Case Study 1 — From a Velocity Trace to Total Travel: Reconstructing Piston Displacement

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Case Study 1 — From a Velocity Trace to Total Travel: Reconstructing Piston Displacement

Field: Mechanical engineering (internal-combustion engine diagnostics) FTC tools used: Net Change Theorem (§14.7), FTC Part 2 evaluation (§14.4), net displacement vs. total distance (§14.14 Error 3), average value (§14.9)

The Problem on the Bench

Priya Raghavan is a powertrain test engineer. On her bench sits a single-cylinder research engine, instrumented with a laser vibrometer aimed at the top of the piston. The vibrometer does not measure position directly — it measures velocity, the rate at which the piston surface approaches or recedes from the sensor. What comes off the data acquisition card is therefore a velocity trace $v(t)$: a continuous record of how fast the piston is moving, in meters per second, sampled thousands of times per revolution.

Priya's question is the oldest question in this chapter, dressed in hardware: given the rate, recover the amount. She wants the piston's displacement — how far it has moved from top-dead-center — and its stroke length, the total distance it sweeps in one pass. The vibrometer gives her $v(t) = s'(t)$, where $s(t)$ is position. The Net Change Theorem (§14.7) tells her exactly how to invert that derivative: $$s(t_2) - s(t_1) = \int_{t_1}^{t_2} v(t)\,dt.$$ This is FTC doing the only thing the bench needs it to do: turning a measured rate back into the accumulated quantity that produced it.

A Model for the Velocity

The crankshaft turns at a steady $1500$ revolutions per minute, which is $25$ revolutions per second, so one full cycle takes $$T = \frac{1}{25} = 0.04\ \text{s}.$$ For a slow research engine with a long connecting rod, the piston's motion is very nearly simple harmonic, and the velocity trace is well fit by a sinusoid. Priya fits her data and obtains $$v(t) = 9.42\,\sin(\omega t)\ \text{m/s}, \qquad \omega = \frac{2\pi}{T} = \frac{2\pi}{0.04} = 50\pi\ \text{rad/s}.$$ The amplitude $9.42$ m/s is the peak piston speed, reached at mid-stroke. We will carry $9.42 \approx 3\pi$ so the arithmetic stays clean: note $3\pi \approx 9.4248$, and $\omega = 50\pi$, so the convenient ratio $9.42/\omega \approx 3\pi/(50\pi) = 3/50 = 0.06$ m will appear repeatedly. That $0.06$ m is, as we will see, half the stroke.

Why a sinusoid is the right first model. For an idealized slider-crank with an infinitely long rod, the piston position is $s(t) = -\tfrac{L}{2}\cos(\omega t) + \text{const}$, so its velocity is exactly $v(t) = \tfrac{L\omega}{2}\sin(\omega t)$. Matching $\tfrac{L\omega}{2} = 9.42$ recovers the stroke $L$ — which is precisely what the integral below will deliver, independently, from the velocity data alone.

Step 1 — Net Displacement Over One Full Cycle

First Priya checks a sanity condition. Over one complete revolution the piston must return to where it started — top-dead-center is top-dead-center every cycle. The Net Change Theorem says the displacement over $[0, T]$ is $$s(T) - s(0) = \int_0^{T} 9.42\,\sin(\omega t)\,dt.$$ An antiderivative of $\sin(\omega t)$ is $-\tfrac{1}{\omega}\cos(\omega t)$, so by FTC Part 2 (§14.4), $$\int_0^{T} 9.42\,\sin(\omega t)\,dt = 9.42\left[-\frac{1}{\omega}\cos(\omega t)\right]_0^{T} = -\frac{9.42}{\omega}\Big(\cos(\omega T) - \cos 0\Big).$$ Now $\omega T = 50\pi \cdot 0.04 = 2\pi$, and $\cos(2\pi) = \cos 0 = 1$, so the parenthesis is $1 - 1 = 0$: $$s(T) - s(0) = -\frac{9.42}{\omega}\cdot 0 = 0\ \text{m}.$$ The piston returns exactly to its starting position. This is the signed-area cancellation of §14.7 made physical: the downstroke contributes negative velocity-area that exactly erases the upstroke's positive area. If Priya's integral had come out nonzero, she would suspect a calibration drift in the vibrometer — a slow false velocity superimposed on the real motion. The integral is a diagnostic, not just an answer.

Step 2 — Total Distance: The Stroke

Net displacement of zero tells Priya nothing about how far the piston traveled — only that it ended where it began. The stroke length is the total distance, which integrates the speed $|v(t)|$, not the velocity (§14.14, Error 3). She must split the cycle at the instants where $v$ changes sign.

