# A Sherlock Holmes Problem **URL:** \ **Category:** Factual Questions\ **Created:** [August 14, 2003, 7:56pm UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219 "2003-08-14T19:56:49Z")\ **Posts on this page:** 18\ **Page:** 1
**Author:** ![CaptainGeordie](https://avatars.discourse-cdn.com/v4/letter/c/bbce88/32.png) [@CaptainGeordie](https://boards.straightdope.com/u/CaptainGeordie)\ **Post date:** [August 14, 2003, 7:56pm UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/1 "2003-08-14T19:56:49Z")
I’ve just been re-reading that fine Sherlock Holmes short story ‘Silver Blaze’. There’s a problem I can’t get my head around - can anyone suggest how the Great Detective did it? From ‘Silver Blaze’ : (Holmes and Watson are travelling on a train): "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 per hour”. “I have not observed the quarter-mile posts” said I. " Nor have I. but the telegraph posts on this line are sixty yards apart, and the calculation is a simple one". Simple to Holmes, maybe, but not to me! How did he do it? ---
**Author:** ![Mort\_Furd](https://avatars.discourse-cdn.com/v4/letter/m/8e8cbc/32.png) [@Mort\_Furd](https://boards.straightdope.com/u/Mort_Furd)\ **Post date:** [August 14, 2003, 8:09pm UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/2 "2003-08-14T20:09:19Z")
Count posts pased in three minutes, subtract one. Divide by 88 - you now have the number of miles passed in one minute. Multiply by 60 for miles per hour - or just multiply by 0.68181818182 without first dividing and be done with it. Simplicity itself - “Elemntary my dear Watson.” Unless I fucked up the math somewhere. 🙂 ---
**Author:** ![Mort\_Furd](https://avatars.discourse-cdn.com/v4/letter/m/8e8cbc/32.png) [@Mort\_Furd](https://boards.straightdope.com/u/Mort_Furd)\ **Post date:** [August 14, 2003, 8:10pm UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/3 "2003-08-14T20:10:25Z")
And if I could spell, too. Sheesh. ---
**Author:** ![Mort\_Furd](https://avatars.discourse-cdn.com/v4/letter/m/8e8cbc/32.png) [@Mort\_Furd](https://boards.straightdope.com/u/Mort_Furd)\ **Post date:** [August 14, 2003, 8:14pm UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/4 "2003-08-14T20:14:57Z")
Or reduce the fraction to 22/15. ---
**Author:** ![pravnik](https://avatars.discourse-cdn.com/v4/letter/p/8edcca/32.png) [@pravnik](https://boards.straightdope.com/u/pravnik)\ **Post date:** [August 14, 2003, 8:32pm UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/5 "2003-08-14T20:32:55Z")
There are about 29 and 1/3 posts in a 1760 yard mile. Count the number of seconds between posts, then mulitply that number by 29 and 1/3. Divide the product by 60. If it’s 110 seconds between posts, you’re going roughly 53 and a half miles per hour. ---
**Author:** ![bryanmcc](https://avatars.discourse-cdn.com/v4/letter/b/eada6e/32.png) [@bryanmcc](https://boards.straightdope.com/u/bryanmcc)\ **Post date:** [August 14, 2003, 8:46pm UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/6 "2003-08-14T20:46:02Z")
Where’d you get that last bit? If the posts are 60 yards apart, and the train is travelling 53.5 miles per hour, then it isn’t going to be 110 seconds between posts. Did you mean merely 10 seconds between posts? No, that doesn’t work out either: my math gives a post every 2.3 seconds (approximately). ---
**Author:** ![CalMeacham](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/calmeacham/32/35_2.png) [@CalMeacham](https://boards.straightdope.com/u/CalMeacham)\ **Post date:** [August 14, 2003, 9:30pm UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/7 "2003-08-14T21:30:16Z")
There’s a note on this – along with a graph! – in **The Annotated Sherlock Holmes** , with nnotations by William S. Baring-Gould ---
**Author:** ![pravnik](https://avatars.discourse-cdn.com/v4/letter/p/8edcca/32.png) [@pravnik](https://boards.straightdope.com/u/pravnik)\ **Post date:** [August 14, 2003, 9:36pm UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/8 "2003-08-14T21:36:07Z")
Whoops. :smack: Actually, if you’re going 110 seconds between posts, it will take you 53 and a half minutes to go a mile. Excuse me for a second, I need to go see if I can fish my income tax return out of the mailbox. ---
