Gear Noise and Vibration Episode 1 Part 2 ppt

Metal Machining - Theory and Applications Episode 1 Part 2 pps

Metal Machining - Theory and Applications Episode 1 Part 2 pps

... 0.85; n s = 2 n s = 2 n s = 2 n s = 3 Mechanical 1 kW 6000 0. 028 Simple CNC 1 kW 20 000 0.0 92 0.060 5 kW 28 000 0 .13 0.086 15 kW 50000 0 .23 0 .15 Turning centre 5 kW 60000 0 .18 0 .16 0 .11 15 kW 12 0000 ... previous figures and tables – Figure 1. 16 (milling machine costs), Table 1. 1 (cutting tool data) and equations (1. 10) and (1. 11) for cost rates. 30 Intro...

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Cutting Tools Episode 1 Part 2 ppt

Cutting Tools Episode 1 Part 2 ppt

... Thickness (2) A= 0.0 01 B = 0.00 02 C = 0.0005 D = 0.0005 E = 0.0 01 G = 0.0 01 (3)M = 0.0 02- 0. 010 (3)U = 0.005-0. 0 12 0.0 01 0.0 01 0.0 01 0.0 01 0.0 01 0.0 01 0.0 02- 0.004 0.005-0. 010 0.0 01 0.005 0.0 01 0.005 0.0 01 0.005 0.005 0.005 R ... segment Grain fragments Workpiece Chip Tool 675 750 850 930 930 11 00 11 00 11 00…F 13 00 12 00 12 00 13 00 FIGURE 2. 7...

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Metal Machining - Theory and Applications Episode 1 Part 10 ppt

Metal Machining - Theory and Applications Episode 1 Part 10 ppt

... (6.39) cos l s 19 0 Advances in mechanics Childs Part 2 28:3 :20 00 3 :13 pm Page 19 0 (x′ 6 – x′ C2 + t 1e,q3–4 sin h′ c ) 2 + (y′ 6 – y′ C2 + t 1e,q3–4 cos h′ c ) 2 = r 2 n (6.43) t 1eq3–4 is obtained ... book, equation (2 .18 ) is modified by a factor c 17 4 Advances in mechanics Childs Part 2 28:3 :20 00 3 : 12 pm Page 17 4 6.4 .2 Tool geometry Figure 6 .16 shows plans...

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Metal Machining - Theory and Applications Episode 1 Part 9 pptx

Metal Machining - Theory and Applications Episode 1 Part 9 pptx

... mild steels, after (a) Palmer and Oxley (19 59), and (b) Roth and Oxley (19 72) Childs Part 2 28:3 :20 00 3 : 12 pm Page 16 9 Harris, A., Hastings, W. F. and Mathew, P. (19 80) The experimental measurement ... face. Annals CIRP 19 (1) , 395–398. References 15 7 Fig. 5 .24 Relation between flank wear VB and amplitude of AE signal, after Miwa et al. (19 81) Childs Part...

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Metal Machining - Theory and Applications Episode 1 Part 7 ppt

Metal Machining - Theory and Applications Episode 1 Part 7 ppt

... min 1 , d = 1. 0 mm, f = 0 .19 mm rev 1 , t = 1 min; (e) tool: cemented carbide P10, v = 15 0 m min 1 , d = 0.5 mm, f = 0 .1 mm rev 1 , t = 2 min. Childs Part 1 28 :3 :20 00 2: 41 pm Page 12 0 takes ... carbon steel with a P10 carbide tool at 23 9 m/min (Yamane and Narutaki, 19 83) Depth of cut: 1. 0 mm Feed rate: 0 .2 mm rev 1 or tooth 1 Childs Part...

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TOEIC Vocabulary Tests Episode 1 part 2 pptx

TOEIC Vocabulary Tests Episode 1 part 2 pptx

... www.english-test.net Test 21 TOEIC Vocabulary / Meaning by Word / Test # 21 Q1 n. infancy (a) a support that consists of a horizontal surface for holding objects; projection; rock ledge; sandbank (b) beginning ... before noon; hours between midnight and noon (d) language used to write pages and sites for the Internet Q9 abbr. H.R. (a) period of about 30 days or 4 weeks; 1/ 12 of a y...

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Writing Skills For GRE-GMAT Episode 1 Part 2 ppt

Writing Skills For GRE-GMAT Episode 1 Part 2 ppt

... nol .a1ql3e11r u() sl)ploq a^urtgatpurrl :no,,i. ;1 . 'tu)rf,U1) ;r:ou eq ,i11nn::r .i.tru s,irss: :no,i fuuu,spuurl .:srcl;il ,r,oJr- ,(y;nrrads: ur a:.no,i 11 . :3u1no11o; aql ... lA NOO ,tr&apos ;2 .1& quot;-> ,r. ,1+ r-relyr4 n. -' ,1 A7. @l fn.lr,nt o4 'aT74 .,oe .11 !' ,. @ .%,__:t_, ii. 1. '1o .1, t1''.' or,, E....

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GMAT reading Episode 1 Part 2 pptx

GMAT reading Episode 1 Part 2 pptx

... 600, and its remaining area was being raided by Arabs and Bulgarians, who at times threatened to take Constantinople and extinguish the (10 ) empire altogether. The wealth of the state and ... on its own territory and had begun to raid and conquer enemy territory, Byzantine resources naturally expanded and more money became available to patronize art and lit-...

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SAT II Math Episode 1 Part 2 ppt

SAT II Math Episode 1 Part 2 ppt

... If log 2 = a and log 3 = b, express log 12 in terms of a and b. Solution: 4. In the formula A = P (1 + r) n , express n in terms of A, P, and r. Solution: 5. If log t 2 = 0.87 62, log 10 0t = Solution: Part ... for y, and y for x. 9. If = Solution: Hence 10 . If the functions f and g are defined as f(x) = x 2 – 2 and g(x) = 2x + 1, what is f [g(x)]? Solution: Math...

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SAT II History Episode 1 Part 2 pptx

SAT II History Episode 1 Part 2 pptx

... and Arizona Year 17 00 17 10 17 20 17 30 17 40 17 50 17 60 17 70 17 75 395,0 21 24 9, 814 468 ,18 8 5 72, 585 718 , 416 814 ,768 7 61, 099 1, 015 ,535 1, 920 ,950 344,3 41 29 3,659 319 ,7 02 536,860 813 ,3 82 1, 313 ,083 2, 611 ,764 1, 925 ,5 71 196 ,16 2 Exports Imports Source: ... B 14 . B 15 . C 16 . B 17 . C 18 . D 19 . D 20 . C 21 . D 22 . A 23...

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