T\u00ednh c\u01b0\u1ee3c xi\u00ean theo k\u00e8o t\u00e0i x\u1ec9u v\u00e0 k\u00e8o Ch\u00e2u \u00c1<\/span><\/h3>\nK\u00e8o t\u00e0i x\u1ec9u v\u00e0 k\u00e8o Ch\u00e2u \u00c1 \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng r\u1ea5t ph\u1ed5 bi\u1ebfn t\u1ea1i Kubet hi\u1ec7n nay. H\u01a1n n\u1eefa \u0111\u00e2y l\u00e0 lo\u1ea1i h\u00ecnh k\u00e8o c\u01b0\u1ee3c c\u1ef1c d\u1ec5 th\u1eafng. V\u00ec v\u1eady anh em c\u1ea7n n\u1eafm \u0111\u01b0\u1ee3c c\u00f4ng th\u1ee9c t\u00ednh c\u01b0\u1ee3c xi\u00ean theo 2 k\u00e8o n\u00e0y nh\u01b0 sau:\u00a0<\/span><\/p>\n\n- Trong tr\u01b0\u1eddng h\u1ee3p k\u00e8o c\u01b0\u1ee3c xi\u00ean th\u1eafng = H\u1ec7 s\u1ed1 c\u01b0\u1ee3c k\u00e8o 1 x H\u1ec7 s\u1ed1 k\u00e8o c\u01b0\u1ee3c 2 x \u2026.x h\u1ec7 s\u1ed1 k\u00e8o c\u01b0\u1ee3c n.<\/span><\/li>\n
- N\u1ebfu t\u1ef7 l\u1ec7 k\u00e8o c\u01b0\u1ee3c xi\u00ean \u0103n 0.5 th\u00ec s\u1ebd b\u1eb1ng (t\u1ef7 l\u1ec7 k\u00e8o -1)\/2 = 1.\u00a0<\/span><\/li>\n
- Trong t\u00ecnh hu\u1ed1ng k\u00e8o c\u01b0\u1ee3c xi\u00ean thua 0.5 = h\u1ec7 s\u1ed1 c\u1ea3 xi\u00ean : 2.\u00a0<\/span><\/li>\n
- N\u1ebfu k\u00e8o c\u01b0\u1ee3c h\u00f2a th\u00ec s\u1ebd b\u1eb1ng\u00a0 = (h\u1ec7 s\u1ed1 x 1).\u00a0\u00a0<\/span><\/li>\n
- Trong tr\u01b0\u1eddng h\u1ee3p c\u00f3 4 k\u00e8o xu\u1ea5t hi\u1ec7n t\u1ea1i m\u1ed9t v\u00e1n c\u01b0\u1ee3c th\u00ec t\u00ednh l\u00e0: H\u1ec7 s\u1ed1 k\u00e8o th\u1eafng x(1 +(t\u1ef7 l\u1ec7 k\u00e8o \u0103n 0.5-1)\/2×1\/2(t\u1ef7 l\u1ec7 k\u00e8o thua 0.5)x1x(t\u1ef7 l\u1ec7 k\u00e8o h\u00f2a)x\u2026x h\u1ec7 s\u1ed1 k\u00e8o th\u1ee9 n.\u00a0<\/span><\/li>\n<\/ul>\n