In *Silver Blaze*, Holmes and Watson are on a train going to Exeter. Holmes has been reading all the papers have to say about the disappearance of the horse, Silver Blaze. Apparently satisfied, he thrusts the papers under the seat and comments to Watson:

*“We are going well,” said he, looking out of the window, and glancing at his watch. “Our rate at present is fifty-three and a half miles an hour.”*

*“I have not observed the quarter-mile posts,” said I.*

*“Nor have I. But the telegraph posts upon this line are sixty yards apart, and the calculation is a simple one.*

So how did Holmes calculate the speed of the train? According to Watson, Holmes was looking out of the window then glanced at his watch so we will assume he counted a set number of poles, then noted elapsed time.

Let’s assume Holmes decided to count, say, 26 posts (or 25 sixty-yard intervals), how many seconds would have elapsed when he again looked at his watch? We know that the rate of the train divided into the distance equals elapsed time (Rate x Time = Distance → Time = Distance ÷ Rate). In this case we know the rate is 53.5 mph and the distance of the 25 posts is 1500 yards (the posts are 60 yards apart). After Holmes counted, the 26th post, he would have consulted his watch and noted 57.35 seconds elapsed time. The calculation would have been: 1500 yards/57.35 seconds = 53.5 mph. Maybe not a simple calculation but for Holmes, apparently it was.

But wait, we’re talking Sherlock Holmes here, surely thee is a better way.

Suppose Holmes noted the time interval between any two posts (60 yards)? The elapsed time would have to be 2.33374 seconds which would be difficult to note on a simple pocket-watch but the distance between any *four* posts (180 yards) would be 7.001 seconds, a much easier determination and 180 yards in 7.001 seconds is a relatively simple calculation and yields: 53.509 miles per hour.

Elementary.

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How could someone, even Sherlock Holmes, count 57.35 seconds on a pocket watch or even 29.33 posts ?!

53.5 mph is one post every 2.2 seconds, or 10 posts every 22 seconds.

Comment by Anonymous — November 24, 2011 @ 6:08 pm |

Unfortunately, although nearly correct and sensical, I cannot agree with your calculation for the 3 posts. Your error lies in the initial assertion that traversing a single gap between two posts must occur over 2.33374 seconds.

The calculation you must use is 53.5 mi/h * (1h/60m) * (1m/60s) = 0.01486111111 mi/s. Converting from yards to miles: 60 yds * (3ft/yd) / (5280 ft/mi) = 0.03409090909 mi

Very simply now, we take .034091 mi / 0.014861 mi/s and you find 2.29 (2.293967715) s per gap.

The actual elapsed time for 4 posts (3 gaps between) is 6.88 seconds (even using 2.29*3 gives 6.87, which, performing the necessary reverse calculation actually results in a measure of 53.6 m/h, so even that is not accurate enough).

All this is to say, Holmes estimated a speed, threw on a “point 5” for drama and impressed Dr. Watson.

Comment by Max PackerBacker Magee — May 24, 2017 @ 8:38 am |

Just to verify, I worked backwards from your 2.33374 seconds/gap number and found a speed of 52.588 m/h, rounded to 52.6. This is close, but no calabash pipe.

Comment by Max PackerBacker Magee — May 24, 2017 @ 9:14 am |

Max, you are correct! I did make a trip – only a little, little trip – but it was more than I could afford with you so close upon me (paraphrased from The Final Problem).

The first part of my calculation is correct – 1500 yards in 57.35 seconds is indeed 53.5 miles/hour – but, alas, the elapsed time between any two posts, at 53.5 mph, is, as you demonstrated, is 2.293967715 seconds.

Possibly Holmes counted 11 posts (600 yards) and did the calculation based on 22.939 seconds. If he rounded 22.939 to 23 seconds, the result would have been 53.59 mph which he may have reported to Watson as, “Fifty-three and a half miles per hour”.

Good catch… a seven percent solution to you!

Comment by Mark Loper — May 26, 2017 @ 8:33 am |