SubPrivate Sub Command6_Click()Timer1.Enabled = TrueEnd SubPrivate Sub Timer1_Timer()Dim yd, xd, yc, xb, yb, a, b, i, ve, vc, ye, ye0, yc0, ve0, vc0, ae, ac, ae0, ac0, x1 As Singlea1.Clsa1.ScaleWidth = 500a1.ScaleHeight = -700a1.ScaleLeft = -250a1.ScaleTop = 350a1.Line (-350, 0)-(350, 0)a1.Line (0, -350)-(0, 350)For i = -7 To 7a1.Line (-2, i * 50)-(2, i * 50)a1.CurrentX = 0a1.CurrentY = i * 50a1.Print i * 50a1.Line (i * 50, -2)-(i * 50, 2)a1.CurrentY = 0a1.CurrentX = i * 50a1.Print i * 50Nextye = 200x1 = 26q = q + 1 / 100xb = 100 * Sin(q)yb = -100 * Cos(q)xd = (3 / 2) * xbyd = yb + Sqr(40000 - xb * xb) / 2ye = yd + Sqr(40000 - xd * xd)yc = yb - Sqr(40000 - xb * xb)yc0 = ycye0 = yea1.Line (0, 0)-(xb, yb)a1.Line (0, yc)-(xd, yd)a1.Line (10, yc + 10)-(-10, yc + 10) a1.Line (10, yc - 10)-(-10, yc - 10) a1.Line (-10, yc + 10)-(-10, yc - 10) a1.Line (10, yc + 10)-(10, yc - 10)a1.Line (0, ye)-(xd, yd)a1.Line (10, ye + 10)-(-10, ye + 10) a1.Line (10, ye - 10)-(-10, ye - 10) a1.Line (-10, ye + 10)-(-10, ye - 10) a1.Line (10, ye + 10)-(10, ye - 10) End Sub得到如下结果动画D点的行程曲线速度曲线位移曲线加速度曲线六、结果分析我们从可以看出计算出的曲线与标准答案一致我们可以看出我们的结果是正确的。
哈工大机械原理大作业-连杆
Harbin Institute of Technology机械原理大作业一课程名称:机械原理设计题目:连杆运动分析院系:机电工程学院班级:设计者:学号:指导教师:明设计时间: 2013年6月25日1、运动分析题目在图1-10中所示的干草压缩机中,已知LAB=150mm,LBC=600mm,LCE=120mm,LCD=500mm,LEF=600mm,XD=400mm,YD=500mm,YF=600mm,曲柄1作等速转动,其转速n1=50r/min。
求在一个运动循环中活塞5的位移、速度和加速度的变化曲线。
图1-102、机构的结构分析(1)基本杆组的划分①AB即杆件1为原动件②DECB即杆件2、3为RRR型II级杆组,其中CE为同一构件上点。
③ EF 和滑块即4、5为RRP 型II 级杆组(2)、建立以点A 为原点的固定平面直角系3、确定已知参数和求解流程(1)原动件1(I 级杆组RR )如图所示,已知原动件1的转角πϕ2~01=原动件杆1的角速度s rad /236.51=ω原动件1的角加速度01=α运动副A 的位置坐标0,0==A A y x运动副A 的速度0,0==yA xA v v运动副A 的加速度0,0==yA xA a a原动件杆I 的长度mm l 1501=可求出B 的位置B 的速度B 的加速度(2)构件2、3(II 级杆组RRR )D 的位置500400==D D y xD 的速度 00==yD xD v vD 的加速度00==yD xD a a杆长mm l l CD j 500==,mm l l BC i 600==由关系j j D i i B C l x l x x ϕϕcos cos +=+=j j D i i B C l y l y y ϕϕsin sin +=+=其中)(20sin cos B D i i i x x l A C B A -==-+ ϕϕ)(2B D i y y l B -=222j BD i l l l C -+=222)()(B D B D BD y y X x l -+-=可解得C A C B A B i +-++=222arctan 2ϕ DC D C j x x y y --=arctan ϕ 由上面两个式子可以得到两杆的角速度 1)()(G y y s x x c w B D j B D j i i -+-==•ϕ132G s G c G ij i i +==ϕα其中i j j i s c s c G -=1,i i i l c ϕcos =,i i i l s ϕsin =,j j j l c ϕcos =,j j j l s ϕsin =可得E 的位置iB E iB E y y x x ϕϕsin 480cos 480+=+=E 的速度 i i yB E yE i i xB E xE v yv v x v ωϕωϕcos 480sin 480+==-==E 的加速度i i i i xB E xE a xa αϕωϕsin 480cos 4802--== i i i i yB E yE a ya ϕαϕωcos 480sin 4802+-== (3)、构件4、5杆组(II 级杆组RRP )在建立的坐标系中取一参考点K 600,0==K K y x则速度0,0==yK xK v v加速度0,0==yK xK a a杆长mm l EF 600=,设F 位移为s由s x l x x K i EF E F +=+='cos ϕ K i EF E F y l y y =+='sin ϕ由上面两个式子可以得到 600600arcsin arcsin'E EF E k i y l y y -=-=ϕ所以:F 点位移 )600600arcsin(cos 600E E F y x x s -+== 速度F F xv = 加速度F F xa =四、编程计算并输出结果(VB 编程)主程序:Private Sub Command1_Click()Dim s5(3600) As DoubleDim v5(3600) As DoubleDim a5(3600) As DoubleDim pi As DoubleDim pa As Doublepi = 3.1415926pa = pi / 180Dim i As LongDim f1(3600) As Double Dim RR1 As RRDim RR2 As RRDim RRR1 As RRRDim RRP1 As RRPSet RR1 = New RRSet RR2 = New RRSet RRR1 = New RRRSet RRP1 = New RRPFor i = 0 To 3600 Step 1 f1(i) = i * pa / 10RR1.delt = 0RR1.f = f1(i)RR1.w = 5.24RR1.e = 0RR1.L = 150RR1.xA = 0RR1.yA = 0RR1.vxA = 0RR1.vyA = 0RR1.axA = 0RR1.ayA = 0RR1.calRRR1.Li = 600RRR1.Lj = 500RRR1.xB = RR1.xBRRR1.yB = RR1.yBRRR1.vxB = RR1.vxBRRR1.vyB = RR1.vyBRRR1.axB = RR1.axBRRR1.ayB = RR1.ayBRRR1.xD = 400RRR1.yD = 500RRR1.vxD = 0RRR1.vyD = 0RRR1.axD = 0RRR1.ayD = 0RRR1.M = 1RRR1.calRRRRR2.delt = 0RR2.f = RRR1.fi RR2.w = RRR1.wi RR2.e = RRR1.ei RR2.L = 480RR2.xA = RR1.xB RR2.yA = RR1.yB RR2.vxA = RR1.vxB RR2.vyA = RR1.vyB RR2.axA = RR1.axB RR2.ayA = RR1.ayB RR2.calRRP1.Li = 600RRP1.Lj = 0RRP1.fj = piRRP1.wj = 0RRP1.ej = 0RRP1.xB = RR2.xB RRP1.yB = RR2.yB RRP1.vxB = RR2.vxB RRP1.vyB = RR2.vyB RRP1.axB = RR2.axB RRP1.ayB = RR2.ayB RRP1.xK = 0RRP1.yK = 600RRP1.vxK = 0RRP1.vyK = 0RRP1.axK = 0RRP1.ayK = 0RRP1.M = 1RRP1.cals5(i) = RRP1.ssv5(i) = RRP1.vssa5(i) = RRP1.assNext iPicture1.Scale (-30, 700)-(360, 580)Picture1.Line (0, 0)-(360, 0) 'XPicture1.Line (0, 580)-(0, 700) 'YFor i = 0 To 360 Step 10 'X轴坐标Picture1.DrawStyle = 2Picture1.Line (i, 700)-(i, 580)Picture1.CurrentX = i - 10: Picture1.CurrentY = 0 Picture1.Print iNext iFor i = 580 To 700 Step 10 'Y轴坐标Picture1.DrawStyle = 2Picture1.Line (0, i)-(360, i)Picture1.CurrentX = -10: Picture1.CurrentY = iPicture1.Print iNext iFor i = 0 To 3600 Step 1Picture1.PSet (i / 10, s5(i))Next iEnd SubRR:Public L As DoublePublic f As DoublePublic delt As DoublePublic w As DoublePublic e As DoublePublic xA As DoublePublic yA As DoublePublic vxA As DoublePublic vyA As DoublePublic axA As DoublePublic ayA As DoublePublic xB As DoublePublic yB As DoublePublic vxB As DoublePublic vyB As DoublePublic axB As DoublePublic ayB As DoublePublic Sub cal()xB = xA + L * Cos(f + delt)yB = yA + L * Sin(f + delt)vxB = vxA - w * L * Sin(f + delt)vyB = vyA + w * L * Cos(f + delt)axB = axA - w ^ 2 * L * Cos(f + delt) - e * L * Sin(f + delt) ayB = ayA - w ^ 2 * L * Sin(f + delt) + e * L * Cos(f + delt) End SubRRR:Public Li As DoublePublic Lj As DoublePublic fi As DoublePublic fj As DoublePublic wi As DoublePublic wj As DoublePublic ei As DoublePublic ej As DoublePublic xB As DoublePublic yB As DoublePublic vxB As DoublePublic vyB As DoublePublic axB As DoublePublic ayB As DoublePublic xC As DoublePublic yC As DoublePublic vxC As DoublePublic vyC As DoublePublic axC As DoublePublic ayC As DoublePublic xD As DoublePublic yD As DoublePublic vxD As DoublePublic vyD As DoublePublic axD As DoublePublic ayD As