Commerce - Energy and Power - Friction and Heat - Strength and Deflection of Beams - Strength of Shafts and Deflection of Springs - Electricity - Building - Surveying - Navigation - Miscellaneous.

WE give in this section a selection of examples and problems which illustrate the various classes of work in which the slide rule may be used. The number of such examples could be increased indefinitely.

We trust that no reader will think that we are suggesting he should search for his particular type of problem and then merely memorise the movements of slide and cursor and repeat them for his own calculations. To follow such a course would be futile. The only way to become proficient with the slide rule is to understand its fundamental principles, and to work out simple exercises. When a practical problem presents itself the relevant numbers should be written down; figures which cancel out completely should be eliminated, and simple factors should be combined mentally to reduce the slide rule operations to a minimum.

Take as a simple case 4/6 x 342. There is no saving in cancelling this to 2/3 x 342. There may be a saving in cancelling the 4/6 to 1.5 , but it is doubtful whether it is worth while doing this. A case such as (3 x 12 x 78.3) / ( 6 x 4 x 5.7) should be cancelled down to 3/2 x (78.3 / 5.7) , or better, to 15 x (78.3 / 5.7). When two or three simple factors appear as in 8 x 3 x 163, they should be combined mentally, and the figures treated as 24 x 163. We would warn the reader against attempting to cancel or combine anything beyond very simple factors.

In some of the examples which follow, the rule best suited for use is mentioned. If scales C and D are to be used, any rule will meet the case. If trigonometrical work is involved, use the Navigational rule every time; the reader will soon see why we recommend this rule.

The time occupied in making acquaintance with the duplicated C and D scales of Dualistic and Electrical rules will be a good investment.

In studying the practical examples we give below, the reader should write down the essential figures arising from the problem. He will find we have cancelled out or combined simple factors when they occur, but only in the obvious cases.

If the reader has worked through the exercises in the earlier sections, and solved some of the problems, probably there will be no need to study all those given in this section, but if he feels he needs still more practice with his slide rule, the following examples are suitable for the purpose.


Example: 56, 8s. 0d. is invested at 5% per annum compound interest. Calculate the value after 8 years.

1 at the end of one year becomes (1 + 055) = 1055; at the end of two years becomes (1055) (1055) = (1055)2, and at the end of 8 years becomes (1055)8.

Use log-log scale, and if 1055 is not within the range of the scale, treat as  2.11/2.
    Set X to 211LU. 10C to X. X to 85C.
    Read in LL under X 570.
    Set X to 2LU. 10C to X. X to 85C.
    Read in LL under X 360.
    Over 57D set 36C.
    Under 1C read 158D.
    Now 56, 8s. 0d. = 564.
Under 564C read 891 in D. 891 = 89, 2s. 0d.

Problem 54. In costing a job it was found that 55 operations in a certain machine took 450 minutes to complete. The operator's rate of pay being 6, 7s. 0d. for 48 hours. Calculate the wages cost per operation.

Example: One gross articles weigh 84 lb. What is the weight of one piece?
Set X to 84K. Under X read 583 lb. in U. No setting of slide is required, the conversion being effected by projecting direct from K to U, using Commercial rule.

Problem 55. A time sheet for a group of workers credits:

A with 6 hours' overtime. Worker's rate 92s. for 48 hrs.
B with 5 hours' overtime. Worker's rate 88s. for 48 hrs.
C with 8 hours' overtime. Worker's rate 85s. for 44 hrs.
D with 2 hours' overtime. Worker's rate 84s. for 44 hrs.
E with 11 hours' overtime. Worker's rate 84s. for 44 hrs.

Calculate the wages due, time and a quarter being paid for all overtime. (Use Commercial rule.)

Example: Material costs 2s. 2d. per cwt., the equivalent price per ton is required. 2s. 2d. = 2209s.- taken from conversion table.
    Set X to 2209D.
Under X in H read 442s. = 2, 2d.

Problem 56. The price of a certain kind of foil is 3d. per sq. foot. Calculate the cost of 100 sheets of foil 21" x 15". (Use scales C and K for multiplication, but read result in U which automatically divides by 144.)

