The Slow Cable
The problem
Mid-1850s. A telegraph cable across the Atlantic would be a single copper conductor wrapped in insulation and laid on the sea floor, thousands of miles long. William Thomson, later Lord Kelvin, worked out in 1854 that a cable like that does not carry a signal the way a wire on poles does. It behaves like a very long resistor feeding a very long capacitor: each puff of current has to charge the whole length before the far end notices, and the delay grows with the square of the length, not with the length.
Here is a cable to try it on. The figures are round numbers chosen for the puzzle, not measurements of any real cable:
- length L = 2,050 nautical miles
- conductor resistance r = 3.0 Ω per nautical mile
- capacitance, conductor to the surrounding sea, c = 0.32 µF per nautical mile
Define the cable’s retardation time as T = R × C, where R and C are the total resistance and capacitance of the whole length.
The telegraph operator sends Morse. Assume that every Morse element (one dot-length of signal or of silence) needs 2T to settle at the far end before the next one can follow, and that a standard word, “PARIS”, is 50 elements long, letter and word gaps included.
How many words per hour can this cable carry?
Notes
- Everything is in nautical miles, so there is no distance conversion. Farads are the only unit that needs care: 1 µF = 10⁻⁶ F.
- R and C are for the whole cable, not per mile. Watch what happens to each when the cable gets longer.
- 3,600 seconds make an hour.
Hint
Click for a hint
Total resistance and total capacitance each grow in proportion to length, so their product does:
T = (r · L) · (c · L)
Time per word is 50 elements at 2T each. Words per hour is 3,600 s divided by that.
Solution
Click to reveal the solution
Total resistance and capacitance:
R = 3.0 Ω/nmi × 2,050 nmi = 6,150 Ω
C = 0.32 µF/nmi × 2,050 nmi = 656 µF
Retardation time:
T = R · C = 6,150 Ω × 656×10⁻⁶ F = 4.0344 s
Each element takes 2T = 8.0688 s, so a 50-element word takes 403.44 s, and:
words per hour = 3600 / (50 × 2 × T) = 36 / T
Answer: ≈ 8.92 words per hour.
The square in the law is the whole story. R and C each scale with L, so T scales with L²: halve the cable and it carries four times as many words, and double it and it carries a quarter as many.
Calculator solutions
HP-42S
Two things make this short. The 3600 / (50 × 2) folds into a constant 36, entered first and left
on the stack while T is worked out on top of it. And the final ÷ divides that 36 by T, so the
last step is just the one key.
36 ENTER
2050 ENTER
× L²
3 × × r
0.32 × × c, in µF
1000000 ÷ µF to F: T in seconds
÷ 36 / T
Display: 8.923259964
Elektronika MK-61
Same sequence, with two habits of this machine: the ENTER key is В↑, and a decimal number must
begin with a digit, so 0.32 is keyed 0 . 3 2. The display holds eight digits.
3 6 ENTER
2 0 5 0 ENTER
×
3 ×
0 . 3 2 ×
1 0 0 0 0 0 0 ÷
÷
Display: 8.9232599
HP-15C
Identical keystrokes to the 42S. The 15C starts in FIX 4, which is enough digits here.
36 ENTER
2050 ENTER
×
3 ×
0.32 ×
1000000 ÷
÷
Result: 8.9233 words per hour.
HP-41C
Identical keystrokes again, and also FIX 4 by default.
36 ENTER↑
2050 ENTER↑
×
3 ×
0.32 ×
1000000 ÷
÷
Display: 8.9233
HP-48GX
This is the one machine where copying the RPN sequence fails, and it fails quietly. On the 48,
ENTER after typing a number just ends the entry and pushes it: it does not copy it the way
classic RPN does. So 2050 ENTER × multiplies 2050 by the 36 underneath, no squaring happens,
and the stack ends up holding a small, plausible-looking, wrong number. Square with the x² key
instead (left-shift √x; 2 y^x gives the same result, but x² is the natural key for a square):
36 ENTER
2050 ENTER
x² 2050²
3 ×
0.32 ×
1000000 ÷
÷
Result: 8.92325996431 words per hour, matching the HP-42S to eleven significant digits.
