60 year old problem

Discussion in 'Aviation' started by b17sam, Nov 20, 2007.

  1. b17sam

    b17sam Member

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    Can anyone come up with an answer to a problem that has that has vexed me since 1944? I've had wild guesses, improbable theories, but zip that rings true.
    Problem: What is the elapsed time for an 88 flak shell to rise to 25000 feet? Educated guesses accepted.
     
  2. HoHun

    HoHun Active Member

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    Hi B17sam,

    >Problem: What is the elapsed time for an 88 flak shell to rise to 25000 feet? Educated guesses accepted.

    I have no definite answer for you, but here is an interesting picture of a stop watch that in my opinion might have been used to time and cross-check projectile flight times and altitudes.

    http://www.museumofworldwarii.com/Images2005/03Tachometerlge.gif

    The numbers are distance in hm (hectometer), so the scale reads from 2 to 11 km.

    The interval between the individual numbers increases because the projectile slows down and needs more time to cover the kilometer from 10 to 11 km than the one from 2 to 3 km.

    The curved lines mean that there time to target should be indexed with a second variable, and the reading not be taken on the inside at the numbers, but at a defined distance out of the centre.

    The curvature indicates that towards the rim, the projectiles slow quicker. That could indicate that the second variable was barrel elevation, as shooting straight up obviously will make the projectile slow down quicker than shooting it in a flat trajectory.

    I don't know why there's a second scale - I could speculate it was etched into the glass, and the watch has a replacement glass lacking this scale. No idea!

    I can't figure out the purpose of this device, either. I'd say it was for determining burst distance from observation of the projectiles and reading the elevation scale on the gun.

    However, the first step in flak shooting probably is range estimation, so the variable this watch could establish should already be established (with greater precision) anyway.

    Not that I have any idea of flak artillery procedures! :)

    Regards,

    Henning (HoHun)
     

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  3. Aussie1001

    Aussie1001 Member

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    would the speed = distance over time be of any use ?
    can modify to equal time = distance over speed i think any way might be speed over distance though....
     
  4. ToughOmbre

    ToughOmbre Active Member

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    Wouldn't you just divide the altitude (in this case 25,000 ft) by the muzzle velocity (in fps) of the 88 in question, taking into consideration the type of ordnance used? Or are there other laws of physics involved :?: :?: :?:

    TO
     
  5. mhuxt

    mhuxt Active Member

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    Well, you'd want to compensate for the cumulative effect of gravity - can't remember how many m/s^2 that is though.

    I actually used to be able to do ballistics - mark of a mis-spent middle age.
     
  6. HoHun

    HoHun Active Member

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    Hi ToughOmbre,

    >Or are there other laws of physics involved :?: :?: :?:

    Gravity :) And air resistance, which always makes the problem much more complicated.

    I have re-checked my thoughts on the above stop watch (which I posted on another forum a while back), and think my time-to-height explanation makes no sense. I guess the watch is meant for checking oblique shots, and the proposed additional variable would probably be something like elevation angle.

    I can work out the gravity bit, but not the air resistance:

    Purely vertical shot, 7.5 cm Flak gun, 616 m/s muzzle velocity, air resistance neglected, read "," as decimal point:

    Height - Time

    1000 m - 1,6 s
    2000 m - 3,3 s
    3000 m - 5,1 s
    4000 m - 6,9 s
    5000 m - 8,7 s
    6000 m - 10,6 s
    7000 m - 12,6 s
    8000 m - 14,7 s
    9000 m - 16,9 s
    10000 m - 19,2 s
    11000 m - 21,6 s
    12000 m - 24,1 s
    13000 m - 26,8 s
    14000 m - 29,8 s
    15000 m - 33 s

    In reality, the 7.5 cm gun is not going to shoot to that height, and the heights it does reach will be reached only in a longer time.

    Shooting obliquely will of course also affect flight times since the distance the projectile has to travel until it reaches a certain altitude will be longer.

    Note that the flight times under perfect conditions are longer than the 18 s for 360° time the above stop watch takes, and yet there are 100 hm marked on the dial - it can't be altitude! :)

    Regards,

    Henning (HoHun)
     
  7. renrich

    renrich Active Member

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    The muzzle velocity is, I believe, about 3000 fps but of course the projectile starts slowing down as soon as it leaves the muzzle. If the projectile maintains it's muzzle velocity(which it won't) it would take slightly more that 8 sec to get to 25000 feet. This is only a guess but I am going to say it will take sightly more than 20 sec to reach 25000 feet and will have a velicity of less than 800 fps at that altitude.
     
  8. mhuxt

    mhuxt Active Member

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    Nope, not going to attempt it - air drag has to be fairly significant if the shell can only go to 30-odd k.

    Has to be longer than 10 seconds to get to 25k feet though.
     
  9. mhuxt

    mhuxt Active Member

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    Oh hell, had originally posted that HoHun would spank me with better math. Heheheh here's the offering I edited out:

    Info In the anti-aircraft role it fired a 20.3-pound shell at a muzzle velocity of 2,600 ft/s to an effective ceiling of 26,000 feet (at maximum 32,000 ft, 10,600 m).
    Gravity -9.8 m/s^2
    Feet/metre 1 feet = 0.304 8 meter
    MuzzVel 822.96 m
    Tgt Alt 7620 m


    Second Velocity Alt
    0 792 0
    1 783 792
    2 773 1575
    3 763 2348
    4 753 3111
    5 743 3864
    6 734 4608
    7 724 5342
    8 714 6065
    9 704 6780
    10 694 7484
    11 685 8178


    So I thought something around 10 seconds, but that's clearly wrong. The shell is still shown to be going 700 m/s at that point. However, if it can only go another 7,000 feet (~2,100 m), it must in reality be travelling far more slowly. Air resistance.

