I'm curious what supersonic aerodynamics phenomenon is seen on this interesting rocket photo? The linked page describes the event, but I'd like to add that this photo is a high-quality colour version of the frame at 0:09:02 (min:sec:frame) of that 1.5 MB greyscale video clip. The exhaust gases seen are from the recent firing of the 8 relatively small solid fuel rockets located at the "tail" of the rocket's first stage whose thrust is directed opposite of the thrust of the propulsion engines in order to quickly jettison the stage.
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What supersonic aerodynamics phenomenon is seen here? 3
- Thread starter luchezar
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Probably all the 'junk' from the rockets. Especially if there is water vapor or similar that is caused to compress by the shock wave.
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btrueblood
Mechanical
Normally the light output from the F1 main engines would swamp any refracted/reflected light from the bow shock wave. This frame is taken at a point in time when the mains have (mostly) shut down, the second stage is not yet ignited (or not yet at full output), and the only light is from (relatively) weak retro engines that are (mostly) firing away from the camera. The retros are also throwing a lot of plume dust into the flow field behind the bow shock to further help reflect/scatter light and help illuminute the shock wave.
I'd bet a few frames later, as the 2nd stage ignites, the bow shock essentially disappears along with the rest of the rocket, and all that can be seen is a huge "fireball" as the light from the 2nd stage engines swamps the film faster than the iris can respond to the light increase.
I'd bet a few frames later, as the 2nd stage ignites, the bow shock essentially disappears along with the rest of the rocket, and all that can be seen is a huge "fireball" as the light from the 2nd stage engines swamps the film faster than the iris can respond to the light increase.
MikeHalloran
Mechanical
The other thing that makes it visible is that the bow shock represents a step change in pressure, and hence refractive index, of the air.
Mike Halloran
Pembroke Pines, FL, USA
Mike Halloran
Pembroke Pines, FL, USA
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- #9
Thank you very much for your answers!
Can we summarise that what we see is not anything else but a conical shock wave, and that the gases from the solid fuel rockets (that stopped working a few film frames ago) help make it visible but don't form its shape?
IRStuff: I assume that you too mean that we see a conical shock wave here.
KENAT: An old description of the photo here confirms your thought about the vapour (probably one of the combustion products from the work of the solid fuel rockets):
(old NASA photo id: 107-KSC-69PC-415):
Can we summarise that what we see is not anything else but a conical shock wave, and that the gases from the solid fuel rockets (that stopped working a few film frames ago) help make it visible but don't form its shape?
IRStuff: I assume that you too mean that we see a conical shock wave here.
KENAT: An old description of the photo here confirms your thought about the vapour (probably one of the combustion products from the work of the solid fuel rockets):
(old NASA photo id: 107-KSC-69PC-415):
That's right.
However I doubt the gases from the retro engives can propagate upstream to the nose of the rocket. At this location, the flow is clearly downstream and at Mach number around 2-3 as indicated by the shock angle.
I would favor an optic explanation, involving the change of density and hence refractive index through the shock and the special lighting due to the firing engines and/or sunlight. A word from a specialist in optics or flow visualisation would be welcome here.
However I doubt the gases from the retro engives can propagate upstream to the nose of the rocket. At this location, the flow is clearly downstream and at Mach number around 2-3 as indicated by the shock angle.
I would favor an optic explanation, involving the change of density and hence refractive index through the shock and the special lighting due to the firing engines and/or sunlight. A word from a specialist in optics or flow visualisation would be welcome here.
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- #11
I agree. As the rocket axis isn't strictly perpendicular to the line of sight so the observed (projected) shock cone angle is slightly larger than the actual one, the Mach number must be closer to 3 than to 2.
As to the visualisation issue, as the Sun was shining from southeast at this time (about 9:35 local time), if the plane that took the film has flown to the northwest of the rocket, the rocket may have been seen from the plane in almost direct sunlight. If so, even a not so big change of the refractive index (e.g. caused by the vapour) may have caused the lighting effect.
If you're still curious about this, a colleague in our university is a specialist in optics - I can ask him for his opinion and write it here.
As to the visualisation issue, as the Sun was shining from southeast at this time (about 9:35 local time), if the plane that took the film has flown to the northwest of the rocket, the rocket may have been seen from the plane in almost direct sunlight. If so, even a not so big change of the refractive index (e.g. caused by the vapour) may have caused the lighting effect.
If you're still curious about this, a colleague in our university is a specialist in optics - I can ask him for his opinion and write it here.
The point at which this is filmed also represents a drastic change in acceleration, and as such, it can cause drops in the bow wave pressure. A sudden pressure drop could cause condensation to appear.
While a refractive index change could also occur, it would be changing in the wrong direction, and would be unlikely to cause the usual refractive mirage effects associated with index transitions, given the viewing angle.
