转矩: Ft = 100 (Nm)
位置矢量: R = 0.2 (m)
连杆长 : L = 1.6 (m)
作用力 : Fc = 500 (N)
设 连杆摆角为 A,曲柄角为B, 角度 A、B 均为弧度。因 R sin B = L sin A , 得 sin A = ( R sin B ) / L ---- (1)
设 切向力为 Fc、 连杆力为 Fg ,
得 Fg = Fc / sin B ------------------- (2)
F = Fg cos A ------------------- (3)
将 (2)代入(3) F = (Fc / sin B) * cos A
年龄 | 转换为弧度 | 余弦值 | Δ 的开方 | 边长 C 的值1 | 边长 C 的值2(负值,舍去) | Fn 的值 | 角度 γ | 力值 |
24 | =PI()*(I10/180) | =COS(J10) | =SQRT(0.4*K10*0.4*K10+4*2.52) | =(0.4+L10)/2 | =(0.4-L10)/2 | =100/M10 | =ASIN(SIN(J10)/8) | =O10/TAN(P10) |
25 | =PI()*(I11/180) | =COS(J11) | =SQRT(0.4*K11*0.4*K11+4*2.52) | =(0.4+L11)/2 | =(0.4-L11)/2 | =100/M11 | =ASIN(SIN(J11)/8) | =O11/TAN(P11) |
26 | =PI()*(I12/180) | =COS(J12) | =SQRT(0.4*K12*0.4*K12+4*2.52) | =(0.4+L12)/2 | =(0.4-L12)/2 | =100/M12 | =ASIN(SIN(J12)/8) | =O12/TAN(P12) |
27 | =PI()*(I13/180) | =COS(J13) | =SQRT(0.4*K13*0.4*K13+4*2.52) | =(0.4+L13)/2 | =(0.4-L13)/2 | =100/M13 | =ASIN(SIN(J13)/8) | =O13/TAN(P13) |
28 | =PI()*(I14/180) | =COS(J14) | =SQRT(0.4*K14*0.4*K14+4*2.52) | =(0.4+L14)/2 | =(0.4-L14)/2 | =100/M14 | =ASIN(SIN(J14)/8) | =O14/TAN(P14) |
29 | =PI()*(I15/180) | =COS(J15) | =SQRT(0.4*K15*0.4*K15+4*2.52) | =(0.4+L15)/2 | =(0.4-L15)/2 | =100/M15 | =ASIN(SIN(J15)/8) | =O15/TAN(P15) |
30 | =PI()*(I16/180) | =COS(J16) | =SQRT(0.4*K16*0.4*K16+4*2.52) | =(0.4+L16)/2 | =(0.4-L16)/2 | =100/M16 | =ASIN(SIN(J16)/8) | =O16/TAN(P16) |
31 | =PI()*(I17/180) | =COS(J17) | =SQRT(0.4*K17*0.4*K17+4*2.52) | =(0.4+L17)/2 | =(0.4-L17)/2 | =100/M17 | =ASIN(SIN(J17)/8) | =O17/TAN(P17) |
32 | =PI()*(I18/180) | =COS(J18) | =SQRT(0.4*K18*0.4*K18+4*2.52) | =(0.4+L18)/2 | =(0.4-L18)/2 | =100/M18 | =ASIN(SIN(J18)/8) | =O18/TAN(P18) |
小鱼人zhangrsl 发表于 2016-1-14 16:46
F=990.813814
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dongdong4834 发表于 2019-4-9 14:10
你好 请问这个是什么软件计算的
wenbotao2004 发表于 2020-3-7 16:29
如果行程是75mm,F是20000N怎么算连杆长度?
子子61961 发表于 2020-3-8 12:01
看你的提问,像是有一个新的机构要设计的样子。
感觉你的条件还不够完全。
一个是,扭矩需要知道。
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