**Author:** ![Mycroft\_H](https://avatars.discourse-cdn.com/v4/letter/m/898d66/32.png) [@Mycroft\_H](https://boards.straightdope.com/u/Mycroft_H)\ **Post date:** [August 15, 2003, 11:01am UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/9 "2003-08-15T11:01:31Z")
This has been discussed for years by Sherlockians, with varying degrees of creativity, complexity and brilliance. As **CalMeacham** said, Baring-Gould’s work, long considered the ‘Bible’ of the Holmesian world, does present a clear solution. IIRC it is a bit convoluted, but my copy is currently boxed up from a recent move so I can’t check. Instead I’ll move onto what I could find in my notes. The simplest solution involves four mathematical operations, two multiplications and two divisions, and knowledge of how many yards are in a mile and how many seconds in an hour. The multiplications even I can do in my head and as for the division operations, while fairly cumbersome, I would be willing to believe that many of the Teeming Millions could do them as well. Part of the assumption is that Holmes supposedly ‘glanced’ at his watch and did not keep track for too long. **Mort Furd** could be right that he viewed for 3 minutes, but that isn’t an elegant solution and does not allow us the precision necessary. The other necessary ability, which is easy enough, is to be able to look at the second sweep hand of a watch (presumably a pocket watch given the date of the story) and using peripheral vision, tell when you pass an object with decent precision. Checking this while driving, I figure I can determine the time of passing to within 0.1 - 0.2 seconds. So, Holmes starts counting while passing a telegraph pole and the second hand ‘click’ coincide and continues to count poles and seconds until they coincide again. He would probably note that they passed 17 poles in 39 seconds. Then for the mathematics - 17 poles x 60 yards/pole = 1020 yards (in 39 seconds) 1020 yards / 1760 yds/mile = 0.58 miles (in 39 seconds) 0.58 miles x 3600 seconds/hour = 2088 2088 / 39 = 53.54 mph I could not manage all the calculations in my head, but the point of the exercise (for the reader) was to illustrate that Holmes was smart and could do it. You can get similar speed answers for 10 poles in 23 seconds, 14 poles in 32 seconds and 24 poles in 55 seconds. Then, for rabid Sherlockians (or ‘Holmesians’ on **CaptainGeordie’s** side of the pond), _ **brilliantly** _ outlined by **Dex** in [http://www.straightdope.com/mailbag/msherlock.html](http://www.straightdope.com/mailbag/msherlock.html), they begin the discussion of why he would subject Watson to this show of brilliance/egotism, where on which railway line it may have occurred, what page of Bradshaw’s timetable the train in question was mentioned, etc. ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [August 15, 2003, 11:23am UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/10 "2003-08-15T11:23:57Z")
> [@](#): > > \*Originally posted by Mycroft H. \* > Checking this while driving, I figure I can determine the time of passing to within 0.1 - 0.2 seconds. > > So, Holmes starts counting while passing a telegraph pole and the second hand ‘click’ coincide and continues to count poles and seconds until they coincide again. He would probably note that they passed 17 poles in 39 seconds. You could be right, but if they are actually travelling at 53.500 MPH, then the first time that a click happens within 0.1 seconds of a pole is at 7 poles in 16 seconds. This gives a speed of 53.69 MPH. ---
**Author:** ![The\_Great\_Unwashed](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/the_great_unwashed/32/4791_2.png) [@The\_Great\_Unwashed](https://boards.straightdope.com/u/The_Great_Unwashed)\ **Post date:** [August 15, 2003, 1:33pm UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/11 "2003-08-15T13:33:54Z")
If Holmes knows that telegraph poles are 60 yards apart then the number of poles passed in 2 minutes and 3 seconds is approximately equal to the number of miles per hour. He could have done the maths before, and merely has to remember the “magic number”, 2’03". Tsk! Those Sherlockians, eh? ---
**Author:** ![CookingWithGas](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cookingwithgas/32/485_2.png) [@CookingWithGas](https://boards.straightdope.com/u/CookingWithGas)\ **Post date:** [August 15, 2003, 2:08pm UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/12 "2003-08-15T14:08:48Z")
How the heck did he know that the poles are 60 yards apart to begin with? ---
