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Do the 50-calibre-machine-gun-bursts reduce the speed of the P-51?

On strafing runs WWII P-51 Mustangs often fired bursts of 1000 rounds from its six - 50 caliber machine guns. Those streams of "fired" lead changed the air speed of the Mustang. The details of strafing are complex. This calculation will include some "guess-work" and approximation. The sketch includes hypothetical data of one near-horizontal (tree-top level) P-51 strafing event.

**Calculate** the P-51's change of air speed after it fires a burst of 1000 rounds.

♦ We take the system to be the mass of the P-51 and the 1000 rounds of 50 cal bullets. The P-51 mass was about 5400 pounds. Its speed about 300 miles per hour (440 ft/s) when firing commenced. The bullet masses we take as 0.1 lbm. They left the machine gun muzzles at about 3000 feet per second. A simple, before-after sketch is helpful.

Assume the drag force of air initially to be “about” the same finally.

We take the airplane and the bullets it carries (then fires in the event) as our system. Since there is "plane" momentum and "bullets" momentum, we place a summation, left of equality, on momentum.

(1)
System consists of the fighter, its remaining ammunition and its fired ammunition. |

**Initial Condition:** An event is a change from some initial condition to a final condition. Initially the plane and bullets move horizontally at same constant speed. The state is called uniform motion. Propeller thrust causes the velocity, the drag (the opposing force of air as the plane moves through it) equals the thrust force. So velocity and momentum are constant for all times prior to firing (**t < 0+**). Stated by equation, this is:

(2) Momentum is constant and the force of the prop equals that of aero-dynamic drag. |

Eqn-2 tells that prior to firing the system momentum is constant in two ways. Left-of equality says the bullets and aircraft speeds are constant. Right-of-equality says "thrust equals drag."

**Firing Duration:** The six guns are triggered in a burst. Each gun fired at about 1000rnds/minute so, the "duration" for 1000 rounds was about 10 seconds. To use the momentum equation we write it in general and consider it at the time: **t = 0+**

(3) System mass remains constant as bullets speed ahead of the P-51. |

All Forces of the event are notated. You might wonder about the forces as the guns recoil. But the recoil effects are not forces since they do not occur at the system boundary. The bullets, powder and aircraft are all part of the system. Forces of the momentum equation are __effects of the surroundings__. To proceed toward solution of the first order differential equation, we separate variables then apply the integration operator:

(4)
Same procedure as always. Let math take the lead. Separate variables and integrate. |

The "burst" time is 10 seconds. Air speed might change over that time span. An exhaustive approach would be to integrate from **t = 0+** to **t = 1 sec**. Determine the conditions after 1 second, then integrate from time, **t = 1sec** to **t = 2** sec... It would be costly in time to go at this one-second-at-a-time. For the sake of expedience, to check the indications and so, we calculate the case for which the firing duration is assumed instanteous.

We like to do easy parts first. The integrand, right of equality (**F**_{thrust} - **F**_{propeller}) was zero before change started, when time was **t < 0+**. We suppose the firing happens very fast so that either "dt" is zero or (**F**_{thrust} - **F**_{propeller}) remains what it was at the start ~ zero. Use either approximation to set "right of equality" equal to zero for the event. Left of equality integrates immediately.

(5)
Integration of an exact differential yields the difference of its quantities. |

The summation is required - our system has two components. Now we expand the sum.

(6)
There are two components to our system. |

The direction of all vectors is to the right, so we scalar multiply or "dot" multiply ("• **I**") the equation then put numbers into it.

(7)
As always, care with units is essential. |

This amounts to a worrisome, 14% decrease in speed. We suppose 14% is the least reduction. In events of "squandered energy," approximations tend to be "rosey."

**History:** Early trials of the P-51 with its original Allison engines showed the plane performance
to be disqualifyingly inadequate. However once fitted with the Spitfire engin e; Rolls-Royce Merlin, perfomance was vastly improved.

The 12-cylinder inline 60-deg engines were superb. At the link, (i) you can see an early Merlin II (ex Hurricane MkI from 1937), the only known MkVIII remaining (Fairey Fulmar). The video (ii) shows an Mk24 (upgraded from Mk22, ex Lancaster).

========= Code development below.

(1) CODE: Open Video New Window at YouTube (use browser to return):

Works: (Click opens Video at YouTube. Use browser "back arrow hyphen" to Close)

(2) CODE: Video First-Frame-Display Inline:

Works: See below. Initiate Video, Pause Video... etc.

(3) CODE: Image via Thickbox (return by "esc or close"):

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To ensure that the video shuts off when the area is hidden, add "?enablejsapi=1" to the end of the iframe src.

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*Note:* to prevent scrollbars on the popup, popup height should be the height
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After Thought (away from above): To use this functionality, make sure that you add the "click"
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**Related Comment:** The son of a Spitfire pilot told me his father never experienced the "machinegun-speed-loss" effect. Being in the seat, he knows what he felt. Maybe the physics is that straffing had a decrease of elevation and that decrease in height caused a speed increase which off-setting the machine gun effect. With bombers, the nose cannon firings (20mm) flying flat and level did cause noticeable slowing.

On strafing runs WWII P-51 Mustangs often fired bursts of 1000 rounds from its six - 50 caliber machine guns. These streams of "fired lead" changed the air speed of the Mustang. This calculation will include some guess-work and approximation.

**Calculate** the P-51's change of air speed after it fires a burst of 1000 rounds.

Premise presently unwritten!