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#51
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EDIT: I would have done even the one to simulate the belts and limit the head movement on the cockpit but...
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![]() A whole generation of pilots learned to treasure the Spitfire for its delightful response to aerobatic manoeuvres and its handiness as a dogfighter. Iit is odd that they had continued to esteem these qualities over those of other fighters in spite of the fact that they were of only secondary importance tactically.Thus it is doubly ironic that the Spitfire’s reputation would habitually be established by reference to archaic, non-tactical criteria. |
#52
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I am referring to the Cliffs of Dover Hurricane FM vs the old 1946 FM.
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#53
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.. ...
![]() Let me say I am flying a 109 or a Hurri or whatever and I have a six. If the enemy behind me is only 20 km/h faster he needs only 10 seconds to reduce the gap by more than 50 meters.... ... In that situation the performence is not the key factor, the reaction of both pilots is important. .. ... ![]() Happy hunting Varrattu Last edited by Varrattu; 08-02-2011 at 02:15 PM. |
#54
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Interesting problem Vara...
Let's put it in nbr ![]() let's say that your speed is 400kph. if the guy in your 6 is 20kph faster it equate to 5% diff in frd speed. When you'll start to maneuver E will be the most important thing. With 95% less speed, your Kin E would be 0.95*0.95=0.90 less than that of the other guy (kin E = 0.5 x m x Speed²) with m the mass of the planes. No you'll say to me that your hurri is heavier than a 109 thx to her heavy plywood and tube structure and that masse at speed equate Kin E. humm let's think about that... Mass = lift at level flight. hence the more mass to carry, the more lift you'll have to produce. As lift = drag in the sense that the more lift at a given speed generate more drag, hence you'll see that given you are at the same speed but with an heavier mass, your plane will generate more drag. And then less E. So let's put this aside as it complicate the prob without giving any advantage o your hurri (but let's keep it in mind when we will need to firce him to bypass our plane ![]() So your hurri has 10% ((1.-0.90)*100 in %) less kin E than the 109 in your 6 and you generate more drag. You'll understand now that even that "little" 20kph disadvantage is uncomfortable whan you'll start to move the plane in the vertical plane or in a high G situation. Why in high G ? let's have a look let's say you pull hard in the vertical at 5g for a brief moment. In a 5G verical turn (not a 360, or a full loop - you'll soon see that it is impossible), your plane will have a corresponding mass of 5 time heavier. Hence it will need 5 time more lift to stay on the same horizontal plane with your wings level. hence your drag will increase, let's say linearly by a factor of 5. What will counteract the drag in that situation ? You'd say so : engine power and you E state. How your Engine power will contribute ? Drag increase basically with the square of the speed. Has the power necessary to sustain a given speed is the product of the forward speed AND the force opposite to the speed vector, the power of the drag forces will be : the fraction of the mass x by the G x by the forward speed. As you see, basically pulling 5 G increase 5 time your drag and necessitate 5 time more power than in a level flight Let's say that at 400kph your hurri will need 80% of it's eng power. Hence at 5g it would have need 0.8 x 5 = 4 time more power. A total amount that your engine can not produce. This is were your kin E will start to play it's role : being a source of E to keep your global E state approximatively cte (your hurri is not a conservative system !). How will that kin E play in this situation ? humm let's have a look. Kin E is a variable of the square of the speed. It has to compensate for the lack of the power of your engine. But the power is the product of a force by a forward speed and the drag force itself evolve in function of the square of the speed ! You'd see : speed x speed for the drag and drag x speed for the power means speed x speed x speed for he drag power Hence your kin E will proportionally contribute only as the square root of itself to compensate for the lack of power.... that is 3 time what your eng can deliver at it's peak power (4 -1 x engPower) Humm let's put it together : you have 95% of the 109 speed hence you'll get only 0.95x0.95x0.95 = 85 % of what the 109 will have for power to balance the 5G pull up during the pul up your 90% less Kin E will transform some E to help you to sustain your forward speed acting only proportionally to the root square of the speed (90%) = 0.95 = 95% SO if we sum the overall lack of power to sustain your frwd speed we will have (1-0.85) + 3x(1-0.95) = 30% ! 30% less power at 5 G means that your speed (considering that drag btw your hurri and the 109 are equal - what we hve seen above is not !) will drop by P = DragForce x Speed => Speed = P/DragForce what means that the drop will be more than 30% of that of the 109 (more since your plane is heavier and have to generate more lift hence face more drag) ! And this is only for the first move ! As you've seen (if you are still there) 5% diff in forward speed means a lot less of manoevrability in term of E and speed after each move (30%). Let's imagine that after your brief 5G move your speed has dropped down to 300kph. The 109 will still hev a speed of 300 x (1+0.30) = 390 kph hence a 90kph difference vs 20 at the begining of the move. ~S! PS : Humm humm now you have said that he was 50m behind you (what is really close for two fighters). 50m at 390kph means (390-300)/3.6 = 25 m/s means that the 109 pilot has 2 seconds to do something to stay behind you. Holly Cow ! you are just ready to make an evasive break ![]() PPS: I kown that Vara know all about the above. I just took his remark as a pretexte for a short refreshing cursus ![]() Last edited by TomcatViP; 08-10-2011 at 11:52 PM. |
#55
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Small correction tomcat, mass does not equal lift in level flight, you are confusing that with weight.
weight = mass * gravity |
#56
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After a small pause due to moving buisness I checked the prop pitch behaviour of the 109 and to my delight the speed is now definitely better. All thumps up for this.
However, I also realized that pushing the lever up will make the pp go up but the pp does not stop increasing when releasing the lever. Basically this will make the pilot guessing how long he has to push the lever in order to achieve a certain pp. I have strong doubts that - considering how important precise setting of proppitch is - this kind of behaviour is accurate history-wise. At least for me it does not make sense. It is more precise to keep the lever up until the pp clock shows the desired value and then one would release it. Bingo, desired pp achieved. With the current behaviour the pilot needs guessing. This simply does not make sense. |
#57
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Hi. I don't know what may be your trouble (axis maybe?), this works at me. I release the lever, the increase/decrease of the propeller-pitch stops. Sometimes a couple of degrees moves longer, but nonsignificant (at the start, it may not respond to the lever until almost a full 6° turn... but there is OK after take-off).
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#58
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#59
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Wrt to Oxy systems. Above 30,000feet (in an unpressurised cockpit) even 100% Oxy is not sufficient. Above 30,000feet an Oxy regulator needs to produce an overpressure at 100% Oxy to keep O2 partial pressures in the blood to acceptable/viable levels. I am not sure if WW2 regulators could automatically produce this required overpressure.
In a pressurised cockpit its not an issue until the Cabin Altitude goes over 30,000feet, then the same arises. B29s were of course pressurised. |
#60
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I do understand that some delay may occur in reality between pushing/releasing the lever and the clock and the proppitch. But this delay should at least be constant and independent of how long the lever is pushed. This really does not make any sense. |
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