The eight calculations

Each procedure has the same five parts, and the fourth is the one that matters.

the competency standard requires a candidate to recognize a wrong answer, not merely to produce a right one. Every calculation below has a check available. Use it every time, and the habit will still be there when the calculation is one you have not seen before.

Procedures 1–3: areas and volumes are one method with three applications

S2, S3 and S4 differ only in shape. The method is the same and so are the mistakes.

Method (all three)

1. Identify the shape and write down which dimension is which.
2. Check for diameters. If a dimension is given as a diameter — bore, hole size, tank radius — halve it before it enters a formula that wants a radius.
3. Apply the formula, keeping every intermediate figure unrounded.
4. Convert the units the answer is wanted in. Volume comes out in cubic inches and is almost never wanted in cubic inches.
5. Round once, at the end.

Procedure 1 — Engine displacement (S2, cylinder volume)

Given — bore 5.125 in, stroke 3.875 in, four cylinders. Sought — total displacement, cubic inches.

Method

1. Radius = bore ÷ 2 = 5.125 ÷ 2 = 2.5625 in.
2. One cylinder: $V=\pi r^{2}h=\pi \times {2.5625}^2\times 3.875=79.94{\text{in}}^3$.
3. Four cylinders: $79.94\times 4=319.7{\text{in}}^3$.
4. ≈ 320 cubic inches.

Check — the engine's designation is the answer. A Lycoming-style O-320 is named for its displacement, so 320 confirms the arithmetic. Any engine you meet in the field carries its displacement in its model number, and that makes this the most self-checking calculation in the unit. Use it.

Failure — forgetting to halve the bore gives 1,279 in³, four times too large, because the radius is squared. An answer four times the expected size is almost always a diameter used as a radius.

Procedure 2 — Wing area (S3)

Given — span 36 ft, root chord 5.5 ft, tip chord 3.5 ft, straight taper. Sought — wing area, square feet.

Method

1. Mean chord = (root + tip) ÷ 2 = (5.5 + 3.5) ÷ 2 = 4.5 ft.
2. Area = span × mean chord = 36 × 4.5 = 162 sq ft.

Check — compute aspect ratio, span² ÷ area = 1296 ÷ 162 = 8.0. Light airplane wings generally fall between about 6 and 10. A result far outside that says the area is wrong.

Failure — using root chord alone gives 198 sq ft, about 22% high. Treating a tapered wing as a rectangle always overstates area.

Procedure 3 — Compartment or tank volume (S4)

Given — a rectangular compartment 30 in long, 18 in wide, 12 in deep. Sought — capacity in gallons, and the weight of avgas it would hold.

Method

1. Volume = 30 × 18 × 12 = 6,480 in³.
2. Gallons = 6,480 ÷ 231 = 28.05 gal. (231 in³ = 1 US gallon, exactly.)
3. Weight = 28.05 × 6 lb/gal = 168 lb.

Check — a US gallon is a bit over 230 cubic inches, so gallons ≈ cubic inches ÷ 230. Roughly 6,500 ÷ 230 is about 28. If your answer is 6,480 gallons you have skipped step 2, and if it is 0.12 you have divided the wrong way.

Failure — reporting cubic inches as gallons. It is a factor of 231 and it does not look wrong on a form.

In service, read the capacity from approved data, not from a tape measure. Real tanks are irregular and the manufacturer publishes both capacity and usable fuel — which are different numbers (GEN-I, AF-J).

Procedure 4 — Square root (S1)

Given — a circular access hole of area 12.57 in². Sought — its diameter.

Method

1. From $A=\pi r^2$, rearrange: $r=\sqrt{A/\pi }$.
2. $r=\sqrt{12.57/3.1416}=\sqrt{4.00}=2.00\text{in}$.
3. Diameter = 2r = 4.00 in.

Check — square the answer's radius and multiply by π. 2² × 3.1416 = 12.57. A root is checked by squaring it, and that check costs seconds.

Failure — taking the root of the area without dividing by π first gives r = 3.55, d = 7.09 — a hole nearly twice the right size.

Procedure 5 — Fraction and decimal conversion (S5)

Given — a drawing calling for a 3/16 in hole; a limits table quoting 0.0625 in. Sought — each in the other form.

Method

• Fraction to decimal: divide. 3 ÷ 16 = 0.1875.
• Decimal to fraction: multiply by the denominator you want and see if it comes out whole. 0.0625 × 16 = 1, so 0.0625 = 1/16.