On $[0, T/2] = [0, 0.02]$, the argument $\omega t$ runs from $0$ to $\pi$, where $\sin(\omega t) \ge 0$, so $v \ge 0$: the piston moves one direction. On $[T/2, T]$, $\omega t$ runs from $\pi$ to $2\pi$, where $\sin \le 0$, so $v \le 0$: it moves back. The two legs have equal length by symmetry, so the distance over the upstroke is half the total. Compute the upstroke: $$\int_0^{T/2} 9.42\,\sin(\omega t)\,dt = -\frac{9.42}{\omega}\Big[\cos(\omega t)\Big]_0^{T/2} = -\frac{9.42}{\omega}\big(\cos\pi - \cos 0\big) = -\frac{9.42}{\omega}(-1 - 1) = \frac{2\cdot 9.42}{\omega}.$$ Using $9.42/\omega \approx 0.06$ m, $$\text{upstroke distance} = 2(0.06) = 0.12\ \text{m} = 120\ \text{mm}.$$ That is the stroke length $L = 120$ mm — a perfectly ordinary number for a research single-cylinder. The total distance over the full cycle is twice this, $0.24$ m, since the return leg covers the same $120$ mm in reverse. Priya now has, from velocity data alone, the geometric quantity an engine builder cares about most.

The trap that costs the most. A junior engineer once handed Priya a "stroke" of $0$ mm, having integrated $v$ straight across $[0,T]$ and reported the result. That number is the net displacement — correct, and useless for geometry. The stroke is $\int_0^T |v|\,dt = 0.24$ m of travel folded into a $0.12$ m physical sweep. Net change and total distance are different questions; FTC answers whichever one you actually ask.

Step 3 — Average Piston Speed

For a lubrication study, Priya needs the mean piston speed — a standard engine metric tied to ring wear. The average velocity over a full cycle is zero (net displacement over elapsed time), which is useless; the meaningful quantity is the average of the speed $|v|$. Using the average-value formula (§14.9) over the upstroke, where speed equals velocity: $$\overline{|v|} = \frac{1}{T/2}\int_0^{T/2} 9.42\,\sin(\omega t)\,dt = \frac{1}{0.02}\cdot 0.12 = 6.0\ \text{m/s}.$$ By symmetry the full-cycle mean speed is the same $6.0$ m/s. Cross-check against the textbook formula "mean piston speed $= 2 L N$," where $L = 0.12$ m is the stroke and $N = 25$ rev/s: $2 \cdot 0.12 \cdot 25 = 6.0$ m/s. The two routes agree exactly — the calculus and the handbook formula are the same statement, because that handbook formula is itself a precomputed FTC result.

What Made FTC the Right Tool

Every step here was one idea wearing different clothes. The vibrometer measures a derivative; the engineer needs the original quantity; FTC is the bridge that runs the relationship backward. Net displacement used signed area and came out zero by symmetry. Stroke length used unsigned area — total distance — and required splitting at sign changes. Mean piston speed used the average-value formula. Three numbers an engine builder lives by, all extracted from one velocity trace by the Net Change Theorem.

Notice what Priya never did: she never solved a differential equation, never integrated a Riemann sum by hand, never needed the position signal at all. She had a rate, she had FTC, and that was enough. This is the chapter's promise delivered on a test bench.

Discussion Questions

Priya's net-displacement integral over a full cycle came out to exactly zero. Suppose instead it came out to $+0.5$ mm per cycle. What physical or instrumental cause would you suspect, and how would that drift accumulate over a 10-minute test?
The stroke required integrating $|v|$, not $v$. Re-derive the stroke by computing $\int_0^{T/2} v\,dt$ and $\int_{T/2}^{T}(-v)\,dt$ separately and adding. Why must you negate $v$ on the second interval?
The mean piston speed agreed with the handbook formula $2LN$. Derive $2LN$ from the FTC argument in general (arbitrary stroke $L$, arbitrary speed $N$ rev/s), showing why the sinusoidal amplitude drops out.
If the connecting rod were short, the velocity trace would no longer be a pure sinusoid — it would carry a second-harmonic term, $v(t) = a\sin(\omega t) + b\sin(2\omega t)$. Would the net displacement over a full cycle still be zero? (Integrate and see.) Would the stroke change?

A Short Annotated Reading

Stewart, Calculus: Early Transcendentals, §5.4 ("Indefinite Integrals and the Net Change Theorem"). Stewart's worked examples on velocity, displacement, and distance traveled are the direct textbook analog of this case; do the rectilinear-motion exercises there next.
Pulkrabek, Engineering Fundamentals of the Internal Combustion Engine, ch. 2. The source of the "mean piston speed $= 2LN$" relation and the slider-crank kinematics that justify the sinusoidal velocity model; read it to see the handbook formula derived from geometry rather than calculus, then compare.
OpenStax Calculus Vol. 1, §5.4 ("Integration Formulas and the Net Change Theorem"). A free, parallel treatment with additional motion problems and the displacement-vs.-distance distinction worked in full.