HP-25
Same keystrokes as the 42S, after setting the display: the 25 powers up in FIX 2, which would show
8.92, so f FIX 4 first gives the digits to compare against the other machines. The multiplies stay
inside the ten-digit range, largest intermediate 12,607,500.
f FIX 4
36 ENTER
2050 ENTER
×
3 ×
0.32 ×
1000000 ÷
÷
Display: 8.9233
TI-84
One expression. The 36 is 3600 ÷ (50 × 2), and the 10⁻⁶ converts µF to F:
36/(3*.32*2050²*10^-6)
Result: ≈ 8.9233 words per hour.
Casio fx-991
Same expression, natural display mode:
36÷(3×0.32×2050²×10^-6)
Result: ≈ 8.9233 words per hour.
Slide Rule
Everything here is multiplication, division and one square, so the rule can do all of it. The two powers of ten also cancel by hand: the 10⁻⁶ of µF against the 10⁶ inside 2050², so the arithmetic is just 3 × 0.32 × 2.05², with no zeros to track.
- Square the length: put the cursor on 2.05 on the D scale and read the A scale under it: 4.20 (4.2025 to a hair more precision than the rule gives).
- Multiply by r and c. Pencil says 3 × 0.32 = 0.96; then 0.96 × 4.20 on C and D gives T ≈ 4.03 s (the 10⁶ from the square and the 10⁻⁶ from µF have cancelled).
- Words per hour is 36 ÷ T. Set 36 on D, bring 4.03 on C over it, and read the result at the index: ≈ 8.9, third digit a 2 or a 3.
Answer: ≈ 8.92 words per hour, hand-checked against the verified 8.9233. Reading the last digit as 8.93 instead is ordinary slide-rule rounding, since rounding T to 4.03 before dividing already moves it that far.
Thomson presented the law of squares to the British Association in 1855, and Wildman Whitehouse, the Atlantic Telegraph Company’s chief electrician, challenged it. The company went with Whitehouse’s cheaper cable, and his attempt to overcome the sluggishness by driving it with ever higher voltage permanently damaged its insulation.
Bonus: program it
For programmable calculators only.
The main problem asked how fast a cable can signal. Turn it around: how much help does a cable need to reach a target speed? That question is worth a program, because it has inputs. Change the cable or the target and the answer changes.
The idea. Suppose, for the puzzle, that the signal could be caught part-way along the cable and sent on afresh by a repeater that takes no time of its own. With n − 1 repeaters spaced evenly, the cable becomes n equal sections in a row. Each section is a cable in its own right, only 1/n as long, and the chain as a whole can go no faster than one of its sections.
The task. Write a program with two inputs:
- T, the retardation time of the whole cable in seconds, as defined in the main problem
- W, the target speed in words per hour
It gives the smallest number of equal sections n for which the chain carries at least W words per hour. The settling time (2T per element) and the 50-element word are the same as before.
Test your program on these three cases. All use the same wire, 3.0 Ω and 0.32 µF per nautical mile.
| Case | Cable length | Target W (words/hour) |
|---|---|---|
| A | 2,050 nmi (the main problem’s cable) | 60 |
| B | 2,050 nmi | 200 |
| C | 4,100 nmi | 60 |
Work out T for each cable first, the way the main problem did. The program starts from T.
Click for a hint
A section 1/n as long has 1/n² of the retardation, so a chain of n sections signals at
words per hour = 36 · n² / T
(the same 36 as in the main problem, where n was 1). Start at n = 1 and add one at a time. Stop at the first n for which 36 · n² / T reaches W. That n is the answer.
Click to reveal the solution
| Case | T (s) | W | Rate with n − 1 sections | Rate with n sections | n |
|---|---|---|---|---|---|
| A | 4.0344 | 60 | 35.7 (n = 2) | 80.3 | 3 |
| B | 4.0344 | 200 | 142.8 (n = 4) | 223.1 | 5 |
| C | 16.1376 | 60 | 55.8 (n = 5) | 80.3 | 6 |
The loop is not needed to get these numbers: n is the smallest whole number with n² ≥ W · T / 36, which is the square root of that quantity rounded up (2.59, 4.73 and 5.19 for A, B and C). The program is still the honest way to see the square law at work.
