The speed on the measured mile

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By

MEASURED

THE SPEED

ON THE

W. Muckle,

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M.Sc.

,

MILE.

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Reprinted

from

THE

SHIPBUILDER

MARINE ENAGNIDNE -BUILDER

July,

1945.

-FOUNDED BY HOOD IN JULY, 1906. :.,\\\\\\-\\\\S:

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VOL. 52. JULY NUMBER, 1945,

THE SPEED ON THE MEASURED MILE.

By W. Muckle, M.Sc.

IN

to ensure that the vessel has acquired a steadycarrying out measured-mile trials, it is important

speed by the time the measured course is reached

and further, at the conclusion of a run, and before the vessel returns to the mile, that the distance run on a straight course is sufficient to ensure that any

speed lost on the turn has been regained. The problem

of ascertaining the necessary distance which a vessel

must run before the speed can be increased by any

given amount has been examined by others (see, for

example, Attwood's " Theoretical Naval Architecture ")

but it appears to the present writer that there are some aspects of the problem worthy of further investigation. In the attempt to investigate a problem of this sort,

several assumptions are necessary at the outset. Firstly,

the engine conditions must be known or assumed ; e.g.,

whether the revolutions per minute are maintained

constant at the estimated number for the measured-mile

speed ; or whether the valve setting is maintained

constant, in which case it would be fair to assume that, with reciprocating machinery, the torque is maintained

constant ; or whether the power is constant.

Secondly, since the motion is accelerated, the question

of the added virtual mass due to the entrained water

enters into the problem. The figure obtained by Froude

from the Greyhound experimentsand which does not

appear to have been disputedwas that, for the

Greyhound, the virtual increase of mass was 20 per cent.

This figure will be used in the present note, though the actual value does not matter a great deal so far as the

analysis is concerned. If more accurate values are

available, these may be readily substituted.

A third assumption made is that the instantaneous

resistance of the ship, at any speed, whether accelerated

I or not, is the same as when the ship is being propelled

uniformly at this speed, i.e., the resistance is independent

of the acceleration.

With these assumptions, the problem may be expressed

as an instance of Newton's second law of motion.

Let A = displacement of the ship

= added virtual weight due to the entrained

water

x= distance run on straight course ;

Rt = the resistance of the ship at time t (including

the resistance augment due to the

pro-peller) ; and

T = the corresponding propeller thrust.

The force causing acceleration is (T R), and the

A+ 8

mass being accelerated is . The equation of

motion may therefore by written :

A + d2x

dtz

The equation in this form does not lend itself very

easily to solution. There are, however, two alternative

forms in which it may be written. The acceleration

dv dv

may be expressed as di or as v .

From the former, we have : A + 8 dv

= T

Rt,

di A + 8

e" dv

g (TRt )

The time necessary to increase the speed from v, to

v, is then given by :

t =

1'2

A +

g (T Rt)

dv (2)

Taking the second alternative, we have

A -I- 8 dv

v =

dx

from which the distance required to increase the speed from v, to v., is given by

x = jv2

A + 8

vi g (T)

v dv

A difficulty arises, necessitating a further assumption

in the evaluation of either of these integrals ; for,

clearly, when the ship has acquired the speed v, at

which it is to be propelled uniformly, T = Rt At this

instant, T Re =0 and the integrand becomes infinite. The integrations to be carried out in equations (2)

and (3) thus involve the determination of the area

underneath a curve whose final ordinate becomes infinite. There are many problems in dynamics where this feature

occurs ; e.g., in the rolling of ships at large angles,

although the ordinate of the curve becomes infinite, the

area is finite, and may be obtained by approximate

methods. In the present connection, however, that is

not so ; a little consideration will show that the

acceler-ation of the ship will not suddenly vanish, but will tend

progressively to zero, becoming zero only at infinite time or infinite distance from the start. It is clear,

however, that, although the results yielded by equations

EDITED BY ALFRED FLETCHER, M.Sc., A.M.Inst.N.A. No. 436. Rt . (1) (3) ; = ;

=

.

(2) and (3) are infinite, it is possible to evaluate the

time or the distance necessary to attain a speed which is

some small fraction less than the upper speed 05.

It

is necessary, therefore, to assume some arbitrary value

for this speed, and we can actually approximate as

closely as we may wish to v2, although it is never possible

to reach it. It may be sufficiently accurate to fix the

upper limit of integration as, say, 0.01 knot less than the top speed 02.

