Chinese missile could shift Pacific power balance

[chauism]

Har...I owe you zilch. Not only that, what you said above has nothing to do with what you claim to exist: a 'non ideal gravity turn'.

Of course it is zilch, because you have none by using your argument for a flat ballistic trajectory.

I already have given you a definition of an ideal gravity turn from you own wiki source, look it up. Let me make it more simple for you, an ideal gravity turn is a maneuver of a rocket or missile that the only(ideal) force used to change the angular moment of inertia(turn) is the gravity force(gravity).

For the example I give to you in my previous post, for the same target, I can achieve it by an ideal gravity turn ascending phase by using minimum pitch, so the missile exit its boost phase around angle of 55 degree, or it can use a less ideal gravity turn ascending phase by using excessive pitch, so that the missile exit the boost phase around 35 degree. Also I can achieve those angles by mixing different set of duration of boost phase and thrust induced steering.

Yes...Really. You misunderstood what was said about the gravity turn, which tells me you never knew about it in the first place.


Which is I why I said that in the lower atmosphere, both behave essentially the same. Now where is that 'non ideal gravity turn' source ?

I never mentioned about gravity turn before because it is irrelevant to my point of how the boost period (acceleration phase) will affect the overall trajectory since the gravity force relatively remains the same throughout the flight path, which at this point you still have not answered how the duration of the boost phase does not affect the trajectory of the missile yet.

Nothing above illustrate how a 'non ideal gravity turn' is created. The fact that you continue to use 'ideal gravity turn' instead of the simple 'gravity turn' tells me you still do not understand how you were in error in the first place. So what we see here is you using a lot of numbers, not to explain anything relevant to your claim, but to hide the fact that you do not know what you are talking about.

Let's see who doesn't understand the concept here.

The moment the vehicle is in a gravity turn, it cannot maneuver for course correction.

IF you did know it, you would know that a 'gravity turn' is essentially a zero angle of attack maneuver...

For tactical missiles, the flight profiles are determined by the thrust and load factor (lift) histories. These histories can be controlled by a number of means including such techniques as using an angle of attack command history or an altitude/downrange schedule that the missile must follow. Each combination of missile design factors, desired missile performance, and system constraints results in a new set of optimal control parameters.

Trajectory optimization - Wikipedia, the free encyclopedia
See the problem here, if the gravity turn as you emphasized is the only factor in your assumption of predicting the flight path of the missile and since the AoA is 0 for any gravity turn as you assumed, and the only the changes of AoA is before the gravity turn with minimum pitch. Tell me why even bother with the history of AoA command.

Also as soon as the missile is not in a vertical flight anymore, it already enters a gravity turn. Your statement of "The moment the vehicle is in a gravity turn, it cannot maneuver for course correction," is dubious.

Until ? That is what radar is for. Radar will tell us target heading, speed, altitude and aspect angle with respect to us. Everything we need for fixed coordinates collision intercept.

Once the missile is detected and tracked while in this turn, its flight path can be estimated straight up to orbit.

Really!!?? Why do we need radar for here? I thought you only need to know the state of missile when it enters a gravity turn to know the trajectory of its path up to the "orbit".

This still does not answer the question of why SHOULD an ICBM be more structurally robust than a satellite launcher. You already admitted that in the boost phase, both are essentially the same. Further, an ICBM is usually launched in home territory, its flight will be mainly for altitude, by the burn-out time of 3-4 min the most important part of the missile will be 6-7 km/s and far above the operating altitude of any fighters, hostile or not. Any horizontal travel that could potentially take the missile into enemy controlled territory would be negated by that burn-out altitude. So what need is there for an ICBM to be any more structurally robust than what is necessary, or more than a satellite launcher, when being so will cost us payload capability ?

Of course it does answer why an ICBM should or shouldn't in this case be built more structurally robust, because it doesn't need to. In the equation, it tells you that for the same material used to construct satellite launcher or missile and assuming both casting has the same material property and has the same endurance for stress, the one with smaller diameter will experience less stress per unit area for the same pressure applied to it. Thus, the one with smaller diameter can endure higher g forces during a maneuver than the one with larger diameter. Of course for missiles and rocket the calculation of each of the stress point during maneuvers is more complicated, but the principle is still the same. Let me give you a demonstration here using the DF-31(ro=1125mm, ri = 975mm) and CZ-2 (ro = 1675mm, ri=1525mm) here since their lifting capacity is more or less the same, if let's say the same pressure 100 MPa is applied to both. The stress at the out surface is
For DF-31, the Hoop stress σc = 704 Mpa, the Axial stress σa = 402 Mpa
For CZ-2, the hoop stress σc = 1069 Mpa, the Axial stress σa = 585 Mpa

What it means here is for DF-31 to experience the same structure stress as CZ-2, more pressure is needed. In here, let's say CZ-2's critical pressure load is 100 Mpa, then DF-31's critical pressure load would be 145 Mpa. So if the maximum g force CZ-2 can sustain is 10, then DF-31's maximum g force is 14.5 g. Thus gives DF-31 more maneuverability. Therefore your statement of "it should not maneuver if it is to retain structural integrity." is false even if it is true for commercial rockets.

Once again...More numbers to obfuscate the fact that you cannot support your argument.

Very well said about yourself.

 

[chauism]

Ok I don't mean to butt in here as there are too many technicalities that I do not know about, but just to dwell a little on the maths here,
let's say x is a [6x1] vector for ICBM as you have explained, y is the [6x1] vector for the intercepting missile , A is a [6x6] configuration matrix that will be used to determine y given x (once it is launched and detected): y = Ax.

However this model only works if the velocity of the ICBM is constant or known as in the case of air-traffic or shipping lane traffic control etc, it doesn't take into account of the ICBM's acceleration as we know it maneuvers itself, so we'd need to have at least x = [X, X-dot, X-dot dot] (X for space vector), but the last three elements of x is unknown and you essentially have an under-determined linear system that doesn't have a definitive solution. Unless the intercepting missile is capable of constantly detecting and calculating the incoming missile's acceleration and adjusting itself on the fly (no pun intended), are the current missiles sophisticated and powerful enough to do this?

Because he has hard time understanding the changes in variables.

I already explained to him about how different value for variable's like duration of the boost phase(acceleration) and steering(exit angle) have effects on the overall trajectory in mathematics, and he is still insist on his notion that those variables does not matter in his prediction of trajectory.

 

[gambit]

Of course it is zilch, because you have none by using your argument for a flat ballistic trajectory.

You ran out of credible arguments for the DF-21 a long time ago. So now you are treading into the absurd. There is no such thing as a 'flat ballistic trajectory'. But there is flatTER, meaning less of an arc, in comparison to others. Even if it is merely one degree off, it can be described as flatTER.

For example...

RT-2UTTH Topol M - Wikipedia, the free encyclopedia

The first stage has three rocket motors developed by the Soyuz Federal Center for Dual-Use Technologies. This gives the missile a much higher acceleration than other ICBM types. It enables the missile to accelerate to the speed of 7,320 m/s and to travel a flatter trajectory to distances of up to 10,000 km.

Just because some ballistic trajectories may be flatter than others does not mean a gravity turn is not required. If there is a rocket motor powerful enough and with sufficient duration, then we can have a missile that will cruise at 1000 m altitude, for example, then there is no need for a gravity turn. So no, I do not owe you any explanation about any 'flat ballistic trajectory' because there really is no such animal.

I already have given you a definition of an ideal gravity turn from you own wiki source, look it up. Let me make it more simple for you, an ideal gravity turn is a maneuver of a rocket or missile that the only(ideal) force used to change the angular moment of inertia(turn) is the gravity force(gravity).

For the example I give to you in my previous post, for the same target, I can achieve it by an ideal gravity turn ascending phase by using minimum pitch, so the missile exit its boost phase around angle of 55 degree, or it can use a less ideal gravity turn ascending phase by using excessive pitch, so that the missile exit the boost phase around 35 degree. Also I can achieve those angles by mixing different set of duration of boost phase and thrust induced steering.

No...What you gave is a hilarious misunderstanding of the relevant paragraph. From what basis can you claim that 55 deg is 'ideal' and others are 'non ideal'? Show us a source that says so.

But here it is again for all to see...

Gravity turn - Wikipedia, the free encyclopedia

The pitch over maneuver consists of the rocket gimbaling its engine slightly to direct some of its thrust to one side. This force creates a net torque on the ship, turning it so that it no longer points vertically. The pitch over angle varies with the launch vehicle and is included in the rocket's initial guidance system,[1] for some vehicles it is only a few degrees while other vehicles use relatively large angles (a few tens of degrees). After the pitch over is complete the engines are reset to point straight down the axis of the rocket again. This small steering maneuver is the only time during an ideal gravity turn ascent that thrust must be used for purposes of steering. This pitch over maneuver serves two purposes. First, it turns the rocket slightly so that its flight path is no longer vertical, and second, it places the rocket on the correct heading for its ascent to orbit. After the pitch over the rocket's angle of attack is adjusted to zero for the remainder of its climb to orbit. This zeroing of the angle of attack reduces lateral aerodynamic loads and produces negligible lift force during the ascent.