Cases A and C are the same problem at two scales. Doubling the cable multiplies T by four, and the square root brings that back to a factor of two, so C needs twice the sections of A. Both end up with sections 683⅓ nmi long, and both signal at the same 80.3 words per hour. Only the length of a section matters.
Programs
Programs follow for the HP-42S, HP-15C, HP-41C, Elektronika MK-61 and HP-25. Open the one for your calculator. Key the program in once, then for each case store T and W in registers 1 and 2 and run it. The text to the right of a listing explains it; do not key it in. Each machine runs at its real speed, and all three cases take under a minute on the slowest of them.
HP-42S
Press PRGM to enter program mode, key in the lines below, and press PRGM again to leave it.
The program clears register 0 (the count) itself.
To run it, store the two inputs, then start the program. For case A:
4.0344 STO 01
60 STO 02
XEQ 01
The answer, 3, is left in X. For another case, store its T and W in the same two
registers and press XEQ 01 again.
Text to paste. In Free42 you can paste this in program mode, without the line numbers. On a real HP-42S, key it in one line at a time.
LBL 01
0
STO 00
LBL 02
1
STO+ 00
RCL 02
RCL 00
RCL 00
×
36
×
RCL 01
÷
X<Y?
GTO 02
RCL 00
RTN
The same program with commentary.
01 LBL 01
02 0
03 STO 00 n = 0
04 LBL 02 start of a try
05 1
06 STO+ 00 n = n + 1
07 RCL 02 W, to be compared with later
08 RCL 00
09 RCL 00
10 × n²
11 36
12 × 36 · n²
13 RCL 01
14 ÷ 36 · n² / T, the rate; W is now in Y
15 X<Y? is the rate still below W?
16 GTO 02 if so, try the next n
17 RCL 00
18 RTN stop, with n in X
19 END
HP-15C
This program uses 18 of the machine’s steps.
Press g P/R (the shifted R/S key) to enter program mode, key in the steps below, and press
g P/R again to leave it. In the listing, LBL 1 is f LBL 1, and x≤y is g x≤y
(the shifted ÷ key).
To run it, clear the registers (the program counts up from the 0 it finds in register 0) and store the two inputs in registers 1 and 2. For case A:
f CLEAR REG
4.0344 STO 1
60 STO 2
GSB 1
The answer, 3.0000, shows when the program stops. For another case, clear the registers again,
store its T and W, and press GSB 1 again.
The stop test x≤y compares X with Y and does the next step only if X is less than or equal to Y.
Here X is W and Y is the rate, so it is true once the rate has reached W. Then GTO 2 leaves the
loop. Otherwise the program skips that step and GTO 1 tries the next n.
Keys to press. Type each number as an ordinary number. The lines column is the line number the calculator shows as you key in; it is there so you can check your place, and you do not enter it.
press lines
f LBL 1 001
1 002
STO + 0 003
RCL 0 004
RCL 0 005
× 006
36 007-008
× 009
RCL 1 010
÷ 011
RCL 2 012
g x≤y 013
GTO 2 014
GTO 1 015
f LBL 2 016
RCL 0 017
R/S 018
The same program with commentary.
001 LBL 1 start of a try
002 1 003 STO + 0 n = n + 1
004 RCL 0 005 RCL 0
006 × n²
007 3 6 009 × 36 · n²
010 RCL 1 011 ÷ 36 · n² / T, the rate
012 RCL 2 W, so X = W and Y = the rate
013 x≤y has the rate reached W?
014 GTO 2 if so, leave the loop
015 GTO 1 otherwise, try the next n
016 LBL 2
017 RCL 0 n
018 R/S stop, with the answer showing
HP-41C
This program is 20 lines long.
Press PRGM to enter program mode, key in the lines below, and press PRGM again to leave it.
LBL is the gold shift of the STO key, GTO is the gold shift of RCL, and x≤y? is the gold
shift of +. Labels are two digits, so LBL 01 is LBL 0 1. A number such as 36 is one line
of the program.
