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E4004

E4004. Offset to Three Straights. A circular curve has clearances to the three straights as shown in the diagram. Calculate R. A. 2.0. 114°16’. 4.5. 180°0’ 37.0. B. 8.0. 106°12’. C. There are two deflection angles D ’ and D ”. EC is a straight line so.

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E4004

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  1. E4004 Offset to Three Straights

  2. A circular curve has clearances to the three straights as shown in the diagram. Calculate R. A 2.0 114°16’ 4.5 180°0’ 37.0 B 8.0 106°12’ C

  3. There are two deflection angles D’ and D” EC is a straight line so... DBE is a straight line so... A 2.0 114°16’ D D’ 4.5 180°0’ 37.0 B E 8.0 D” 106°12’ C

  4. There is a formula to solve for R if the curve is tangential to the three straights so construct lines parallel to the three straights and tangential to the curve The deflection angles between the three parallel lines are the same 114°16’ A 2.0 114°16’ D A’ D’ D’ D’ 4.5 180°0’ 37.0 B E D” E’ C’ Require the distance D’E’ 8.0 D” 106°12’ 253°48’ C

  5. Construct D’N perpendicular to DE Construct E’P perpendicular to DE 114°16’ A 2.0 114°16’ D A’ D’E’ = NP = DE - ( DN + PE ) D’ D’ N D’ 4.5 180°0’ 37.0 B P E D” E’ C’ 8.0 D” 106°12’ 253°48’ C

  6. Construct D’M perpendicular to AD Close D’MDN D’M 24°16’ 2.0 MD 114°16’ ( ) DN 180°0’ ( ) ND’ 270°0’ 4.5 4.0351 114°16’ A 0.16517 M 2.0 114°16’ D A’ D’ D’ N D’ 4.5 180°0’ 37.0 B P E D” E’ C’ 8.0 D” 106°12’ 253°48’ C

  7. Construct E’Q perpendicular to EC Close E’PEQ E’P 90°0’ 4.5 PE 180°0’ ( ) EQ 253°48’ ( ) QE’ 343°48’ 8.0 7.02341 114°16’ A 2.36185 M 2.0 114°16’ D A’ D’ D’ N D’ 4.5 180°0’ 37.0 B P E D” E’ C’ 8.0 D” 106°12’ Q 253°48’ C

  8. D’E’ = NP = DE - ( DN + PE ) D’E’ = NP = 37.0 - ( 7.02341 + 0.16517 ) = 29.81142 114°16’ A M 2.0 114°16’ D A’ D’ D’ N D’ 4.5 180°0’ 37.0 B P E D” E’ C’ 8.0 D” 106°12’ Q 253°48’ C

  9. Check Calculate AD and EC by closing OADEC OA 24°16’ 21.34074+2.0 AD 114°16’ ( ) DE 180°0’ 37.0 EC 253°48’ ( ) CO 343°48’ 21.34074+8.0 114°16’ 17.82285 A M 2.0 114°16’ D A’ 18.38493 D’ D’ N D’ 4.5 180°0’ 37.0 B AO = 23.34074 OB = 25.84074 OC = 29.34074 P E D” E’ C’ 8.0 D” 106°12’ Q 253°48’ C

  10. Check OA 24°16’ 21.34074+2.0 AD 114°16’ 17.82285 AO = 23.34074 OB = 25.84074 OC = 29.34074 REF L 114°16’ E N -23.34074 114°16’ 17.82285 A M 2.0 114°16’ REF L 180°0’ E N 25.84074 D A’ D’ D’ -13.95350 N D’ B?D? DE 180°0’ 37.0 REF L 253°48’ E N 4.5 180°0’ 37.0 B -29.34074 P E D” E’ -18.38493 C’ 8.0 D” 106°12’ Q 253°48’ C

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