DoublePublic M As DoublePublic Sub calRRR()Dim fDB As DoubleDim Ci As DoubleDim Cj As DoubleDim Si As DoubleDim Sj As DoubleDim G1 As DoubleDim G2 As DoubleDim G3 As DoubleDim LBD As DoubleDim JCBD As DoubleDim val As Doublepi = 3.1415926LBD = Sqr((xB - xD) ^ 2 + (yD - yB) ^ 2)If LBD < Li + Lj And LBD > Abs(Li - Lj) Thenval = (Li ^ 2 + LBD ^ 2 - Lj ^ 2) / (2 * Li * LBD) JCBD = Atn(-val / Sqr(-val * val + 1)) + 2 * Atn(1) End IfRRP:Public Li As DoublePublic Lj As DoublePublic fi As DoublePublic fj As DoublePublic wi As DoublePublic wj As DoublePublic ei As DoublePublic ej As DoublePublic xB As DoublePublic yB As DoublePublic vxB As DoublePublic vyB As DoublePublic axB As DoublePublic ayB As DoublePublic xK As DoublePublic yK As DoublePublic vxK As DoublePublic vyK As DoublePublic axK As DoublePublic ayK As DoublePublic xC As DoublePublic yC As DoublePublic vxC As DoublePublic vyC As DoublePublic axC As DoublePublic ayC As DoublePublic xD As DoublePublic yD As DoublePublic vxD As DoublePublic vyD As DoublePublic axD As DoublePublic ayD As DoublePublic M As SinglePublic ss As DoublePublic vss As DoublePublic ass As DoublePublic Sub cal()Dim A0 As DoubleDim Q1 As DoubleDim Q2 As DoubleDim Q3 As DoubleDim Q4 As DoubleDim Q5 As DoubleDim val As DoubleDim pi As Doublepi = 3.14159216A0 = Lj + ((yK - yB) * Cos(fj) - (xK - xB) * Sin(fj)) val = A0 / Lifi = M * Atn(val / Sqr(-val * val + 1)) + fjxC = xB + Li * Cos(fi)yC = yB + Li * Sin(fi)ss = (xC - xK) * Cos(fj) + (yC - yK) * Sin(fj)xD = xK + ss * Cos(fj)yD = yK + ss * Sin(fj)Q1 = vxK - vxB - wj * (ss * Sin(fj) + Lj * Cos(fj))Q2 = vyK - vyB + wj * (ss * Cos(fj) - Lj * Sin(fj))Q3 = Li * Sin(fi) * Sin(fj) + Li * Cos(fi) * Cos(fj) wi = (-Q1 * Sin(fj) + Q2 * Cos(fj)) / Q3vss = -(Q1 * Li * Cos(fi) + Q2 * Li * Sin(fi)) / Q3vxC = vxB - wi * Li * Sin(fi)vyC = vyB + wi * Li * Cos(fi)vxD = vxK + vss * Cos(fj) - ss * wj * Sin(fj)vyD = vyK + vss * Sin(fj) + ss * wj * Cos(fj)Q4 = axK - axB + wi ^ 2 * Li * Cos(fi) - ej * (ss * Sin(fj) + Lj * Cos(fj)) - wj ^ 2 * (ss * Cos(fj) - Lj * Sin(fj)) - 2 * vss * wj * Sin(fj)Q5 = ayK - ayB + wi ^ 2 * Li * Sin(fi) + ej * (ss * Cos(fj) - Lj * Sin(fj)) - wj ^ 2 * (ss * Sin(fj) + Lj * Cos(fj)) + 2 * vss * wj * Cos(fj)ei = (-Q4 * Sin(fj) + Q5 * Cos(fj)) / Q3ass = (-Q4 * Li * Cos(fi) - Q5 * Li * Sin(fi)) / Q3axC = axB - ei * Li * Sin(fi) - wi ^ 2 * Li * Cos(fi)ayC = ayB + ei * Li * Cos(fi) - wi ^ 2 * Li * Sin(fi)axD = axK + ass * Cos(fj) - ss * ej * Sin(fj) - ss * wj ^ 2 * Cos(fj) - 2 * vss * wj * Sin(fj) ayD = ayK + ass * Sin(fj) + ss * ej * Cos(fj) - ss * wj ^ 2 * Sin(fj) + 2 * vss * wj * Cos(fj) End Sub五、计算结果数据如图:位移曲线:速度曲线:加速度曲线:六、计算结果分析主动件转角为0时,滑块的位移为628mm,随着转角的匀速增加,滑块位移先上升,速度为负快速下降,加速度为负且开始值较小并逐渐下降,到达最低点-2200左右,此时速度为0, 然后位移开始继续下降,速度继续下降,然后到达最小值-100左右,此时加速度为0,又开始上升(向左运动),速度正向增大,在后面一段时期速度继续增大,加速度也正向增大,然后速度下降,加速度下降.滑块分别在4.2s,11.4s速度达到正向最大,1.7s,7.7s速度达到负向最大,0.8s,2.7s,5.8s,9.6s加速度达到极值,滑块就是这样周期性的左右运动。
机械原理大作业连杆机构.
Harbin Institute of Technology大作业设计说明书课程名称:机械原理设计题目:连杆机构设计院系:班级:设计者:学号:指导教师:设计时间:2013-6-13哈尔滨工业大学1.连杆题目(16):如图所示机构,已知机构各构件的尺寸为==100AC CE l l mm ,==200BC CD l l mm ,90BCD ∠=︒,构件1的角速度为1=10/w rad s ,试求构件5的角位移、角速度和角加速度,并对计算结果进行分析。
1.1 机构的运动分析AB 为原动件,AB 转动通过转动导杆机构带动杆BCD 转动,BCD 转动通过转动导杆机构带动杆DE 摆动。
1.2 机构的结构分析杆组可以划分为一个RR I级杆组(杆1)、RRPII级杆组(滑块2,杆3)、RPRII 级杆组(滑块4,杆5)1)RRI级杆组1:2)RRPII级杆组2,3:3)RPRII级杆组4,5:1.3 机构各杆组的运动分析数学模型1)RRI 级杆组1:B 点位移方程:abab ·cos ·sin B A ab B A ab x x l y y l ϕϕ=+⎧⎨=+⎩B 点速度方程: xB xA ab ab ab yB yA ab ab abv v l Sin v v l Cos ωϕωϕ=-⎧⎨=+⎩B 点加速度方程:22 cos sin xB xA ab ab AB yB yA ab ab AB a a l a a l ωϕωϕ⎧=-⎪⎨=+⎪⎩2)RRPII 级杆组2,3: 以A 点为参考点-y -y Cos C A ab C A abAO ϕϕ=(x -x )Sin ()0.5ab MO Cos ϕ=2=tan ()++1bc ab arc MO ϕϕ-+B C bc bc l Cos ϕ=x x+B C bc bc l Sin ϕ=y y-=B Aabs Cos ϕx x1xA xC ab ab Q v v w sSin ϕ=--2=+yA yC ab ab Q v v w sCos ϕ-3=+bc bc ab bc bc ab Q l Sin Sin l Cos Cos ϕϕϕϕ123-+=ab abbc Q Sin Q Cos w Q ϕϕ123()+()=-bc bc bc bc s Q l Cos Q l Sin v Q ϕϕ224=--2bc ab bc bc ab s ab ab Q w l Cos w sCos v w Sin ϕϕϕ 225=--2bc ab bc bc ab s ab ab Q w l Sin w sSin v w COs ϕϕϕ45=-+bc ab ab Q Sin Q COs αϕϕ3)RPRII 级杆组4,5:de l =arctan arctan arctan arctan 2 0.5 y D Ede D E D Ede D E D Ede D E D Ede D E de dey y x x y y x x y y x x y y x x ϕϕπϕπϕπϕπϕ-=--=+--=+--=+-=第一象限第二象限第三象限第四象限轴正半轴1.5 y π⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪=⎪⎩轴负半轴()yDde xD dededev Cos v Sin l ϕϕω-=() 2yD de de de de dede xD xD yD dea Cos a Sin v Cos v Sin w l ϕϕϕϕα--+=1.4 机构各杆组的编程1)对RRI级杆组1::xB = lab * Cos(fab)yB = lab * Sin(fab)vxB = -wab * lab * Sin(fab)vyB = wab * lab * Cos(fab)axB= -wab ^ 2 * lab * Cos(fab) - eab * lab * Sin(fab )ayB = -wab ^ 2 * lab * Sin(fab) + eab * lab * Cos(fab)1)对RRPII级杆组2,3:A0 = (xC - xA) * Sin(fab) - (yC - yA) * Cos(fab)M0 = 0.5 * Cos(fab)fi = Atn(M0 / Sqr(-M0 * M0 + 1)) + fabxB = xC + lbc * Cos(fbc)yB = yC + lbc * Sin(fbc)s = (xB - xA) / Cos(fab)Q1 = vxA - vxC - wab * s * Sin(fab)Q2 = vyA - vyC + wab * s * Cos(fab)Q3 = lbc * Sin(fbc) * Sin(fab) + lbc * Cos(fbc) * Cos(fab)wbc = (-Q1 * Sin(fab) + Q2 * Cos(fab)) / Q3vs = -(Q1 * lbc * Cos(fbc) + Q2 * lbc * Sin(fbc)) / Q3Q4 = wbc ^ 2 * lbc * Cos(fbc) - wab ^ 2 * s * Cos(fab) - 2 * vs * wab * Sin(fab) Q5 = wbc ^ 2 * lbc * Sin(fbc) - wab ^ 2 * s * Sin(fab) + 2 * vs * wab * Cos(fab) ebc = (-Q4 * Sin(fab) + Q5 * Cos(fab))2)对RPRII级杆组4,5:lde = Sqr((xD - xE) ^ 2 + (yD - yE) ^ 2)If xD>xE And yD>yE Then '第一象限fde = Atn((yD - yE) / (xD - xE))ElseEnd IfIf xD<xE And yD>= yE Then '第二象限fde = Atn((yD - yE) / (xD - xE)) + piElseEnd IfIf xD<xE And yD<yE Then '第三象限fde = Atn((yD - yE) / (xD - xE)) + piElseEnd IfIf xD>xE And yD<= yE Then '第四象限fde = Atn((yD - yE) / (xD - xE)) + 2 * piElseEnd IfIf xD = xE And yD>yE Then 'y轴正向fde = 0.5 * piElseEnd IfIf xD = xE And yD<yE Then 'y轴负向fde = 1.5 * piElseEnd Ifwde = (vyD * Cos(fde) - vxD * Sin(fde)) / ldeede = ((ayD * Cos(fde) - axD * Sin(fde) - 2 * (vxd * Cos(fde) + vyd * Sin(fde)) * wde) / lde2.