A Problem with Spitfires

This problem stems from an interesting exchange on Twitter last night, prompted by a tweet from the Reverend Richard Coles:

I think his clerical vocation may be responsible for the spelling mistake. The answer to his question doesn’t require any physics beyond GCSE but it does require data that I didn’t have access to last night.

Here’s a version for you to try at home with all the necessary numbers (though not necessarily in the right units):

A model of a Mark VI Spitfire showing its two 20mm cannons.

A Supermarine  Mark VI (Type 350) Spitfire fighter aircraft weighing 6740 lb is initially travelling at its top speed of 354 mph. The aircraft is armed with two Hispano-Suiza HS.404 20mm cannons, one on each wing, each of which is fed by a drum magazine containing 60 rounds. Each projectile  fired from  the cannon weighs 130 grams, the rate fire of each cannon is 700 rounds per minute and the muzzle velocity of each shell is 860 m/s.

(a) Calculate the reduction in the aircraft’s speed if the pilot fires both cannon simultaneously until the magazines are empty, if the pilot does nothing to compensate for the recoil. Express your answer in kilometres per hour.

(b) Calculate the average deceleration of the aircraft while the cannons are being fired, and express your result as a fraction of g, the acceleration due to gravity at the Earth’s surface which you can take to be 9.8 ms-2.

(c) A Mark 24 Spitfire – which is somewhat heavier than the Mark VI, at 9,900 lb (4,490 kg) – is armed with 4×20mm cannons, two on each wing. The inboard cannon on each wing has a magazine containing 175 rounds; the outboard one has 150 rounds to fire. Repeat the above  analysis for these new parameters and comment on your  answer.

Answers through the comments box please!




28 Responses to “A Problem with Spitfires”

  1. Bryn Jones Says:

    Oh. That’s quite a mix of systems of units: imperial, cgs, SI. Presumably to make the problem more complex.

    I can believe Rev. Coles has witnessed the firing of canons, though for what transgressions I can’t imagine.

  2. andyinkuwait Says:

    For the first problem, assuming the magazine cases weigh nothing, I make it:

    557 km/hr with an average deceleration of 0.071g

    For the second, I make it:

    487 km/hr with an average deceleration of 0.156g

    In the second problem, one of the pairs of MGs runs out slightly before the other but I ignored this as 25 rounds takes 2s to fire compared to the 15s I used. Deceleration is about twice as much with Mark 24.

    Changed everything to SI first. I always make some stupid arithmetic error when doing these things and I can’t see this would be much different. Conservation of momentum for first part of the question and Newton II for the second part.

    • telescoper Says:

      I think that in early spitfires the spent cartridges remained in the wing; in later ones they fell out of a hole underneath. In any case the information I gave is actually for the projectile part only.

    • telescoper Says:

      I’m guessing that diving or climbing in a dogfight would produce much larger effects relative to g than firing the guns!

    • telescoper Says:

      I’ve just looked up the Mark 24 data again and discovered that the plane is rather heavier (thanks to a bigger engine). Its mass is
      9,900 lb (4,490 kg). This slightly mitigates the effect of increased weight of fire.

  3. andyinkuwait Says:

    For comparison, initial speed was 570 km/hr so, yes, it slows down but not as much as the Mark 24.

    • telescoper Says:

      Pilots of the latter aircraft frequently found that the guns on one wing would jam while the others continued firing. It’s not hard to see that this could lead to a considerable torque around the vertical axis making the plane hard to steer.

      • Anton Garrett Says:

        The Spitfire was well known to be easier to turn one way than the other because of the angular momentum of the propeller and shaft.

      • telescoper Says:

        That poses an interesting problem. Not sure what the moment of inertia of a Merlin engine is though!

  4. 354 mph = 158.2 m/s
    each shell is 130 g = 0.130 kg
    plane weighs 3057.2 kg
    shell v = 860 m/s

    Each shell’s momentum is 0.130 * 860 m/s = 111.8 kg*m/s

    So each time the shell gets fired one can subtract the momentum of the shell from that of the plane.

    The plane: p = mv so p = (3057.2)(158.2) = (483,649.04) kg*m/s

    There are 120 rounds total fired at 700 per minute per gun, so that’s really 1400 per minute total, and that means the magazines (of 60 rounds) will be empty in 5.14 seconds.

    So the total added momentum of the bullets is (111.8)(120) = (13146)kg*m/s

    (483,649.04) kg*m/s – (13146)kg*m/s = 470,233.04 kg*m/s

    The plane’s velocity is reduced then to:

    (470,233.04 kg*m/s)/(3057.2)kg = 153.8 m/s

    This happens over the course of 5.14 seconds, so the deceleration is
    158.2 m/s – 153.8 m/s = 4.4 m/s / 5.14s = 0.86 m/s/s

    which means the deceleration of the plane is ~0.08 g

    The Spitfire’s speed reduces from 569 kph to 553 kph, or about 2%, which isn’t very much at all in the scheme of things.

    I am curious if I approached this correctly. I could probably have used a version of the rocket equation, but this seemed a simpler method. I realize that the deceleration would actually be a bit less because one would want to account for the mass of the shells added to that of the plane initially, but the difference is pretty small (< 1%)

    For the second set of parameters you'd get more deceleration, though the difference would be much greater in the first few seconds until the bigger magazine runs out. Then you'd get less deceleration (since the firing wouldn't go on for very long after the first 150 rounds went out — you'd get what, 1 second of fire? A bit less)

  5. I like the idea of the gun being synchronized to the propeller, so it shoots “through” the propeller, but of course doesn’t hit it.

