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Pipefitting Math and Calculations Flashcards

6 cards from real PIPEFITTER practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

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  1. A pipe must travel 18 inches horizontally and 18 inches vertically. Using 45° elbows, what is the travel distance (center-to-center) along the offset pipe?

    Answer: 25.46 inches

    For a 45° offset, travel = offset × 1.414. Since both legs are 18 inches, the offset is 18 inches, so travel = 18 × 1.414 = 25.46 inches.

  2. What is the formula to calculate the circumference of a pipe for layout work?

    Answer: C = π × D (3.1416 × outside diameter)

    Circumference = π × D, where D is the outside diameter of the pipe. This is used to lay out saddle cuts, branch miter cuts, and wrap-around marks.

  3. If a pipe run must drop 6 inches over a horizontal distance of 10 feet to achieve proper drainage slope, what is the slope expressed as a fraction per foot?

    Answer: 1/2 inch per foot

    6 inches ÷ 10 feet = 0.6 inches/foot. Expressed as a fraction, this rounds to approximately 1/2 inch per foot (0.5"/ft).

  4. What is the 'take-out' of a 90° long-radius elbow on a 4-inch NPS pipe (LR elbow center-to-face dimension)?

    Answer: 6 inches

    For a long-radius (LR) 90° elbow, the center-to-face dimension = 1.5 × NPS. For 4-inch NPS: 1.5 × 4 = 6 inches.

  5. A pipe flange has 8 bolt holes on a 9.5-inch bolt circle diameter (BCD). What is the spacing between adjacent bolt holes?

    Answer: 3.74 inches

    Bolt hole spacing = π × BCD ÷ number of bolts = 3.1416 × 9.5 ÷ 8 = 29.85 ÷ 8 = 3.73 inches.

  6. What is the developed length of a 90° long-radius elbow fitting for a 3-inch NPS pipe?

    Answer: 7.07 inches

    Developed length of a 90° LR elbow = (π/2) × CLR = 1.5708 × (1.5 × NPS) = 1.5708 × 4.5 = 7.07 inches.