History
The method of checking up torpedo ranges hereinafter described has the following advantages over any other method:
(a) It is independent of navigational marks on shore—hence of charts.
(b) The entire range can be checked with the ship in one location.
(c) I he observations to check the entire range can be taken in 15 minutes.
(d) The method is accurate to within 10 yards for ranges up to 4000 yards.
The method is entirely geometrical, and can be used in any locality. .
Material Required
In order to use the method with the greatest accuracy and facility, the following are required:
(a) One vessel of known water-line length.
(b) One small boat to occupy end buoy of range.
(c) Three sextants of known I. C.
(d) One ship’s pelorus, set with 0° at the ship’s head. The position of the pelorus should be accurately marked by a white strip of bunting over the wing of the bridge.
(e) Ship’s whistle.
Observers
In the motor boat:
Observer D—with most accurate sextant.
Observer B—with sextant.
On the bridge of vessel:
Observer A—with sextant.
Observer C—with pelorus and to operate ship’s whistle.
Stations
Small boat occupies end buoy of range (firing point buoy). This should be done as accurately as possible. If the buoy is well anchored, secure directly to it. Vessel steams off from small boat at, approximately, right angles to the range, to a distance somewhere between 960 yards and 1500 yards, stops, and heads “up range.” Set alidade of pelorus to point directly on beam toward small boat. Vessel gives “stand by ” signal, and maneuvers so as to bring small boat to bear on alidade without shifting ship’s position much.
Signals
“Stand by”—toots on whistle.
“Make observations”—one long blast.
“Make next observation on number — buoy” (indicating buoy from the small boat by serial number, the firing buoy being number zero), one long blast followed by toots to the number of the buoy.
“Discontinue and return to ship”—three long blasts.
Observations
When vessel attains position with small boat bearing on alidade, distant about 1200 yards give signal for “make observations.”
Observer A measures the angle “A” subtended between the firing buoy (#0) and the buoy being observed. Records angle to least count.
Observer B measures the angle "B ” subtended between the buoy being observed and the position of the alidade on the bridge, which should be marked. Records angle to least count.
Observer C gives signal to “make observations” when small boat bears on alidade.
Observer D measures very accurately the angle "D” subtended by the ship’s water-line length. Records to least count of his sextant. He should use sextant telescope in making observations.
Record for each observation the following data:
(a) Date and time.
(b) Number of observation.
(c) Title letter of angle observed.
(d) Amount of angle.
(e) I. C. of sextant.
(f) Corrected angle.
(g) Range buoy being checked by that observation.
(h) Name of observer.
Stand by for next observation. Three should be taken for each buoy distant over 2000 yards, worked out independently, the average of three results being accepted.
Computations
I. BASE-LINE
The base-line (x) is calculated from the angle “D” and known characteristics of the ship. To derive the formula, we first refer to the accompanying sketch (Fig. 1).
L = length between perpendiculars of the vessel, in feet. m=distance from pelorus to forward perpendicular, measured parallel to the keel, in feet.
We have:
x = Mcot0=(L-M)cot(D-0)feet
cot0=L-m/m cot(D-0)
By trigonometry:
cot(D-0) = cot0cotD+1/cot0-cotD
Then:
cot0 = (L-m)(cot0cotD+1)/m(cot0-cotD)
mcot20-mcot0cotD = (L-m)cot0cotD+(L-m)
mcot20+cot0cotD(-m-L+m) = L-m
mcot20-cot0(LcotD) = L-m/m
cot20-2cot0(LcotD/2m) = L-m/m
cot20-2cot0(LcotD/2m)+(LcotD/2m)2 = L-m/m + (LcotD/2m)2
cot0-(LcotD/2m) = /(L-m/m+(LcotD/2m)2)
cot0 = (LcotD/2m) + /(L-m/m + (LcotD/2m)2)
Calling the expression
LcotD/2m = C
and substituting, we have
cot0 = C+ /(L-m/m + C2)
Referring back to (a), we find,
x = mcot0 feet
Then:
x = mC/3 + m/3 /(L-m/m + C2) yards
The data for a 750-ton destroyer are not at hand, but for a vessel of the 1000-ton class, the following characteristics obtain:
L = 300 feet
m = 70
C = 300cotD/2x70 = 2.143cotD
C2 = 4.59cot2D
and the formula becomes:
x = 50cotD+23.33 /(3.286+4.59cot2D yards
This formula lends itself readily to the construction of a curve with values of (x) as abscissae, and values of the angle (D) as ordinates. See graph at end of article.
II. POSITION OF BUOY
The position of the buoy, being observed, is readily determined by solving the simple triangle shown in the sketch (Fig. 2).
We have, by trigonometry,
sinA/sin(180-(A+B)) = y/x
and
sin(180-(A+B)) = sin(A+B)
Then:
Y = xsinA/sin(A+B)
Accuracy
In considering the use of the method an important point is its probable accuracy. This accuracy depends on the accuracy of observing the angles (AB) and especially on the accuracy with which the base-line (x) is determined, as the error of the base-line is multiplied by the factor (y/x) in its effect on the determination of the range (y). We will consider first the base-line (x).
1. base-line
This is affected by two factors:
(a) Error in measuring angle (D),
(b) Error obtaining ship’s length (L).
A simple method of considering error in the angle (£>) is to obtain the equivalent error in (L) necessary to make the same error in the base-line (x). The sketch below (Fig. 3) is not strictly accurate, as the triangle of the base-line is not right angled (see Fig. 1), but the assumption will not interfere with a practical determination of the relative errors in (D) and (L).