A gravity turn is not the only mean to ascent to orbit. If the American Space Shuttle could take off like an aircraft and is equiped with powerful enough engines, it could achieve orbit gradually through aerodynamic exploitation until the atmosphere is too thin. And if you think that is not possible: SR-71. That aircraft's maximum altitude is still secret but all knows that figure is high enough that a rocket motor would propel the aircraft into orbit. Another mean is to simply power straight up. But the reason why the Space Shuttle, satellite launchers and ICBM uses a gravity turn is to reduce gravity loss...

RSE 24

A helicopter hovering over one spot is a perfect example of a mass being held up by raw horsepower. This is gravity loss. Gravity loss is the thrust of a rocket that is expended to counteract the force of gravity. It is maximum at liftoff, and zero as the payload enters orbit.

At moment of lift-off, immediately there is gravity loss. An inclined trajectory coupled with a high power-to-weight ratio and we have an 'ideal' method to achieve orbit. The duration of our stay in orbit depends on our mission, of course. That mean this phrase 'ideal gravity turn ascent' is translated as: the gravity turn is the ideal method of orbital ascent. Not that there are animals like 'ideal gravity turn' and 'non ideal gravity turn' as you tried to pass off here. So if you are confident that there is such a thing as 'non ideal gravity turn' then show us a source that says so. We are still waiting and am willing to be corrected by a credible external and non-bias source.

I never mentioned about gravity turn before because it is irrelevant to my point of how the boost period (acceleration phase) will affect the overall trajectory since the gravity force relatively remains the same throughout the flight path, which at this point you still have not answered how the duration of the boost phase does not affect the trajectory of the missile yet.

I never said it does not. But I have no problems schooling you...

The longer the VERTICAL part of the boost phase, the less physical stress the vehicle has to endure because the vehicle would leave the lower/denser atmosphere sooner. The downside to a PURELY vertical ascent is the aformentioned 'gravity loss', which cost us in terms of large fuel load which naturally contribute to overall vehicle size, also because of 'gravity loss', such a purely vertical climb DOES NOT contribute to orbital velocity. Improved guidance and control technology allow us to create that 'gravity turn' where lateral stresses on the off-vertical vehicle are minimized. As time and experience goes by, we have a decreasing vertical portion of the boost phase and a sooner entry into the gravity turn, which IS a part of the boost phase. In multistage vehicles, the shedding of empty fuel tanks, aka continual mass reduction, and a reduction of gravity loss thanks to an inclined vehicle, allow the vehicle to gain orbital velocity more rapidly. So does the duration of the boost phase affect the trajectory of a missile? Yes, what else but because it is a segment of the overall travel? There is a side benefit in that the gravity turn is best indicator of a trajectory profile and the sooner the entry into the maneuver the better the predictive process of that trajectory.

And do not tell me that you never 'mentioned' about the gravity turn. You never 'mentioned' it because you cannot mention it because never knew about it as part of the boost phase in the first place as evident by that 'non ideal' version you are still trying to pass.

Let's see who doesn't understand the concept here.

Trajectory optimization - Wikipedia, the free encyclopedia
See the problem here, if the gravity turn as you emphasized is the only factor in your assumption of predicting the flight path of the missile and since the AoA is 0 for any gravity turn as you assumed, and the only the changes of AoA is before the gravity turn with minimum pitch. Tell me why even bother with the history of AoA command.

First...I never claimed that the gravity turn is the only factor in estimating flight path.

Second...Here are the relevant paragraphs...

Trajectory optimization is the process of designing a trajectory that minimizes or maximizes some measure of performance within prescribed constraint boundaries. While not exactly the same, the goal of solving a trajectory optimization problem is essentially the same as solving an optimal control problem.

The selection of flight profiles that yield the greatest performance plays a substantial role in the preliminary design of flight vehicles, since the use of ad-hoc profile or control policies to evaluate competing configurations may inappropriately penalize the performance of one configuration over another. Thus, to guarantee the selection of the best vehicle design, it is important to optimize the profile and control policy for each configuration early in the design process.

Consider this example. For tactical missiles, the flight profiles are determined by the thrust and load factor (lift) histories. These histories can be controlled by a number of means including such techniques as using an angle of attack command history or an altitude/downrange schedule that the missile must follow. Each combination of missile design factors, desired missile performance, and system constraints results in a new set of optimal control parameters.

This is not about trajectory optimization BY a missile but about weapons development, specifically missile development in creating more accurate missiles via detailed flight records such as AoA degrees, thrust, duration...etc...etc...Notice the plurals: 'flight profiles', 'flight vehicles' and 'control policies'. This means for any specific flight, the missile is not making several 'trajectory optimization' processes but that the missile's programed flight path(s) are ALREADY optimized based upon historical data from past missiles. In other words, your wiki source is not about A particular missile in flight but about THE missile as a weapons development over time and experience.

Just like the gravity turn gaff, you keep making these interpretation mistakes because you have no relevant experience and are trying to salvage some face for a losing argument.

Also as soon as the missile is not in a vertical flight anymore, it already enters a gravity turn. Your statement of "The moment the vehicle is in a gravity turn, it cannot maneuver for course correction," is dubious.

No...It is not dubious. Any maneuvers outside of what is programmed affect range. Fuel is finite. I was not speaking technically because there are no technical limitations on creating maneuverable ICBMs if we wanted to. I was speaking of 'policy' and it is that we should maneuver no more than what is necessary.

Here is the definition of 'policy'...

Policy - Wikipedia, the free encyclopedia

A policy is typically described as a principle or rule to guide decisions and achieve rational outcome(s). The term is not normally used to denote what is actually done, this is normally referred to as either procedure or protocol. Whereas a policy will contain the 'what' and the 'why', procedures or protocols contain the 'what', the 'how', the 'where', and the 'when'. Policies are generally adopted by the Board of or senior governance body within an organisation where as procedures or protocols would be developed and adopted by senior executive officers.

See that? The 'rational outcome' we desire is to hit a ground point at so-and-so time within a so-and-so diameter circle (CEP). So the 'principle' or 'rule' is that we should not maneuver beyond necessities. That mean once inside a gravity turn, the missile should not, or in accordance to 'policy', it CANNOT maneuver. Course corrections are usually at terminal. But again, what I said should not be construed to be from a technical perspective.

Really!!?? Why do we need radar for here? I thought you only need to know the state of missile when it enters a gravity turn to know the trajectory of its path up to the "orbit".

And you thought wrong. There are two moving bodies here: the missile and the interceptor. Since we are using fixed coordinate system to create collision intercept, real time updates of BOTH bodies are necessary.

NWS JetStream - Exactly how does radar work?

By measuring the shift in phase between a transmitted pulse and a received echo, the target's radial velocity (the movement of the target directly toward or away from the radar) can be calculated. A positive phase shift implies motion toward the radar and a negative shift suggests motion away from the radar.

Gaps between echoes compresses as the two bodies nears each other. Conversely they widens if the bodies are moving away from each other. Phase (shift) Derived Range Acceleration (PDRA) estimations are created for trajectory predictions and, believe it or not, RECONSTRUCTION. Target state contain (x y z) position, velocity and acceleration.

Previously, I showed how fixed coordinate system sees a target state:

X = [x y z Vx Vy Vz]T or [x y z x-dot y-dot z-dot]T
Where X = state vector, T = time, V = velocity and the standard xyz axes.

For target range acceleration estimate, basic target state is seen as:

X = [R(RRC) V(RRC) A(VTC)]
Where R(RRC) = [x y z]T is the target position the east-north-up (ENU) radar reference coordinates.

Coordinate Systems, Basis Sets, Reference Frames, Axes, etc.

Local East-North-Up (ENU) system

In any ENU system, {dX, dY, dZ} is a right-handed orthogonal system.

Where V(RRC) = [x-dot y-dot z-dot]T is the target's velocity inside RRC.
Finally, A(VTC) = [a(v) a(t) a(c)]T is target's acceleration in the right-handed velocity-turn-climb (VTC) coordinates system.

The radar echo phase shift information from all three elements -- R V A -- can be used for both prediction and reconstruction and if the radar is sophisticated enough it can predict as well as reconstruct at the same time regardless of its position in respect to the target.

I still do not want to engage in a mathematical one-upmanship game here but what I do want to show the readers is that your simplistic assertions in the previous pages regarding flight path prediction bears no resemblance to what has been in place for decades. If we go back to this illustration...