THE PROPELLER THRUST AND THE SHIP RESISTANCE. In order to perform the integration of the equations

in the foregoing section, it is necessary to have both

the ship resistance and the propeller thrust for a series

of speeds. The data in regard to the resistance will be

available if a model test has been carried outas is

general nowadays. If not, an estimate of the tow-rope

resistance may be made from such data as those of

Ayre, or from Taylor's resistance contours. The

resistance required for the present problem is the tow-rope resistance plus the augment of resistance due to

the propeller, i.e., allowing for thrust deduction. These quantities are related by the well-known equation

Rt (1 t) R, where is the thrust-deduction

fraction.

Actually, one would expect t to vary with the speed

of the vessel, but, in the absence of definite data to this effect, it will probably be ad visible to take t as

constant.

The next step is to obtain a curve of propeller thrust

for a series of ship speeds. It is unlikely that such

data will be available from a model experiment unless

a detailed experiment has been carried out. In such

cases, the data for the standard series of Taylor or

Troost may be used for the purpose. To apply these

data, the wake fraction will have to be estimated at

various speeds. In the ordinary, moderate-speed vessel,

this can often be taken as constant, and the work of

Baker, Telfer, Bragg and others will be helpful in this

respect. Where high-speed vessels are concerned, it

will not be sufficiently accurate to make this assumption,

since very often vessels such as destroyers, at their top speeds, show considerable variations in wake fraction, due to wave formation, as compared with the value of

this fraction at their cruising speeds. The only

satis-factory way to tackle this problem will be by model

experiment.

It will be assumed that by one of these means the

speeds of advance corresponding to a number of ship

speeds have been determined. To determine the

pro-peller thrusts corresponding to these speeds, some assumptions concerning the engines must be made. We

may assume, for example, that the number of revolutions

per minute is constant, or, alternatively, that the torque

is constant. A third alternative is that the shaft horse-power is constant.

(a) Constant Revolutions per Minute.

The known data consist of the revolutions per minute,

the speed of advance, the propeller diameter and the

ND

pitch ratio. The quantity 8 =-A can thus be deter_

mined for a number of ship speeds. Since a and pitch

ratio are fixed, the position on the propeller chart is fixed and, in the case of Taylor's data, the Be value

can be read off. Now, Be = 131,2 X VA5 so that U N. and 33,000 IT Thrust, T 6± 080 x 60

If Troost's data are employed, the BL, value will have

to be used, since these data are given only in terms of this quantity. x VA. _{, and} In this case, S.H.P. = N2 B,. X VA. U = X V, N2

n being the propeller efficiency obtained from the

diagrams.

Constant Torque.

In this case, the problem is more complicated, since

we do not know the speed of revolution. The known

data are the propeller pitch and diameter, together with the torque which will give the trial speed.

A suitable method of approach is as follows. Assum-ing the wake fraction as before, determine the speeds

of advance for a number of ship speeds. For each of

these speeds, determine the 3 and B, values for each

of, say, five different r.p.m., N1, N2, N2, 1\14 and N2. It is then possible to derive the torque corresponding

to each of these speeds of advance and r.p.m., and curves of torque at various r.p.m. may be plotted, as

shown in Fig. 1. /vs _{Torque corresponding to} trial speed 4/e p. m fOr ConStant r°rque

.Ship speed and speed' ofadvance

Fig. 1.

If, in this diagram, is plotted a horizontal line

repre-senting the constant torque for the trial speed, it is

possible to read off the speed of advance at the various r.p.m. corresponding to this torque, and hence to plot a curve of r.p.m. for constant torque to a base of speed

ND

of advance and ship speed.

Values of 8 =for

_VA

constant torque may then be found, and Be or B,

values and efficiency can be lifted from the Taylor or Troost data, from which it is possible to evaluate

the thrust horse-power and hence thrust.

Constant power.

If the shaft horse-power is to be constant, the known

data are the power, propeller diameter and pitch. As

before, various ship speeds are assumed, and, knowing

the wake fraction, we may determine the corresponding speeds of advance.

One method of determining the thrusts at constant

power is to use the Taylor A, charts,

1,000 a P

where A, =

D. VA3

a = pitch ratio

P = shaft horse-power ;

D = propeller diameter ; and

VA = speed of advance.