计算编程以A为坐标原点,建立坐标系计算编程,源代码如下:Option Explicit '定义自变量Dim xA As Double '点A的坐标,速度,加速度Dim yA As DoubleDim vxA As DoubleDim vyA As DoubleDim axA As DoubleDim ayA As DoubleDim xB As Double '点B的坐标,速度,加速度Dim yB As DoubleDim vxB As DoubleDim vyB As DoubleDim axB As DoubleDim ayB As DoubleDim xC As Double '点C的坐标,速度,加速度Dim yC As DoubleDim vxC As DoubleDim vyC As DoubleDim axC As DoubleDim ayC As DoubleDim xD As Double '点D的坐标,速度,加速度Dim yD As DoubleDim vxD As DoubleDim vyD As DoubleDim axD As DoubleDim ayD As DoubleDim atd As DoubleDim adn As DoubleDim xE As Double '点E的坐标,速度,加速度Dim yE As DoubleDim vxE As DoubleDim vyE As DoubleDim axE As DoubleDim ayE As DoubleDim lbc As Double '杆BC的长度Dim lcd As Double '杆CD的长度Dim fab As Double '杆AB的角位移Dim fbc As Double '杆BC的角位移Dim fde As Double '杆DE的角位移Dim fj1 As Double '循环变量Dim wab As Double '杆AB的角速度Dim wbc As Double '杆BC的角速度Dim wde As Double '杆DE的角速度Dim eab As Double '杆AB的角加速度Dim ebc As Double '杆BC的角加速度Dim ede As Double '杆DE的角加速度Dim lab As Double 'AB的距离Dim lde As Double 'DE的距离Dim s As Double 's的长度Dim vs As Double 's的速度Dim Q1 As Double 'RRP中的Q1Dim Q2 As Double 'RRP中的Q2Dim Q3 As Double 'RRP中的Q3Dim Q4 As Double 'RRP中的Q4Dim Q5 As Double 'RRP中的Q5Dim A0 As Double '杆组的中间变量Dim M0 As DoubleDim pi As Double '圆周率Dim pa As Double '角度与弧度转换的系数Dim i As Double '循环变量Private Sub Form_Load() '赋值Form1.WindowState = 2lbc = 200lcd = 200wab = 10eab = 0xA = 0yA = 0vxA = 0vyA = 0axA = 0ayA = 0xC = 0yC = -100vxC = 0vyC = 0axC = 0ayC = 0xE = 0yE = -200vxE = 0vyE = 0axE = 0ayE = 0pi = 4 * Atn(1)pa = pi / 180fj1 = 0End SubPrivate Sub Command1_Click()Set Picture1.Picture = NothingPicture1.Scale (-20, 8)-(400, -2)Picture1.Line (-20, 0)-(400, 0) 'XPicture1.Line (0, 8)-(0, -2) 'YFor i = -20 To 400 Step 50 'X轴坐标Picture1.DrawStyle = 2Picture1.Line (i, 8)-(i, -2)Picture1.CurrentX = i - 10: Picture1.CurrentY = 0 Picture1.Print iNext iFor i = -2 To 8 Step 1 'Y轴坐标Picture1.DrawStyle = 2Picture1.Line (-20, i)-(400, i)Picture1.CurrentX = -10: Picture1.CurrentY = iPicture1.Print iNext iFor fj1 = 0 To 360 Step 0.01fab = fj1 * paCall RRPCall RPRPicture1.PSet (fj1, fde), vbRedNext fj1End SubPrivate Sub Command2_Click()Set Picture2.Picture = NothingPicture2.Scale (-20, 30)-(400, -4)Picture2.Line (-20, 0)-(400, 0) 'XPicture2.Line (0, 30)-(0, -4) 'YFor i = 0 To 360 Step 30 'X轴坐标Picture2.DrawStyle = 2Picture2.Line (i, 30)-(i, -4)Picture2.CurrentX = i - 10: Picture2.CurrentY = 0 Picture2.Print iNext iFor i = -4 To 30 Step 4 'Y轴坐标Picture2.Line (0, i)-(400, i)Picture2.CurrentX = -20: Picture2.CurrentY = iPicture2.Print iNext iFor fj1 = 0 To 360 Step 0.01fab = fj1 * paCall RRPCall RPRPicture2.PSet (fj1, wde), vbRedNext fj1End SubPrivate Sub Command3_Click() '杆5的角加速度Set Picture3.Picture = NothingPicture3.Scale (-20, 10000)-(400, -20000)Picture3.Line (-20, 0)-(400, 0) 'XPicture3.Line (0, 10000)-(0, -20000) 'YFor i = 0 To 360 Step 30 'X轴坐标Picture3.DrawStyle = 2Picture3.Line (i, 10000)-(i, -20000)Picture3.CurrentX = i - 10: Picture3.CurrentY = 0Picture3.Print iNext iFor i = -20000 To 10000 Step 2500 'Y轴坐标Picture3.Line (0, i)-(400, i)Picture3.CurrentX = -25: Picture3.CurrentY = i + 5 Picture3.Print iNext iFor fj1 = 0 To 360 Step 0.01fab = fj1 * paCall RRPCall RPRPicture3.PSet (fj1, ede), vbRedNext fj1End SubPrivate Sub RRP() 'Ⅱ级杆组RRP(滑块2、杆3)A0 = (xC - xA) * Sin(fab) - (yC - yA) * Cos(fab)M0 = 0.5 * Cos(fab)fbc = Atn(M0 / Sqr(-M0 * M0 + 1)) + fabxB = xC + lbc * Cos(fbc)yB = yC + lbc * Sin(fbc)If fab = pi / 2 Thens = lbc - (yA - yC)ElseIf fab = 3 * pi / 2 Thens = lbc + (yA - yC)Elses = (xB - xA) / Cos(fab)End IfQ1 = vxA - vxC - wab * s * Sin(fab)Q2 = vyA - vyC + wab * s * Cos(fab)Q3 = lbc * Sin(fbc) * Sin(fab) + lbc * Cos(fbc) * Cos(fab)wbc = (-Q1 * Sin(fab) + Q2 * Cos(fab)) / Q3vs = -(Q1 * lbc * Cos(fbc) + Q2 * lbc * Sin(fbc)) / Q3Q4 = wbc ^ 2 * lbc * Cos(fbc) - wab ^ 2 * s * Cos(fab) - 2 * vs * wab * Sin(fab) Q5 = wbc ^ 2 * lbc * Sin(fbc) - wab ^ 2 * s * Sin(fab) + 2 * vs * wab * Cos(fab) ebc = (-Q4 * Sin(fab) + Q5 * Cos(fab))atd = lbc * ebcadn = wbc ^ 2 * lbcxD = xC + lbc * Sin(fbc)yD = yC - lbc * Cos(fbc)vxD = wbc * lbc * Cos(fbc)vyD = wbc * lbc * Sin(fbc)axD = -adn * Sin(fbc) + atd * Cos(fbc)ayD = adn * Cos(fbc) + atd * Sin(fbc)End SubPrivate Sub RPR() 'Ⅱ级杆组RPR(滑块4、杆5)lde = Sqr((xD - xE) ^ 2 + (yD - yE) ^ 2)If xD>xE And yD>yE Then '第一象限fde = Atn((yD - yE) / (xD - xE))ElseEnd IfIf xD<xE And yD>= yE Then '第二象限fde = Atn((yD - yE) / (xD - xE)) + piElseEnd IfIf xD<xE And yD<yE Then '第三象限fde = Atn((yD - yE) / (xD - xE)) + piElseEnd IfIf xD>xE And yD<= yE Then '第四象限fde = Atn((yD - yE) / (xD - xE)) + 2 * piElseEnd IfIf xD = xE And yD>yE Then 'y轴正向fde = 0.5 * piElseEnd IfIf xD = xE And yD<yE Then 'y轴负向fde = 1.5 * piElseEnd Ifwde = (vyD * Cos(fde) - vxD * Sin(fde)) / ldeede = (ayD * Cos(fde) - axD * Sin(fde) - 2 * (vxD * Cos(fde) + vyD * Sin(fde)) * wde) / ldeEnd Sub3.计算结果(绘制构件运动图线):构件5角位移构件5角速度图像构件5角加速度图像4. 