The Deepest Dive

Pressure at depth h in a fluid of density ρ:

P = ρ · g · h

Stress in a thin-walled sphere of radius r and wall thickness t:

σ = P · r / (2 · t)

With r = D/2, that becomes σ = P · D / (4 · t), which is a single chain of multiplications and divisions.

Pressure at the bottom, from the weight of the water column:

P = ρ · g · h = 1025 × 9.80665 × 10916 ≈ 109.7 MPa

Stress in the wall, from the thin-shell formula:

σ = P · r / (2t) = P · D / (4t)
  = 109.7 MPa × 2.16 / (4 × 0.127)

Answer: ≈ 466.5 MPa

Two honest caveats about the model. First, real seawater compresses under that much water above it, so the true pressure is not exactly what a constant density gives. Second, the wall is thick: its thickness is about an eighth of its radius, past where the thin-shell formula is at its best, and a full thick-shell analysis gives a different stress distribution through the wall. For a problem you can do on a slide rule, it is the right first estimate.

For scale: the sphere was calculated to withstand 1,250 kg/cm² (about 123 MPa) at the bottom of the Challenger Deep, with a margin of safety above the pressure worked out here.

The whole calculation is one chain of multiplications and divisions, so RPN handles it with no memory registers. The last step divides by a million to turn pascals into megapascals.

1025    ENTER
9.80665  *
10916   *
2.16    *
4       /
0.127   /
1000000 /

Display: ≈ 466.5499 MPa (the 42S carries more digits than it shows).

The intermediate value here is around 10⁸, which is well inside the MK-61’s range, so a plain chain works with no memory register. It shows eight digits.

Enter every number starting with a digit, never a bare decimal point: 0.127 must be typed 0 . 1 2 7.

1025    ENTER
9.80665  ×
10916   ×
2.16    ×
4       ÷
0.127   ÷
1000000 ÷

Display: 466.54985

The last digits differ a little from the HP values because the MK-61 rounds to eight digits at every step.

Same structure as the 42S:

1025    ENTER
9.80665  ×
10916   ×
2.16    ×
4       ÷
0.127   ÷
1000000 ÷

Result: ≈ 466.55 MPa.

1025    ENTER
9.80665  *
10916   *
2.16    *
4       /
0.127   /
1000000 /

Display: ≈ 466.55 MPa.

A plain stack entry, the same shape as the 42S. There is no π in this one, so the symbolic-constant trap from challenge 3 does not come up, and no →NUM is needed.

1025    ENTER
9.80665  ×
10916   ×
2.16    ×
4       ÷
0.127   ÷
1000000 ÷

Result: 466.549906614 MPa.

Ten digits are held internally, but the display shows what the mode allows. Set FIX 2 so the result reads out as decimals rather than a power of ten:

f FIX 2
1025    ENTER
9.80665  ×
10916   ×
2.16    ×
4       ÷
0.127   ÷
1000000 ÷

Display: 466.55

1025*9.80665*10916*2.16/(4*0.127)/1000000

Result: ≈ 466.5499 MPa.

Same single expression, natural display mode:

1025×9.80665×10916×2.16÷(4×0.127)÷1000000

Result: ≈ 466.5499 MPa.

Every step is multiplication or division, so no pencil arithmetic is needed. The exponents are tracked separately.

1. Pressure. Multiply 1.025 × 9.807 on C and D: set the left index of C over 1.025 on D, move the cursor to 9.807 on C, and read ≈ 1.005 on D (that is 10.05, tracking the exponent separately). Multiply by 1.0916 the same way: set the index of C at 1.005 on D, move the cursor to 1.0916 on C, and read ≈ 1.097 on D. The pressure is ≈ 1.097 × 10⁸ Pa.
2. Geometry factor. Divide 2.16 by (4 × 0.127) = 0.508. Set 2.16 on D, bring 5.08 on C above it, and read at the index: ≈ 4.25.
3. Stress. Multiply the pressure by the geometry factor: 1.097 × 4.25 ≈ 4.66, so 4.66 × 10⁸ Pa.
4. Divide by 10⁶ to get megapascals.

Answer: ≈ 467 MPa — three significant figures is a slide rule’s honest precision; hand-checked against the verified 466.55 result above.

The main problem gave the stress in the wall of a thin-walled sphere as σ = P · D / (4t). Turn that around to get the most pressure the wall can take:

P_max = 4 · t · σ / D

With D = 2.16 m and t in millimetres, this comes to P_max = t · σ / 540 in megapascals.