...Collision navigation by the missile does not care where the bogey came from. Pure pursuit navigation (PPuN) require a true tail chase situation and air-air interceptions seldom allow us that luxury. That leave us the much more complex proportional navigation (PN) based hybrid intercept laws that require the missile to constantly (re)calculate the closing distance between bogey and itself. Even if the missile is behind the target, the laws are still PN based. In the PN based scheme, the missile does not care where the bogey came from or its thrust. The missile cares only for the closing velocity between the bogey and itself at any point in space/time. No different for ICBM collision intercept. The incorporation of phase shift data rendered origination data quite -- pointless.

So yes...Really.

Of course it does answer why an ICBM should or shouldn't in this case be built more structurally robust, because it doesn't need to.

If the ICBM does not need to be more structurally robust than a satellite launcher, then why did you criticized me for legitimately comparing the two? Remember this...

You are comparing the rigidness of a commercial rocket to a missile here????

In the equation, it tells you that for the same material used to construct satellite launcher or missile and assuming both casting has the same material property and has the same endurance for stress, the one with smaller diameter will experience less stress per unit area for the same pressure applied to it. Thus, the one with smaller diameter can endure higher g forces during a maneuver than the one with larger diameter. Of course for missiles and rocket the calculation of each of the stress point during maneuvers is more complicated, but the principle is still the same. Let me give you a demonstration here using the DF-31(ro=1125mm, ri = 975mm) and CZ-2 (ro = 1675mm, ri=1525mm) here since their lifting capacity is more or less the same, if let's say the same pressure 100 MPa is applied to both. The stress at the out surface is
For DF-31, the Hoop stress σc = 704 Mpa, the Axial stress σa = 402 Mpa
For CZ-2, the hoop stress σc = 1069 Mpa, the Axial stress σa = 585 Mpa

What it means here is for DF-31 to experience the same structure stress as CZ-2, more pressure is needed. In here, let's say CZ-2's critical pressure load is 100 Mpa, then DF-31's critical pressure load would be 145 Mpa. So if the maximum g force CZ-2 can sustain is 10, then DF-31's maximum g force is 14.5 g. Thus gives DF-31 more maneuverability. Therefore your statement of "it should not maneuver if it is to retain structural integrity." is false even if it is true for commercial rockets.

You argued that a missile is more physically robust than a 'commercial rocket'...

Commercial rockets are not designed to endure high stress manuvers that modern missiles have to withstand.

...But you show no source to support your claim, and that a missile should be so because it IS more maneuverable, but again you show no source to support your claim. The math you presented is irrelevant and it serves nothing more than to distract from the fact that you consistently failed to support your claims with nonbias sources.

So according to you...

Q - Why should a satellite launcher be less physically robust than a missile?
A - Because the missile is more maneuverable.
Q - Why should the missile be more maneuverable?
A - Because the missile is not a satellite launcher.

What the hell kind of argument is that? In road racing, your opponents are threats so it behooves you to have a highly responsive and maneuverable car. In air combat, someone is trying to kill you so it behooves you to have a highly maneuverable aircraft. How is a satellite launcher a threat to a missile?

Here is the real logical thought process...

An ICBM is usually launched from home territory, so what is the need to make the thing maneuverable? None other than one slight pitch over. So why should it be any more physically robust than a satellite launcher? Not at all. So what if the fuel used allows for a smaller diameter tank? If anything, I would keep the same dimensions but pack it with more fuel for higher speed and for greater range. So if I build a vehicle that has just enough structural strength to carry a payload to a location, then what I said is very correct -- that 'it should not maneuver if it is to retain structural integrity'.

This is why you guys and your desperation to exaggerate anything Chinese are soooo entertaining.

 

[chauism]

You ran out of credible arguments for the DF-21 a long time ago. So now you are treading into the absurd. There is no such thing as a 'flat ballistic trajectory'. But there is flatTER, meaning less of an arc, in comparison to others. Even if it is merely one degree off, it can be described as flatTER.

First of all, I don't think that I ever argued about DF-21 since it has not been tested yet.

For example...

RT-2UTTH Topol M - Wikipedia, the free encyclopedia

That is the example I gave to you.

One of the Topol-M's most notable features is its short engine burn time following take-off, intended to minimize satellite detection of launches and thereby complicate both early warning and interception by missile defense systems during boost phase. The missile also has a relatively flat ballistic trajectory, complicating defense acquisition and interception.

Just because some ballistic trajectories may be flatter than others does not mean a gravity turn is not required. If there is a rocket motor powerful enough and with sufficient duration, then we can have a missile that will cruise at 1000 m altitude, for example, then there is no need for a gravity turn. So no, I do not owe you any explanation about any 'flat ballistic trajectory' because there really is no such animal.

Since when did I say that gravity turn is not required for ballistic missile?

No...What you gave is a hilarious misunderstanding of the relevant paragraph. From what basis can you claim that 55 deg is 'ideal' and others are 'non ideal'? Show us a source that says so.

The example already showed you that for the same intended target, there are at least two trajectories for the same missile. For the one with flatter trajectory, the pitch maneuver can and should happen slowly maybe even over the whole boost phase to reduce the lateral stress of the missile.

But here it is again for all to see...

Gravity turn - Wikipedia, the free encyclopedia

A gravity turn is not the only mean to ascent to orbit. If the American Space Shuttle could take off like an aircraft and is equiped with powerful enough engines, it could achieve orbit gradually through aerodynamic exploitation until the atmosphere is too thin. And if you think that is not possible: SR-71. That aircraft's maximum altitude is still secret but all knows that figure is high enough that a rocket motor would propel the aircraft into orbit. Another mean is to simply power straight up. But the reason why the Space Shuttle, satellite launchers and ICBM uses a gravity turn is to reduce gravity loss...

RSE 24

At moment of lift-off, immediately there is gravity loss. An inclined trajectory coupled with a high power-to-weight ratio and we have an 'ideal' method to achieve orbit. The duration of our stay in orbit depends on our mission, of course. That mean this phrase 'ideal gravity turn ascent' is translated as: the gravity turn is the ideal method of orbital ascent. Not that there are animals like 'ideal gravity turn' and 'non ideal gravity turn' as you tried to pass off here. So if you are confident that there is such a thing as 'non ideal gravity turn' then show us a source that says so. We are still waiting and am willing to be corrected by a credible external and non-bias source.

Gravity turn happens once the missile is not in a vertical ascend or descend, and it happens throughout missiles flightpath. The other steering whether it is by thrust or by control surface can also add to the turn. An ideal gravity turn is a turn where gravity turn is the only turning of missile experience, or maybe I should say pure gravity turn then so you can understand. In your wiki source it said ideal gravity turn ascending phase, then tell me what is a non-ideal gravity turn ascending phase then.

I never said it does not. But I have no problems schooling you...

The longer the VERTICAL part of the boost phase, the less physical stress the vehicle has to endure because the vehicle would leave the lower/denser atmosphere sooner.

Wrong, it is to reduce the aerodynamic drag/friction force that causes a negative acceleration force to the missile. Also if the missile enter the thinner atmosphere too early during its boost phase, it will experience more longitude stress because thinner atmosphere will result in less aerodynamic drag, thus with the reduction of mass due to fuel consumption, this gives the missile over all higher net acceleration.

Maximum payload acceleration — Another limitation related to engine thrust is the maximum acceleration that can be safely sustained by the crew and/or the payload. Near main engine cut off (MECO) when the launch vehicle has consumed most of its fuel it will be much lighter than it was at launch. If the engines are still producing the same amount of thrust the acceleration will grow as a result of the decreasing vehicle mass. If this acceleration is not kept in check by throttling back the engines injury to the crew or damage to the payload could occur. This forces the vehicle to spend more time gaining horizontal velocity, increasing gravity drag.

In order to minimize gravity drag the vehicle should begin gaining horizontal speed as soon as possible On an airless body such as the Moon this presents no problem, however on a planet with a dense atmosphere this is not possible. A trade off exists between flying higher before starting downrange acceleration, thus increasing gravity drag losses; or starting downrange acceleration earlier, reducing gravity drag but increasing the aerodynamic drag experienced during launch.

The downside to a PURELY vertical ascent is the aformentioned 'gravity loss', which cost us in terms of large fuel load which naturally contribute to overall vehicle size, also because of 'gravity loss', such a purely vertical climb DOES NOT contribute to orbital velocity.

Wrong again here, if T(thrust) > mg, it still can produce an acceleration (T/m-g) that contribute to its speed (T/m-g)*t in a vertical climb. How do you suppose rocket or missile gain its speed in the vertical climb in the first place. Its lift only become zero(vertical velocity become constant) when T = mg + 1/2 p(density of the atmosphere )* v(velocity of the missile)^2 *C(drag coefficient of the this particuluar missile) *A(reference area of the missile).

Improved guidance and control technology allow us to create that 'gravity turn' where lateral stresses on the off-vertical vehicle are minimized.

As if V-2 does not use gravity turn. Any projectile motion has to use a gravity turn unless its motion is pure vertical.