构件运动属性随主动件的转动变化列举(角度变化为Δθ=1°):角构件5角位移(rad) 构件5角速度(rad/s) 构件5角加速度(rad/s2) 度0 5.651270995 14.76627109 -14893.44451 5.676722398 14.40057723 -14666.11492 5.701544405 14.04504413 -14437.389243 5.72575477 13.69968448 -14207.738354 5.749371219 13.36445541 -13977.573885 5.772411362 13.03926746 -13747.254436 5.794892617 12.72399249 -13517.091297 5.81683215 12.41847071 -13287.353848 5.838246824 12.12251691 -13058.274499 5.859153164 11.83592579 -12830.0532810 5.879567318 11.55847668 -12602.8619311 5.899505042 11.28993752 -12376.8475912 5.918981679 11.0300683 -12152.1361113 5.938012146 10.77862402 -11928.8350114 5.956610932 10.53535705 -11707.0360415 5.974792092 10.30001929 -11486.8174616 5.992569251 10.07236376 -11268.2460217 6.009955602 9.85214605 -11051.3786618 6.026963915 9.639125439 -10836.2639619 6.043606546 9.433065763 -10622.9434620 6.059895442 9.23373613 -10411.4526421 6.07584215 9.040911451 -10201.8219122 6.091457833 8.854372832 -9994.07727823 6.106753275 8.673907846 -9788.24104524 6.121738897 8.499310711 -9584.3322825 6.136424764 8.330382385 -9382.36725426 6.150820603 8.166930588 -9182.35977527 6.164935808 8.008769777 -8984.32146228 6.178779457 7.855721072 -8788.26195529 6.192360321 7.707612149 -8594.18907930 6.205686876 7.564277103 -8402.10897231 6.218767314 7.425556294 -8212.02617732 6.231609553 7.291296171 -8023.94371133 6.244221248 7.161349095 -7837.86311234 6.256609803 7.035573143 -7653.78447335 6.268782379 6.913831914 -7471.70646336 6.2807459 6.795994337 -7291.62634137 9.32E-03 6.68193447 -7113.53996638 2.09E-02 6.571531308 -6937.44180339 0.032262796 6.464668594 -6763.32492540 4.35E-02 6.361234628 -6591.18102541 5.45E-02 6.26112209 -6421.00041842 6.53E-02 6.164227865 -6252.77205243 0.075988655 6.070452866 -6086.48352644 8.65E-02 5.979701878 -5922.121145 9.69E-02 5.891883397 -5759.66971946 0.107072173 5.806909481 -5599.11303747 0.117135002 5.724695602 -5440.43344848 0.127056689 5.645160514 -5283.6121149 0.136841842 5.568226113 -5128.62898950 0.146494934 5.493817317 -4975.46289451 0.15602031 5.421861945 -4824.09151752 0.16542219 5.352290598 -4674.4914853 0.174704676 5.285036556 -4526.63838254 0.183871758 5.220035668 -4380.50684555 0.192927313 5.157226257 -4236.07056556 0.201875114 5.096549022 -4093.30236857 0.210718833 5.037946952 -3952.17425558 0.219462042 4.981365236 -3812.65746359 0.228108222 4.926751184 -3674.72251360 2.37E-01 4.874054148 -3538.3392761 2.45E-01 4.823225447 -3403.4769962 2.53E-01 4.774218296 -3270.10438263 2.62E-01 4.726987743 -3138.18965364 2.70E-01 4.681490596 -3007.70056765 2.78E-01 4.637685368 -2878.60449466 0.286188901 4.595532216 -2750.86846467 0.294174009 4.554992884 -2624.45921368 0.302089743 4.51603065 -2499.34323469 0.309938826 4.478610274 -2375.48682770 0.31772392 4.442697951 -2252.85614471 0.325447626 4.40826126 -2131.41723472 0.333112495 4.375269121 -2011.13608873 0.34072102 4.343691751 -1891.97868474 0.348275646 4.313500625 -1773.91102575 0.355778768 4.284668431 -1656.89918776 0.363232735 4.257169038 -1540.9093577 0.370639852 4.230977453 -1425.90784578 0.37800238 4.206069791 -1311.86118679 0.385322539 4.182423239 -1198.73611280 0.392602512 4.160016021 -1086.49961881 0.399844443 4.13882737 -975.118993782 0.407050442 4.118837498 -864.561859383 0.414222585 4.100027562 -754.796196684 0.421362914 4.08237964 -645.790384785 0.428473444 4.065876704 -537.51323386 0.435556158 4.050502591 -429.934014487 0.442613013 4.036241977 -323.022498488 0.44964594 4.023080356 -216.748983889 0.456656844 4.011004013 -111.08433290 0.463647609 4 -691 0.470620095 3.990056116 98.5319262392 0.477576142 3.981160883 202.538697793 0.484517572 3.973303522 306.046868294 0.491446186 3.966473937 409.082259795 0.49836377 3.960662689 511.669926496 0.505272094 3.955860979 613.834118797 0.512172913 3.952060626 715.598246298 0.519067966 3.949254046 816.984839299 0.525958984 3.947434235 918.0155099 100 0.532847682 3.946594748 1018.710912 101 0.539735765 3.946729679 1119.0907 102 0.546624931 3.947833642 1219.173486 103 0.553516865 3.94990175 1318.976795 104 0.560413247 3.9529296 1418.51702 105 0.567315749 3.956913248 1517.809379 106 0.574226034 3.961849192 1616.867862107 0.581145763 3.967734353 1715.705182 108 0.588076591 3.97456605 1814.332729 109 0.595020166 3.982341987 1912.760512 110 0.601978135 3.991060224 2010.997111 111 0.608952141 4.000719164 2109.049616 112 0.615943826 4.011317523 2206.923577 113 0.622954827 4.022854314 2304.622942 114 0.629986782 4.035328821 2402.150002 115 0.637041327 4.048740578 2499.505327 116 0.644120098 4.063089343 2596.687713 117 0.651224729 4.078375073 2693.694114 118 0.658356857 4.094597902 2790.519586 119 0.665518118 4.111758112 2887.157222 120 0.672710146 4.129856105 2983.598094 121 0.679934581 4.148892379 3079.831188 122 0.687193059 4.168867498 3175.843346 123 0.69448722 4.189782061 3271.619206 124 0.701818704 4.211636674 3367.141145 125 0.709189153 4.234431917 3462.389221 126 0.716600208 4.258168314 3557.34112 127 0.724053512 4.282846299 3651.972103 128 0.73155071 4.308466182 3746.254959 129 0.739093444 4.335028115 3840.159964 130 0.74668336 4.362532057 3933.654834 131 0.754322102 4.390977737 4026.704696 132 0.762011312 4.420364617 4119.272056 133 0.769752632 4.450691856 4211.316778 134 0.777547703 4.481958271 4302.796065 135 0.785398163 4.514162296 4393.664455 136 0.793305647 4.547301949 4483.873819 137 0.801271785 4.581374785 4573.373368 138 0.809298203 4.616377864 4662.10968 139 0.817386522 4.652307705 4750.026723 140 0.825538357 4.689160253 4837.065902 141 0.833755314 4.726930833 4923.166108 142 0.842038989 4.765614119 5008.263792 143 0.850390973 4.805204089 5092.293039 144 0.85881284 4.845693993 5175.185667 145 0.867306155 4.887076317 5256.871335 146 0.875872468 4.929342744 5337.277666 147 0.884513316 4.972484126 5416.330391 148 0.893230216 5.016490453 5493.953504 149 0.902024668 5.06135082 5570.069429 150 0.910898153 5.107053405 5644.599218151 0.91985213 5.153585444 5717.462746 152 0.928888034 5.200933207 5788.578936 153 0.938007278 