Path: utzoo!utgpu!news-server.csri.toronto.edu!rpi!zaphod.mps.ohio-state.edu!swrinde!elroy.jpl.nasa.gov!lll-winken!sun-barr!newstop!sun!amdcad!amdcad!military From: henry@zoo.toronto.edu (Henry Spencer) Newsgroups: sci.military Subject: Re: 16" Battleship gun info wanted Message-ID: <1991Apr6.020906.14151@amd.com> Date: 5 Apr 91 18:34:39 GMT References: <1991Apr4.041931.11567@amd.com> Sender: military@amd.com Organization: U of Toronto Zoology Lines: 26 Approved: military@amd.com 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