JULY, 1945 THE SHIPBUILDER AND MARINE ENGINE-BUILDER PAGE 3

=

VA (b) =

=

; ; =

PAGE 4 THE SHIPBUILDER AND MARINE ENGINE-BUILDER JULY, 1945

Since power is constant, for a given propeller and a

given speed of advance, Ap may be calculated. Contours

of Ap are plotted against B, as abscissa and pitch ratio

as ordinate. Pitch ratio being known, and Ap having been calculated, the corresponding Bts can be read off.

This fixes the position on the ordinary Bppitch ratio

efficiency charts.

The efficiency being thus determined, the thrust

horse-power is given by

propeller efficiency

The thrust horse-power being known, the thrust may be ascertained, since the speed of advance is known.

When the thrust has been determined by the methods

outlined in the foregoing, a curve of propeller thrust

can be plotted, as shown in Fig. 2.

Prope

thpt

EVALUATION OF THE INTEGRALS.

The problem is reduced to that of finding the area

beneath a curve whose ordinate approaches infinity at

the trial speed. It is clear that the time which must

elapse, or the distance which must be run, to raise the

speed from, say, one knot below the trial speed to

almost the trial speed will be very large.

This is shown in Fig. 3. If we fix the upper limit

rial

0speeCa)knots

of integration as 0.01 knot short of the trial speed, the

important area is ABCD. To attempt to calculate

this by a graphical method will undoubtedly involve

considerable error ; nor will Simpson's rules be of

much use here. It is better, therefore, to attempt a

mathematical method which approximates to the

correct result.

N study of curves such as those in

Fig. 2 suggests a procedure for the purpose. The curve of thrust necessary to propel the ship, to a base of speed,

is concave upwards ; while, in general, the curve of

propeller thrust to a base of speed is concave

down-wards. Hence, the curve of T Rt to a base of speed

as in Fig. 4, will be concave downwards. Over the last

4-,

c V+099

1

under-estimated so that lJy is over-estimated; and, hence,

1

the area under the

_{T Rt}

curve will be over-estimated.

The error is, therefore, on the safe side. Making this straight-line assumption, we have

1 1

TI21

y , and y

K (1 x) ;

1 1

Re K (1 x)

The speed, v, corresponding to any x value is V+x.

V+ x

K(lx)

111

v

.1

= (1 )i

whence, JT:Ri dv =

[

x (V+ I) loge (1 x)]

The limits of integration in this case will be v= V,

i.e., x=0, for the lower limit, and, as previously assumed,

v =V+ 0 99, i.e., x=0.99, for the upper limit. Thus,

T Rt

dv h.,

[ 0.99 (V+1) loge 0.01]

.

Hence, from equation (3), the distance to be run to

Fig. 3. increase speed from V to V + 0.99 is given I y :

g K

A + 8 [

_{0-99+(V + 1) loge 100}

-t0

proPe

tp Trial

Speed (knots) V+I

Fig. 4.

knot, this curve will not depart very far from a straight

speed Speed.

Fig. 2.

line having the ordinate K at speed V, and a zero ordinate

at speed (V+ 1), the trial speed. The effect of this straight-line assumption is that the ordinates, y, are

(Trial spee_c)knots

; ;

=

+ 1

,

-JULY, 1945 THE SHIPBUILDER AND MARINE ENGINEIBUILDER PAGE 5

Similarly, from equation ,(2),, the time requited is

given by V 099 A + 8 dv g g K 0 I 8.fi) 99 1 xd -

lg

o 100.. A ± g K e 1

The integration of the curve of T -R1 or

as the case may be, from the lower limit of speed up to the speed V (one knot less than the top speed), should

present no difficulty and should be capable of being

performed sufficiently accurately by Simpson's first rule.

or by planimeter.

It is of some importance to decide the lower I'mit of

speed. When first coming on to the mile, the ship may

start from a quite low speed, and, for the purposes or

the calculation, it may be as well to assume that the

lower limit of speed is half-speed. 'When the vessel

turns at the end of a run and is about to come back for the next, an estimate must be made of the loss of

seed, so that the distance the vessel must run to regain

this speed may be found. It may be thought that this

loss of speed is difficult to determine, but some recent

work of Professor Kenneth S. M. Davidson, of the United

States of America, throws some light on the problem. In his paper " On the Turning and Steering of Ships,"t

he has shown that a major factor governing the ratio of the speed in steady turning to the approach speed

is the ratio of the tactical diameter to the length of ship.. American Society of Naval Architects and Marine Engineers,

Nov-ember, 1944; and No. 433, Vol. 52, of this journal,

J6,000-4,000

,F.c.111

Cr!

hittal speed (knots)9

The accompanying table has been derived from the

curve given in the paper mentioned.