**Author:** ![The\_Great\_Unwashed](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/the_great_unwashed/32/4791_2.png) [@The\_Great\_Unwashed](https://boards.straightdope.com/u/The_Great_Unwashed)\ **Post date:** [August 15, 2003, 2:14pm UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/13 "2003-08-15T14:14:58Z")
Elementary, my dear **CookingWithGas** , at 53.5mph he would pass one pole every 2.3 seconds indicating the poles were 60yds apart. ---
**Author:** ![CookingWithGas](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cookingwithgas/32/485_2.png) [@CookingWithGas](https://boards.straightdope.com/u/CookingWithGas)\ **Post date:** [August 15, 2003, 2:56pm UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/14 "2003-08-15T14:56:23Z")
::rolleyes: Droll. Very droll. ---
**Author:** ![Mycroft\_H](https://avatars.discourse-cdn.com/v4/letter/m/898d66/32.png) [@Mycroft\_H](https://boards.straightdope.com/u/Mycroft_H)\ **Post date:** [August 15, 2003, 5:16pm UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/15 "2003-08-15T17:16:56Z")
**CookingWithGas** , you have to know these things when you are king… of the detectives. It was an illustration of the level of knowledge that Holmes had of everything. If he knows the spacing of telegraph poles along different railway lines, what all else does he know? **The Great Unwashed** , your point is exactly the sort of thing Sherlockians enjoy raising. It is a good suggestion as an answer. One line of discussion, similar to yours, is that Holmes was just pulling Watson’s leg, and Watson was too much in awe (or too polite) to call BS on him. In discussing ‘Silver Blaze’ with friends some people have stated just that – How do we know the train was _really_ going 53-1/2 mph? Holmes could have said anything without any way to verify it. But this depends on how one wishes to interpret Holmes and his relationship with Watson. Is Holmes a genius detective in a great friendship with Watson, or is he a sham artist in a sycophantic relationship? These are the sorts of discussions Sherlockians enjoy. (Yes, we are a strange lot :rolleyes:, but it can be a fun hobby.) I have seen very convoluted methods whereby one can come up with an answer using fractions, approximations and the like, but it gets away from the illustration of Holmes just using his brain power which is the image I prefer. ---
**Author:** ![Mort\_Furd](https://avatars.discourse-cdn.com/v4/letter/m/8e8cbc/32.png) [@Mort\_Furd](https://boards.straightdope.com/u/Mort_Furd)\ **Post date:** [August 15, 2003, 7:33pm UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/16 "2003-08-15T19:33:43Z")
> [@](#): > > \*Originally posted by Mort Furd \* > \*\*Count posts pased in three minutes, subtract one. Divide by 88 - you now have the number of miles passed in one minute. Multiply by 60 for miles per hour - or just multiply by 0.68181818182 without first dividing and be done with it. Simplicity itself - “Elemntary my dear Watson.” > > Unless I fucked up the math somewhere. 🙂 \*\* Which I did. Count posts passed in three minutes, subtract one. Divide by 88 - you now have the number of miles passed in **three minutes.** Multiply by \*\*20[/] for miles per hour - or just multiply by 0.22727272727 without first dividing and be done with it. Or use the fraction 5/22 Musta gotta hold of some a Sherlock’s 3% solution. ---
**Author:** ![gazpacho](https://avatars.discourse-cdn.com/v4/letter/g/6f9a4e/32.png) [@gazpacho](https://boards.straightdope.com/u/gazpacho)\ **Post date:** [August 15, 2003, 9:56pm UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/17 "2003-08-15T21:56:48Z")
Sherlock used a 7% solution. ---
**Author:** ![Exapno\_Mapcase](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/exapno_mapcase/32/1051_2.png) [@Exapno\_Mapcase](https://boards.straightdope.com/u/Exapno_Mapcase)\ **Post date:** [August 16, 2003, 12:02am UTC](https://boards.straightdope.com/t/a-sherlock-holmes-problem/195219/18 "2003-08-16T00:02:49Z")
What is means in reality was that Doyle was one of the worst hack writers who ever lived. It also means that an entire industry has been formed of people taking Doyle’s ten million idiocies and trying to make them coherent.\* Think of it as what people used to do before Star Trek. 😃 \*I have shelves of this stuff. It’s better reading than the stories these days.