As time and experience goes by, we have a decreasing vertical portion of the boost phase and a sooner entry into the gravity turn, which IS a part of the boost phase. In multistage vehicles, the shedding of empty fuel tanks, aka continual mass reduction, and a reduction of gravity loss thanks to an inclined vehicle, allow the vehicle to gain orbital velocity more rapidly.

You never took physics before, didn't you. The jettison of empty fuel tanks has nothing to do with reducing the gravity loss because the downrange acceleration caused by gravity drag is the same regardless of the mass. The downrange acceleration g = 9.81 m/s^2 = 32.2 ft/s^2. The reason for jettison is to increase the acceleration by the thrust T, since T = ma. So the less the m, the bigger the a. That is why most of jettison happens before the ignition of the second or third stage.

So does the duration of the boost phase affect the trajectory of a missile? Yes, what else but because it is a segment of the overall travel? There is a side benefit in that the gravity turn is best indicator of a trajectory profile and the sooner the entry into the maneuver the better the predictive process of that trajectory.

Since you have proved your knowledge of simple physics above, let me give you a lesson.

We can agree that after the boost phase, the missile follows a classic projectile motion assuming there is no air friction. The exit velocity is v and angle θ.
It motion in relation to time(t)
in the x axis is x = v*(cos&#952*t.
in the y axis is y= v*(sin&#952*t - 1/2*g*t^2

In the boost phase, the duration will cause the difference in the missile exit speed because T = ma, V(x,y) = a*t1. b(angle) here is also a variable caused by gravity turn or steering, now let assume there is now steering involved, and the speed of missile exit the vertical flight is v0 with a tilt angle b0.

In the x axis, Vx = a*cos(b)*t1
In the y axis, Vy = v0 + [a*sin(b)-g]t1
the D(x,y) is the integral of v(x,y) with respect of t1.
tan(b) = Dy/Dx = sin(b)/cos(b)

from the formula, we the see the duration of the boost phase will not only cause the difference in missile's exit speed, but also the angle which it exit. So it is not only a segment of overall trajectory, but also the determining fact of missile's trajectory. There are hugh difference in its trajectorybetween 60 sec and 120 sec of its duration

And do not tell me that you never 'mentioned' about the gravity turn. You never 'mentioned' it because you cannot mention it because never knew about it as part of the boost phase in the first place as evident by that 'non ideal' version you are still trying to pass.

The reason I don't mention it is because all the effects gravity have on the missile is already included in my equation that predict the flightpath, so there is no need to single it out especially when g is constant.

First...I never claimed that the gravity turn is the only factor in estimating flight path.

However you implied that once the missile is not in the vertical flight, then its trajectory is pretty much determined, which I just proved you wrong.

The moment the vehicle is in a gravity turn, it cannot maneuver for course correction, or it should not maneuver if it is to retain structural integrity. The boost phase include this gravity turn maneuver. Once the missile is detected and tracked while in this turn, its flight path can be estimated straight up to orbit.

Second...Here are the relevant paragraphs...

This is not about trajectory optimization BY a missile but about weapons development, specifically missile development in creating more accurate missiles via detailed flight records such as AoA degrees, thrust, duration...etc...etc...Notice the plurals: 'flight profiles', 'flight vehicles' and 'control policies'. This means for any specific flight, the missile is not making several 'trajectory optimization' processes but that the missile's programed flight path(s) are ALREADY optimized based upon historical data from past missiles. In other words, your wiki source is not about A particular missile in flight but about THE missile as a weapons development over time and experience.

Just like the gravity turn gaff, you keep making these interpretation mistakes because you have no relevant experience and are trying to salvage some face for a losing argument.


No...It is not dubious. Any maneuvers outside of what is programmed affect range. Fuel is finite. I was not speaking technically because there are no technical limitations on creating maneuverable ICBMs if we wanted to. I was speaking of 'policy' and it is that we should maneuver no more than what is necessary.

You better read the third paragraph again.

And you thought wrong. There are two moving bodies here: the missile and the interceptor. Since we are using fixed coordinate system to create collision intercept, real time updates of BOTH bodies are necessary.

NWS JetStream - Exactly how does radar work?

Gaps between echoes compresses as the two bodies nears each other. Conversely they widens if the bodies are moving away from each other. Phase (shift) Derived Range Acceleration (PDRA) estimations are created for trajectory predictions and, believe it or not, RECONSTRUCTION. Target state contain (x y z) position, velocity and acceleration.

Previously, I showed how fixed coordinate system sees a target state:

X = [x y z Vx Vy Vz]T or [x y z x-dot y-dot z-dot]T
Where X = state vector, T = time, V = velocity and the standard xyz axes.

For target range acceleration estimate, basic target state is seen as:

X = [R(RRC) V(RRC) A(VTC)]
Where R(RRC) = [x y z]T is the target position the east-north-up (ENU) radar reference coordinates.

Coordinate Systems, Basis Sets, Reference Frames, Axes, etc.

Where V(RRC) = [x-dot y-dot z-dot]T is the target's velocity inside RRC.
Finally, A(VTC) = [a(v) a(t) a(c)]T is target's acceleration in the right-handed velocity-turn-climb (VTC) coordinates system.

The equation of still applies in the with vectors
1.D(t) = 1/2 * a * t^2, if 0 <= t <= t1
2.D(t) = 1/2 * a * t1^2 + [v(t-t1) - 1/2*d*(t-t1)^2]
= 1/2 * a * t1^2 + [a*t1(t-t1) -1/2 * d * (t-t1)^2] if t1 < t <= (a/d + 1) * t1

1.D(x,y,z)(t) = 1/2 *a (x,y,z) * t^2 if 0<= t <= t1
2.D(x,y,z)(t) = 1/2 * a(x,y,z) * t^2 + [a(x,y,z)*(t-t1) - 1/2 * d(x,y,z)*(t-t1)^2] if t1 < t <= (a(x,y,z)/d(x,y,Z) + 1) * t1

In case you don't know, distance, velocity and acceleration are all vectors.

The radar echo phase shift information from all three elements -- R V A -- can be used for both prediction and reconstruction and if the radar is sophisticated enough it can predict as well as reconstruct at the same time regardless of its position in respect to the target.

I still do not want to engage in a mathematical one-upmanship game here but what I do want to show the readers is that your simplistic assertions in the previous pages regarding flight path prediction bears no resemblance to what has been in place for decades. If we go back to this illustration...

...Collision navigation by the missile does not care where the bogey came from. Pure pursuit navigation (PPuN) require a true tail chase situation and air-air interceptions seldom allow us that luxury. That leave us the much more complex proportional navigation (PN) based hybrid intercept laws that require the missile to constantly (re)calculate the closing distance between bogey and itself. Even if the missile is behind the target, the laws are still PN based. In the PN based scheme, the missile does not care where the bogey came from or its thrust. The missile cares only for the closing velocity between the bogey and itself at any point in space/time. No different for ICBM collision intercept. The incorporation of phase shift data rendered origination data quite -- pointless.

That is why we need to track missile with radars, there are so many variables that do change in its flight path that is not predictable. The thing you can know about the missile's flight path is after it enters a gravity turn is the direction where the missile is going to fly, not where its target is going to be.

So yes...Really.


If the ICBM does not need to be more structurally robust than a satellite launcher, then why did you criticized me for legitimately comparing the two? Remember this...




You argued that a missile is more physically robust than a 'commercial rocket'...

By reducing its diameter because of using solid fuel, the missile is already more structurally robust than commercial rockets without any engineering modification to enhance construction material and manufacturing.

...But you show no source to support your claim, and that a missile should be so because it IS more maneuverable, but again you show no source to support your claim. The math you presented is irrelevant and it serves nothing more than to distract from the fact that you consistently failed to support your claims with nonbias sources.

I already give you a scientific proof from a material mechanics' prospective. Now you are calling science biased?

So according to you...

Q - Why should a satellite launcher be less physically robust than a missile?
A - Because the missile is more maneuverable.
Q - Why should the missile be more maneuverable?
A - Because the missile is not a satellite launcher.

What the hell kind of argument is that? In road racing, your opponents are threats so it behooves you to have a highly responsive and maneuverable car. In air combat, someone is trying to kill you so it behooves you to have a highly maneuverable aircraft. How is a satellite launcher a threat to a missile?

Getting desperate, aren't you. Now you are picking on words.
Missile is physically robust comparing to rockets because it has smaller diameter by using solid fuel as its propellant. Thus it can be more maneuverable than a commercial rocket.

Here is the real logical thought process...

An ICBM is usually launched from home territory, so what is the need to make the thing maneuverable? None other than one slight pitch over. So why should it be any more physically robust than a satellite launcher? Not at all. So what if the fuel used allows for a smaller diameter tank? If anything, I would keep the same dimensions but pack it with more fuel for higher speed and for greater range. So if I build a vehicle that has just enough structural strength to carry a payload to a location, then what I said is very correct -- that 'it should not maneuver if it is to retain structural integrity'.

