SECTION SIXTEEN

EXERCISES

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 34·2. There is no saving in cancelling this to 2/3 x 34·2. 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 1·5 x (78.3 / 5.7). When two or three simple factors appear as in 8 x 3 x 16·3, they should be combined mentally, and the figures treated as 24 x 16·3. 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.

Commerce

Example: £56, 8s. 0d. is invested at 5½% per annum compound interest. Calculate the value after 8½ years.

£1 at the end of one year becomes £(1 + ·055) = 1·055; at the end of two years becomes £(1·055) (1·055) = £(1·055)2, and at the end of 8½ years becomes £(1·055).

Use log-log scale, and if 1·055 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.
Now £56, 8s. 0d. = £56·4.
Under 564C read 891 in D. £89·1 = £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 6½ hours' overtime. Worker's rate 92s. for 48 hrs. B with 5¾ hours' overtime. Worker's rate 88s. for 48 hrs. C with 8¼ hours' overtime. Worker's rate 85s. for 44 hrs. D with 2¾ hours' overtime. Worker's rate 84s. for 44 hrs. E with 11½ 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. 2½d. per cwt., the equivalent price per ton is required. 2s. 2½d. = 2·209s.- taken from conversion table.
Set X to 2209D.
Under X in H read 44·2s. = £2, ¼ 2½d.

Problem 56. The price of a certain kind of foil is 3¾d. per sq. foot. Calculate the cost of 100 sheets of foil 21" x 15½". (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: 1·52" and 1·42,,. 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 1·6 lb. of coal per h.p. per hour. (Calorific value of coal 12,200 B.T.U. per lb.)

Friction

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. = 7·1.

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 1¼ 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 = 2·7183).
Set 1C to pD. Under 15C read 4.72 in D. (4.72 is mq)
Set 1C to 2·72LL (log-log scale). Under 472C read in LL, 111.
To 224D set 111C. Under 1C read 20·2 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 ¼" 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: 5·2 grm. of ice added to 54·8 grm. of water at 20° C. give a final temperature of 12° C. Calculate the latent heat of fusion (water equivalent of calorimeter 5·1 grm.).
Over 599D set 52C. Under 8C read 92·1 in D.
Latent heat = 92·1 - 12 = 80·1 C.H.U.

Problem 62. 152·5 grm, of Hg. at 100°C. were mixed with 87·5 grm, of water (including water equivalent of calorimeter) at 10·1° C. The final temperature being 15·2° 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: 5·15" under 1A. 7·3" under 2A. 8·94" under 3A.
Now traverse slide and read in D: 10·3" under 4A. 11·5" under 5A.

Problem 63. A beam of uniform cross-section, breadth = 2·2", depth = 4·6", 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 2·5". Modulus of rigidity = C = 11 x 106 lb. per sq. inch.
Formula:
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 7·15 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.

Electricity

Example: The specific resistance of platinum is 8·96 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, 5·73 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 5·2. (Specific resistance of copper 1·7 microhms.)

Building

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 3·75s. = 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 10½" x 6½" 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 = £4·85 = £4, 17s. 0d.

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

Surveying

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 34·36 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 10·9 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 15½ knots will dock in 11¾ hours; it is imperative to dock earlier. What must be the speed to dock in 9½ hours?
To 155D set 95c.
Set X to 10C. 1C to X.
Result: 19·1 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 10·5 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.

Miscellaneous

Example: 5¾% stock is purchased at (i) a premium of 8%; (ii) a discount of 8%. Calculate the interest yields in these two cases.
Over 5·75D set 108C.
Read 5·32D under 1C. Result £5, 6s. ¾
Set 92C over 5·75D.
Read 6·25D under 10C. Result £6, 5s. 0d.%.
Set 92C over 5·75D.
Read 6·25D 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. 453·6 grammes = 1 lb.)
Over 245D set 96C.
Move X to 4536C.
Set 68C to X.
Move X to 10C.
Set 1C to X.
£38·2 = £38, 4s. 0d. per ton.

Example: Calculate the combined resistance of a divided circuit in which the four branches have separate resistances of 18·6, 24·2, 8·7 and 2·56 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 = ·60110 Over 1D set 6011G. Read 1664D under 10C.

Example: Calculate with one setting of the slide rule the percentages of the various elements in potassium chlorate (KClO3).
KClO3 = 39 + 35·5 + 48 = 122·5
Over 1D set 1225C.
Under 39C read 318D i.e. K = 31.8%
Under 355C read 29D i.e. Cl = 29·0%
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 62·3 lb. Calculate the weight of a sphere of copper 8½" dia, Specific gravity of Cu 8·8. Repeat for lead, cast iron and silver, (Sp. gravities 11·4, 7·2 and 10·5 respectively.)
Over 425D set l0C
Move X to 425C
Set 1C to X
Move X to 425C  (This completes the cubing of 4·25)
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.
Answers 102, 132·5, 83·6 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 4¼ 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,

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 = 2·64H5/2
First find the value of 1·75 using any slide rule equipped with a log-log scale.
Set X over 1·75 LU.
Set 10C to X.
Move X to 2·5C.
Under X read 4·05 in LL.
Now multiply 4·05 by 2·64, using C and D scales.
Answer 10·7 cu. ft. per sec.

Example: Calculate the overall efficiency of a steam engine and boiler which consumes 1·6 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.