The values of ? and ?' derived in the following note are given in Major Sladen's Principles of Gunnery, pages 74 and 75, London, 1879, with a reference to Mr. W. D. Niven's paper On the Trajectories of Shot, published in the Proceedings of the Royal Society, No. 181, Vol. XXVI.
The expressions for ? and ?' however, as obtained in Mr. Niven's paper, differ slightly from those quoted by Major Sladen and derived here.
The method employed in deriving these formulas is similar to that of Mr. Niven.
Putting ? = ? – ?, equation (4), page 70, becomes
The values of ? and ?' derived in the following note are given in Major Sladen's Principles of Gunnery, pages 74 and 75, London, 1879, with a reference to Mr. W. D. Niven's paper On the Trajectories of Shot, published in the Proceedings of the Royal Society, No. 181, Vol. XXVI.
The expressions for ? and ?' however, as obtained in Mr. Niven's paper, differ slightly from those quoted by Major Sladen and derived here.
The method employed in deriving these formulas is similar to that of Mr. Niven.
Putting ? = ? – ?, equation (4), page 70, becomes
1/u3 – 1/p3 = c/g (P? - P? – ?) = c/g [3tan? + tan3? – 3tan(?-?) – tan3(?-?)]
Expanding the last two terms as functions of ? by Maclaurin’s Theorem, we have
1/u3 – 1/p3 = c/g [3sec2?.? + …
+ 3tan2?.sec2?.? + terms involving ?2],
Or
1/u3 – 1/p3 = c/g [3sec4?.?
+ terms involving squares and higher powers of ?]
To obtain an approximate value of ?, we omit all terms containing its square and higher powers; whence
? = gcos4?/3c (1/u3-1/p3)
Again, since q is the value of u when ? = ß [see equation (5) page 70], we have from (?)
? – ß = gcos4?/3c (1/q3 – 1/p3);
whence ? = (1/u3 – 1/p3)/(1/q3 – 1/p3) (? – ß)
The X Integral
In equation (c), page 73, putting r=cv3=cu3sec3?, we have
X = 1/c ?pq cos2?(du/u2)
By Maclaurin’s Theorem we have
cos2? = cos2(? – ?) = cos2? + sin2?.? + terms in ?2;
omitting the terms containing powers of ? higher than the first and substituting in (?),
X = cos2?/c ?pq du/u2 + sin2?/c ?pq ? du/u2
Introducing the value of ? from (r), we have
X = cos2?/c ?pq du/u2 + (? – ß)sin2?/c(1/q3 – 1/p3) ?pq (du/u5 – 1/p3 du/u2)
Whence, integrating,
X = cos2?/c (1/q – 1/p) + (? – ß)sin2?/c(1/q3 – 1/p3) [1/4(1/q4 – 1/p4) – 1/p3(1/q – 1/p)],
and, putting 1/q – 1/p = Q,
X = Q/c cos2? + sin2?.(? – ß)[(1/q3 + 1/q2p + 1/qp2 – 3/p3)/4(1/q3 – 1/p3)]
It is now necessary to obtain an approximate value of the fraction
f = (1/q3 + 1/q2p + 1/qp2 – 3/p3)/4(1/q3 – 1/p3) = (1/q2 + 2/qp + 3/p2)/4(1/q2 + 1/qp + 1/p2) = (p2 + 2pg + 3q2)/4(p2 + pq + q2)
Put l = p-q/p+q, whence p = (1+l/1-l)q. We now substitute this value of p in the expression for f and omit l2 in the result, since l is a small quantity.
F = [(1+l/1-l)2q2 + 2(1+l/1-l)q2 + 3q2]/4[(1+l/1-l)2q2 + (1+l/1-l)q2 + q2]
= [(1+l)2 + 2(1-l2) + 3(1-l)2]/4[(1+l)2 + (1-l2) + (1-l)2]
and, omitting l2, we have approximately
f = (6–4l+2l2)/[4(3+l2)] = (3-2l)/6
Whence X = Q/c [cos2? + sin2?.(? – ß)(3-2l)/6]
Comparing the equations (?) and (n), it is evident that the expression in braces may be put equal to cos2?, if
? = ? – (3-2l/6)(? – ß) = ? – (? – ß/2) + l/3(? – ß) = (? + ß)/2 + (p-q)/[3(p+q)] (? – ß)
That is, we have
X = cos2?/c Q, Q = 1/q – 1/p, and ? = (? + ß)/2 + (p-q)/[3(p+q)] (? – ß)
The Y Integral
In equation (d), page 73, putting r = cu3 sec3? we have
Y = 1/c ?pq cos? du/u2
Expanding by Maclaurin’s Theorem, we have
Sin?cos? = sin(?-?)cos(?-?) = sin?cos? – cos2?.? + terms in ?2
Applying to this integral the same process we applied to (?), we have
Y = (sin?cos?)/c Q,
in which ? and Q have the same values as in the X intergral.
The Time Integral
Putting r = cu3sec3? in equation (i), page 72, we derive
dt = du/(cu3sec3?cos?); whence T = 1/c ?pq cos2? du/u3
Substituting the value of cos2? from equation (?)
T = cos2?/c ?q du/u3 + sin2?/c ?pq ?.du/u3
Integrating and introducing the value of ? from the equation (r)
T = cos2?/c ½ (1/q2 – 1/p2) + [(? – ß)sin2?]/[c(1/q2 – 1/p2)] [1/5(1/q5 – 1/p5) – 1/2p3(1/q2 – 1/p2)];
Whence, putting Q’ = ½(1/q2 – 1/p2), we have
T = Q’/c [cos2?+sin2?.(?-ß) (1/5(1/q5 – 1/p5) – 1/2p3(1/q2 – 1/p2)/½(1/q2 – 1/p2)(1/q3 – 1/p3)]
Denoting by f’ the fraction in the right hand term, and reducing
F’ = 2(1/q5 - 1/p5) – 5/p3(1/q2 – 1/p3)/5(1/q2 – 1/p2)(1/q3 – 1/p3)
= 2(1/q4 + 1/q3p + 1/q2p2 + 1/qp3 + 1/p4) – 5/p3(1/q + 1/p) / 5(1/q +1/p)(1/q3 – 1/p3)
= 2p3/q3 + 4p2/q2 + 6p/q + 3 / 5(p/q + 1)(p2/q2 + p/q + 1)
= 2p3+4p2q+6pq2+3q3 / 5(p+q)(p2+pq+q2)
Putting l = p-q/p+q, whence p = (1+l/1-l)q, we have
f’ = 2(1+l/1-l)8q3+4(1+l/1-l)2q3+6(1+l/1-l)q3+3q3 / 5(2q/1-l)[(1+l/1-l)2q2+q2]
= 2(1+l)3+4(1+l)2(1-l)+6(1+l)(1-l)2+(1-l)3 / 10[(1+l)2+1-l2+(1-l)2]
Whence f’ = 15-5l+5l2+l3 / 10(3+l2) = 3-l/6 (omitting l2 and l3)
Hence T = Q’/c [cos2?+sin2?.(? – ß).3-l/6]
On comparing this equation with (?), it is obvious that we can express T in the form
T = Q’/c cos2?’, if ?’ = ? – 3-l/6 (? – ß) = ? – (? – ß)/2 + l/6 (? – ß),
Or ?’ = (?+ß)/2 + 1/6 p-q/p+q (? – ß)