Energy and Power

Example: A gas engine uses 93 1 cu. ft. of Dowson gas per i.h.p. per hour. Calorific value of this gas is 123,000 ft. lb. per cu. ft. Calculate the efficiency of the engine.
    Over 33D set 931C.
    Set X to 6C. 123C to X.
Result: 17.25% in D under 1C.

Problem 57. The average heights of indicator diagrams taken from both ends of a cylinder are: 152" and 142,,. Spring 1" = 60 lb. weight. Piston 8" dia. Stroke 16". Speed 220 r.p.m. Calculate i.h.p.

Example: A cut of depth 11" is being made in a lathe. Feed is 03" per rev. Speed 50 r.p.m. Dia. of bar 4"; pressure on tool 910 lb. weight. Find the h.p. expended at the tool, and weight of metal removed per minute. (Density of steel 28 lb. per cu. inch.)
    To 91D set 3C.
    Set X to pC. 33C to X. X to 1C. 10C to X.
    Read in D under 5C 144 h.p.
    To 44D set 10C.
    Set X to pC. 1C to X. X to 15C. 1C to X.
Result in D under 28C = 58 lb.

Problem 58. Calculate the overall efficiency of a steam engine and boiler using 16 lb. of coal per h.p. per hour. (Calorific value of coal 12,200 B.T.U. per lb.)


Example: A horizontal shaft, 8" dia., carries a load of 5 tons. Calculate the h.p. absorbed in friction at 200 r.p.m. m. = 05.
    To 25D set 33C.
    Set X to 224C. 12C to X. X to pC. 1C to X.
In D under 16C read h.p. = 71.

Problem 59. Calculate the h.p. required to drive a motor car at 60 miles per hour along a level road. Rolling resistance 30 lbs. per ton. Assume air resistance accounts of four-fifths of the total h.p. at this speed. Weight of car 1 tons.

Example: To lower a load of 1 ton, a rope is wrapped three times round a horizontal post. Diameter of post 4". m . = 25. Calculate the pull necessary at the free end of the rope to lower steadily.
    Formula: T1 / T2 = emq   (e = 27183).
    Set 1C to pD. Under 15C read 4.72 in D. (4.72 is mq)
    Set 1C to 272LL (log-log scale). Under 472C read in LL, 111.
To 224D set 111C. Under 1C read 202 lb. in D.
(Provided the post is strong enough its diameter does not enter into the calculation.)

Problem 60. A belt laps 180 round a flat pulley, and is just on the point of slipping. Tension in the slack side is 400 lb. Calculate tension in the tight side. m. = 3.

Example: A railway line is laid in 40-ft. lengths on a day when the temperature is 10 C., and gaps of " are left between adjacent lengths. What will be the gaps when the temperature is 40 C. and -10 C.?
    Coefficient of linear expansion of steel 000012 per degree C.
    Set l0C to 48D.
    Read in D under 36C increase in length 173.
    Read in D under 24C decrease in length 115.
    Gap at 40 C. = 25 -  173 = 077".
    Gap at -10 C. = 25 -  115 = 365".

Problem 61. Gas has a volume of 188 c.c. at a pressure of 76 cm. of Hg. and temperature 20 C. What will be the volume at 40 cm. of Hg. and 80 C.?

Example: 52 grm. of ice added to 548 grm. of water at 20 C. give a final temperature of 12 C. Calculate the latent heat of fusion (water equivalent of calorimeter 51 grm.).
    Over 599D set 52C. Under 8C read 921 in D.
Latent heat = 921 - 12 = 801 C.H.U.

Problem 62. 1525 grm, of Hg. at 100C. were mixed with 875 grm, of water (including water equivalent of calorimeter) at 101 C. The final temperature being 152 C. Calculate the specific heat of Hg.

Strength and Deflection of Beams

Example: A cantilever, 50" long, carries a load of 4000 lb. at its free end. The section is rectangular of breadth 3". Calculate the depth at distances of 10", 20", 30", 40" and 50" from the free end if the maximum stress is to be 3000 lb. per square inch.
   Formula: f / y = M / I
    Under 80A set 3B.
Read results in D: 515" under 1A. 73" under 2A. 894" under 3A.
Now traverse slide and read in D: 103" under 4A. 115" under 5A.

