戰列艦齊射會倒退嗎？比如一米這樣？

先說結論：不能。（下文摘自戰列艦論壇）

IOWA級為例

16寸炮開火時的總機械能

Given:

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

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

Projectile Kinetic Energy （彈丸動能）= 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.

計算髮射葯燃氣動能

Average outflow velocity of propellant gases（平均燃氣噴射速度）: w = 3,937 fps

Gas Kinetic Energy（燃氣動能） = 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:

計算後座部件動能

w = B*Vo, therefore B = 1.5748

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

Recoil Energy （後座能量）= ((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.

彈丸的旋轉能量極小可以忽視。

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.

總機械能只是發射葯化學能的40%－50%，其餘部分作為熱能被燃氣帶走，或傳遞給炮管。

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:

單一彈丸的動量

Projectile momentum （彈丸動量）= (Weight of projectile/g) * Muzzle velocity of projectile （彈丸重量/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 （燃氣動量）= (Wc/g) * w

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

Summing these:

兩者的和：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.

後座動量應該和彈丸加燃氣的動量大致相等。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發齊射的動量：

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:

計算IOWA在齊射後的速度：

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

齊射質量×速度＝船質量×船速度

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

因為齊射質量×速度等於之前計算的齊射動量，推得：

Broadside Momentum = Mass of ship * Velocity of ship

齊射動量＝船質量×船速度（橫移速度）

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

計算：Velocity of ship （船橫移速度）= Broadside Momentum / [Mass of ship] （齊射動量/船質量）= 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.

結論：IOWA在冰面上以0角度齊射造成到船體橫移速度為每秒7.7英寸。考慮到橫移會被船體在水中部分阻擋，而船體的水線長度為860英尺，喫水38英尺，具體橫移距離應該不到一毫米。

瀉藥

不會

原貼地址：

【圖片】最近經常有人水戰列艦齊射會不會平移甚至翻船的奇怪問題，回復的也五花八門，其實navweaps上早有這樣的文章_戰列艦吧_百度貼吧?

tieba.baidu.com

原文地址：

Do battleships move sideways when they fire？ -NavWeaps?

www.navweaps.com

如果戰列艦齊射會造成後退一米這麼大誤差，當炮彈飛到40000碼開外的時候，那誤差估計連濺起來的水花都揚不到目標臉上。

設計出這種戰列艦的人會被陸軍馬鹿三天內暗殺掉，骨灰都給揚了那種。

你要知道戰列艦本來就是海上會移動的超遠程炮臺，這麼大的變數怎麼打炮？？

不會。以衣阿華級戰列艦為例，全裝葯發射9枚Mark8Mod1型穿甲彈，該炮彈射彈全重1224.7千克，則9枚炮彈全重11022.3千克即11.0223噸，炮彈初速為762米/秒，以衣阿華級戰列艦標準排水量45000噸計，僅考慮動量定理，可得一次齊射後船體獲得的速度為0.187米/秒，這個速度較小且因為海水的阻力還會迅速下降，因此船體的位移幾乎可以忽略不計。

會。但是到不了1m。

這張照片是衣阿華級的威斯康辛號右舷齊射的情景，應該是在海灣戰爭的時候拍的。你注意看艦艏處的白色浪花。當然了，這可不是船橫移時產生的，確實有橫移但是絕逼橫移不了1m，這是艦在繞著穩心搖擺的時候產生的。在艦靜止的時候朝前射擊，艦一定會後退的，除非用的是無後坐力炮或者物理定律出現了變化。但是1m就有點誇張了，至於準確的是多少米，沒有數據，也沒有靠譜的模型。。