    • Anton Garrett Says:

      That was an issue in WW1 and a system to ensure you didn’t hit your own propeller was created. One such design was by the teenage Andrew Irvine who was lost on Everest in the 1920s with George Mallory.

  6. andyinkuwait Says:

    I included the mass of the shells in the calculation which is why my calculation gives a slightly lower answer.

    25 rounds, which is the excess once one of the MGs runs out, takes 2s to discharge on 700 rounds/minute.

    Momentum of the bullets is 13416 Ns not 13146 Ns.

    All minor issues.

  7. Chris Brunt Says:

    According to Wikipedia, the bombs in a Lancaster could be up to 50% of its empty weight. I think you may need to write a bonus-points Part B to this question… e.g. by what angle would the plane’s trajectory change during bomb release, assuming the pilot makes no adjustments?

    • andyinkuwait Says:

      But we aren’t talking about bombing. We are talking about horizontal shooting with MGs.

    • telescoper Says:

      Some versions of the Spitfire were equipped to carry bombs, including the Mark 24 which could carry up to 2 × 250 lb (110 kg) bombs (wing racks), plus 1 × 500 lb (230 kg) bomb (centre-section rack). These would increase the weight of the plane by 450kg so dropping them would make a significant but not huge difference.

  8. jim Dunlop Says:

    Hmmm – depends what you mean by the pilot does nothing to compensate. If the Spitfire is at constant speed, then presumably forward force is exactly compensated by the air resistance. When the cannon shell leaves, sure a simple momentum conservation will mean plane slows slightly, but then air resistance will drop slightly, and if the pilot doesnt change the throttle the plane will accelerate back up a bit (indeed, helped by the fact it will be slightly lighter).

  9. jim Dunlop Says:

    Or, to put it another way, modulo some juddering, the speed will be unchanged, since keeping the throttle in the same position will rapidly recover the velocity to reach the same air ram pressure that the Rolls Royce engine can match.

    • andyinkuwait Says:

      I did some ‘back of fag packet’ calculations for the Mark VI.

      I figured out the air resistance going at 354 mph with the engine flat out at 1054 kW. After the firing of the guns (assuming air resistance is proportional to v^2) the net force forward is roughly 275 N. This gives an initial acceleration of 0.09 m/s^2. If it were to stay at that value, while the Spitfire got back up to its maximum speed, (it won’t as air resistance will increase and the net forward force and acceleration will decrease) it would take roughly 35 s to get back to full speed. As the actual speed change is roughly 7.5 mph I’m not sure it would be noticed but >35 s is not an insignificant time.

    • telescoper Says:

      It’s a question of timescales.

      You can get the notional change in velocity impulsively by assuming that instead of 60 shells over time from each cannon you fire the total weight forward as one big lump. That would slow the plane down, but it would naturally speed up again as you describe. The change in mass of the plane is a very slight effect.

      However, if you fired one round at a time over a very long period, the plane would adjust in time for the notional change in velocity resulting from the cannon never to happen.

      You can set this up as a differential equation for dv/dt with three terms on the RHS: the drag force (depends on v), the cannon fire recoil and the thrust of the engines. In the steady state without cannons being fired dv/dt=0 so you can get a useful relation between the first and last terms at the known v at maximum speed. When you fire the guns dv/dt becomes negative until the drag force changes sufficiently to compensate. For the parameters given that timescale is longer than the maximum duration of fire as defined by the magazine capacity.

      • Jim Dunlop Says:

        Yep – maybe set this up as a question for early undergrads! What’s interesting is that if the shooting could go on long enough to make a big dent in velocity – e.g. to drop speed by ~10% (~15 seconds) then the catchup effect can start to bite, because reduction in air resistance is more like 1300 N and the acceleration ~0.5 m/s – so, yes, it depends on the timescales and what change in speed you are interested in as to whether the change in air resistance needs to be factored in. Obviously change in mass negligible.

  10. Jim Dunlop Says:

    Sorry, 25 seconds I meant to type.

    Frightening how little firing time they had actually – hence the short burst tactic described by Bader, rather than letting everything go in 5s.

  11. Jim Dunlop Says:

    It does seem that the tiny velocity drop was the least of the pilot’s problems – here, from an interview with Bader:

    “He recalled the first time he used the cannons while engaging a Messerschmitt BF-109 fighter over the English Channel, saying that its use was so “bloody noisy, as to make one go deaf.” Aside from the noise associated with the cannons, he also recalled the vibrations felt in his airplane every time he depressed the button to fire off a few 20mm rounds, “it very nearly shook you apart”.”

    • telescoper Says:

      The velocity drop is indeed probably too small to be noticeable for early Spitfires but quite large for later, more heavily armed, versions – the cannons of which were presumably even louder!

      When I was at School I was in the Navy Section of the CCF and I went for a week on a Guns and Missiles course, during which I fired a few rounds from an old 20mm anti-aircraft cannon similar to those adapted for air-air combat. I can confirm that it was very noisy.

      This gives you an idea:

  12. I must say that altitude was not taken into consideration in the responses. I knew a P-47D pilot who stated that at 35,000ft firing all 8 .50s for 3-4 seconds slowed the aircraft by about 100mph

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