tavarish palkovnik

Base drag is something else and actually not related to this topic which is more about internal ballistic. Here I made some simple text about base drag or better to say, about difference of drag coefficients in active and passive stages if someone finds interesting -> But to return on equation, it is empirical equation in form Ical=Th*Inom+190,3+76*pk-3,058*pk^2-7000*pa+25484*pa^2 pk- chamber pressure (MPa) pa- ambient pressure (MPa) Inom- nominal impulse, theoretical impulse, maximal impulse. Fuels of USA origin are usually given in ratio of 70:1 while of Russian origin in ratio 40:1 (40 bars in chamber vs 1 bar ambient) And Th are coefficients of losses which are always present. I took some maximal values, losses of unburned fuel, losses of friction in nozzle and losses due to effect of condensation phases in exhaust

I will later today...in the meantime just to make one correction, it was with typing mistake -> I=245*9,81*0,96*0,97*0,98+190,3+76*5-3,058*5^2-7000*0,1+25484*0,1^2=2242

This is how it looks in my ''shots'', AIM-54C ballistic flight, 45000ft, 1,2M, 0;30;45 deg, constant thrust of 13595N in 27 seconds But, thrust of 13595 N is something about what I would like to talk. 13595*27/163=2252 kg*s/kg ... it means impuls is 2252 or 229,5 s what seams is quite less then regulary used 250 or 245 or 260 etc One example for start, rocket motor of Neva 5V27 rocket Fuel in this motor is known, composite with maximal theoretical impulse of 245s at pressure ratio pk:pa=40:1 Diagram says 345300/151,4=2280 kg*s/kg or 232 s what is 5% less then theoretical maximum at 40:1. It can be seen that average pressure in chamber is 50 bar and test is performed at pa=1 bar. So mathematic will say for this -> I=245*9,81*0,96*0,97*0,98+190,3+76*5-3,058*5^2-7000*0,1+25484*0,1^2=2242 or 228,5 s In practise, motor with fuel with maximal impulse 245s at 40:1 will give real 228,5 s at 50:1 with combined losses in burning process. What about AIM-54C and these ''shots'' with 13595N of thrust. I have one fuel composition which I believe is or it is close to what is inside. It is fuel with maximal theoretical impulse of 247 s at ratio pk:pa=70:1, density is 1,62 kg/dm^3 and burning rate at 70 bar is 5,3 mm/s. Let's say nozzle throat is 55mm, chamber pressure should be around 4 bar -> 13595/1,4*4/3,14/55^2 I=247*9,81*0,96*0,97*0,98+190,3+76*4-3,058*4^2-7000*0,1+25484*0,1^2=2211 or 225 s So, this motor in practise at sea level will have thrust as result of impulse of cca 225-230s. But these shots are up there at high. Configuration of nozzle is such to allow expansion of gases to the pressure quite less then 1 bar, based on some calculations I'm getting 25000 Pa what is in line of ambient pressure up there. So what exactly should be impulse of motor burning there -> I=247*9,81*0,96*0,97*0,98+190,3+76*4-3,058*4^2-7000*0,025*25484*0,025^2=2498 or 255 s 2498*163/27=15080 N or 10% more then at sea level Higher altitude -> higher impuls Higher altitude -> higher Cx In some cases what impuls gives drag takes in roughly same percentage but here taking in consideration nozzle exit and expansion, I belive impuls will give a little more then what drag will take and by that, mach numbers should be more then as presented at the beginning, by my opinion...unless flight is not ballistic but with kinematic overload

...

I have something else in addition to ask. This could be more tougher. This is the only one cutaway model that can be found on the net. I’m the most interested in motors and it would be fantastic to be able to see it in real in close look…but…not possible So what I wanted to ask, did anyone perhaps visited this museum and have pictures not blurred like this. This motor could be MK60, seems like cylindrical grain inhibited on outside in 2/3 length, burning from inside, from up and back heads and from outside on uninhibited surface. From geometry it should be roughly 200kg of fuel and with some very regular burning rate of 7mm/s that gives 20 seconds of active

Thanks The_Tau, it has sense and it is more correct than many fairytales that can be found on Internet

Can anyone help with translating this?

Today for fun I modeled this flight, for fun and relaxation. Are we close with other variables of this flight? Just from curiosity

Thank you Маэстро, understood and accepted. Although for physics and mathematics all times are good, I (we) will leave it till some better time comes

These are not random stuffs, these are sequences in order to get answers. I would like, if there is no in free sources full description, and there is no, to get answer how this rocket actually flys. Also I would like to find someone with same enthusiasm (I’m not rocket scientist) to share thoughts. I was reading here lot about R-27, mostly complaints, so I guessed there will be someone, however. If there is no any single photo and drawing of R-27ER than why not to make one by itself. Collecting visual segments and textual descriptions in assembly is possible isn’t it. With knowing how it looks and what is inside it’s easy to describe how it works. Than it is hard to convince me in something else. Further sequences are ballistic characteristics, all that can be calculated, just regular physics and mathematics, some books, some tables, calculator and there it is…source. Next sequence is guiding method together with all previous (not random) and there it is…answer is there. Conclusion is there.

It is very hard to animate anybody in this very interesting stuff, even here nobody want to talk I thought R-27ER is rocket of interest here, at least I got such impression. Anyway I will continue with some more words, at least it will be at one place. So now when having Cx f(M ; H) at zero angle of attack and Cy f(M) also at zero angle of attack and some allowable angles, it is good time to use these numbers for cases when rocket is under overload, when angle of attack starts to be involved. This is simplified mathematic but sufficient enough. So let’s take one time moment, 1km and 2M. Cx0-drag coefficient at zero angle of attack Cy0-lift coefficient at zero angle of attack, 1/deg Cx1-drag coefficient with angle of attack Cy1-lift coefficient with angle of attack, 1/deg A-angle of attack, deg Cx0=0,713 ; Cy0=0,5079 ; and for example A=5 deg Cy1=Cy0-Cx0/57,3=0,5079-0,713/57,3=0,495 (1/deg) Cx1=Cx0+Cy0*A^2/57,3=0,713+0,5079*5^2/57,3=0,935

...

But…all that how-how…however inertial navigation with R-27R/ER is missed completely

It would be indeed (for this and these), but that and those probably you don’t have extracted. However when I see that, for example H-58, is powered with lousy 210s in march, and this rocket is filled with highly energetic composite fuel, and plus to that with 195s in buster , everything can be expected.

...

...

...

...

...

...

...

...

...

...

Uffff…should we now about available reserved overload capacity after 60 seconds of flight. Or about what “blues” like to call flight with loft what “reds” have also and call it inertial flight with weight compensation