Following your logic, then a solid fuel missile should be built with material that has less strength.
Let me give you another physics lesson about why missile is more maneuverable here. Since missile's trajectory is much lower than commercial rocket, which means it exit boost phase with a much lower angle. Assuming the duration of boost phase is the same, then the missile will experience more angular acceleration during that period. From F = ma, the missile will experience more torque than commercial rocket.

This is why you guys and your desperation to exaggerate anything Chinese are soooo entertaining.

Your pretentious attitude without any really scientific knowledge besides hearsays is soooo annoying.

 

[TopCat]

^^^^^^^^^^^^^^^
Guys you are making the whole thread a boring by copy and paste Physics' theory which has no relevance to the thread here... We know the missile going to hit the target and let the designer do all the maths....

 

[lcloo]

I feel so bore by these debates on academic mathematics equations, I fall asleep.

They are simply beyond us laymen who only knows simple newton's laws.

 

[gambit]

First of all, I don't think that I ever argued about DF-21 since it has not been tested yet.

May be so. But you are defending its purported claims indirectly. So far I have never said those claims are technically impossible. I only challenge their efficacy and no Chinese member here like that. Here is the definition of 'efficacy'...

Efficacy - Wikipedia, the free encyclopedia

Efficacy is the capacity to produce an effect.

All this running around by you is simply because you do not like anything Chinese challenged, even on technically legitimate grounds. But it is fun to watch, though.

That is the example I gave to you.

Yes...And your demand of me to explain a 'flat ballistic trajectory' was meaningless in that example.

Since when did I say that gravity turn is not required for ballistic missile?

You never knew about it in the first place.

The example already showed you that for the same intended target, there are at least two trajectories for the same missile. For the one with flatter trajectory, the pitch maneuver can and should happen slowly maybe even over the whole boost phase to reduce the lateral stress of the missile.

Fine...Now which one is a 'non-ideal gravity turn' and do show the readers a credible nonbias source on it. We want to see the source say that given a pitch over angle after X altitude and given vehicle velocity there are X available trajectories and this 'one' here is the 'ideal gravity turn and the rest are not and here are the reasons why. Remember, the 'non ideal gravity turn' are YOUR words by implications.

Gravity turn happens once the missile is not in a vertical ascend or descend, and it happens throughout missiles flightpath. The other steering whether it is by thrust or by control surface can also add to the turn. An ideal gravity turn is a turn where gravity turn is the only turning of missile experience, or maybe I should say pure gravity turn then so you can understand. In your wiki source it said ideal gravity turn ascending phase, then tell me what is a non-ideal gravity turn ascending phase then.

Now that is dishonest. The creature 'non-ideal gravity turn' is YOURS, not mine...

YOf course its AoA is zero if it is an ideal gravity turn. I doubt you even understand what AoA means in this matter. However even with an ideal gravity turn, the vector velocity of missile trajectory is still determined by the duration of the boost phrase when rocket is firing.

...even if it is an ideal gravity turn.

Besides not all the missile has to follow an ideal gravity turn for its trajectory during its boost phase, that is why it is called ideal gravity turn.

Only if it is an ideal gravity turn.

You misunderstood the gravity turn wiki source. If there is such a thing as an 'ideal gravity turn' then its opposite also must exist. I have twice now explained how you came to that false conclusion. So for you to ask me about what is a 'non-ideal gravity turn' is outright dishonest. That is expected of someone who is exposed to be a pretender. But I have no problem exposing you again for the readers to see...

Gravity turn - Wikipedia, the free encyclopedia

This small steering maneuver is the only time during an ideal gravity turn ascent that thrust must be used for purposes of steering. This pitch over maneuver serves two purposes. First, it turns the rocket slightly so that its flight path is no longer vertical, and second, it places the rocket on the correct heading for its ascent to orbit.

There are other ways to ascent to orbit. Aerodynamics exploitation like the SR-71 where that aircraft's altitude is very near space. Simple thrust if we have sufficient power. But because we are still working on the previous two methods, a 'gravity turn' give us the best solution. That mean the 'gravity turn' is 'ideal' for what we want to do, not that there is such an animal called 'ideal gravity turn'. Asking me to explain something that I never even implied and that YOU misunderstood is dishonest. As we Americans say -- Grab your balls and Man Up.

===

The longer the VERTICAL part of the boost phase, the less physical stress the vehicle has to endure because the vehicle would leave the lower/denser atmosphere sooner.

Wrong,...

Wrong...??? The shortest path to orbit is straight up. Unfortunately, we do not have that kind of power yet so we created the gravity turn.

Your response looked like it was hobbled together from disparate sources in trying to make anything stick...

...it is to reduce the aerodynamic drag/friction force that causes a negative acceleration force to the missile.

Say what? Going vertical reduces aerodynamic drag/friction? You sure about that? Because the sources I have says something else...

Gravity turn - Wikipedia, the free encyclopedia

The gravity turn is commonly utilized with launch vehicles such as a rocket or the Space Shuttle which launch vertically. The rocket begins by flying straight up, gaining both vertical speed and altitude. During this portion of the launch gravity acts directly against the thrust of the rocket, lowering its vertical acceleration. Losses associated with this slowing are known as gravity drag, and can be minimized by executing the next phase of the launch, the pitch over maneuver, as soon as possible. The pitch over should also be carried out while the vertical velocity is small to avoid large aerodynamic loads on the vehicle during the maneuver.

In atmosphere, there will be aerodynamic drag, even when going vertical. But the greatest retardant force, for a vertically launched vehicle, is gravity drag, not aerodynamics. If I shoot a bullet straight up, the distance between ground and its maximum altitude will be shorter than if I had shoot the bullet horizontal and until its fall to the ground. Aerodynamic drag exist for both. If you mean something else -- rephrase.

Also if the missile enter the thinner atmosphere too early during its boost phase,...

Too early? Based upon what standards? Show me a credible source that say there is a time that a vehicle must not violate upon entering a certain atmospheric layer.

...it will experience more longitude stress because thinner atmosphere will result in less aerodynamic drag, thus with the reduction of mass due to fuel consumption, this gives the missile higher acceleration.

News for you, kid. Rockets are designed to withstand longitudinal stresses better than those created by lateral acceleration. As far as structural stresses goes, denser atmosphere increases stresses everywhere, even if the vehicle is standing straight up, and especially when the vehicle is in flight where the possibility of high winds could overstress the body, induce vibrations and course deviations, and all three could be too great for the guidance system to compensate. Whenever the USAF abort a satellite launch due to high wind, sometimes it was because of altitude wind, sometimes it was because of high wind at ground level. The rocket would be swaying too much for comfort just standing there and upon lift off when there is nothing to secure the rocket at its most vulnerable, a gust could send the rocket unpredictably sideways.

When you said this: '...experience more longitude stress because thinner atmosphere will result in less aerodynamic drag...' May be what you were looking for is -- maximum aerodynamic pressure (Qmax). At high atmospheric density, if velocity is low, then aerodynamic drag is low. Initial velocity at lift off will be low. As we gain vertical velocity, which also increase our altitude, aerodynamic pressure will increases.

Here is Qmax and examples of different Qmax altitudes for different vehicles...

Max Q - Wikipedia, the free encyclopedia

...max Q is the point of maximum dynamic pressure, the point at which aerodynamic stress on a spacecraft in atmospheric flight is maximized.

During a normal Space Shuttle launch, for example, max Q is at an altitude of approximately 11 km (35,000 ft).[1] During a typical Apollo mission, max Q occurred between 13 and 14 km of altitude (43,000–46,000 ft).

The fact that you say 'longitude stress' tells me you do not know about Qmax, its relationship to altitude and vehicle design.

So what I said is not 'wrong'. The longer we stay in vertical flight, the sooner we leave the lower/denser atmosphere and the less dense atmosphere will decrease Qmax. Just in case you have a reading comprehension problem, the word 'longer' here does not mean distance but time duration. If it is possible that we can vertically launch all the way to orbit -- we would have. But we cannot, so we must devise other ways and we are still working on that.

 

[gambit]

The downside to a PURELY vertical ascent is the aformentioned 'gravity loss', which cost us in terms of large fuel load which naturally contribute to overall vehicle size, also because of 'gravity loss', such a purely vertical climb DOES NOT contribute to orbital velocity.

Wrong again here, if T(thrust) > mg, it still can produce an acceleration (T/m-g) that contribute to its speed (T/m-g)*t in a vertical climb. How do you suppose rocket or missile gain its speed in the vertical climb in the first place. Its lift only become zero(vertical velocity become constant) when T = mg + 1/2 p * v^2 *C *A(aerodynamic drag)

How? There is no disputing that a vertical launch is possible, but you are wrong in saying that aerodynamic drag is the cause of lift loss. No surprise here but you are confused by the context of the word 'lift'. For a vehicle that relies mostly on thrust, the word 'lift' has nothing to do with the phrase 'aerodynamic lift'. From the moment of lift off in a vertical launch, it is gravity loss that eventually will give zero lift and the context of that word mean rise above the ground via direct thrust against gravity.