TTFN
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While a refractive index change could also occur, it would be changing in the wrong direction, and would be unlikely to cause the usual refractive mirage effects associated with index transitions, given the viewing angle.
TTFN
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sll, good point bout the vapor being behind the cone. The image wasn't real clear when I looked at it the other day, guess I'll look again when I have a decent internet connection.
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I meant my suggestion that the vapor was from the rockets. There could of course be vapor naturally in the atmosphere. Though not sure what altitude this is at etc.
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Dynamic pressure is dependent on velocity or Mach number, not on acceleration.
According to the NASA press kit of the time, first stage burnout occurred at an altitude of 38 miles (about 200,000ft). According to the standard density at that altitude would be 0.0053 x 10^-4 slugs/ft^3 or 0.02 per cent of sea level density. There isn't enough air or water vapour at that altitude to create the effect seen in that film frame.
The vehicle velocity at Stage I burnout was about 6000mph, according to NASA, which is 8800 ft/sec. Based on propellant mass and burn time (to give mass flow rate) and stated thrust, the exhaust velocity of the second stage rockets would have been around 13,500 ft/sec.
This indicates that the velocity of the separation thrusters could well have been greater than the velocity of the vehicle at first stage separation.
I consider that the vapour seen within the shock cone would have been the exhaust of the separation thrusters.
According to the NASA press kit of the time, first stage burnout occurred at an altitude of 38 miles (about 200,000ft). According to the standard density at that altitude would be 0.0053 x 10^-4 slugs/ft^3 or 0.02 per cent of sea level density. There isn't enough air or water vapour at that altitude to create the effect seen in that film frame.
The vehicle velocity at Stage I burnout was about 6000mph, according to NASA, which is 8800 ft/sec. Based on propellant mass and burn time (to give mass flow rate) and stated thrust, the exhaust velocity of the second stage rockets would have been around 13,500 ft/sec.
This indicates that the velocity of the separation thrusters could well have been greater than the velocity of the vehicle at first stage separation.
I consider that the vapour seen within the shock cone would have been the exhaust of the separation thrusters.
Well there you go, do your homework and you get your answer.
Thanks Glenn.
Thanks Glenn.
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- #18
So, the vapour was really a part of the separation thruster (retro-motor) exhaust gases, and their velocity must have been significantly greater than the velocity of the vehicle at the first stage separation in order for the gases to reach the rocket "nose" and light up the shock wave cone.
The composition of the propellant used in the retro-motors is basically known. The Saturn V flight manual, page 4-28, last paragraph, states:
By the way, the film whose frame is the photo in question indicates that the gases do reach the nose of the rocket.
But if you re-read the whole thread again, wouldn't you see something strange?
The composition of the propellant used in the retro-motors is basically known. The Saturn V flight manual, page 4-28, last paragraph, states:
What velocity do the exhaust gases of this propellant reach? Does this velocity depend on something else?
By the way, the film whose frame is the photo in question indicates that the gases do reach the nose of the rocket.
But if you re-read the whole thread again, wouldn't you see something strange?
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- #19
The Saturn V flight manual (SA-503), page 4-28, last paragraph, says:
where I[sub]sp[/sub] is the effective impulse, F is the average effective thrust [kgf], m is the propellant mass [kg], and t is the burn (firing) time . And then multiplied the specific impulse I[sub]sp[/sub] to the gravitational acceleration constant g[sub]0[/sub] to get the effective retro-motor exhaust velocity v:
The Saturn V flight evaluation report (AS-503), page 12-3, gives the 0.658 seconds for the average effective retro-motor burn time, and 375,470 N for the average effective thrust. I asked a more knowledgeable friend to calculate the effective exhaust velocity from this data. He first calculated the specific impulse by the following formula:
I[sub]sp[/sub] = F / (m/t) = (375470 / 9.80665) / (278 * 0.45359237 / 0.658) = 199.788241 s
where I[sub]sp[/sub] is the effective impulse
v = I[sub]sp[/sub] * g[sub]0[/sub] = 199.788241 * 9.80665 = 1 959 m/s
I haven't read all the replies in this thread, if I'm repeating someone else's reply I apologise.
What it is seen is an oblique shock, that is correct. I advise you to do some research on the atmosphere chemical composition at that altitude, and what speed is the rocket travelling at. We might be seeing a rocket at hypersonic flight, i.e. the break up of the air on its ions, and the actual "ignition" of this air due to the high temperatures.
Regards
What it is seen is an oblique shock, that is correct. I advise you to do some research on the atmosphere chemical composition at that altitude, and what speed is the rocket travelling at. We might be seeing a rocket at hypersonic flight, i.e. the break up of the air on its ions, and the actual "ignition" of this air due to the high temperatures.
Regards
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