Check — the common ones should become recognition rather than calculation:

Failure — a #30 drill is not a 1/8 hole. Number, letter and fractional drill sizes are separate series and they interleave. Converting a fraction to a decimal does not tell you which drill to pick; the drawing does (AF-A).

Procedure 6 — Comparing two values with a ratio (S6)

Given — a bushing with a new inside diameter of 0.500 in and a maximum allowable of 0.5250 in. Measured: 0.5312 in. Sought — how the wear compares with the allowance.

Method

1. Actual wear = 0.5312 − 0.500 = 0.0312 in.
2. Allowable wear = 0.5250 − 0.500 = 0.0250 in.
3. Ratio = 0.0312 ÷ 0.0250 = 1.25.

The part is at 125% of its allowable wear. Reject.

Check — a ratio above 1 means beyond limit and below 1 means within it, so the answer should agree with a direct comparison of the two diameters. 0.5312 is larger than 0.5250, so a ratio below 1 would mean you have divided the wrong way.

Failure — comparing the measured diameter to the allowable diameter as a ratio (0.5312 ÷ 0.5250 = 1.012) and concluding the part is only 1% over. The ratio must be of the wear, not of the dimension, or a badly worn part looks nearly new.

Procedure 7 — Compression ratio (S7)

Given — swept volume 79.9 in³ per cylinder; clearance volume 13.3 in³. Sought — compression ratio.

Method

1. Total volume at bottom of stroke = swept + clearance = 79.9 + 13.3 = 93.2 in³.
2. Volume at top of stroke = clearance = 13.3 in³.
3. Ratio = 93.2 ÷ 13.3 = 7.0, written 7.0:1.

Check — compression ratios for normally aspirated piston aircraft engines sit roughly between 6:1 and 10:1. A ratio below 1 means you have divided the wrong way; a ratio near 6 without the clearance volume added on top means you used swept ÷ clearance and forgot step 1.

Failure — swept ÷ clearance alone gives 6.0:1. It is a plausible number, it is wrong, and nothing about it looks wrong. This is the unit's central lesson in one calculation.

Procedure 8 — Torque unit conversion (S8)

Given — a maintenance manual value of 450 in-lb; a wrench reading foot-pounds. Sought — the setting.

Method

1. Inch-pounds to foot-pounds: divide by 12. 450 ÷ 12 = 37.5 ft-lb.
2. Foot-pounds to inch-pounds: multiply by 12. 25 ft-lb × 12 = 300 in-lb.

Check — inch-pounds are always the bigger number. A foot is twelve inches, so the same torque expressed in inch-pounds is twelve times the foot-pound figure. If your converted number went up when moving to foot-pounds, you have multiplied instead of divided.

Failure — setting 450 on a foot-pound wrench applies twelve times the specified torque. The fastener fails, or the thread does, or the part distorts and is found weeks later. There is no outward sign at the moment it happens.

The extension bar

If you use an extension in line with the handle, the wrench reads less than what reaches the fastener. Set it lower:

Given — target 300 in-lb; wrench length L = 12 in; extension E = 3 in. Method — $T_W=300\times \frac{12}{15}=240\text{in-lb}$. Set the wrench to 240.

Check — with an extension in line with the handle, the setting is lower than the target. If your answer is higher, you have inverted the fraction. This is the error to expect, because the intuition — the extension makes it harder, so set it higher — points the wrong way.

The correction depends on the geometry, and "always lower" is wrong. The formula assumes the adapter lengthens the lever arm along the handle's axis. Two common cases do not:

• An adapter at 90° to the handle — a crowfoot fitted across, rather than in line — does not change the effective length, so no correction applies.
• An adapter that offsets back toward the handle shortens the arm, and the setting is higher than the target.

Establish the adapter's orientation before applying any correction. Applying the in-line formula to a crowfoot at 90° under-torques the fastener by the same proportion it would correctly have adjusted.

Failure — setting 300 with the extension fitted delivers 375 in-lb, 25% over. GEN-E explains what over-torque does and why nothing about the fastener will tell you.

The habit this unit is actually teaching

Eight calculations, and every one of them has a check that takes seconds:

None of these calculations is difficult. All of them are checkable. The only way to get one wrong and not know is to skip the check — and every wrong answer in this unit looks exactly like a right one.

License

CC BY-SA 4.0 (Licensing).