emery@aries.mitre.org (David Emery) (04/04/91)
From: emery@aries.mitre.org (David Emery) To resolve a debate some friends of mine are having: Does anyone know the muzzle velocity, time of flight, or max ordinate for a 16" gun firing at max range? Apparently USN&WR said in an article that a 16" gun firing at a range of 25 miles, the projectile would travel 50,000 feet into the air. Some people asserted this was correct, others said it sounded much too high (assuming low angle fire). With either time-of-flight or muzzle velocity, we can approximate the max ordinate (assuming a vacuum, which gives an upper bound on the problem). thanks in advance dave emery emery@aries.mitre.org
jmc@DEC-Lite.Stanford.EDU (John McCarthy) (04/04/91)
From: jmc@DEC-Lite.Stanford.EDU (John McCarthy) Assuming a vacuum, the longest range is obtained when a gun is fired at an elevation angle of 45 degrees. In this case, the altitude reached is 1/4 the range. This would be 6 1/4 miles if the range is 25 miles, which is 33,000 feet. If one takes air resistance into account, the maximum range would be obtained at a somewhat higher elevation angle, because that would make the shell spend more of its trajectory where the air is less dense. If 25 miles is not the maximum range, then the elevation angle can be higher or lower. It is not obvious to me which would be more accurate.
chidsey@smoke.brl.mil (Irving Chidsey) (04/04/91)
From: Irving Chidsey <chidsey@smoke.brl.mil>
In article <1991Apr4.041931.11567@amd.com> emery@aries.mitre.org (David Emery) writes:
< Does anyone know the muzzle velocity, time of flight, or max
<ordinate for a 16" gun firing at max range?
< Apparently USN&WR said in an article that a 16" gun firing at
<a range of 25 miles, the projectile would travel 50,000 feet into the
<air. Some people asserted this was correct, others said it sounded
<much too high (assuming low angle fire).
Max range and low angle fire are incompatible! Max range means
an elevation angle of around 45 degrees. Air resistance and possibly
the curvature of the earth affect the angle. 50,000 feet is about 9.7
miles, which should not be too far from the maximum ordinate for a near
45 degre elevation angle. I see no reason to object to the figures
given in the article.
--
I do not have signature authority. I am not authorized to sign anything.
I am not authorized to commit the BRL, the DA, the DOD, or the US Government
to anything, not even by implication. They do not tell me what their policy
is. They may not have one. Irving L. Chidsey <chidsey@brl.mil>chidsey@smoke.brl.mil (Irving Chidsey) (04/06/91)
From: Irving Chidsey <chidsey@smoke.brl.mil> In article <1991Apr5.092149.8384@amd.com> jmc@DEC-Lite.Stanford.EDU (John McCarthy) writes: >If 25 miles is not the maximum range, then the elevation angle can >be higher or lower. It is not obvious to me which would be more >accurate. Against a target that was as tall as it was deep ( front to back ) the lower trajectory would be most accurate because it would be shorter. Against a point target, it would depend. The higher trajectory Might have a smaller ground footprint. Or, it might not. -- I do not have signature authority. I am not authorized to sign anything. I am not authorized to commit the BRL, the DA, the DOD, or the US Government to anything, not even by implication. They do not tell me what their policy is. They may not have one. Irving L. Chidsey <chidsey@brl.mil>
henry@zoo.toronto.edu (Henry Spencer) (04/06/91)
From: henry@zoo.toronto.edu (Henry Spencer) >From: emery@aries.mitre.org (David Emery) > Does anyone know the muzzle velocity, time of flight, or max >ordinate for a 16" gun firing at max range? > > Apparently USN&WR said in an article that a 16" gun firing at >a range of 25 miles, the projectile would travel 50,000 feet into the >air. Some people asserted this was correct, others said it sounded >much too high (assuming low angle fire). A reasonable rule of thumb is that high-velocity artillery is circa 1000m/s at the muzzle. The old ballistics rule of thumb is that horizontal range is double vertical range, max altitude of the max-range trajectory is half vertical range, and for max range you fire at 45-degree elevation. (You do not use "low angle fire" for long range!) The usefulness of this is that vertical range is easy to calculate: v^2 = 2ag, so 1000m/s in a 9.81m/s^2 gravitational field gives a vertical range of a bit over 50km. So we have a 100km range at 45 degrees, with a trajectory reaching 25km altitude, circa 75000ft. This is a wee bit optimistic because it ignores air resistance, of course, but clearly 50000ft is not implausible. -- "The stories one hears about putting up | Henry Spencer @ U of Toronto Zoology SunOS 4.1.1 are all true." -D. Harrison| henry@zoo.toronto.edu utzoo!henry
henry@zoo.toronto.edu (Henry Spencer) (04/07/91)
From: henry@zoo.toronto.edu (Henry Spencer)
I wrote:
>... (You do not use "low angle fire" for long range!)
I got mail about this from Jon W. Meyer, an artillery type who can't post
from his machine:
Sorry, Henry, but "low angle fire" for artillery has a particular
definition. Essentially, any tube elevation less than or equal
to 1600 mils or 45 degrees is "low angle". Anything above this
is "high angle". Actually, the firing tables for most indirect
fire weapons don't provide data within two hundred or so mils of
1600 becase the CEP becomes unacceptably large. (This is more
true when firing with larger propellant charges than with smaller
ones).
--
"The stories one hears about putting up | Henry Spencer @ U of Toronto Zoology
SunOS 4.1.1 are all true." -D. Harrison| henry@zoo.toronto.edu utzoo!henrywbt@cbema.att.com (William B Thacker) (04/08/91)
From: wbt@cbema.att.com (William B Thacker) In article <1991Apr5.092149.8384@amd.com> jmc@DEC-Lite.Stanford.EDU (John McCarthy) writes: >Assuming a vacuum, the longest range is obtained when a gun is fired >at an elevation angle of 45 degrees. [...] >If one takes air resistance into account, the >maximum range would be obtained at a somewhat higher elevation angle, >because that would make the shell spend more of its trajectory where >the air is less dense. Correct; I think about 48 degrees is optimal for large guns. The turrets of the Iowas, however, only allow elevation to 45 degrees, so it's a moot point. A bit of data for the 16"/50: Projectile -> AP Mk 8 HC Mk 14 shell wt. (lb) 2700 1900 muz.vel(ft/sec) 2500 2690 range@45deg 42345yds 41622yds 38720m 38049m -- Bill Thacker AT&T Network Systems - Columbus wbt@cbnews.att.com
nash@cs.nps.navy.mil (David Nash) (04/10/91)
From: nash@cs.nps.navy.mil (David Nash) In his reply, Mr. Meyer mistakenly equated 1600 mils with 45 degrees. Lest there be any doubt, low angle fire is any fire where the quadrant elevation does not exceed 800 mils, or 45 degrees.