Problem 63. A beam of uniform cross-section, breadth = 22", depth = 46", is simply supported at its ends. It is 30" long and carries a load of 1000 lb. 18" from one end. Calculate the maximum stress induced by the load.

Example: A uniform beam, 3" diameter, 4' 0" long, simply supported at ends, carries a load of 4000 lb. uniformly distributed. Find the maximum deflection.
E = 30 x 106 lb. per sq. inch. (Use Electrical or Dualistic rule.)
    Formula: (5 W L3) / (384 E I )
    Under 48d set 81c
    Set X to 64c.  pC to X.
Result 484 in d or D coincident with 4 in c or C.
Approximation gives .04. Result .0484".

Problem 64. Calculate the maximum stress induced in the beam of the above example.

Strength of Shafts-Deflection of Springs

Problem 65. Calculate the h.p. which may be safely transmitted by a circular shaft 4" diameter at 200 r.p.m. Stress to be limited to 9000 lb. per sq. inch.

Example: Calculate the angle of twist per foot of length of the shaft in Problem 65. Modulus of rigidity = C = 13 x 106 lb. per sq. inch.
    Formula: Angle in radians  = (2 f l ) / C. Dia = 32 T l / (pC dia.4 )
    After cancelling, over 54D set 13C.
    Under 1C read in D 416.
    Approximation gives 004. Result: 00416 radians.
X to 416D. pC to X. Under 180C read 238 in D.

Example: Calculate the number of coils necessary in a helical spring to give an extension of 5" for an axial load of 12 lb. Dia. of wire 21", dia. of coil 25". Modulus of rigidity = C = 11 x 106 lb. per sq. inch.
    Number of coils = (C x (dia. of wire)4 x extension) / (8 x Load x (dia. of coil)3 )
    Over 5D set 25C. Set X to 21C. 25C to X. X to 21C.
    25C to X. X to 21C. 1C to X. X to 21C. 96C to X.
    X to 10C. 1C to X.
Result in D under 11 C 715 coils.

Problem 66. A coiled spring has 50 turns of wire 11" dia. Dia, of coil 1.5". C = 10 x 106 lb. per sq. inch. Find the extension caused by a load of 2 lb.


Example: The specific resistance of platinum is 896 microhms at 0 C., and its temperature coefficient is 0034. What length of platinum wire of 32 s.w.g. (dia. 0274 cm.) will have a resistance of 5 ohms, at 50 C.? What will be its resistance at 100 C.?
    Over 5D set 4C.
    Set X to pC. 117C to X. X to 274C. 896C to X.
Read in D under 274C, the length, 282 cm.
    Over 5D set 117C.
Read in D under 134C, 573 ohms, at 100 C.

Problem 67. A generator feeds 1500 75-watt lamps at 230 volts; find the current supplied.

Example: Calculate the h.p. required to drive a dynamo generating 40 kw. Efficiency 85%. (Use Electrical rule.)
    Set N in c to 85d.
Above 4D read 63 in C. Result: 63 h.p.

Problem 68. The output of an electric motor is 65 h.p. and its efficiency is 82%. Find the power required to drive it.

Problem 69. A copper conductor is 500 yards long and carries a current of 21 amps. Calculate the diameter of wire if the volt drop is to be limited to 52. (Specific resistance of copper 17 microhms.)


Example: Imported scantlings cost 25 per standard (165 ft. cube) to which must be added 2 per standard for delivery and 4 per standard for planing. Find the cost per foot cube wrought.
    25 + 2 + 4 = 31.
    Over 31D set 165C.
Under 1C read in D 188 or under 2C read 375s. = 3s. 9d.

Problem 70. Calculate the cost per foot run to excavate, fill and ram a trench for a 4" drain; depth of trench 3' 0". Concrete bed 18" wide, 6" thick, Benching up and displacement of pipe taken as half volume of bed. Excavating at 2s. 3d. per yard cube. Returning, filling and ramming at 4s. per yard cube.

Example: Calculate the cost of tiles per square (100' super), tiles 10" x 6" gauge (i.e. exposed part of tile) 4"; tiles at 8, 15s. 0d. per 1000.
    Over 144D set 4G.
    Set X to 875C. 65C to X.
Result in D under l0D = 485 = 4, 17s. 0d.

