The physics calculations

Same five-part form as GEN-H Procedures, and the same rule: every procedure ends with a check, and the check is how you know the answer is wrong.

Six of this unit's eight skills are calculations. S5 is a bench demonstration and S7 is an identification; both are covered in §7.

Procedure 1 — Temperature conversion (S1)

Given — an outside air temperature of 30 °C. Sought — the same in °F.

Method — °F = °C × 9/5 + 32 = 30 × 1.8 + 32 = 86 °F. The reverse: °C = (°F − 32) × 5/9 = (86 − 32) × 5/9 = 30 °C.

Check — −40 is the same in both scales. Anchor to it: any Celsius value above −40 must convert to a Fahrenheit value that is higher, and the gap widens as you go up. If your answer is lower than the Celsius figure and you are above −40, you have subtracted 32 before multiplying.

Failure — applying the 32 in the wrong order. Order of operations, arriving from GEN-H Concepts §5, and it produces a number that is wrong by about 26 degrees and looks reasonable.

Procedure 2 — Pressure altitude (S3)

Given — field elevation 3,000 ft, altimeter setting 29.42 in Hg. Sought — pressure altitude.

Method

1. Difference from standard: 29.92 − 29.42 = 0.50 in Hg.
2. Each inch is about 1,000 ft, so 0.50 × 1,000 = 500 ft.
3. Pressure altitude = 3,000 + 500 = 3,500 ft.

Check — a low altimeter setting means high pressure altitude. Low pressure, thin air, the aircraft behaves as though higher up. If your answer came out below field elevation with a setting under 29.92, you have subtracted the wrong way.

Failure — reversing the subtraction gives 2,500 ft, and every performance figure derived from it is then optimistic. Optimistic is the dangerous direction.

Procedure 3 — Density altitude (S2)

Given — pressure altitude 3,500 ft (from Procedure 2), OAT 30 °C. Sought — density altitude.

Method

1. Standard temperature at that altitude: 15 − (2 × 3.5) = 8 °C.
2. Temperature deviation: 30 − 8 = +22 °C, warmer than standard.
3. Density altitude = 3,500 + (120 × 22) = 3,500 + 2,640 = 6,140 ft.

Check — warmer than standard always raises density altitude; colder always lowers it. Confirm the sign of the deviation matches the direction the answer moved. On a hot day at a modest field elevation, a density altitude two or three thousand feet above pressure altitude is normal and expected.

Failure — using field elevation instead of pressure altitude at step 1, or forgetting that the standard temperature falls with altitude. Both give a density altitude that is too low, and an engine judged against it looks better than it is.

This calculation is a rule of thumb. It is the right tool for deciding whether a ground run looks reasonable. The aircraft's performance section governs any actual determination (GEN-J Concepts §5).

Procedure 4 — Force, area and pressure (S4)

Given — a piston of 5.125 in diameter with 100 psi applied. Sought — the force produced.

Method

1. Radius = 5.125 ÷ 2 = 2.5625 in. Halve the diameter — the same trap as GEN-H Procedures Procedure 1.
2. Area = π r² = π × 2.5625² = 20.63 in².
3. Force = pressure × area = 100 × 20.63 = 2,063 lb.

Check — pressure is force per unit area, so the three quantities must divide back. 2,063 ÷ 20.63 = 100 psi. A force smaller than the pressure figure means area was omitted; a force four times too large means the diameter was used as a radius.

Failure — this is the calculation behind every hydraulic actuator and every brake. Getting the area wrong here understates or overstates the force a system can produce, which is AF-F's subject.

Procedure 5 — Mechanical advantage of an inclined plane (S6)

Given — a ramp 12 ft long rising 3 ft. Sought — its mechanical advantage, and the effort to raise 800 lb.

Method

1. Mechanical advantage = length ÷ height = 12 ÷ 3 = 4.
2. Effort = load ÷ MA = 800 ÷ 4 = 200 lb.

Check — mechanical advantage is always greater than 1 for a useful inclined plane, and the effort is always less than the load. An answer where effort exceeds load means the ratio is inverted.

Where it matters — a screw thread is an inclined plane wrapped round a cylinder, which is why a modest torque on a bolt produces a large clamping force (GEN-E).

Procedure 6 — Mechanical advantage of a lever (K3, supporting S5)

Given — a bellcrank with a 30 in effort arm and a 6 in resistance arm.

Method — MA = effort arm ÷ resistance arm = 30 ÷ 6 = 5.

Check — force and distance trade against each other. A mechanical advantage of 5 means one-fifth the force and five times the travel at the effort end. If your answer promises more force and more travel, it is wrong.

Where it matters — moving a control attachment one hole changes force and travel together, which is why rigging is done to a procedure rather than adjusted by feel (AF-C).

Procedure 7 — Horsepower (S8)

Given — an engine producing 300 ft-lb of torque at 2,400 RPM. Sought — horsepower.

Method — HP = torque (ft-lb) × RPM ÷ 5,252 = 300 × 2,400 ÷ 5,252 = 137 hp.

Check — torque and horsepower are equal at 5,252 RPM, by definition of the constant. Below that speed horsepower is the smaller number; above it, the larger. At 2,400 RPM the horsepower figure must be well below the torque figure, and 137 against 300 satisfies that.

Failure — using inch-pounds instead of foot-pounds gives an answer twelve times too large. GEN-H Procedures Procedure 8 again, in a different subject.

The two non-calculations

S5 — demonstrate the mechanical advantage of various types of levers. A bench task with first-, second- and third-class levers and a spring balance. It needs no rig and no supervision beyond ordinary shop practice; the point is to feel the trade in Procedure 6 rather than compute it.

S7 — identify changes in pressure and velocity as a fluid passes through a venturi. Delivered as instrumented imagery: a venturi with pressure taps, so the fall in static pressure at the throat is visible alongside the rise in velocity. This is Bernoulli as a measurement rather than an assertion, and it is the cleanest demonstration of K5 available (the delivery-mode analysis).

The checks, collected

License

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