战列舰齐射会倒退吗？比如一米这样？

先说结论：不能。（下文摘自战列舰论坛）

IOWA级为例

16寸炮开火时的总机械能

Given:

Projectile Weight （弹重）: Wp = 2,700 lbs.

Charge Weight（药包重）: Wc = 650 lbs.Muzzle Velocity（炮口初速）: Vo = 2,500 fps.Weight of Recoiling Parts（后座部件重）: Wr = 250,000 lbs.g （重力加速度）= 32.174 fps^2

Projectile Kinetic Energy （弹丸动能）= 0.5*((Wp/g)/2)*Vo^2 = 2.622*10^8 ft-lb.

To compute the kinetic energy of the propellant gases we must know the average velocity of the gases as they escape the muzzle. Experiments have shown that this velocity varies between 1,200 and 1,400 mps. depending on the muzzle velocity of the weapon. For purposes of these calculations we will use 1,200 mps or 3,937 fps.

计算发射药燃气动能

Average outflow velocity of propellant gases（平均燃气喷射速度）: w = 3,937 fps

Gas Kinetic Energy（燃气动能） = 0.5*((Wc/g)/2)*w^2 = 78.29*10^6 ft-lb

To compute the Kinetic energy of the recoiling parts, we must determine the velocity that they would achieve if allowed to recoil with no retarding force. This is commonly referred to as the free recoil velocity. To account for the difference between the velocity of the projectile and that of the propellant gases, we will use the aftereffect coefficient B which is defined by the relationship:

计算后座部件动能

w = B*Vo, therefore B = 1.5748

Free Recoil Velocity（自由后座速度）: Vre = (((Wp/g)+B*(Wc/g))/(Wr/g))*Vo = 37.236 fps

Recoil Energy （后座能量）= ((Wr/g)/2)*Vre^2 = 5.387*10^6 ft-lb

The rotational energy of the projectile is small by comparison and can be neglected.

弹丸的旋转能量极小可以忽视。

The overall mechanical energy is only a part (40 to 50%) of the chemical energy of the propellant, since a considerable portion of the energy is carried off as heat by the propellant gases, or transmitted to the gun barrel.

总机械能只是发射药化学能的40%－50%，其余部分作为热能被燃气带走，或传递给炮管。

Ref. Rheinmetall Handbook on Weaponry, 1982, chapter 9.

Now, with these factors in mind, here are some additional calculations.

The momentum of a single projectile can be calculated as follows:

单一弹丸的动量

Projectile momentum （弹丸动量）= (Weight of projectile/g) * Muzzle velocity of projectile （弹丸重量/重力）×炮口初速

= (2,700 / 32.174) * 2,500= 209.80 x 10^3

Momentum of the propellant gasses can be calculated from Leos numbers as follows:

Propellant gas momentum （燃气动量）= (Wc/g) * w

= (650 / 32.174) * 3,937= 79.54 x 10^3

Summing these:

两者的和：Projectile momentum + Propellant gas momentum = 209.80 x 10^3 + 79.54 x 10^3= 289.34 x 10^3

The Free Recoil momentum calculation can be used as a check, as it should be about equal to the sum of the momentums of the projectile and propellant gasses.

后座动量应该和弹丸加燃气的动量大致相等。Free Recoil momentum = (Wr/g) * Vre= (250,000 / 32.174) * 37.236= 289.33 x 10^3

The Broadside Momentum for 9 projectiles can now be calculated as follows:

9发齐射的动量：

Broadside Momentum = 9 * (momentum of projectile + momentum of propellant gasses)= 9 * (209.80 x 10^3 + 79.54 x 10^3)= 2.60 x 10^6

Using Gregs formula, the velocity of an Iowa firing a 9-gun broadside can be recalculated as follows:

计算IOWA在齐射后的速度：

Mass of broadside * Velocity of broadside = Mass of ship * Velocity of ship

齐射质量×速度＝船质量×船速度

As the Mass of broadside * Velocity of broadside term is equivalent to Broadside Momentum, this formula can be restated as follows:

因为齐射质量×速度等于之前计算的齐射动量，推得：

Broadside Momentum = Mass of ship * Velocity of ship

齐射动量＝船质量×船速度（横移速度）

Solving for the velocity of the ship and using the above calculated momentum figures:

计算：Velocity of ship （船横移速度）= Broadside Momentum / [Mass of ship] （齐射动量/船质量）= 2.60 x 10^6 / [58,000 * (2,240 / 32.174)]= 0.64 fps

So, the ships velocity ON ICE with the guns firing at zero degrees elevation would be about 7.7 inches per second rather than the 6 inches per second calculated above. When one considers that any sideways motion of the ship through water is actually resisted by the wall created by the hull of the ship, whose wetted surface is about 860 feet long and 38 feet deep, then it can be easily understood that Dick Landgraffs comment above, "theoretically, a fraction of a millimeter," is closer to the truth.

结论：IOWA在冰面上以0角度齐射造成到船体横移速度为每秒7.7英寸。考虑到横移会被船体在水中部分阻挡，而船体的水线长度为860英尺，吃水38英尺，具体横移距离应该不到一毫米。

泻药

不会

原贴地址：

【图片】最近经常有人水战列舰齐射会不会平移甚至翻船的奇怪问题，回复的也五花八门，其实navweaps上早有这样的文章_战列舰吧_百度贴吧?

tieba.baidu.com

原文地址：

Do battleships move sideways when they fire？ -NavWeaps?

www.navweaps.com

如果战列舰齐射会造成后退一米这么大误差，当炮弹飞到40000码开外的时候，那误差估计连溅起来的水花都扬不到目标脸上。

设计出这种战列舰的人会被陆军马鹿三天内暗杀掉，骨灰都给扬了那种。

你要知道战列舰本来就是海上会移动的超远程炮台，这么大的变数怎么打炮？？

不会。以衣阿华级战列舰为例，全装药发射9枚Mark8Mod1型穿甲弹，该炮弹射弹全重1224.7千克，则9枚炮弹全重11022.3千克即11.0223吨，炮弹初速为762米/秒，以衣阿华级战列舰标准排水量45000吨计，仅考虑动量定理，可得一次齐射后船体获得的速度为0.187米/秒，这个速度较小且因为海水的阻力还会迅速下降，因此船体的位移几乎可以忽略不计。

会。但是到不了1m。

这张照片是衣阿华级的威斯康辛号右舷齐射的情景，应该是在海湾战争的时候拍的。你注意看舰艏处的白色浪花。当然了，这可不是船横移时产生的，确实有横移但是绝逼横移不了1m，这是舰在绕著稳心摇摆的时候产生的。在舰静止的时候朝前射击，舰一定会后退的，除非用的是无后坐力炮或者物理定律出现了变化。但是1m就有点夸张了，至于准确的是多少米，没有数据，也没有靠谱的模型。。