Problem 71. Find the cost of 450 ft. run of timber 9" x 6" at 84, 7s. 6d. per standard.


Problem 72. A tower subtends an angle of 10 4' at a point 627 ft. horizontally from its base. Calculate the height of the tower.

Example: A hill subtends an angle of 8 30' at a point A. At B, 1750 ft. nearer along a horizontal line, the hill subtends 12 50'. Find the height of the summit above the points of observation.
    12 50' - 8 30' = 4 20'.
    To 1750A set 4 20' S2.
    Set X to 12 50' S2. 90 S2 to X.
Result: 760 ft. in A over 8 30' S2.

Problem 73. A plot of land measured by a planimeter on a plan gave an area of 3436 sq. inches. The scale of the plan being 1 chain = 1 inch; find the actual area of the plot.

Example: A survey line, XYZ, crosses a river too wide to be chained. X is on one bank, Y on the opposite bank, and Z is 100 ft. beyond Y. At a point P 200 ft. from Y, and on a line through Y at right angles to XYZ, sights are taken on X and Z. The angle XPZ was observed to be 78. Calculate the width of the river.
    Set X to 5D. In T1 read 26 40' under X.
    78 - 26 40' = 51 20'.
    90 - 51 20' = 38 40'.
    Set X to 2D. 38 40' T2 to X.
Result: 250 ft. in D under 10C.


Example: A ship steaming at 12 knots runs into a 2-knot current setting S. 10 W. Ship's true course is to be N. 50 W. Calculate course to steer, and find true speed.
    Angle between true course and current is 120 sin 120 = sin 60.
    Set X to 60S1. 12B to X. Over 2B read 8 20' in S1. 60 - 8 20' = 51 40'.
Course to steer is N. 41 40' W.
    Set X to 12A. 60S2 to X. X to 51 40' S2.
In A over X read 109 knots, the true speed.

Problem 74. From a vessel a lightship was observed 65 forward of the beam on the port side; after steaming 12 miles the light was abeam. Calculate the distance at which the vessel passes the light, and the distance from the light when the first observation was made, If it was desired to pass the light at 2 miles, what alteration to course would have been necessary at the time of first observation?

Example: A ship steaming at 15 knots will dock in 11 hours; it is imperative to dock earlier. What must be the speed to dock in 9 hours?
    To 155D set 95c.
    Set X to 10C. 1C to X.
Result: 191 knots in D under 1175C.
(This result could have been obtained in one setting of the slide with the duplicated C and D scales-see Section 10.)

Problem 75. From a ship steering N. 60 E., two observations on a lightship were made, First bearing was due E. After steaming 105 miles, the second bearing was 5. 28 W. Find the distances from the light when the observations were made.
(This problem is: Given K= 30, L = 118, l = 10.5. Find k and m.) A diagram should be drawn.


Example: 5% stock is purchased at (i) a premium of 8%; (ii) a discount of 8%. Calculate the interest yields in these two cases.
    Over 575D set 108C.
    Read 532D under 1C. Result 5, 6s.
    Set 92C over 575D.
Read 625D under 10C. Result 6, 5s. 0d.%.
    Set 92C over 575D.
Read 625D under 10C. Result 6, 5s. 0d.%.

Example: 68 kilos of material cost 2450 francs. Calculate the cost per lb. and per ton in sterling. (Rate of exchange 960 francs = 1. 4536 grammes = 1 lb.)
    Over 245D set 96C.
    Move X to 4536C.
    Set 68C to X.
    Move X to 10C.
    Set 1C to X.
    Under 24C read 408D.
    Under 224C read 382D.
Answer 408 pence per lb.
    382 = 38, 4s. 0d. per ton.

Example: Calculate the combined resistance of a divided circuit in which the four branches have separate resistances of 186, 242, 87 and 256 ohms.
    1 / R = 1 / r1 +1 / r2 +1 / r3 +1 / r4 + ...

  Over 1D set 186C.  
  Read 5375D under 10C   05375
  Over 1D set 242G.  
  Read 4135D under 10C.   04135
  Over 1D set 87G.  
  Read 115D under 10C.   115
  Over 1D set 256G.  
  Read 391C under 10   391

1 / R =

  Over 1D set 6011G.  
  Read 1664D under 10C.  