You say: if T(thrust) > mg. Here is what we know IN A GRAVITY FIELD AND AN ATMOSPHERE: Increasing thrust reduces gravity loss but increases aerodynamic loss. Unfortunately, the engines, liquid or solid, are at full power at lift off. That mean we have finite thrust. Velocity gain cannot be infinite with finite thrust in a gravity field and an atmosphere. Or is that 'Chinese physics' talking again? So IF we can maintain increasing thrust to overcome all losses in a gravity field and an atmosphere, THEN WHAT IS THE NEED FOR A GRAVITY TURN ANYWAY ? We can just go vertical all the way to orbit.

===

As if V-2 does not use gravity turn. Any projectile has to use a gravity turn unless its motion is pure vertical.

Now this is hilarious. We know that gravity is a constant vector. But a 'gravity turn' is a voluntary maneuver and is designed to achieve orbit. The V-2 rocket was nothing more than a powered bullet. Both are affected by gravity into a parabollic trajectory but that hardly qualify that trajectory to be FROM a gravity turn.

You never took physics before, don't you. The jettison of empty fuel tanks has nothing to do with reducing the gravity loss because the downrange acceleration caused by gravity drag is the same regardless of the mass. The downrange acceleration g = 9.81 m/s^2 = 32.2 ft/s^2. The reason for jettison is to increase the acceleration by the thrust T, since T = ma. So the less the m, the bigger the a. That is why most of jettison happens before the ignition of the second or third stage.

And you have a serious reading comprehension problem. Here is what I actually said...

In multistage vehicles, the shedding of empty fuel tanks, aka continual mass reduction, and a reduction of gravity loss thanks to an inclined vehicle, allow the vehicle to gain orbital velocity more rapidly.

Got that? See how what I actually said, as highlighted, is different than what you claimed I said? Let us take a look at the wiki source again...

Gravity turn - Wikipedia, the free encyclopedia

During this portion of the launch gravity acts directly against the thrust of the rocket, lowering its vertical acceleration. Losses associated with this slowing are known as gravity drag, and can be minimized by executing the next phase of the launch, the pitch over maneuver, as soon as possible.

I will summarize the highlighted: 'gravity drag can be minimized by executing the pitch over maneuver.' The 'pitch over maneuver' produce an 'inclined vehicle', correct? And what I actually said was: reduction of gravity loss thanks to an inclined vehicle. I did not say shedding empty tanks reduces gravity loss. You are so eager to throw in distracting textbook numbers that you ended up making a fool out of yourself.

Since you have prove your knowledge of simple physics here, let me give you a lesson.

No thanks. Not interested in 'Chinese physics'.

 

[gambit]

The reason I don't mention it is because all the effects gravity have on the missile is already included in my equation that predict the flightpath, so there is no need to single it out especially when g is constant.

You did not mention it because you did not know about it. This is a tissue thin cover up. If you did know about it, you would not have said 'ideal gravity turn' in the first place and that you still do. No such animal -- 'ideal gravity turn'. Make me suspicious if you even understand how the concept works.

However you implied that once the missile is not in the vertical flight, then its trajectory is pretty much determined, which I just proved you wrong.

And reality proved you wrong. Ever wonder how professional athletes are that good? Try watching them closely. If the sports broadcast is showing the event just right, often you will see the ball receiver begin to move even BEFORE the ball is at apogee. American football are full of this example when one side kick the ball to the other end. The designated receiver will already begin to move to where he guess the ball may fall. You can see some of that in baseball as well when the batter hit to the outfielders. That mean barring any inflight interference, like a wind gust, the ball's trajectory is quite determined. Professional athletes are that good because they are better than us at subconsciously calculating and recalculating the ball's trajectory throughout. Less capable people have to wait later in time to make positional adjustments but by then it is often too late.

You better read the third paragraph again.

This feeble response mean you know you got caught misunderstanding your own source. I have no problem posting all the relevant paragraphs to show everyone that third paragraph in proper context.

Trajectory optimization - Wikipedia, the free encyclopedia

Trajectory optimization is the process of designing a trajectory that minimizes or maximizes some measure of performance within prescribed constraint boundaries. While not exactly the same, the goal of solving a trajectory optimization problem is essentially the same as solving an optimal control problem.

The selection of flight profiles that yield the greatest performance plays a substantial role in the preliminary design of flight vehicles, since the use of ad-hoc profile or control policies to evaluate competing configurations may inappropriately penalize the performance of one configuration over another. Thus, to guarantee the selection of the best vehicle design, it is important to optimize the profile and control policy for each configuration early in the design process.

Consider this example. For tactical missiles, the flight profiles are determined by the thrust and load factor (lift) histories. These histories can be controlled by a number of means including such techniques as using an angle of attack command history or an altitude/downrange schedule that the missile must follow. Each combination of missile design factors, desired missile performance, and system constraints results in a new set of optimal control parameters.

You brought this source on to show that A missile performs its own trajectory optimization. That is not true. This source explain how the process works for development purposes and that it is from performance histories of past designs.

 

[gambit]

That is why we need to track missile with radars, there are so many variables that do change in its flight path that is not predictable. The thing you can know about the missile's flight path is after it enters a gravity turn is the direction where the missile is going to fly, not where its target is going to be.

If I launch an ICBM, why would I want to allow variables to change its flight path? Are we talking about externally induced course deviations such as wind that are automatically corrected? If so, then it is still reasonable that even though we know there are variables, just like the ball receiver, we can still focus our attention on the general area that the parabollic arc seems to end.

I already give you a scientific proof from a material mechanics' prospective. Now you are calling science biased?

Nope...I am calling for a credible nonbias source. Got any?

Getting desperate, aren't you. Now you are picking on words.
Missile is physically robust comparing to rockets because it has smaller diameter by using solid fuel as its propellant. Thus it can be more maneuverable than a commercial rocket.

Me desperate? Hardly. And the picking here is picking apart your argument: 'A missile is more robust than a rocket because a missile uses solid fuel.' This is an interesting argument because it presume that ALL rockets use liquid fuel and ALL missiles use solid fuel. Is there a credible nonbias source that say so? That is a repeated gross assumption and sounds pretty desperate to me. Here is a satellite launcher from India's DRDO that is solid fuel...

Polar Satellite Launch Vehicle - Wikipedia, the free encyclopedia

The PSLV has four stages using solid and liquid propulsion systems alternately. The first stage is one of the largest solid-fuel rocket boosters in the world and carries 138 tonnes of Hydroxyl-terminated polybutadiene (HTPB) bound propellant with a diameter of 2.8 m.

Yes it does use both types. But the fact that it does use both types debunked you.

Here is another satellite launcher from Japan that uses solid fuel...

M-V

M-V-1 Solid propellant rocket stage. Loaded/empty mass 83,560/12,070 kg. Thrust 3,780.35 kN. Vacuum specific impulse 276 seconds.

So when you say this...

Commercial rockets are not designed to endure high stress manuvers that modern missiles have to withstand.

You effectively told us that there are distinguishing features between civilian and military launch vehicles and that fuel type is a major part of that distinction with 'commercial' being liquid and 'military' be solid. I presented sources that say you are wrong. Now you need to present at least a couple sources to prove your claim.

Following your logic, then a solid fuel missile should be built with material that has less strength.
Let me give you another physics lesson about why missile is more maneuverable here. Since missile's trajectory is much lower than commercial rocket, which means it exit boost phase with a much lower angle. Assuming the duration of boost phase is the same, then the missile will experience more angular acceleration during that period. From F = ma, the missile will experience more force than commercial rocket.

Absolutely. All the available sources will show that if there is anything 'maneuverable' on an ICBM, it is on the warhead(s), not the vehicle itself. You can declare that the ICBM 'can' be more maneuverable than the satellite launcher and no one can argue against it. But in the real world, not the one you live in, we do not want to build the thing any more robust than necessary and if that mean build it to the same specs as the satellite launcher to save weight to carry more payload, so be it. So you can throw up all the textbook numbers you want and all it shows is someone who has no real world experience in the subject he is talking about.

Your pretentious attitude without any really scientific knowledge besides hearsays is soooo annoying.

My 'hearsays' are sources that so far you could not challenge while you continue to produce nothing to support your arguments. My personal experience in the military and later in the civilian life in field weapons testing and development give me the necessary background to provide people here with sources, along with my comments, that they can read and judge for themselves. Some may be satisfied with what they see. Some may have researched further. Some may take it to their military friends. So far, no one came back to a discussion and accuse me of giving out misleading information or lied to them.

 

[chauism]

Wrong...??? The shortest path to orbit is straight up. Unfortunately, we do not have that kind of power yet so we created the gravity turn.