Work in megapascals throughout, so K = 2340. Keep two numbers: the pressure P and the depth h. Both start at 0. Repeat these two lines:

P ← P + 1025 · (1 + P / 2340) · 9.80665 · 100 / 10⁶
h ← h + 100

After each repeat, compare P with P_max. Stop at the first repeat that brings P up to P_max or more. That repeat has just added a layer that goes past the limit, so the answer is the value of h minus 100.

This model also has an exact solution, h = K / (ρ₀ · g) · ln(1 + P / K). It gives 11,414 m, 4,380 m and 1,652 m. The program counts only whole 100 m layers, so in each case it gives the exact depth rounded down to a multiple of 100 m.

Sphere A has a wall about as thick as the Trieste’s (127 mm). With steel that strong it would reach deeper than the 10,916 m of the Challenger Deep.

Press PRGM to enter program mode, key in the lines below, and press PRGM again to leave it. The listing is shown the way the calculator itself displays it. Pressure is in megapascals.

To run it, store the two inputs, then start the program. For sphere A:

127  STO 02
500  STO 03
XEQ 01

For another sphere, store its t and σ 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
STO 01
LBL 02
RCL 00
2340
÷
1
+
1025
×
9.80665
×
10000
÷
RCL+ 00
STO 00
100
STO+ 01
RCL 02
RCL 03
×
540
÷
RCL 00
X<Y?
GTO 02
RCL 01
100
-
RTN

The same program with commentary. The numbers at the left are the line numbers the calculator shows.

01 LBL 01
02 0
03 STO 00           P = 0
04 STO 01           h = 0
05 LBL 02           start of a layer
06 RCL 00
07 2340
08 ÷                P / K
09 1
10 +                1 + P / K
11 1025
12 ×
13 9.80665
14 ×
15 10000
16 ÷                the pressure this layer adds (MPa)
17 RCL+ 00
18 STO 00           add it to P
19 100
20 STO+ 01          h = h + 100
21 RCL 02
22 RCL 03
23 ×
24 540
25 ÷                P_max = t · σ / 540
26 RCL 00
27 X<Y?             is P below P_max?
28 GTO 02           if so, do another layer
29 RCL 01
30 100
31 -                h − 100
32 RTN              stop, with the answer in X
33 END

This program uses 53 of the machine’s steps. Pressure is in megapascals.

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 and store the two inputs in registers 2 and 3. For sphere A:

f CLEAR REG
127  STO 2
500  STO 3
GSB 1

For another sphere, clear the registers again, store its t and σ, and press GSB 1 again. The 15C runs at the machine’s real speed, about three and a half seconds per layer, so sphere A takes around six minutes.

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 P_max and Y is P, so it is true when P has reached P_max. Then GTO 2 leaves the loop. Otherwise the program skips that step and GTO 1 starts another layer.

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
RCL 0          002
2340           003-006
÷              007
1              008
+              009
1025           010-013
×              014
9.80665        015-021
×              022
10000          023-027
÷              028
RCL 0          029
+              030
STO 0          031
100            032-034
STO + 1        035
RCL 0          036
RCL 2          037
RCL 3          038
×              039
540            040-042
÷              043
g x≤y          044
GTO 2          045
GTO 1          046
f LBL 2        047
RCL 1          048
100            049-051
-              052
R/S            053

The same program with commentary. The numbers at the left are the line numbers again. A number such as 2 3 4 0 takes one line per digit, so the operator after it appears further along the row.

001  LBL 1                     start of a layer
002  RCL 0                     P
003  2 3 4 0        007  ÷     P / K
008  1              009  +     1 + P / K
010  1 0 2 5        014  ×     × 1025
015  9 . 8 0 6 6 5  022  ×     × g
023  1 0 0 0 0      028  ÷     the pressure this layer adds (MPa)
029  RCL 0          030  +     add it to P
031  STO 0                     store it as the new P
032  1 0 0          035  STO + 1   h = h + 100
036  RCL 0                     P
037  RCL 2          038  RCL 3     t, then σ
039  ×                         t · σ
040  5 4 0          043  ÷     P_max = t · σ / 540
044  x≤y                       has P reached P_max?
045  GTO 2                     if so, leave the loop
046  GTO 1                     otherwise, another layer
047  LBL 2
048  RCL 1                     h
049  1 0 0          052  -     h − 100
053  R/S                       stop, with the answer showing

This program is 35 lines long. Pressure is in megapascals.

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 2340 is one line of the program.

