Time — Question 5

Question 5

What a video can say about an event between frames

An ideal camera samples a shutter instantaneously at exactly $240$ frames per second, with frame $k$ taken at $t_k=k/240\ \mathrm{s}$. No frames are dropped. The shutter opens once and then closes once, with observations shown below.

Use conservative closed intervals: the opening occurred between frames 100 and 101, and the closing between frames 137 and 138. There are no other timing errors in this model.

1. Find the earliest and latest allowed opening and closing times as fractions of a second. Then find the smallest and largest allowed open duration.

2. Give the midpoint estimate of the duration and its half-width. A student counts 37 open frames and reports exactly $37/240\ \mathrm{s}$. Explain what is useful and what is unjustified in that report.

3. For the same type of two-transition bracketing at a new exact frame rate $f$, derive the worst-case duration half-width. What minimum frame rate would make it no greater than $0.0020\ \mathrm{s}$?

4. Why would finite exposure time, missing frames, or an uncertain frame rate require revisiting this result?

Question 5 - Solution

What a video can say about an event between frames

(a) Bound both transition times, then subtract.

$$t_{\mathrm{open}}\in[100/240,101/240]\ \mathrm{s},\qquad t_{\mathrm{close}}\in[137/240,138/240]\ \mathrm{s}.$$ The shortest duration pairs the latest opening with the earliest closing; the longest pairs the earliest opening with the latest closing: $$\boxed{\Delta t\in[36/240,38/240]\ \mathrm{s}} = [0.150000,0.158333\ldots]\ \mathrm{s}.$$

(b) Keep the unresolved fractions of a frame.

The midpoint is $\boxed{37/240\ \mathrm{s}\approx0.154167\ \mathrm{s}}$ and the half-width is $\boxed{1/240\ \mathrm{s}\approx0.004167\ \mathrm{s}}$.

The 37 open frames suggest this midpoint under the symmetric bracketing convention, but do not locate either transition exactly. Many durations inside the allowed interval give the same frame states. The midpoint is a representative estimate, not a uniquely measured duration or a statistical expectation supplied by the data.

(c) Two one-frame brackets determine the duration bound.

$$\delta(\Delta t)=\frac12\left(\frac1f+\frac1f\right)=\frac1f,\qquad f_{\min}=\frac1{0.0020\ \mathrm s}=500\ \mathrm{s^{-1}}.$$

Each transition bracket has width $1/f$. Their difference interval has width $2/f$, hence half-width $1/f$. Requiring $1/f\le0.0020\ \mathrm{s}$ gives $\boxed{f\ge500\ \mathrm{s^{-1}}}$, or 500 frames per second.

(d) The sampling model matters.

A finite exposure can blur when a state changed; dropped frames enlarge gaps; an uncertain timebase changes the mapping from frame index to SI time. Each affects the stated brackets or their scale. A nominal frame-rate label alone does not justify the ideal bound.