5.249081987 5857.865994 154 0.947211244 5.298016082 5925.241659 155 0.95650129 5.347718782 5990.623467 156 0.965878741 5.398172371 6053.929029 157 0.975344892 5.449358117 6115.076322 158 0.984901004 5.50125628 6173.983991 159 0.994548301 5.553846118 6230.571658 160 1.004287974 5.607105897 6284.760239 161 1.01412117 5.661012913 6336.472274 162 1.024048999 5.715543509 6385.63225 163 1.034072529 5.770673107 6432.166939 164 1.044192782 5.826376242 6476.005722 165 1.054410737 5.882626595 6517.080922 166 1.064727325 5.939397042 6555.328128 167 1.07514343 5.9966597 6590.686511 168 1.085659886 6.054385983 6623.099128 169 1.096277477 6.112546661 6652.513223 170 1.106996934 6.171111921 6678.880498 171 1.117818939 6.230051439 6702.157385 172 1.128744116 6.289334449 6722.305279 173 1.139773039 6.348929824 6739.290767 174 1.150906226 6.408806148 6753.085818 175 1.162144138 6.468931809 6763.66796 176 1.173487184 6.529275074 6771.020416 177 1.184935716 6.589804184 6775.132227 178 1.19649003 6.650487441 6775.998326 179 1.208150368 6.711293297 6773.619598 180 1.219916916 6.772190444 6768.002899 181 1.231789807 6.83314791 6759.161041 182 1.243769119 6.894135142 6747.112752 183 1.255854878 6.955122103 6731.882601 184 1.268047058 7.016079353 6713.500888 185 1.280345581 7.076978139 6692.00351 186 1.292750322 7.137790478 6667.431793 187 1.305261104 7.198489235 6639.832297 188 1.317877708 7.2590482 6609.256601 189 1.330599866 7.319442166 6575.761054 190 1.343427271 7.379646989 6539.406511 191 1.356359571 7.439639663 6500.258049 192 1.369396376 7.49939837 6458.384664 193 1.382537261 7.558902541 6413.858952 194 1.395781764 7.618132907 6366.756784195 1.409129392 7.677071537 6317.156965 196 1.42257962 7.735701887 6265.14089 197 1.436131897 7.794008824 6210.792195 198 1.449785647 7.851978666 6154.196404 199 1.46354027 7.909599198 6095.440585 200 1.477395147 7.966859697 6034.612997 201 1.491349642 8.023750943 5971.802751 202 1.505403102 8.080265231 5907.099479 203 1.519554865 8.136396376 5840.593006 204 1.533804258 8.192139715 5772.373037 205 1.5481506 8.247492105 5702.528856 206 1.562593208 8.302451918 5631.149038 207 1.577131396 8.357019033 5558.321178 208 1.59176448 8.411194821 5484.131631 209 1.606491779 8.464982136 5408.665274 210 1.621312618 8.518385297 5332.005281 211 1.636226333 8.571410069 5254.232914 212 1.651232268 8.624063647 5175.427339 213 1.666329783 8.676354634 5095.665446 214 1.681518254 8.728293022 5015.021703 215 1.696797074 8.779890169 4933.568012 216 1.712165661 8.831158781 4851.373588 217 1.727623451 8.882112888 4768.504855 218 1.743169909 8.932767829 4685.025357 219 1.758804528 8.983140227 4600.995675 220 1.77452683 9.033247978 4516.473375 221 1.790336369 9.083110231 4431.512952 222 1.806232735 9.132747374 4346.165793 223 1.822215553 9.182181025 4260.480158 224 1.838284488 9.231434017 4174.501155 225 1.854439246 9.280530396 4088.270741 226 1.870679574 9.329495413 4001.827724 227 1.887005267 9.378355524 3915.207766 228 1.903416166 9.427138389 3828.443409 229 1.91991216 9.475872882 3741.56409 230 1.936493191 9.524589094 3654.596168 231 1.953159255 9.573318347 3567.56296 232 1.969910403 9.622093209 3480.484768 233 1.986746743 9.670947517 3393.378917 234 2.003668445 9.719916394 3306.259795 235 2.02067574 9.769036285 3219.138886 236 2.037768924 9.818344983 3132.024812 237 2.054948362 9.867881672 3044.923365 238 2.072214486 9.917686961 2957.837547239 2.089567801 9.967802939 2870.767604 240 2.107008888 10.01827322 2783.71105 241 2.124538404 10.06914301 2696.662703 242 2.142157087 10.12045915 2609.614703 243 2.159865759 10.17227021 2522.556533 244 2.177665326 10.22462654 2435.475033 245 2.195556786 10.27758035 2348.354411 246 2.213541228 10.33118583 2261.176243 247 2.231619839 10.38549918 2173.919471 248 2.249793905 10.44057878 2086.560394 249 2.268064815 10.49648522 1999.072648 250 2.286434068 10.55328149 1911.427181 251 2.304903273 10.61103303 1823.59222 252 2.323474156 10.66980789 1735.533225 253 2.342148565 10.72967686 1647.212842 254 2.360928473 10.79071361 1558.590832 255 2.379815985 10.85299484 1469.624007 256 2.398813342 10.91660043 1380.266139 257 2.417922928 10.9816136 1290.467866 258 2.437147274 11.0481211 1200.176582 259 2.456489068 11.11621339 1109.336319 260 2.475951156 11.18598479 1017.887606 261 2.495536555 11.25753373 925.7673213 262 2.515248456 11.33096292 832.9085287 263 2.535090233 11.40637958 739.2402931 264 2.555065453 11.48389563 644.687483 265 2.57517788 11.56362796 549.1705546 266 2.595431489 11.64569862 452.605317 267 2.615830472 11.73023508 354.9026776 268 2.636379246 11.81737043 255.9683687 269 2.657082469 11.9072437 155.7026516 270 2.677945045 12 54 271 2.698972136 12.09579084 -49.25123962 272 2.720169177 12.19477433 -154.16921 273 2.741541882 12.29711538 -260.8789293 274 2.763096261 12.40298597 -369.5127 275 2.784838628 12.5125653 -480.2105427 276 2.806775619 12.62603996 -593.1206523 277 2.828914201 12.7436041 -708.3998776 278 2.851261686 12.86545951 -826.2142221 279 2.873825745 12.99181568 -946.7393647 280 2.896614422 13.1228898 -1070.161198 281 2.919636147 13.25890668 -1196.676381 282 2.942899748 13.40009858 -1326.492899283 2.966414465 13.54670498 -1459.830631 284 2.990189962 13.69897215 -1596.92191 285 3.014236339 13.85715265 -1738.012069 286 3.03856414 14.02150458 -1883.359973 287 3.063184366 14.19229075 -2033.238495 288 3.088108479 14.36977744 -2187.934953 289 3.113348406 14.55423308 -2347.751462 290 3.138916547 14.74592648 -2513.005186 291 3.164825766 14.94512472 -2684.028462 292 3.191089393 15.15209073 -2861.168755 293 3.217721209 15.36708025 -3044.788411 294 3.244735433 15.59033836 -3235.264152 295 3.272146701 15.82209544 -3432.986263 296 3.299970033 16.06256236 -3638.357411 297 3.328220797 16.31192507 -3851.791015 298 3.356914659 16.57033834 -4073.709092 299 3.386067523 16.83791857 -4304.539493 300 3.415695457 17.11473584 -4544.712426 301 3.445814605 17.40080484 -4794.656164 302 3.476441079 17.69607489 -5054.791817 303 3.50759084 18.00041893 -5325.527077 304 3.53927955 18.31362154 -5607.248798 305 3.571522411 18.63536601 -5900.314316 306 3.60433397 18.96522071 -6205.041424 307 3.637727914 19.30262471 -6521.696914 308 3.671716826 19.64687324 -6850.483668 309 3.706311923 19.99710298 -7191.526309 310 3.741522764 20.35227795 -7544.855479 311 3.777356945 20.71117638 -7910.390911 312 3.813819753 21.07237927 -8287.923564 313 3.850913824 21.43426153 -8677.097172 314 3.888638771 21.79498642 -9077.389742 315 3.926990817 22.15250437 -9488.095643 316 3.965962423 22.50455722 -9908.309074 317 4.005541936 22.84868866 -10336.90988 318 4.045713253 23.18226199 -10772.55272 319 4.086455532 23.50248579 -11213.66079 320 4.127742946 23.80644791 -11658.42518 321 4.169544512 24.09115807 -12104.81101 322 4.211823995 24.35359852 -12550.57119 323 4.254539912 24.59078188 -12993.26869 324 4.297645637 24.79981492 -13430.30738 325 4.341089626 24.97796606 -13858.97143 326 4.384815759 25.12273412 -14276.47267327 4.428763801 25.23191553 -14680.00441 328 4.472869976 25.30366675 -15066.80024 329 4.517067644 25.33655883 -15434.19516 330 4.561288069 25.32962121 -15779.68675 331 4.60546124 25.28237224 -16100.99329 332 4.649516748 25.19483483 -16396.10616 333 4.693384663 25.06753614 -16663.334 334 4.736996405 24.90149146 -16901.33649 335 4.780285573 24.69817319 -17109.14645 336 4.823188699 24.45946661 -17286.17927 337 4.865645929 24.18761506 -17432.23011 338 4.907601592 23.88515729 -17547.45927 339 4.949004657 23.55486017 -17632.36742 340 4.98980907 23.19964997 -17687.7623 341 5.029973974 22.82254493 -17714.71923 342 5.069463816 22.42659175 -17714.53761 343 5.108248339 22.01480796 -17688.69562 344 5.146302494 21.5901315 -17638.805 345 5.183606256 21.15537851 -17566.56784 346 5.220144385 20.71320962 -17473.73648 347 5.255906129 20.2661045 -17362.07765 348 5.290884894 19.81634447 -17233.34134 349 5.325077886 19.36600218 -17089.23479 350 5.358485742 18.91693765 -16931.40148 351 5.391112162 18.47079954 -16761.40501 352 5.422963541 18.0290308 -16580.7174 353 5.454048622 17.59287767 -16390.71142 354 5.484378157 17.16340106 -16192.65635 355 5.513964598 16.74148976 -15987.71668 356 5.54282181 16.32787449 -15776.95316 357 5.570964803 15.92314255 -15561.32571 358 5.5984095 15.52775234 -15341.69776 359 5.62517252 15.14204762 -15118.84162 360 5.651270995 14.76627109 -14893.44455.计算结果分析:1、原动件AB转动一周,构件5随之转动一周。
哈工大机械原理大作业(连杆机构)
建立坐标系:以C为原点,水平方向为X轴,CA所在直线为Y轴建立直角坐标系(如图4)。
取曲柄1水平且位于A点右侧为初始时刻,设曲柄1角速度为w,由题意知w= =8.5π rad/s………………(1)
设曲柄1转角为θ,则B点坐标:
xB=ιABcosθ=ιABcoswt
yB=H1+ιABsinθ=H1+ιABsinwt………………(2)
form=1:length(t)-1
ddxF(m)=(dxF(m+1)-dxF(m))/0.0001;
end
ddxF(length(t))=ddxF(length(t)-1);
figure
plot(t,ddxF)
title('¼ÓËÙ¶ÈͼÏñ');
xlabel('t /s'),ylabel('v /(m/s^2)');
输出图像:
xE(m)=yE(m)/k(m);
xF(m)=xE(m)-(-H^2+lEF^2-yE(m)^2+2*yE(m)*H)^(1/2)+0.1142;
end
form=1:length(t)-1
dxF(m)=(xF(m+1)-xF(m))/0.0001;
end
dxF(length(t))=dxF(length(t)-1);
∵ιEF+ιCE>H且ιCE<H
∴E点始终在F点的右下方
∴xF<xE,所以x2舍去,只取xF=x1……………(8)
∴点F坐标为(xF,H)
当t=0时,可得F点初始位置坐标,不妨设为(xo,H)。
则F点位移(通过计算,t=0时,得xo=-0.1142)
哈工大机械原理大作业连杆14-2015版
Harbin Institute of Technology机械原理大作业连杆课程名称:机械原理设计题目:连杆机构运动分析院系:xx学院班级:xxxxxxxx姓名:xxxxxxxxx学号:xxxxxxxxxxx指导教师:焦映厚陈照波设计时间:2015年xxxxxxxxxx哈尔滨工业大学题目14如图所示机构,已知各机构尺寸为,,,,构件l的角速度为,试求构件2上点D的轨迹及构建5的角位移、角速度和角加速度,并对计算结果进行分析。
(题中构建尺寸满足)对机构进行结构分析该机构由原动件AB(Ⅰ级组),CD(RPRⅡ级杆组)和DE(RPRⅡ级杆组)组成,拆分如下:RR杆组RPR杆组RPR杆组数学模型计算对B点:对D点(∠BAC=,∠BCA=)流程框图开始编程及代码根据以上计算采用vb编程编程代码如下:Option ExplicitPrivate lab, lac, lbd, lce, lbc, lae, pi, sita As DoubleFunction arcsin(x As Double) As DoubleIf x >= -1 And x < -0.5 Then arcsin = -Atn(Sqr(1 - x * x) / x) - 2 * Atn(1)If x >= -0.5 And x <= 0.5 Then arcsin = Atn(x / Sqr(1 - x * x))If x > 0.5 And x <= 1 Then arcsin = -Atn(Sqr(1 - x * x) / x) + 2 * Atn(1) End FunctionFunction arccos(x As Double) As DoubleIf x >= -1 And x < -0.5 Then arccos = Atn(Sqr(1 - x * x) / x) + 4 * Atn(1)If x >= -0.5 And x <= 0.5 Then arccos = -Atn(x / Sqr(1 - x * x)) + 2 * Atn(1) If x > 0.5 And x <= 1 Then arccos = Atn(Sqr(1 - x * x) / x)End FunctionPrivate Sub Command1_Click()Line1.Visible = FalseLine2.Visible = FalseLine3.Visible = False'画坐标系lab = 200lac = 400lbd = 700lae = 800lce = 400pi = 3.1416Dim i, j As IntegerDim beta, xb, yb, xd, yd As DoublePicture1.BackColor = vbWhitePicture1.FillStyle = 0Picture1.Scale (-400, 440)-(1500, -440)Picture1.Line (-400, 0)-(1500, 0)Picture1.Line (0, -440)-(0, 440)For i = 0 To 22Picture1.CurrentX = -68Picture1.CurrentY = 440 - 40 * iPicture1.Print " " & 440 - 40 * i & " "Picture1.DrawStyle = 2Picture1.Line (-400, 440 - 40 * i)-(1500, 440 - 40 * i)NextFor j = 0 To 70Picture1.CurrentX = 1520 - 80 * jPicture1.CurrentY = -5Picture1.Print 1520 - 80 * jPicture1.DrawStyle = 2Picture1.Line (1520 - 80 * j, 440)-(1520 - 80 * j, -440) Next'画D点位移图线For sita = 0 To 2 * pi Step 0.001xb = lab * Cos(sita)yb = lab * Sin(sita)lbc = Sqr(lab ^ 2 + lac ^ 2 - 2 * lab * lac * Cos(sita)) beta = arcsin(lab * Sin(sita) / lbc)xd = xb + lbd * Cos(beta)yd = yb - lbd * Sin(beta)Picture1.PSet (xd, yd)NextEnd SubPrivate Sub Command2_Click()‘D点角位移图Line1.Visible = FalseLine2.Visible = FalseLine3.Visible = Falselab = 200lac = 400lbd = 700lae = 800lce = 400pi = 3.1416Dim i, j As IntegerDim