NUMERICAL EXAMPLES,

Knowing approximately the turning qualities of the ship under consideration, one is then able to determine the distance to be run after the turn in order to regain speed.

Results have been worked out, for both constant r.p.m. and constant torque, for a yes3e1 having the

following characteristics :

Dimensions 410ft. ()in. x 57ft. 6in. x 35ft. 6in.

` Displacement 12,500 tons.;

Draught '7in. ; -Q)170 h-r"

Trial speed 11 knots 'Corresponding r.p.m..60'.

The results are shoWn in Fig. 5, where distance to be run to achieve a speed of 10.99 knots is plotted against

starting speed. In the same diagram are drawn vertical

lines, each labelled with the size of the turning circle

based on Davidson's work. It is fairly clear that loss

of speed on the turn has not a great influence on the

distance v, hich must be run to recover the original speeda conclusion arrived at by Davidson. The most impartant stage is when the trial speed is being

;approached. It appears that the distance required to

attain a speed of 10 99 knots from an initial speed of

10 knots is much greater than to reach a speed of 10 knots from a speed of 7 knots. It is also seen that the distance

to be run is greater with constant torque than with

constant revolutions.

During the turn at the end of a run on the mile, it'

is fairly accurate to assume that the torque is constant,

since the valve setting remains constant. Thus, for

_ 151;1-a-c."1,c1 16

tjt- rei.M

to -to sij I. Tactical Diameter

- .

Length of Ship ... 2 4 5 6 7 . 8 '9 110 11 12 , 13 Steady-turning speed ± approach speed 0.39 0.58 0.71 01.80 0-85 0.88 0-90 0.92 0,93, 0.94 0.945 0-95 Distance Ti'me I I

v

-1 E - P. 7 ri-5 an oraue constant_revolution) constant torq-ue El !I 111 1 El it II ai) cs1 lhI EA t z. r co Et Iss! "0" I .0 I 15

Fig. 5,Curves of- Distance and Time necessary to Attain a Speed of 10 99 knots, from Varioui

Initial Speeds, for a 12,500-ton Cargo Vessel. (Top Speed, 11 knots.)

+ (T - Rt ) p. 202. 24ft. 7c 7 8 1 E

8,000-the vessel considered, 8,000-the straight run required before

the vessel comes on to the mile again is about 12,600ft.,

or a little over two nautical miles, if the diameter of

the turn made is four lengths.

Also included in Fig. 5 is a curve showing the time

necessary to reach the speed of 10.99 knots.

At a

tactical diameter of four lengths, this time is about

12.5 minutes.

The time on the mile at a speed of

11 knots is about 5.5 minutes, so that the total time for one run is 30.5 minutes. This gives a time of

122 minutes for a series of four runs, which illustrates the importance of the tidal correction to the measured

speeds, when the measured course is subject to tidal

effects. In a period of 2 hours, appreciable variation

of the tide speed may occur.

The second example chosen is that of a destroyer, particulars of which are as follows :

Dimensions 320ft. Oin. x 33ft. Oin. ;

Displacement 1,400 tons

Draught 81t. 8fin. ;

-Maximum speed 36 knots.

The distances which must be run to attain a speed

of 35.99 knots have been worked out for two conditions,

viz., constant r.p.m. and constant S.H.P. The results are shown in Fig. 6.

In this case, considering a turn having a tactical

diameter of seven lengths,: and, accepting Davidson's

results, one finds the distance which must be run (at

constant power) before coming on to the mile again is

about 7,100ft.

It would seem at first sight that, for the destroyer, a much greater run would be required to recover the speed lost on the turn than is required for the cargo

vessel. The results indicate that this is not the case.

It must be remembered that the mass of the cargo

vessel is many times that of the destroyer. (The ratio

t This appears to be about the minimum diameter of turning circle which can be achieved by a destroyer at this speed. (See " Destroyer Turning Circles" by A. P. Cole, R.C.N.C., Institution of Naval Architects,

1938; and No. 342, Vol. 45, of this journal, p. 229.)

C,000

/6

Distance

Distance

Tirne

Fig. 6.Curves of Distance

Initial

20

is actually. 12,500/1,400, i.e., about 9.) A further factor which would favour the destroyer is that (as is almost

certain) the added virtual mass for the cargo vessel

will be a greater percentage of the total displacement

than for the destroyer. In the calculations, a uniform

figure of 20 per cent, has been employed for both ships. CONCLUSION.

