For what it’s worth, 100mm is roughly 4 inches, which was a WW1 destroyer armament. 
I think that chasing splashes is as silly as jumping into a newly-formed shell crater.
The crater might provide you with some cover from other incoming fire. At least against the shrapnel.
I disagree. Chasing splashes is based on the assumption that your enemy is going to be adjusting his range on account on his last salvo having been too long or too short, and if you make your change of course after he has corrected his range and before the shells arrive, his next salvo will also miss. That’s not the usual situation on a battlefield, where jumping into a shellhole is based on a fallacious appeal to the law of averages.
“Chasing splashes” is one choice for making course corrections to avoid (hopefully) being struck by the next salvo. A captain/helmsman could make arbitrary course corrections instead. Is one more effective than the other? Maybe, maybe not. The object of constantly changing course is to keep the opposition from guessing where you’ll be when they fire the next round.
Shell craters, foxholes, serpentine trenches would all provide some cover from flying debris and shrapnel of subsequent incoming rounds but not from direct hits. Using them would keep troop losses to a minimum.
Not all that fallacious : firing causes the gun to shift a bit. Assuming identical wind and pressure conditions, it’s really unlikely that gun will land two shells in the exact same spot (plus the artillery guys are usually trying to carpet the area, not pinpoint strike anyway).
Of course, you’ve still got the other 15+ guns in the battery to worry about :). But as said, if the shrapnel if flying over your head because you’re in a shell hole, you know it’s not going *through *it.
I was wondering what the time on target would be for a 14" gun at combat range.
It’s really amazing that they hit at all with those ranges and shell flight times, and non-digital rangefinding and mechanical targeting computers.
Oh, and the point of jumping into a shellhole wasn’t that shells wouldn’t hit the same place twice, it’s that it’s a HOLE, and you’ll be sheltered from fragments from other shells, since you were clearly caught in the open when the barrage started.
I’ve read several personal accounts of WWII destroyers altering course to “chase splashes” in order to avoid incoming rounds. The USS Johnson DD-557 and USS Hoel (DD-533) of Taffy 3, the Battle off Samar, fame to name a few. The Imperial Navy didn’t have radar to assist targeting.
It’s important to note that “chasing splashes” didn’t prevent either the Johnson or Hoel from being repeatedly hit by incoming rounds but may have reduced the number of hits they received. Maybe. Either way, both the Johnson and Hoel were sunk but not before damaging, sinking, and turning back a much, much, much larger force.
Not sure what “combat range” is, but for the max effective range of the main battery of the WWI dreadnought USS Texas (BB-35), it comes out to about 42 seconds for the 14"/45 caliber gun to send a 1,400 pound armor piercing shell at some poor sap 20 miles away. Assuming the shell never loses any velocity, anyways.
Considering that I barely passed high school Algebra, I’m gonna BS the calculations and call it just under a minute for the shell to travel that far. Maybe 20-25 seconds for a target 10 miles away. IANAArtillerist.
An abridged range table for the 2700 pound Mark 8 AP projectile out of the 16 in./50 rifles on the BB Iowa class. The table assumes a MV of 2425 fps. Max range from it is 40,185 yds, and time of flight is 92.72 seconds. Changing the MV gives slightly different figures for TOF and range.
The thing that got me when I first heard about it in connection with large artillery like this, is how high the shells go. The previous range table I linked to didn’t have the maximum ordinate (highest point of the trajectory), but this one does. At 2500 fps MV, and a max range of 42,345 yds, the shell will get 36,610 feet in the air at one point. For most of the ranges listed, the maximum ordinate is a much smaller percentage of the total range. E.g., for a shot at 10,000 yds., the TOF is 13.24 seconds, and the max ordinate is only 703 feet.
Thanks, Gray Ghost!
I rounded my numbers, to make the math easy. I assume that a shell loses velocity from the initial firing blast, but at 36,000 feet, I imagine that a shell might actually gain some of that back as it plunges back to the surface.
Well they didn’t hit all that often - take Jutland for example:
A few things just for grins: That info about shell trajectory heights is why helicopter pilots insist on liaison with the artillery to work out not only impact areas, but fire corridors between the guns and the targets, so they can literally fly around those lines of fire. Most artillerists hold to the “Big sky little bullet” theory, but pilots do NOT agree with that way of thinking.
And, the USS Texas has the distinction of being the only US battleship to be struck by enemy gunfire during WW II (gunfire, not bombs or torpedoes). She was hit by a German 9-inch shell while bombarding the Normandy coast during the June 6th landings. The shell lodged in the main armor belt on her main deck and failed to explode. And calling her a “WWI dreadnought” is a little misleading. Her design was indeed begun in the 19th century when HMS Dreadnought was state of the art, but she fought all the way to the end of WWII. She was steaming to the Pacific to take part in Operation Olympic when Japan surrendered. She’s on permanent display in Buffalo Bayou, just east of Houston, and to anyone interested in naval architecture or history, the price of a ticket to go aboard is well worth it.
For a “flat-earth, no-atmosphere” flight, the maximum range is obtained by firing at a 45 degree angle, which leads to a pretty high flight. For round-earth flight in atmosphere the best angle is going to be somewhat different, but 45 degrees is still going to be a pretty good approximation.
Hitting at long ranges was very difficult, and as a result, infrequent, as described above. Occasionally there were spectacular shots:
Added to all this is that when the splash lands, the guns on the firing ship may well have already loosed of the next salvo, and in fact at longer ranges and higher elevations there could be up to three rounds per gun in the air at the same time, what it means is that by the time you have noticed splashes around you, the next lot of shells are incoming and will not be able to correct for your course movements, and assuming those in flight rounds have already been corrected for your predicted course it makes sense to assume that misses will have had correction added and they will not be directed to the same place they already missed.
Unless they’ve guessed that you’re going to chase splashes. And even if there’s only, say, a 1 in 3 chance that the enemy captain decides to chase splashes, that still makes that one spot a lot likelier place for the target to be than any other location you might care to choose.
The optimum defensive tactic would be to choose spots to chase completely at random, without regard to where the splashes are.
Maybe you’re thinking of German gunfire or land-based gunfire, because the USS South Dakota got pretty jacked up by gunfire in the naval battles around Guadalcanal by IJN cruisers and destroyers.
All ballistic (non-guided, non-powered and wingless) shells absolutely gain speed after apogee—in the vertical direction. That’s because their vertical speed is 0 at apogee. If a ballistic shell didn’t gain vertical speed after apogee, it would be orbiting the earth.
Horizontal speed decreases continually (but not linearly) as soon as the shell leaves the barrel. Of course, the horizontal speed never increases no matter how fast the shell falls vertically.
The range table Gray Ghost linked to shows that a 16-inch shell’s impact velocity reaches a minimum at about 33,000 yards and an apogee of about 13,500 feet. Beyond that range, the shell’s vertical speed adds to its net velocity at impact. At ranges less than 33,000 yards, the impact velocity is dominated by the shell’s horizontal speed.
At about 40,000 yards, the “angle of fall” is 45 degrees, which suggests that the horizontal speed and the vertical speed are equal.
(I used quotes and cautious wording there because the phrase “angle of fall” is ambiguous. The shell falls in a nearly parabolic arc, so there is no one “angle of fall.” But I bet that range chart is indicating the angle of fall at impact).