How do you reach a orbit in a straight up fashion without turning since a orbit in space flight is the gravitationally curved path of one object around a point or another body.

Rather, the objective is achieve the position/velocity combination for the desired orbit. For instance, the way to maximize acceleration is to thrust straight downward; however, thrusting downward is clearly not a viable course of action for a rocket intending to reach orbit.

Orbit - Wikipedia, the free encyclopedia

Your response looked like it was hobbled together from disparate sources in trying to make anything stick...

My response is derived from simple physics principles which you have difficulty to understand.

Say what? Going vertical reduces aerodynamic drag/friction? You sure about that? Because the sources I have says something else...

Going vertically to reach upper atmosphere will
1. reduce the vertical acceleration due to gravity drag which reduce the increase in the velocity.
2. reach the thinner atmosphere fastest, thus reduce the density p.

D(aerodynamic drag) = 1/2p*v^2*C*A. Go figure yourself.

Gravity turn - Wikipedia, the free encyclopedia

In atmosphere, there will be aerodynamic drag, even when going vertical. But the greatest retardant force, for a vertically launched vehicle, is gravity drag, not aerodynamics. If I shoot a bullet straight up, the distance between ground and its maximum altitude will be shorter than if I had shoot the bullet horizontal and until its fall to the ground. Aerodynamic drag exist for both. If you mean something else -- rephrase.

That just showed you how much you know about aerodynamic drag.
Drag(D) = 1/2p*v^2*C*A, the drag force is proportional to the velocity to its second power, which means the increase of aerodynamic drag is much fast than the increase of speed. In part of the vertically flight, the missile or the rocket's velocity remains constant is because that the velocity has reached the point where aerodynamic drag is so big that it = T-mg and neglect any lift that the thrust produces.

Don't make me laugh about the bullet example. The distance between the bullet and the ground(vertical distance) is the greatest if its is shot straight up. However the overall distance D(x,y) traveled by the bullet is the greatest if it is shot at 45 degree angle, not horizontally.

Assuming without air drag. v is the initial velocity, and &#952; is the angle it shoots.
Its motion in the air with respect to time(t) are
in the x axis is x = v*(cos&#952*t.
in the y axis is y= v*(sin&#952*t - 1/2*g*t^2

xmax = v^2*sin(2&#952/g, xmax is the greatest when sin(2&#952 = 1, &#952; = 45
ymax = v^2*[sin(&#952]^2/2g, ymax is the greatest when sin(&#952 = 1, &#952; = 90.

For the rest of the angles, the projectile can reach its intended distance with two different angle with the same initial velocity, which also applies to missile here after it finishes boost phase.

That just shows that how much you know about physics, and this is the basic of basic of basic Newtonian physics here.
Effect of Gravity on an Artillery Projectile - Succeed in Understanding Physics: School for Champions

On a planet with an atmosphere, the objective is further complicated by the need to achieve the necessary altitude to escape the atmosphere, and to minimize the losses due to atmospheric drag during the launch itself.

Too early? Based upon what standards? Show me a credible source that say there is a time that a vehicle must not violate upon entering a certain atmospheric layer.

That is the problem with commercial rocket with payloads especially the payload is human. For missile it does not matter. If the rocket is in the thinner atmosphere with a lot of overall thrust left, it will increase the acceleration of the rocket, so the the overall g force experienced by the payload will increase.

News for you, kid. Rockets are designed to withstand longitudinal stresses better than those created by lateral acceleration. As far as structural stresses goes, denser atmosphere increases stresses everywhere, even if the vehicle is standing straight up, and especially when the vehicle is in flight where the possibility of high winds could overstress the body, induce vibrations and course deviations, and all three could be too great for the guidance system to compensate. Whenever the USAF abort a satellite launch due to high wind, sometimes it was because of altitude wind, sometimes it was because of high wind at ground level. The rocket would be swaying too much for comfort just standing there and upon lift off when there is nothing to secure the rocket at its most vulnerable, a gust could send the rocket unpredictably sideways.

When you said this: '...experience more longitude stress because thinner atmosphere will result in less aerodynamic drag...' May be what you were looking for is -- maximum aerodynamic pressure (Qmax). At high atmospheric density, if velocity is low, then aerodynamic drag is low. Initial velocity at lift off will be low. As we gain vertical velocity, which also increase our altitude, aerodynamic pressure will increases.

Here is Qmax and examples of different Qmax altitudes for different vehicles...

Max Q - Wikipedia, the free encyclopedia

The fact that you say 'longitude stress' tells me you do not know about Qmax, its relationship to altitude and vehicle design.

That is interesting, now you want to talk about the aerodynamic pressure while saying that aerodynamic drag is not important.

The formula for aerodynamic pressure is 1/2p*v^2, does it looks familiar to you? Right, it is exactly the pressure caused by the aerodynamic drag. Q = D/(C*A). While density here is important to the aerodynamic pressure here, the effects caused by velocity is much greater. Aerodynamic pressures acts on all its surface, the stress the missile body experience is everywhere, not just longitudinally or laterally. So, NO, it is not what I am looking for, you can find what I am looking for in the formulas I have provided you with before.

So what I said is not 'wrong'. The longer we stay in vertical flight, the sooner we leave the lower/denser atmosphere and the less dense atmosphere will decrease Qmax. Just in case you have a reading comprehension problem, the word 'longer' here does not mean distance but time duration. If it is possible that we can vertically launch all the way to orbit -- we would have. But we cannot, so we must devise other ways and we are still working on that.

You are wrong to the second power according to 1/2p*v^2.

If vehicle's velocity is still increasing after it enters a low density atmosphere, which means the thrust is still available, then the aerodynamic pressure will increase faster with the increasing speed than the reduction in atmospheric density can decrease the dynamic pressure until it reaches a new equilibrium point, which will result in a higher Max Q than the one in the lower atmosphere if the velocity is left unchecked. However since we do have a limited thrust available, the velocity will not reach that level in the upper atmosphere.

How? There is no disputing that a vertical launch is possible, but you are wrong in saying that aerodynamic drag is the cause of lift loss. No surprise here but you are confused by the context of the word 'lift'. For a vehicle that relies mostly on thrust, the word 'lift' has nothing to do with the phrase 'aerodynamic lift'. From the moment of lift off in a vertical launch, it is gravity loss that eventually will give zero lift and the context of that word mean rise above the ground via direct thrust against gravity.

Confused??? Let's see who, since I already gave the formula of lift in the vertical phase ages ago, according to your logic that since you did not mention it before first, you have no knowledge of what it is then. L= T-mg-1/2p*v^2*C*A.
Gravity drag is constant here, the only thing that contributes to the decrease of lift here is the increase of the velocity where it increases the aerodynamic drag to square of its value.

You say: if T(thrust) > mg. Here is what we know IN A GRAVITY FIELD AND AN ATMOSPHERE: Increasing thrust reduces gravity loss but increases aerodynamic loss. Unfortunately, the engines, liquid or solid, are at full power at lift off. That mean we have finite thrust. Velocity gain cannot be infinite with finite thrust in a gravity field and an atmosphere. Or is that 'Chinese physics' talking again? So IF we can maintain increasing thrust to overcome all losses in a gravity field and an atmosphere, THEN WHAT IS THE NEED FOR A GRAVITY TURN ANYWAY ? We can just go vertical all the way to orbit.

T is a constant. It does not increase or decrease in our assumption here. The duration of T however is limited. If let's say we have a nuclear powered rocket where its thrust is unlimited, and yes it can reach from planet A to planet B in a straight up fashion. However if it wants to enter an orbital flight path or an orbital descend around any planet, it has to use a gravity turn or any form of steering. That is how much you know about orbit. To enter an orbital flight, one must enter tangentially to this particular orbit at its orbital speed. Orbital speed v^2 = m2^2*G/[r*(m1+m2)].

Now this is hilarious. We know that gravity is a constant vector. But a 'gravity turn' is a voluntary maneuver and is designed to achieve orbit. The V-2 rocket was nothing more than a powered bullet. Both are affected by gravity into a parabollic trajectory but that hardly qualify that trajectory to be FROM a gravity turn.

That is hilarious indeed. You are saying V-2 is not a ballistic missile, and does not have a sub-orbital flight path now? Show me how V2 does not have to use gravity turn in its flight path.

The rocket was the world's first long-range combat-ballistic missile and first human artifact to achieve sub-orbital spaceflight.

http://en.wikipedia.org/wiki/V-2

Your knowledge of V-2 and its physics behind it is no more than what Hitler knew about it.

At the time, Adolf Hitler was not particularly impressed by the V-2; he pointed out that it was merely an artillery shell with a longer range and much higher cost.

No thanks. Not interested in 'Chinese physics'.

It is physics in any text books, if you ever read one.

You did not mention it because you did not know about it. This is a tissue thin cover up. If you did know about it, you would not have said 'ideal gravity turn' in the first place and that you still do. No such animal -- 'ideal gravity turn'. Make me suspicious if you even understand how the concept works.