The upper limit of speed assumed in the calculations,

viz., 0.01 knot short of the required speed, is a purely arbitrary choice, and obviously the results will differ

if a different limit is chosen. For example, if the limit were chosen as 0001 knot below the required speed,

the distance to be travelled to raise the speed from II knot short of the trial speed to this limit would be

log 1.000

increased approximately in the ratio of

log 106-'

i.e., as 3 : 2.

If it is desired actually to reach the full trial speed, the vessel must be over-propelled before coming on to the mile, by slightly increasing the r.p.m., which can be reduced to the correct number for the trial run.

The effect of increasing the r.p.m. from 70 to 71 has been worked out for the cargo ship already dealt with, and it appears that the distance the vessel must travel

to reach 11 knots is 4,360ft. This method of attaining

the trial speed cannot, of course, be used at the end

of a run before the vessel comes back on to the mile. In this case, the speed is never really fully recovered,

although, as shown, it is possible to determine the distance which must be run to approximate very closely

to the desired speed.

From the two examples which have been presented, it is seen that the most important aspect of the question is that of raising the speed from, say, 1 knot below the

trial speed to a speed just below the trial speed. With

the assumptions made in these notes, this distance is

given by

g K x

[

99+ (trial speed) loge 100]

A + a 30 4-20 10 r_

and Time necessary to Attain a Speed of 35.99 knots, from various

Speeds, for a Destroyer. (Top Speed, 36 knots.)

PAGE 6 THE SHIPBUILDER AND MARINE ENGINE-BUILDER JULY, 1945

( constant P owe,) VI -G) I CT, 0.1 Pl (constant revolutior75)n I 1 . a - a I .I I

(constant' -plower) -. 51 il III111-iilli

-e ! I _...1 J j -i5 --,3 .P4--rd II _{,-:4 I} 5'1 (81%1 -,`.-='I S14;3,1 k. 1--1--- 1---1 I 1 Is) 24 28 32

1,->itial speed (kns)ot

#.6.,000

L,4,000

2,000

:

This may be reduced to .:

a,eti,ttir-q>

\z:

-JULY, 1945 THE SHIPBUILDER AND MARINE ENGINE-BUILDER PAGE

A 8 _{x trial speed x loge 100, as 0,99 is small :in}

gK

-

-comparison with the second term in the square brackets.

A coefficient which may therefore be used for

com-parison would be :

(A x trial speed

(propeller

thrustship resistance)

at 1 knot below trial speed

If a constant relation between 8 and A is assumed

as might well be done in the absence of definite

knowledgethis coefficient could be replaced by :,

A x

TRt

where VT -= trial speed

T = propeller thrust ; and

Rt = resistance at a speed I knot below trial speed.

The values of this coefficient for the destroyer and

the cargo, vessel at constant r.p.m are as follows :

1,400 x 36

Destroyer _20,400 = 2 47

Cargo vessel 12,500 x 11

18,100

. 7.6.

The distance which the destroyer must run to increase

the speed from 35.0 to 35.99 knots is 2,700ft. ; and,

if the cargo-ship distance, is estimated from this result,

we find

:-76

6

,177 x 2,700.= 8,3001t.

The distance found for the cargo vessel by the detailed

method of calculation is about 8,000ft.

As a final example, we may estimate the distance for

a large liner of 30 knots trial speed. The particulars chosen are those for the Normandie, viz..

Displacement 66,400 tons ;

Speed 30 knots.

_SI

It is estimated that, at the speed of 30 knots, the

thrust necessary to propel the ship is 811,000 rb, and

at 29 knots 756,000 /b. The propeller thrusts at these

ship speeds, and at constant r.p.m., are estimated as

811,000 /b and 915,000 lb, the difference at 29 knots

being 159,000 lb.. Hence, distance to be run to raise the speed from 29 to 29.99 knots

66,400 x 1.2

32.2 x 159,009[0- 99 + 30 loge 100

6,080 \

(3,600)

x 2,2401 x 13,700ft.

The value of the suggested coefficient Itt this case is 66,400 x 30 159,000 (Lei) C C)t..P

Li 0-0

/25-7-11) ill

i/vv,,,OLC.Z.:

IQ, :;Zeito

P/

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Newcastle-on-Tyne, 6, England.

+

;

;