In the sketch,
L=ship’s length
D=angle(D)
dD=error in measuring angle(D)
dL=equivalent error in assuming length(L)
X=base-line, in yards
To get a relation we must express dD in the same terms as is dL (in feet). The angle (dD) subtends a small length of the circumference of a circle whose radius is x yards = 3x feet. The circumference is 2nx yards = 6nx feet, and 1° subtends = 6nx/360 = nx/60 feet (h')
Then:
dD(feet) = nxdDo/60
From the small triangle of differentials, we have dL = dD sec D(feet)
=nxsecDodDo/60, in which dDo=error in the angle (Do)
Let us determine the error in (D°) necessary to equal an error in (L) of 1 foot. That is, let dL=1.
Then:
1 = nxsecD0dDo/60
dDo = 60/nxsecDo degrees
= 60x60/nxsecDo minutes
Let us use a value of D = 7°30'. Substitution in Formula 1 gives us x = 761.9 yards.
log 60 = 1.77815 log 3.1416 = 0.49715
log 60 = 1.77815 log 761.9 = 2.88190
log sec 7o30’ = 0.00373 3.37905
3.56003
3.37905
.18098 = log 1.517 minutes
60
91.020 seconds =
Equivalent error in angle (D) to error of 1 foot in length (L) 90 seconds
In considering the effect on the base-line (x) of an error in length (L) we find the following sketch of use (Fig. 4):
Symbols as in Fig. 3, so far as they apply. dx = error in base-line (x) caused by an error (dL) in length (L) We have:
dx=dLcotD
Let
dL = 1 foot, and angle D = 7.5o
Then:
dx = cot(7o30’) feet
= cot 7o30’/3 yards
log cot7o30’ = 0.88057
log 3 = 0.47712
0.40345 = log 2.532 yards
Then an error of 1 foot in the length (L) will cause an error of 2.532 yards in the base-line(x).
Recapitulating, it should be possible to observe the angle (D) to within 30" by using the sextant telescope. This error then would be equivalent to x 1 foot = 0.333 foot error in determining the length (A). The actual error in determining this length for computation should never exceed two inches, making a combined possible error, if both were in the same direction, of foot in the length (L) and therefore of 1.25 yards in the base-line (x).
II. final determination
In the triangle of the range (see Fig. 2) the error produced in the range determination by an error (dx) in the base-line (x), is in accordance with relation:
dy/dx = yx
dy = y/x dx
Considering the extreme case of checking up the point-of-aim buoy (4000 yards):
Y = 4000 yards
x = 761.9 yards
dx = 1.25 yards
Then:
Dy = 4000 x 1.25/761.9
Log 4000 = 3.60206
log 1.25 = 0.09691
3.69897
log 761.9 = 2.88190
0.81707 = log 6.5625 yards
Maximum error in final determination due to base-line is 6.5625 yards.
Angles A and B lend themselves readily to accurate measurement. The error in either one should not exceed 30". As the angle (B) is approximately 90°, it is evident that error in observing the angle (A) will have the greater effect. If we consider the error due to errors in both (A) and (B) as double that due to (A) we will then be on the safe side. The sketch (Fig. 5) indicates an error of (dA°) in measuring the angle (A) and the equivalent error (dc) in the base-line (.tr) to produce the same error in the determination of (y).
dA = dxsinA
dAo = subtends part of a circumference whose length is 2nk yards.
The part subtended by 1o is 2nk/360 = nk/180 yards, and the part subtended by dAo = dA = nkdAo/180
To eliminate (K), we have by trigonometry:
k/x = sinB/sin(180-A-B) = sinB/sin(A+B)
k = xinB/sin(A+B)
dA = nxsinBdAo/18-sin(A+B) = dxsinA
dx = nxsinBdAo18-sinAsin(A+B)
Letting dAo=30” = 1/120 degree, and x=761.9 yards, we have:
dx = 1x761.9nsinB/120x18-sinAsin(A+B)
By letting B=80o and substituting in Formula 3 along with values of (x)=761.9 and (y)=4000, we get
A=89o 01’ A+B=169o 01’
Log120=2.07918 log761.9=2.88190
log 180=2.25527 log n = 0.49715
log sin A=9.99994 log sinB=9.99335
long sin(A+B)=9.27995 3.27240
3.61434 3.61434
log dx = log 0.45505 = 9.65806
=0.46 yard
Then the equivalent error in the base-line (*) to an error of 30" in observing the angle (A) is 0.46 yard. Double this is 0.9 yard, the total error that can result from the angles (A) and (B). The effect of 0.9 yard error in (x) on the final determination of (y) is readily determined from (5) and (7) as follows:
dy/dx = 6.5625/2.5
dy = 6.5625dx/2.5 } dx = 0.9 yard
= 6.5625x0.9/2.5
log 6.525 = 0.81707
log 0.9 = 9.95425
0.77131
log 2.5 = 0.39794
0.37337 = 2.3625 yards
Total error due to errors in observing angles (A) and (B) should not exceed 2.4 yards.
Recapitulating, we have:
Error caused by (dx) 5.6 yards
Error caused by (dAo) and (dBo) 2.4 yards
Total error, all causes 8.0 yards
It must be realized that these errors exist in any one set of observations. If three sets are made by any one set of observers and each is worked out separately, the mean of their results should be very close and well within the limit of 10 yards.
Note.—The appended graph may not be used for vessels which have not the following characteristics:
Horizontal distance – pelorus to water-line at bow / Water-line length = 0.2335
For vessels of these characteristics, but with a water-line length other than 300 feet, the curve may be used if the result obtained is multiplied by the expression:
Vessel’s actual water-line length, in feet / 300 feet
The vessels of the McDougal, Benham and Nicholson classes may use this curve without modification.