To run it, store the two inputs in registers 02 and 03, then start the program. The program clears registers 00 and 01 itself. For sphere A:

127  STO 02
500  STO 03
XEQ 01

For another sphere, store its t and σ 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 P_max and Y is P, so it is true when P has reached P_max, and then GTO 03 leaves the loop. Otherwise the program skips that line and GTO 02 starts another layer.

01 LBL 01
02 0
03 STO 00           P = 0
04 STO 01           h = 0
05 LBL 02           start of a layer
06 RCL 00
07 2340
08 ÷                P / K
09 1
10 +                1 + P / K
11 1025
12 ×
13 9.80665
14 ×
15 10000
16 ÷                the pressure this layer adds (MPa)
17 RCL 00
18 +
19 STO 00           add it to P
20 100
21 STO+ 01          h = h + 100
22 RCL 00           P
23 RCL 02
24 RCL 03
25 ×
26 540
27 ÷                P_max = t · σ / 540
28 X<=Y?            has P reached P_max?
29 GTO 03           if so, leave the loop
30 GTO 02           otherwise, another layer
31 LBL 03
32 RCL 01
33 100
34 -                h − 100
35 R/S              stop, with the answer showing

This program uses 61 of the machine’s 105 steps. Pressure is in megapascals.

1. Press F ПРГ (the shifted ВП key) to enter program mode.
2. Key in the steps below.
3. Press F АВТ (the shifted /-/ key) to return to normal mode.

To run it, clear registers 0 and 1 and store the two inputs in registers 2 and 3. For sphere A:

0     x→П 0
0     x→П 1
127   x→П 2
500   x→П 3
БП 0 0
С/П

For another sphere, clear registers 0 and 1, store its t and σ in registers 2 and 3, then press БП 0 0 and С/П again.

A few notes on the keys:

• П→x 0 recalls register 0 and x→П 0 stores into it. Each is a single step.
• Type every number starting with a digit, so 0.5 and never .5.
• The test F x≥0 (the shifted В/О key) 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↑   2340   ÷                    P / K
07    В↑   1   +                               1 + P / K
10    В↑   1025   ×                            × 1025
16    В↑   9.80665   ×                         × g
25    В↑   10000   ÷                           the pressure this layer adds (MPa)
32    П→x 0   +   x→П 0                        add it to P
35    П→x 1   В↑   100   +   x→П 1             h = h + 100
42    П→x 0   П→x 2   П→x 3   ×   В↑   540   ÷   -
                                               P − P_max, where P_max = t · σ / 540
52    F x≥0   00                               P still below P_max: go back to step 00.
                                               Otherwise carry on.
54    П→x 1   В↑   100   -   С/П               h − 100, and stop

The HP-25 has room for 49 program steps, and this program uses 43. To save steps it folds three constants into one: 1025 × 9.80665 × 100 ÷ 10⁶ = 1.005181625, the pressure in megapascals that the first layer adds. Pressure is in megapascals.

Two things are 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.
• The program runs at the calculator’s real speed, about three seconds per layer. Sphere A is 115 layers, about five minutes. Sphere C is 16 layers, under a minute.

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, and store the two inputs in registers 2 and 3. For sphere A:

127  STO 2
500  STO 3
GTO 0 0
R/S

For another sphere, clear the registers again, store its t and σ, 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
RCL 0          01
2340           02-05
÷              06
1              07
+              08
1.005181625    09-19
×              20
RCL 0          21
+              22
STO 0          23
100            24-26
STO + 1        27
RCL 2          28
RCL 3          29
×              30
540            31-33
÷              34
RCL 0          35
f x<y          36
GTO 0 1        37
RCL 1          38
100            39-41
-              42
R/S            43

The same program with commentary. A number such as 2 3 4 0 takes one step per digit, so the operator after it appears further along the row.

01  RCL 0                       P
02  2 3 4 0        06  ÷        P / K
07  1              08  +        1 + P / K
09  1 . 0 0 5 1 8 1 6 2 5   20  ×    × 1.005181625: the pressure the first layer adds
21  RCL 0          22  +        add it to P
23  STO 0                       store it as the new P
24  1 0 0          27  STO + 1  h = h + 100
28  RCL 2          29  RCL 3    t, then σ
30  ×                           t · σ
31  5 4 0          34  ÷        P_max = t · σ / 540
35  RCL 0                       P
36  f x<y                       is P below P_max?
37  GTO 01                      if so, do another layer
38  RCL 1                       h
39  1 0 0          42  -        h − 100
43  R/S                         stop, with the answer showing