To run it, store the two inputs in registers 01 and 02, then start the program. The program clears register 00 (the count) itself. For case A:
4.0344 STO 01
60 STO 02
XEQ 01
The answer, 3.0000, shows when the program stops. For another case, store its T and W in the
same two registers and press XEQ 01 again.
The test x≤y? does the next line only if X is less than or equal to Y. Here X is W and Y is the
rate, so it is true once the rate has reached W, and then GTO 03 leaves the loop. Otherwise the
program skips that line and GTO 02 tries the next n.
01 LBL 01
02 0
03 STO 00 n = 0
04 LBL 02 start of a try
05 1
06 STO+ 00 n = n + 1
07 RCL 00
08 RCL 00
09 × n²
10 36
11 × 36 · n²
12 RCL 01
13 ÷ 36 · n² / T, the rate
14 RCL 02 W, so X = W and Y = the rate
15 X<=Y? has the rate reached W?
16 GTO 03 if so, leave the loop
17 GTO 02 otherwise, try the next n
18 LBL 03
19 RCL 00 n
20 R/S stop, with the answer showing
Elektronika MK-61
This program uses 20 of the machine’s 105 steps.
- Press
FПРГ(the shiftedВПkey) to enter program mode. - Key in the steps below.
- Press
FАВТ(the shifted/-/key) to return to normal mode.
To run it, clear register 0 (the count) and store the two inputs in registers 1 and 2. For case A:
0 x→П 0
4.0344 x→П 1
60 x→П 2
БП 0 0
С/П
The answer, 3., shows when it stops. For another case, clear register 0, store its T and
W in registers 1 and 2, then press БП 0 0 and С/П again.
A few notes on the keys:
П→x 0recalls register 0 andx→П 0stores into it. Each is a single step.- Type every number starting with a digit, so
0.5and never.5. - The test
Fx≥0(the shiftedВ/Оkey) is followed by an address. If x is zero or positive, the program carries on past the address. If x is negative, it jumps to that address.
addr keys what it does
00 П→x 0 В↑ 1 + x→П 0 n = n + 1
05 П→x 0 П→x 0 × n²
08 В↑ 3 6 × 36 · n²
12 П→x 1 ÷ 36 · n² / T, the rate
14 П→x 2 - rate − W
16 F x≥0 00 rate still below W (x negative): back to step 00.
Otherwise carry on.
18 П→x 0 С/П n, and stop
HP-25
The HP-25 has room for 49 program steps, and this program uses 15.
One thing is specific to this calculator: the loop goes back with GTO 01, not GTO 00. On the
HP-25 a jump to line 00 stops the program.
To enter the program, set the switch to PRGM, press f PRGM to clear program memory, and key in
the steps below.
To run it, set the switch to RUN, press f CLEAR REG to clear the registers (the program counts
up from the 0 it finds in register 0), and store the two inputs in registers 1 and 2. For case A:
4.0344 STO 1
60 STO 2
GTO 0 0
R/S
The answer, 3.00, shows when the program stops. For another case, clear the registers again,
store its T and W, and press GTO 0 0 and R/S again.
Keys to press. Type each number as an ordinary number. The steps column is the step number the calculator shows as you key in; it is there so you can check your place, and you do not enter it.
press steps
1 01
STO + 0 02
RCL 2 03
RCL 0 04
RCL 0 05
× 06
36 07-08
× 09
RCL 1 10
÷ 11
f x<y 12
GTO 0 1 13
RCL 0 14
R/S 15
The same program with commentary. W is recalled first so that it is already in Y when the rate
comes out on top of it, which is what x<y compares.
01 1 02 STO + 0 n = n + 1
03 RCL 2 W
04 RCL 0 05 RCL 0
06 × n²
07 3 6 09 × 36 · n²
10 RCL 1 11 ÷ 36 · n² / T, the rate, with W in Y
12 f x<y is the rate still below W?
13 GTO 01 if so, try the next n
14 RCL 0 n
15 R/S stop, with the answer showing
Solvable on: HP-42S, HP-15C, HP-41C, HP-48GX, HP-25, Elektronika MK-61, TI-84, Casio fx-991, Slide Rule