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CORRECTED MATHEMATICAL FORMULA, Summaries of Motor Vehicle Design

University of Washington (UW) - SeattleMotor Vehicle Design

45. 1.41. 22. 2.67. 46. 1.39. 23. 2.56. 47. 1.37. 24. 2.46. 48. 1.35. DISTANCE BETWEEN BENDS = OFFSET MULTIPLIER FOR. DEGREE X OFFSET HEIGHT. EXAMPLE:.

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CORRECTED MATHEMATICAL FORMULA
FOR HYDRAULIC TUBE BENDING
The following formula should be used to determine the start point for each required bend.
The example will be for a Double Bevel Bend
L1 = 26 25 degree angle
L2 = 15 65 degree angle
L3 = 39 65 degree angle
L4 = 15 25 degree angle
L5 = 26
Bend #1 Start Point:
L1 – ½ developed length (DL25) – ½ (Gain)
26” – ½ (3.064”) – ½ (.050)
26” – 1.532” - .025” = 24.448
Bend #2 Start Point:
L1 + L2 – Gain 1 – ½ (DL65) – ½ (Gain 2)
26 + 15 - .050 –1/2 (7.941) – ½ (.973”)
26 + 15 - .050 – 3.9705 - .48895 = 36.491
Bend #3 Start Point:
L1 + L2 + L3 – Gain 1 – Gain 2 – ½ (DL65) – ½ (Gain 3)
26 + 15 + 39 - .050” - .978 – ½(7.941) – ½ (.972”)
26 + 15 + 39 - .050” -.978 – 3.9705 - 0.486 = 74.515
Bend #4 Start Point:
L1 + L2 + L3 + L4 – Gain 1 – Gain 2 – Gain 3 – ½ (DL25) – ½ (Gain 4)
26 + 15 + 39 + 15 -.050 - .978 - .978 –1/2 (3.054) – ½ (0.497)
26 + 15 + 39 + 15 - .050 - .978 - .978 – 1.527 – 0.025 = 91.442
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CORRECTED MATHEMATICAL FORMULA

FOR HYDRAULIC TUBE BENDING

The following formula should be used to determine the start point for each required bend. The example will be for a Double Bevel Bend L1 = 26 L2 = 15 25 degree angle65 degree angle L3 = 39 65 degree angle L4 = 15 25 degree angle L5 = 26

Bend #1 Start Point: L1 – ½ developed length (DL25) – ½ (Gain) 26” – ½ (3.064”) – ½ (.050) 26” – 1.532” - .025” = 24. Bend #2 Start Point: L1 + L2 – Gain 1 – ½ (DL65) – ½ (Gain 2) 26 + 15 - .050 –1/2 (7.941) – ½ (.973”) 26 + 15 - .050 – 3.9705 - .48895 = 36. Bend #3 Start Point: L1 + L2 + L3 – Gain 1 – Gain 2 – ½ (DL65) – ½ (Gain 3) 26 + 15 + 39 - .050” - .978 – ½(7.941) – ½ (.972”) 26 + 15 + 39 - .050” -.978 – 3.9705 - 0.486 = 74. Bend #4 Start Point: L1 + L2 + L3 + L4 – Gain 1 – Gain 2 – Gain 3 – ½ (DL25) – ½ (Gain 4) 26 + 15 + 39 + 15 -.050 - .978 - .978 –1/2 (3.054) – ½ (0.497) 26 + 15 + 39 + 15 - .050 - .978 - .978 – 1.527 – 0.025 = 91.

GAIN FACTORS

GAIN = GAIN FACTOR FOR DEGREE OF BEND X RADIUS EXAMPLE: FIND THE GAIN FOR AN 85 DEGREE BEND

  • 1 .0000 31 .0136 61. Degree ofBend Multiplier of BendDegree Multiplier of BendDegree Multiplier
  • 2 .0000 32 .0150 62.
  • 3 .0 000 33 .0165 63.
  • 4 .0000 34 .0181 64.
  • 5 .0000 35 .0197 65.
  • 6 .0001 36 .0215 66.
  • 7 .0001 37 .0234 67.
  • 8 .0003 38 .0254 68.
  • 9 .0003 39 .0276 69.
  • 10 .0005 40 .0298 70.
  • 11 .0006 41 .0322 71 .187
  • 12 .0008 42 .0347 72.
  • 13 .0010 43 .0373 73.
  • 14 .0013 44 .0400 74.
  • 15 .0015 45 .0430 75.
  • 16 .0018 46 .0461 76.
  • 17 .0022 47 .0493 77.
  • 18 .0026 48 .0527 78.
  • 19 .0031 49 .0562 79.
  • 20 .0036 50 .0600 80.
  • 21 .0042 51 .0637 81.
  • 22 .0048 52 .0679 82.
  • 23 .0055 53 .0721 83.
  • 24 .0062 54 .0766 84.
  • 25 .0071 55 .0812 85.
  • 26 .0079 56 .0860 86.
  • 27 .0090 57 .0911 87.
  • 28 .0100 58 .0963 88.
  • 29 .0111 59 .1018 89.
  • 30 .0126 60 .1075 90. - GAIN = .3491 X 7 = 2.4437 OR 2-7/ USING A 7 INCH RADIUS

TABLE FOR OFFSET MULTIPLIER

Degree of

Bend^ Multiplier^

Degree of

Bend^ Multiplier

DISTANCE BETWEEN BENDS = OFFSET MULTIPLIER FOR

DEGREE X OFFSET HEIGHT

EXAMPLE:

FIND THE DISTANCE BETWEEN BENDS FOR A

15 INCH OFFSET USING 25 DEGREE BENDS.

DISTANCE BETWEEN BENDS = 2.37 X 15 = 35.55 OR 35