tavarish palkovnik

Check of mathematical model (with used like they wrote ''publicly available data'' , meaning here I used 13595 N / 27 sec), and comparison with famous NASA ''shots'' 45000ft ; 1,2M ; 0 / 30 / 45 deg 45000ft ; 45deg ; 1,2 / 0,8 / 0,5 M It fits nice, only this thrust envelope has nothing related with real condition

This is how I see flight as per Figure No.7 could looks like, targeting high speed/high altitude ''Mig-25'' Missile usually after releasing is with small decline relatively to airplane axis, I gave it as negative 2 degeess. Then first 4 seconds after motor started, missile released starts its flight, after that I overloaded missile with constant 5g. That is red part of trajectory, and than missile is again released. For velocity this overload has significant influence, 5g in 10 seconds is flight with angle of attack from 25 to 13 deg -> For comparison, three cases, this one, horizontal constantly overloaded flight and ballistic shot with start elevation 30 deg

I would take it as 0,4 M

AIM-54 zone puska.pdf

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Except, it looks nice

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By the way, could be coincidence or maybe not…sustaining stage and 20,8 bar of chamber pressure together with expansion ratio 6,2 for heat ratio 1,2 suggest pressure at nozzle exit of 47300 Pa (0,473 bar) what is ambient pressure at 6000 meters or 20000 ft

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Yes that is Omega, and yes orbital speed how I call it, is angular rate of the line of sight. However, it is speed and speed can be constant, with acceleration or with retardation. Only here it is not m/s but deg/s. In circular (but really circular) motion target with constant velocity around point is with constant orbital speed. If angle like in case of this last declined 45deg trajectory is with every new second bigger and bigger, then it is orbital speed which is with acceleration, positive sign. It is not about clockwise or against clockwise when speed is matter, only about kind of speed, acceleration or retardation

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Can you make some sketches, for example on these 3 cases, it will be easier to follow and to understand. How you would position missile’s inertial navigation trajectories chasing these targets

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Great, comrade for communication is here For start, let’s forget where I pointed point for SARH start, is it on line of launch altitude or not, above or below, doesn’t matter for now. Like said it is unscaled imaginary case. It is more important where it is relatively to target because I see it always or almost always with IN, above. Principle is to hold missile to have in very end of terminal approach maximally released position, with no kinematic overload. By the way, negative sign on angle is always in clockwise side, it is mechanical principle and rule. We will continue for sure, I hope so, momentarily I’m out battery

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Aha...I thought deleting attachments will mean that all will be gone, all published will not be visible anymore...it is about deleting which is not deleting That’s better, it wouldn't be good to delete all and not have chronology. In any case, I think I'm on very end of finding answers about motors of Phoenix, actually I'm very sure. I've had some dilemmas about slots in this finishing, how long they are and how transition looks like, although it was in front of nose, instead of looking picture more patiently and precisely I was struggling with textual description. This is for me final, roughly this is Phoenix's motor Output can’t be much different to this Pressure pick is due to erosion in start, and like Mk60 document says max is 1000 psi (69 bar), and it's true. Thrust also is so close to what document gives, from 5000 lbf to 1000 lbf. Geometry and burning rate data give 25 seconds, everything is in line to what is given to Mk60. This is of course for sea level, but not in converted style but actual, true figures, values which motor actually gives. Because I don't have doubts that FWind’s sketch of cross section is Mk47 and because output is same as given to Mk60, these two motors are same, they give same and that's why they are or better to say, they can be exchangeable. Otherwise it would be difficult to have same guided weapon with two different dynamics Mk47 Mod.0=Mk60 Mod.0

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Motor No.C from document that @FWind shared with us is described in patent No.US4015427A, author W.T.Brooks One more step not to have doubts that cross section of motor No.B is Mk47 (and/or not Mk60)

Few more words (thoughts) about numbers behind Phoenix. Let’s take these 5000 to 1000 lbf and 25 seconds as situation on the ground, at sea level. That makes total impulse 75000 lbs and average thrust 3000 lbf. Few documents say 97000 lbs and 4000 lbf so let’s try to understand these numbers. It is easier to me in metrics so I want to find connection between 333600 Ns (75000 lbs) and 431456 Ns (97000 lbs) and/or 13340 N (3000 lbf) and 17792 N (4000 lbf). This motor operates with average chamber pressure of pk=48 bar (documented) and my calculations are in line with that. Propellant is with heat ratio k=1,2 and nozzle expansion ratio is A=18,5. These three values give what should be pressure at nozzle exit -> E=pk/pe=4,3591*A^1,2666=175,55 -> 48/pe=175,55 -> pe=0,273 bar (0,0273 MPa) Pressure at sea level is 1 bar (0,1 MPa) so full expansion can’t be achieved, motor is choked by ambient (atmospheric) pressure. Exit diameter of nozzle is 238mm so area of choking is 44466mm2. Pressure differential is 0,0237-0,1=-0,0727 Mpa what makes force differential of -3323N. This is what motor “lost” down there because of choking. Now let’s add it to what motor actually gives, 13340+3323=16663N (3746 lbf) Now it is closer to 4000 lbf. The rest is differential between 48bar and 69bar (1000 psi) what is US standard for chamber pressure when presenting internal ballistic. So it is all about ratio 69:1, ratio between chamber pressure and atmospheric pressure. Russian standard is 40:1 As much as I understand American principles, they give motor’s characteristics converted to standard sea level conditions 69:1. If I’m not wrong they just take thrust coefficient Ct for optimal expansion at 69:1 for specific heat ratio, reduces it for losses and multiply it with real chamber pressure and throat area. F=ß1*ß2*Ct*pk*Ak Ct=1,6 for k=1,2 and 69:1, pk=4,8 MPa, Ak=2407mm2 and losses are 0,98 (diverging part of nozzle) and one more 0,98 for other losses (heat, friction etc etc) F=0,98*0,98*1,6*4,8*2407=17754N (3991 lbf) Why they do it that way I don’t know, it doesn’t have sense to me but all right. Conclusion, should be careful with numbers, behind can be everything. Btw, perhaps AIM-7F could be taken in consideration as well, I think it’s numbers on internet are also on same principle. It is not same as Phoenix because nozzle exit is not such big but motors Mk58 and Mk65 indeed operate quite of time much under 69 bar

Configuration with these two slots and this burning rate gives nearly linear drop from 5000 to 1000 lbf in 25seconds at see level. Exactly what is given to Mk60. Author of this text, W.T.Brooks, was from Rocketdyne, motor A is Sidewinder's Mk36, motor of Rocketdyne, and there is no reason not to believe that motor B is description of Mk47. What I want to say, Mk47 and Mk60 were (should be) with same characteristics. I don't know when and how and who started all this on Internet with different data for Mk47 and Mk60, for me these were with same output. 0km: 5000-1000 lbf 5km: 5565-1480 lbf 10km: 5840-1755 lbf 15km: 5985-1900 lbf 20km: 6050-1965 lbf

@FWind you are not much about words but man you can hit great “Restrictor divider located in the cylindrical portion of the grain that separates the grain into two segments” this is new direction and now with fixed number of slots to just 2 it will be easier We will continue for sure