Answer 1 664 ohms.

Example: Calculate with one setting of the slide rule the percentages of the various elements in potassium chlorate (KClO3).
    KClO3 = 39 + 355 + 48 = 1225
    Over 1D set 1225C.
    Under 39C read 318D i.e. K = 31.8%
    Under 355C read 29D i.e. Cl = 290%
    Under 48C read 392D i.e. O = 39.2%
                                Total       = 100%

Problem 76. Repeat the foregoing exercise for (i) sodium bicarbonate (NaHCO3); (ii) sulphuric acid (H2S04).

Problem 77. A body falls freely under the action of gravity. Calculate how far it will fall in the 1st second, the 5th second and the 10th second after release. Also find the velocity after falling 200 feet and its velocity after 10 seconds from rest.

Example: A cubic foot of water weighs 623 lb. Calculate the weight of a sphere of copper 8" dia, Specific gravity of Cu 88. Repeat for lead, cast iron and silver, (Sp. gravities 114, 72 and 105 respectively.)
    Over 425D set l0C
    Move X to 425C
    Set 1C to X
    Move X to 425C  (This completes the cubing of 425)
    Set 3C to X.
    Move X to pC.
    Set 1728C to X,
    Move X to 1C.
    Set l0C to X.
    Move X to 4C.
    Set l0C to X
    Move X to 623C.
    Set 1C to X.
    Under 88C read 102D.
    Under 114C read 1325D.
    Under 72C read 836D.
    Under 105C read 122D.
Answers 102, 1325, 836 and 122 lb.

This exercise involves an unusual number of operations, all of which have been effected with the C and D scales of the 10" rule. The cubing of 4 at the outset might have been carried out by starting in D and moving across to the A and B scales and a slight reduction in operations would have resulted, but the remaining factors would then have been dealt with in A and B which are 5" scales.

There is also one slide traverse, which increases the manipulation of the rule.

The slide traverse would be eliminated and the cubing operation reduced by using a Dualistic slide rule. If the reader possesses this type of rule we invite him to repeat the exercise, which should present no difficulty.

In case some assistance is necessary, we give the slide rule movement below:
    Set X to 425P2.
    Set 1C1 to X.
    Move X to 425C2.
    Set 3C1 to X.
    Move X to pC1.
    Set 1728C2 to X.
    Move X to 4C1.
    Set 1C2 to X.
    Move X to 623C1.
    Set 1C2 to X,
    Over 88C2 read 102D2.
    Over 114C2 read 1325D2.
    Over 72C2 read 836D2.
    Over 105C2 read 122D2.

The reader will notice that the results could have been obtained in D and there are other alternatives.

Problem 78. Calculate the water pressure in lb. per square inch at depths of 400 feet and 10,000 feet in sea water (1 cu. foot sea water weighs 64 lb.)

Example: The still-water surface-level is 21 inches above the bottom of a 90 V gauge notch. Calculate the discharge in cubic feet per sec.
    Formula: Q = 264H5/2
    First find the value of 175 using any slide rule equipped with a log-log scale.
    Set X over 175 LU.
    Set 10C to X.
    Move X to 25C.
    Under X read 405 in LL.
    Now multiply 405 by 264, using C and D scales.
Answer 107 cu. ft. per sec.

Example: Calculate the overall efficiency of a steam engine and boiler which consumes 16 lb. of coal per horse power hour. Calorific value of coal 12,400 B.T.U.
    Over 33D set 778G.
    Move X to 6C.
    Set 16C to X.
    Move X to 1C.
    Set 124C to X.
    Under 1C read 128D.
Answer 12.8%.

Problem 79. Calculate the efficiency of a hydraulic crane which uses 60 gallons of water at a pressure of 900 lb. per square inch in lifting a load of 5 tons to a height of 60 feet.

Problem 80. In an electric circuit in which no mechanical work is done a current of 56 amps is flowing against a resistance of 25 ohms. Calculate the voltage and the heat loss in watts and in horse power.

Hodder Stoughton, reproduced with permission.