ELDING LAYOU d the pipe fo a chalk line around the UCERS 110

ELDING LAYOU d the pipe fo a chalk line around the UCERS 110 draw a chalk line a line C. (See Fig. 67 (a),he ipe draw a chalk lin Note: D is not const D is not constant; it Two-Cut Eccentric Reducer with the size o the ic,ar with the size of the eccent Formulas A=OD. of the larger pipe B=2in. hin auently, the aboveforao only approximate 3. Solve for E and la tance one-half a low line 4. Draw ne.is tance 4. Draw straigt chalk back to line C and numbet halkbe o n sigchalk line frow 5 and 6. Then, using the these liey n. (approximately) -half above and one-ha circumferences distance from Y to L E=½ the difference between the two = as a gu point guidelidrawing htt : Hdistance from L to P locating point p ocatingpe into four equal parts near the end starting at the top. Draw straig lines from the end of the pipe at cach around as a quarter line making them equa in and P length to 1 times.the.diameter of the larger pipe Number these lines 1 2, 3 and 4, as shown in Fig. 67 (B) 4. Point L is located at th section of lines 4 and C. Usithe inte t L. is located at the inte Po lines e, connect poimts 1. Divide the surface of the pipe o of les ocatedon to around as a guide, connect g (B) END VIEew 5. At the intersection of c a 1, lay off B, one-half on each ind (A) SIDE VIEW lay off B, one-half on each nd line line 1 locating points Y, each 6. Lay off distance G f sie r Fig. 67. Two-cut eccentric reducer off dsta Connect Y and M . ne 4, locating M. Connect fr 2 Lay off distances A and D on Y a oh with a straight chalk line, nd M sing the wrap around as a gude, with a

Solution

The formula for calculating G when E is Known is

G=E×3

Given E= 6 inches therefore G= E×3 = 6×3 = 18 inches