beta, xb, yb, xd, yd, gama, gama1 As Double Picture1.BackColor = vbWhitePicture1.FillStyle = 0Picture1.Scale (-20, 360)-(360, -80)Picture1.Line (-0, 0)-(360, 0)Picture1.Line (0, 360)-(0, -80)For i = 0 To 22Picture1.CurrentX = -15Picture1.CurrentY = 360 - 20 * iPicture1.Print " " & 360 - 20 * i & " "Picture1.DrawStyle = 2Picture1.Line (0, 360 - 20 * i)-(360, 360 - 20 * i)NextFor j = 0 To 18Picture1.CurrentX = 360 - 20 * jPicture1.CurrentY = -5Picture1.Print 360 - 20 * jPicture1.DrawStyle = 2Picture1.Line (360 - 20 * j, 360)-(360 - 20 * j, -80)NextFor sita = 0 To 360 Step 0.01xb = lab * Cos(sita * pi / 180)yb = lab * Sin(sita * pi / 180)lbc = Sqr(lab ^ 2 + lac ^ 2 - 2 * lab * lac * Cos(sita * pi / 180)) beta = arcsin(lab * Sin(sita * pi / 180) / lbc)xd = xb + lbd * Cos(beta)yd = yb - lbd * Sin(beta)gama = arccos((xd - 800) / Sqr((xd - 800) ^ 2 + yd ^ 2))If yd <= 0 Thengama1 = 2 * pi - gamaElsegama1 = gamaEnd IfPicture1.PSet (sita, gama1 * 180 / pi)NextEnd SubPrivate Sub Command3_Click()‘画角角速度Line1.Visible = FalseLine2.Visible = FalseLine3.Visible = Falselab = 200lac = 400lbd = 700lae = 800lce = 400pi = 3.1416Dim beta, xb, yb, xd, yd, gama, gama1, beta0, xb0, yb0, xd0, yd0, gama0, gama10, lbc0, w, sita0 As DoubleDim i, j As IntegerPicture1.BackColor = vbWhitePicture1.FillStyle = 0Picture1.Scale (-20, 90)-(360, -540)Picture1.Line (-0, 0)-(360, 0)Picture1.Line (0, 360)-(0, -540)For i = 0 To 22Picture1.CurrentX = -15Picture1.CurrentY = 90 - 30 * iPicture1.Print " " & 90 - 30 * i & " "Picture1.DrawStyle = 2Picture1.Line (0, 90 - 30 * i)-(360, 90 - 30 * i)NextFor j = 0 To 18Picture1.CurrentX = 360 - 20 * jPicture1.CurrentY = -5Picture1.Print 360 - 20 * jPicture1.DrawStyle = 2Picture1.Line (360 - 20 * j, 360)-(360 - 20 * j, -540)NextFor sita = 0 To 360 Step 0.01xb = lab * Cos(sita * pi / 180)yb = lab * Sin(sita * pi / 180)lbc = Sqr(lab ^ 2 + lac ^ 2 - 2 * lab * lac * Cos(sita * pi / 180))beta = arcsin(lab * Sin(sita * pi / 180) / lbc)xd = xb + lbd * Cos(beta)yd = yb - lbd * Sin(beta)gama = arccos((xd - 800) / Sqr((xd - 800) ^ 2 + yd ^ 2))If yd <= 0 Thengama1 = 2 * pi - gamaElsegama1 = gamaEnd Ifsita0 = sita + 0.001xb0 = lab * Cos(sita0 * pi / 180)yb0 = lab * Sin(sita0 * pi / 180)lbc0 = Sqr(lab ^ 2 + lac ^ 2 - 2 * lab * lac * Cos(sita0 * pi / 180)) beta0 = arcsin(lab * Sin(sita0 * pi / 180) / lbc0)xd0 = xb + lbd * Cos(beta0)yd0 = yb - lbd * Sin(beta0)gama0 = arccos((xd0 - 800) / Sqr((xd0 - 800) ^ 2 + yd0 ^ 2))If yd0 <= 0 Thengama10 = 2 * pi - gama0Elsegama10 = gama0End Ifw = (gama10 - gama1) / (pi / 180000)Picture1.PSet (sita, w * 180 / pi)NextEnd SubPrivate Sub Command4_Click()‘画运动图Timer1.Enabled = TrueEnd SubPrivate Sub Command5_Click()‘画角加速度Line1.Visible = FalseLine2.Visible = FalseLine3.Visible = Falselab = 200lac = 400lbd = 700lae = 800lce = 400pi = 3.1416Dim w2, a, sita0, sita02, gama102, gama12, sita2, beta, xb, yb, xd, yd, gama, gama1, beta0, xb0, yb0, xd0, yd0, gama0, gama10, lbc0, w1, beta2, xb2, yb2, xd2, yd2, lbc2, lbc02, gama2, beta02, xb02, yb02, xd02, yd02, gama02 As DoubleDim i, j As IntegerPicture1.BackColor = vbWhitePicture1.FillStyle = 0Picture1.Scale (-20, 140000)-(360, -140000)Picture1.Line (-0, 0)-(360, 0)Picture1.Line (0, 140000)-(0, -140000)For i = 0 To 22Picture1.CurrentX = -15Picture1.CurrentY = 140000 - 10000 * iPicture1.Print " " & 140000 - 10000 * i & " "Picture1.DrawStyle = 2Picture1.Line (0, 140000 - 10000 * i)-(360, 140000 - 10000 * i)NextFor j = 0 To 18Picture1.CurrentX = 360 - 20 * jPicture1.CurrentY = -5Picture1.Print 360 - 20 * jPicture1.DrawStyle = 2Picture1.Line (360 - 20 * j, 360)-(360 - 20 * j, -540)NextFor sita = 0 To 360 Step 0.01xb = lab * Cos(sita * pi / 180)yb = lab * Sin(sita * pi / 180)lbc = Sqr(lab ^ 2 + lac ^ 2 - 2 * lab * lac * Cos(sita * pi / 180))beta = arcsin(lab * Sin(sita * pi / 180) / lbc)xd = xb + lbd * Cos(beta)yd = yb - lbd * Sin(beta)gama = arccos((xd - 800) / Sqr((xd - 800) ^ 2 + yd ^ 2))If yd <= 0 Thengama1 = 2 * pi - gamaElsegama1 = gamaEnd Ifsita0 = sita + 0.001xb0 = lab * Cos(sita0 * pi / 180)yb0 = lab * Sin(sita0 * pi / 180)lbc0 = Sqr(lab ^ 2 + lac ^ 2 - 2 * lab * lac * Cos(sita0 * pi / 180)) beta0 = arcsin(lab * Sin(sita0 * pi / 180) / lbc0)xd0 = xb + lbd * Cos(beta0)yd0 = yb - lbd * Sin(beta0)gama0 = arccos((xd0 - 800) / Sqr((xd0 - 800) ^ 2 + yd0 ^ 2))If yd0 <= 0 Thengama10 = 2 * pi - gama0Elsegama10 = gama0End Ifw1 = (gama10 - gama1) / (pi / 180000)sita2 = sita + 0.01xb2 = lab * Cos(sita2 * pi / 180)yb2 = lab * Sin(sita2 * pi / 180)lbc2 = Sqr(lab ^ 2 + lac ^ 2 - 2 * lab * lac * Cos(sita2 * pi / 180)) beta2 = arcsin(lab * Sin(sita2 * pi / 180) / lbc2)xd2 = xb + lbd * Cos(beta2)yd2 = yb - lbd * Sin(beta2)gama2 = arccos((xd2 - 800) / Sqr((xd2 - 800) ^ 2 + yd2 ^ 2))If yd <= 0 Thengama12 = 2 * pi - gama2Elsegama12 = gamaEnd Ifsita02 = sita2 + 0.001xb02 = lab * Cos(sita02 * pi / 180)yb02 = lab * Sin(sita02 * pi / 180)lbc02 = Sqr(lab ^ 2 + lac ^ 2 - 2 * lab * lac * Cos(sita02 * pi / 180)) beta02 = arcsin(lab * Sin(sita02 * pi / 180) / lbc02)xd02 = xb02 + lbd * Cos(beta02)yd02 = yb02 - lbd * Sin(beta02)gama02 = arccos((xd02 - 800) / Sqr((xd02 - 800) ^ 2 + yd02 ^ 2)) If yd02 <= 0 Thengama102 = 2 * pi - gama02Elsegama102 = gama02End Ifw2 = (gama102 - gama12) / (pi / 180000)a = (w2 - w1) / 0.01Picture1.PSet (sita, a * 180 / pi)NextEnd SubPrivate Sub Form_Load()Timer1.Enabled = FalseEnd SubPrivate Sub Timer1_Timer()‘动画Dim k As IntegerDim a As IntegerDim beta As Doublelab = 200lac = 400lbd = 700lae = 800lce = 400pi = 3.1416Dim i, j As IntegerDim xb, yb, xd, yd As Double'画坐标系Picture1.BackColor = vbWhitePicture1.FillStyle = 