In the equations I have provided, g gravity is always included. The concept of "gravity turn" has no any real application in the calculation beyond a conceptual description of its maneuver.

And reality proved you wrong. Ever wonder how professional athletes are that good? Try watching them closely. If the sports broadcast is showing the event just right, often you will see the ball receiver begin to move even BEFORE the ball is at apogee. American football are full of this example when one side kick the ball to the other end. The designated receiver will already begin to move to where he guess the ball may fall. You can see some of that in baseball as well when the batter hit to the outfielders. That mean barring any inflight interference, like a wind gust, the ball's trajectory is quite determined. Professional athletes are that good because they are better than us at subconsciously calculating and recalculating the ball's trajectory throughout. Less capable people have to wait later in time to make positional adjustments but by then it is often too late.

As if the football is able to produce thrust and maneuver itself in the midair by itself.

You brought this source on to show that A missile performs its own trajectory optimization. That is not true. This source explain how the process works for development purposes and that it is from performance histories of past designs.

you better read it again.

It shows for missile to reach its intended target, it has a set of optimal control parameters resulting from design factors, desired missile performance, and system constraints. The history here is not the histories of past design, but it is the schedules of this particular missile's AoA commands and downrange accelerations during its flight path.

I will answer the rest later.

 

[below_freezing]

why bother. gambit can't do math. vietnam never had science in the modern term. even today most vietnam overseas open restraunts and don't do science.

you're wasting your aerospace engineering degree arguing with him.

 

[dbc]

why bother. gambit can't do math. vietnam never had science in the modern term. even today most vietnam overseas open restraunts and don't do science.

you're wasting your aerospace engineering degree arguing with him.

That sort of racial stereotyping is uncalled for - if you don't have the mental faculty to debunk him stay out of the discussion.

 

[gambit]

That just showed you how much you know about aerodynamic drag.
Drag(D) = 1/2p*v^2*C*A, the drag force is proportional to the velocity to its second power, which means the increase of aerodynamic drag is much fast than the increase of speed. In part of the vertically flight, the missile or the rocket's velocity remains constant is because that the velocity has reached the point where aerodynamic drag is so big that it = T-mg and neglect any lift that the thrust produces.

Not just me but apparently the rest of the world as well. Here is what I originally said...

In atmosphere, there will be aerodynamic drag, even when going vertical. But the greatest retardant force, for a vertically launched vehicle, is gravity drag, not aerodynamics.

Now here is what other people interested in the subject also said...

RSE 24

When I first investigated launch profiles, I knew that drag was a factor in the launch process. Drag was mentioned in several texts, and but nothing specific was stated, I felt that a balance must be struck between drag loss and gravity loss to obtain the best thrust profile. For a high acceleration vehicle like a sounding rocket, drag loss should be a major factor. For a slow liftoff vehicle like the Saturn V, tons of propellant go into the fight with gravity. Factoring in the atmospheric drag was the last step in determining the optimum thrust profile for a launch vehicle. And the answer is: ----- Well, I'm no math genius, but I do have a little computer program that will simulate a rocket flight, and with a few lines of code added, ( not a big problem in basic), I let the computer do the work. I set up the program to keep the thrust at a level where the vehicle drag is equal to the vertical loss to gravity. A limit was placed on the maximum acceleration to keep liftoff thrust below infinite, and thrust was allowed to increase as the drag component decreases with altitude. Only the first stage was examined, as other stages operate in vacuum.

The results - Drag is not a factor in the thrust profile of a launch vehicle. (No Problema!) Well, it's not nearly as great a factor as gravity. You have to design the nose of the rocket to withstand the force of drag, and the heat of air friction, and as a negative force to thrust, but drag can be largely ignored for any practical thrust profile. You have to get up into the acceleration levels that start to tear components off circuit boards before drag becomes a factor in determining launch thrust profiles. The only factors that limit the thrust profile of a launch vehicle are, the strength of the payload and vehicle structure, and the comfort of the passengers. So, why did NASA spend millions of dollars to develop rocket engines for the Space Shuttle that could be throttled down to limit the effects of drag? Because the Space Shuttle in it's launch configuration is an aerodynamic abomination that would be damaged by severe aerodynamic forces, that's why.

So what I said about gravity being the greatest retardant in a VERTICAL launch profile is correct. There are plenty more sources out there that to varying degrees says the same thing, that in a vertical launch profile, gravity loss is the more dominant factor over aerodynamic loss.

You want another smack...???

Drag: Loss in Ascent, Gain in Descent, and What It Means for Scalability Gravity Loss

* Ariane A-44L: Gravity Loss: 1576 m/s Drag Loss: 135 m/s
* Atlas I: Gravity Loss: 1395 m/s Drag Loss: 110 m/s
* Delta 7925: Gravity Loss: 1150 m/s Drag Loss: 136 m/s
* Shuttle: Gravity Loss: 1222 m/s Drag Loss: 107 m/s
* Saturn V: Gravity Loss: 1534 m/s Drag Loss: 40 m/s (!!)
* Titan IV/Centaur: Gravity Loss: 1442 m/s Drag Loss: 156 m/s

As we can see, gravity loss is far greater than aerodynamic loss over a reasonably wide range of vehicles.

If I shoot a bullet straight up, the distance between ground and its maximum altitude will be shorter than if I had shoot the bullet horizontal and until its fall to the ground.

Don't make me laugh about the bullet example. The distance between the bullet and the ground(vertical distance) is the greatest if its is shot straight up. However the overall distance D(x,y) traveled by the bullet is the greatest if it is shot at 45 degree angle, not horizontally.

Assuming without air drag. v is the initial velocity, and &#952; is the angle it shoots.
Its motion in the air with respect to time(t) are
in the x axis is x = v*(cos&#952*t.
in the y axis is y= v*(sin&#952*t - 1/2*g*t^2

xmax = v^2*sin(2&#952/g, xmax is the greatest when sin(2&#952 = 1, &#952; = 45
ymax = v^2*[sin(&#952]^2/2g, ymax is the greatest when sin(&#952 = 1, &#952; = 90.

For the rest of the angles, the projectile can reach its intended distance with two different angle with the same initial velocity, which also applies to missile here after it finishes boost phase.

That just shows that how much you know about physics, and this is the basic of basic of basic Newtonian physics here.

Assuming without air drag? Sorry...That is not how it works. You cannot insist on aerodynamic drag when it is convenient for you and discard it when it inconvenience you. The issue here is gravity loss versus aerodynamic loss. I will admit that using the word 'horizontal' alone was not accurate and should have include 'relative' but are we talking about ground distance covered or is this about total distance the bullet covered inflight? If you insist on talking about ground coverage then there is nothing more to say because it strays from the issue of gravity loss versus aerodynamic loss.

Anyway...This is the US where firearms are plenty and I have several firearms myself as well as no shortage of sources to support my argument.

Maximum Altitude For Bullets Fired Vertically | Close Focus Research - Ballistic Testing Services

22 LR 40 383 1,257 1,179 3,868

From the above source, the 22 Long Rifle bullet, can reach its maximum altitude of 3800 ft.

Here is a testimony of ground coverage of the same caliber...

Google Answers: physics of gunshots

A .22 caliber rifle, for example has a maximum range of about 1 mile...

One mile = 5280 ft and is greater than 3800 ft, correct?

The 7.62 bullet can reach max altitude of 7874 ft. The AK-47's stock sight is graduated out to 1000 meters and the bullet can travel thrice that if the shooter does not care about accuracy.

From the same source above, the 30-06 bullet can reach up to 10,100 ft.

.30-06 180 823 2,700 3,080 10,105

Here is a source for maximum horizontal range for the same caliber...

30-06 Springfield

The return to the lighter bullet came about, at least in part, because of difficulties adapting the new Garand semi-automatic rifle to handle the 172-grain load. The heavier boattail bullet was superior for machine gun use because of its greater maximum range of nearly 6000 yards, compared to about 3500 yards for the 150-grain loading.

Six thousand yards = 18,000 ft. Both sources has similar grain, 172 and 180. Can we agree that 18k ft horizontal is greater than 10k ft vertical?

Just like the rockets example where gravity loss is greater than aerodynamic loss, for my bullet example that you laughed at, consistently over several calibers, we see that the maximum horizontal travel exceeds vertical.

So when you say this: 'The distance between the bullet and the ground(vertical distance) is the greatest if its is shot straight up.' Where did you get it from? I doubt that it is from personal experience because if it is from personal experience you would not argue about it in the first place. You would know that what I said is true. Two items that you claim that I do not know about and two items I brought REAL WORLD sources to support my arguments.

 

[gambit]

That sort of racial stereotyping is uncalled for - if you don't have the mental faculty to debunk him stay out of the discussion.

For that loser, that is all he can do since he now knows all the fantasies about Chinese military hardwares have been debunked.