    Bugga.
     
  10. drgondog

    drgondog Well-Known Member

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    If you know the Ballistic coefficient of the 88mm round, and the muzzle velocity - and assume shortest time to 25,000 feet is a vertical (90degree) shot, you can calculate the drag combined with gravitational acceleration for 'slowing forces' and plot velocity loss from muzzle as a function of heighth.

    When you gradually angle the gun toward horizon, the gravity contribution as a retarding force is a SIN function where the value of zero degrees from vertical is 1.00 and 90 degrees (level shot) is Zero.

    In other words gravity has no effect on drag for a level shot.

    Having said that I don't have the data
     
  11. mhuxt

    mhuxt Active Member

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    "And he puzzled three hours
    'til his puzzler was sore.
    Then the Grinch thought of something
    he hadn't before."

    I don't have the ballistic coefficient, but as dragondog says, there must be a way to "back out" the drag effect, if we know the shell can only get to 32,000 feet, at which point its velocity would be zero.

    Mnnnnnnnnnnnoooooooougggggggggggghhhh this is gonna hurt...
     
  12. drgondog

    drgondog Well-Known Member

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    a= g = 32 feet per second per second..
     
  13. mhuxt

    mhuxt Active Member

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    Hey man, I've been metric since 1974.

    9.8 m/s/s or 9.8m/s^2
     
  14. mhuxt

    mhuxt Active Member

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    meh

    trying to solve

    Vtime=Vlauch-gravity*time

    When I only know Vtime and Vlaunch (with gravity being equal to gravity and drag)

    I knew this would hurt.

    Well, so far as I can see, I have to add -56.2 m/s^2 to get a shell launched at 792m/s to be at 0m/s at 9,500m, instead of at 675m/s.
     
  15. HoHun

    HoHun Active Member

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    Hi B17sam,

    >Problem: What is the elapsed time for an 88 flak shell to rise to 25000 feet?

    Here is a pretty good site with a lot of information on Flak:

    FLAK Geschütze - Flugabwehrkanone

    (It's in German, but follow the blue links, and you'll find actual wartime reports with a lot of drawings and diagrams for which you don't need to understand German.)

    This page has drawings of a Flak fuse:

    Aktiver Luftschutz bekannt unter der Bezeichnung FLAK - Flugabwehrkanonen Eruierung der Flak in Bochum

    The fuses would detonate the projectile before it could reach the highest point of its trajectory - probably because it got so slow up there that hitting was impossible anyway :)

    The average amount of rounds necessary for one kill in WW2:

    8.8 cm Flak 36/37: 16000 rounds
    8.8 cm Flak 41: 8500 rounds
    10.5 cm Flak 39: 6000 rounds
    12.8 cm Flak 40: 3000 rounds

    (From "Die schwere Flak" by Werner Müller)

    Regards,

    Henning (HoHun)
     
  16. mhuxt

    mhuxt Active Member

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    double meh


    BS in, BS out.

    (Newton's Fourth Law)

    Back to teh drawing board.
     
  17. mhuxt

    mhuxt Active Member

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    This site:

    Trajectories

    Has a formula for calculation of peak height, which uses only launch velocity and gravity variables.

    We know peak height, we know launch velocity, assume firing straight up, solve for g, then reapply into time, deceleration and distance?
     
  18. mhuxt

    mhuxt Active Member

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    So, if I work backwards from the above to solve for g (assuming g represents the effect of both gravity and air resistance), and using 9754m as peak height, I get:

    g=(Vo^2)/2h

    which gives me a value for g of 32.15 m/s/s.

    A rough spreadsheet effort as I origninally tried tells me it takes a little more than 24 seconds to slow to 0 m/s, with the shell taking 14 secs to 25,000 feet or around 7750 m.

    Of course, it's bollocks to the extent that I'm assuming air resistance is dependent on time, just like gravity, when of course it's dependent on surface area, start velocity, air thickness, etc.

    I know a guy on another board who does highly-detailed naval ballistics simulations. I'll see what he says.

    Time for some alcohol.
     
  19. HoHun

    HoHun Active Member

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    Hi Mhuxt,

    >Of course, it's bollocks to the extent that I'm assuming air resistance is dependent on time, just like gravity, when of course it's dependent on surface area, start velocity, air thickness, etc.

    It's a good estimate anyway since we know it will give a minimum time. (The deceleration of the round is largest at the highest speed in the high-density air near sea level, so a real round will take longer than your estimate.)

    I googled a bit and found this article on a German Flak director:

    Lone Sentry: German Predictor 40 (WWII Tactical and Technical Trends, No. 40, December 16, 1943)

    Note that it has a limit for "time of flight" of 30 s - in that time the shell might not have reached its absolute apex, but it will be detonated anyhow because it is not expected to hit any more.

    (It will be close to the apex, however. I've got figure of 14.7 km absolute height, 12.35 km maximum fuse height for the - rare - 8.8 cm Flak 41, and it seems reasonable to assume that the 12.35 km fuse height were determined by the 30 s limit of the Flak director.)

    This is also mentioned here:

    Employment of German AAA, WWII Tactical and Technical Trends, No. 35, October 7, 1943 (Lone Sentry)

    Here is a slightly weird article about "skywriting" by Flak units, which however, illustrates how the time fuses were set:

    Flak Directional Arrows to Guide German Fighter Planes, WWII Tactical and Technical Trends, No. 22: April 8, 1943 (Lone Sentry)

    Nice fuse pictures and an animation:

    dekomunition24.de

    Regards,

    Henning (HoHun)
     
  20. syscom3

    syscom3 Pacific Historian

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    Great info Hohun
     
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