0Picture1.Scale (-400, 440)-(1500, -440) Picture1.Line (-400, 0)-(1500, 0) Picture1.Line (0, -440)-(0, 440)For i = 0 To 22Picture1.CurrentX = -68Picture1.CurrentY = 440 - 40 * i Picture1.Print " " & 440 - 40 * i & " " Picture1.DrawStyle = 2Picture1.Line (-400, 440 - 40 * i)-(1500, 440 - 40 * i) NextFor j = 0 To 70Picture1.CurrentX = 1520 - 80 * jPicture1.CurrentY = -5Picture1.Print 1520 - 80 * jPicture1.DrawStyle = 2Picture1.Line (1520 - 80 * j, 440)-(1520 - 80 * j, -440) NextText1.Text = Val(Text1.Text) + 1a = Text1.Textlbc = Sqr(lab ^ 2 + lac ^ 2 - 2 * lab * lac * Cos(sita)) beta = arcsin(lab * Sin(sita) / lbc)Line1.X1 = 0Line1.Y1 = 0Line1.X2 = lab * Cos(a)Line1.Y2 = lab * Sin(a)Line2.X1 = lab * Cos(a)Line2.Y1 = lab * Sin(a)Line2.X2 = Line2.X1 + lbd * Cos(beta)Line2.Y2 = Line2.Y1 - lbd * Sin(beta)Line3.Y1 = Line2.Y2Line3.X1 = Line2.X2Line3.X2 = 800Line3.Y2 = 0Line1.Visible = TrueLine2.Visible = TrueLine3.Visible = TrueFor sita = 0 To 2 * pi Step 0.001xb = lab * Cos(sita)yb = lab * Sin(sita)lbc = Sqr(lab ^ 2 + lac ^ 2 - 2 * lab * lac * Cos(sita)) beta = arcsin(lab * Sin(sita) / lbc)xd = xb + lbd * Cos(beta)yd = yb - lbd * Sin(beta)Picture1.PSet (xd, yd)NextEnd Sub运行上述程序点击“运动”按钮得运动动画,截图如下:D点的轨迹如下:角位移如下:角速度如下:角加速度如下:运动结果分析D点做整周的回转运动,若以向右为角度的起始处,在D转角位180度时角速度最小,转角位0处角速度最大。
- 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
连杆机构设计1设计题目2对机构进行结构分析,找出基本杆组①AB即杆件1为原动件②DECB即杆件2、3为RRR型II级杆组③其中CE为同一构件上点,④EF和滑块即4、5为RRP型II级杆组B<9)在图1-9所示的机构中,已知l AB=60mm , bc=180mm , l DE=200mm , l cD=120mm , l EF=300mm , h=80mm , h i=85mm , h2=225mm,构件1以等角速度W i=100rad/s转动。
求在一个运动循环中,滑块5的位移、速度和加速度曲线。
3各基本杆组的运动分析数学模型② RRR 杆组运动分析的数学模型1. 位置分析设两个构件长度R ,氏及外运动副N i , N2的位置已知,求两个构件的位置 角色及内运动副N 3的位置。
选定坐标系及相应的标号如下图,构件的位置角R 约定从响应构件的外运 动副Ni引x 轴的方向线,按逆时针量取。
设外运动副Nl ,N2的位置坐标分别为Nl <R x , py ),N 2<P 2x ,P 2y ),则2 21d =[( P x - Ex )2" P y -P 2y )2『cos : =(d 2R 2R ;)/(2 Rd)申=arctan((P 2y - P y )/( Bx - P x )哥八-: 内运动副N 3点坐标为:P 3x =Px + RcosqP 3y= R y + R SinQ构件K 2的位置角:arctan[(F 3y - By )/(忠 - Bx )]位置分析过程中应注意两个问题:<1)因为Nl,N2的位置及杆长 Rl,&都是给定的,这就可能出现 d >Ri +民或d< R -R2的情况。
在这 两种情况下实际上不可能形成 RRR 杆组,计算过程中应及时验算上述条件,如满足上述条件应中止运算并给出相应信息。
<2)在给定N1,N 2,R1,&的条件下,N3可能有两个位置如上图中的 N3和N3,相应的弓二'*和弓二'-■,我们称为杆组的两种工作状态。
对于 实际构件而言,杆组只可能在一种工作状态下运动,而且在机构运动过程中只 要不出现d=R -尺 的情况<这种情况下,机构处于瞬时运动不确定状态,设 计时应避免)杆组就不会从一种工作状态变为另一种工作状态,所以运动分析 时可预先按机构的实际工作位置,指明杆组是哪一种工作状态。
约定状态参数 M :叫怡弘为逆时针读取时M =1,叫2“3为顺时钟时M =-1。
2. 速度分析%M设外运动副N1,N2点的速度V1x,V1y及V2x,V2y已知,求N3点的速度V3x,V3y及构件Kl, K 2的角速度11「。
因为 P 3x = P1x RCOS^i = l~2x R2C0S“2Ry =P y + Rsi nq =F 2y + &si n^2将上式对时间t 微分:V 3x =比乂 一 R i sin 齐二 V 2x _ &「2 sin 丁2 v 3y 二My R “cos 冃二v 2y R 2,2COSJ 2 注意到:R iCOS^i = Ex -Pix ,R 1 sin4 = Ey — p y1二-[(V2x -V 1x)( Rx -Bx) (V 2y-£)( By -P2y )]/Q国2 = -|XV 2y-灯)(Ry - R y ) +(V 2x -V 1x)(P3x - R x)]/ Q将J ,' '2值代入式<1)即可求得V3x , V3y 3. 加速度分析设外运动副N1, N 2点的加速度a1x ,a ^y,a2x ,a2y已知,求N3点的加速度a3x ,a3y,及构件K1,K2的角加速度 "2。
将式(1>对时间t 微分得:*-(R y -P y ) (R y - By)严 1 1=「E A 1 l(P 3x - P 1x ) -(P 3x - Px )」[名2 - ]E B _式中.EA = a2x " a1x (V3y - My)'" "(V3y " V 2y^ 2E B- a 2y - a 1y -(V 3x - V 1x )1 (V 3x - V 2x ) 2「-[E A ( Bx —Bx ) E B ( F 3y -By )]/Q 客2= -[E A ( F 3x-R X ) + E B ( F 3y-R y )]/Q 内运动副N3点的加速度a3x ,a 3y可由微分式V1)求得。
③ 平面运动构件V 单杆)的运动分析已知构件K 上的N1点的位置Px ,Py,速度为V1x, V1Y ,加速度为a1x, a1y 及 过点的N1点的线段N^2的位置角二,构件的角速度3,角加速度&,求构件上 点N 2和任意指定点N3<位置参数N 1N3= &,- “2叫“3=)的位置、速度、 加速度。
<1PBy -Py)(B y- -P 2y )'◎ V2x一V 1x=l(P3x — P1x ) 4P3x-P2x)> 5 一 》2y - V1y _Q = ( By - R y )( Bx -巳x ) -( py - P2y )(P3x -P1x)则: R2COS*? = Ex - p>x ,R2sin日2 = py - P2y 式<1)可写为令:④ RRP 杆组运动分析的数学模型1. 位置分析设已知外运动副点Nl 及移动副导路上任意 一选定参考点N2的位置,构件Kl的长度Rl 及 导路的位置角-,求构件&的位置角円及内运 动副N3点的位置 <如右图)。
:角从水平线到即-arctan[(R y - 氐)/ (R x - 氐)]由Nl 向导路作垂线,垂足为 A ,令NlA=u , N2A=e , N3A = f则e = d cos( : _ -)> u =dsin(--)2 2 1f =(R 2-u 2)2N3点相对于导路上参考点N2的滑移距离:R 2 =e _ fNl, N 3点的位置为:忠=Rx + R cos 日P 2y =Ry +R sin 日 R 3x = Rx 十 R 2 cos (日十了)N ia 3yN3点的速度,加速度为:V 2x =V ix-R 佃sin 。
= V ix-叭Ey -R y ) V 2y =V iy _R i 国 sin 8 = My _co (F 2x _ R x ) V 3x % - R^sin(8 +Y )=见-aQ y - R y ) V 3y =% - R 2C0 cos® +?)=知-国傀-R x ) a2x = Cx --R y ) -国(R 2x-R x )2a2y = aiy十 E (R x — R x) — 国(B y —R y )2a 3x =4x - ^(B y - R y )(R3x - R x )2N 2N 3 度量21d =[(R 叽)2+亿七)显然,当R <|u|时无解。
当R |u 1时有两个解,对应于杆组的不同位置状态。
若/R2 =e f,约定状态参数M =1 ;若/ NIN2N3>2,则金=e- f,则约定状态参数M =-1 0 内运动副N3的位置坐标:P3x=P2x R2 COS , Ry 二F2y R z Sin 一:构件K i的位置角:q =arctan[(P3y-R y)/ (R x-R x)]2.速度分析Nl, N2点的速度为Vlx, Vly及v2x, V2y已知,导路的角速度':,求构件Ki 的角速度1,点的速度%x,V3y及N3点相对于导路上重合点的相对速度构件%R3x= P x +R cosq = P2x+ R2 cos P珀=R y +Rsinq =R2y +&sin P<2)上式对时间t微分,可解出:3= ‘i=(-E v Sin :F v COS:)/QR2 =%2 =TE v(R3x -R x) + F v(P3y -R y)]/ Q式中:E v =V2x -V ix 「sin 1F v =V2y -Wy - -cos :Q =(F3y _Py)sin P +(F3x _P x) cosPN3点的速度为:V3x =V ix — R i 1 si n^iJV3y 严y R i cosq3.加速度分析Ni,N2点的加速度aix,aiy,a2x,a2y及移动副导路的角加速度Y已知,求构件Ki的角加速度;i,N3点的加速度a3x,a3y,及N3点相对于移动副导路上重合点的相对角速度a「2 0对式<2)进行两次微分可得:“ =(-E A sin 一:F A cos J/Qa「2 =—(E A Q X一P x) + F A(F3y - R y))/ Q式中:E A =a2x -Qx Wi2(P3x -P x) -晞R2cosP -2沖V r2Sin p -邛(P jy - F2y)F A =a2y _a iy +^i2(P3y _P y) _⑷;R2Sin P +2⑷輕边cosP _g0(P3x _Bx)N1N2N3Q =(F3y _R y)sin 0 +(F3x _P x) cosPN3点的加速度:2a3x =a!X - R i ‘1 cos哥一R “ sin y2a3y =q y—Rc^ sin d+R 岛cosq4建立坐标系,程序设计及画图以D点为坐标原点,自然方向为坐标xy轴<1)滑块5的位移曲线<使用matlab编程画图,详见附录1)<2)滑块5的速度曲线<使用matlab编程画图,详见附录2)8<3)滑块5的加速度曲线<使用matlab编程画图,详见附录3)附录附录 1t=0:0.0002.*pi:0.04.*pi 。
xd=225+60.*cos(100.*t> 。
yd=80+60.*sin(100.*t> 。
A0=2.*120.*xd 。
B0=2.*120.*yd 。
C0=120.A2+xd.A2+yd.A2-180.A2 。
ai=2.*ata n((B0+sqrt(A0.A2+B0A2-C0A2»./(A0+C0>> xe=200.*cos(ai>。
ye=200.*sin(ai> 。
xf=xe-sqrt(300.A2-(165-ye>.A2> 。
plot(t,xf>附录 2t=0:0.0002.*pi:0.04.*pi 。
xd=225+60.*cos(100.*t> 。
yd=80+60.*sin(100.*t> 。