Chinese missile could shift Pacific power balance

[gambit]

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.

This feeble response tells me that YOU have done exactly as I said......Sometime in your past someone threw a ball at you and even BEFORE the ball reached apogee, you moved to intercept. Whether you were successful or not is besides the point, which is that it is eminently possible to successfully predict a ballistic trajectory BEFORE said object inflight reached apogee. The issue is not thrust or maneuverability but about ballistic trajectory predictions and successful intercepts. In effect...You debunked yourself, sarcasm notwithstanding. And if humans can predict ballistic trajectories before apogees, according to you, radars and computers cannot.

 

[huzihaidao12]

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

Haha, that was fun, you can suggest the United States Department of Defense, not to mention the China threat, go, really do.

 

[gambit]

That is hilarious indeed. You are saying V-2 is not a ballistic missile,

Never said it was not.

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.


V-2 - Wikipedia, the free encyclopedia

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

Actually, given how militarily ineffective the V-2 really was, Hitler was right in his criticisms of the V-2.

Wright-Patterson AFB is headquarters for USAF Systems Command. This is the USAF conceptual center. Many years ago, before public access of the Internet and probably before you were borned, some USAF engineers in their off duty time replicated the V2's guidance system with items from their spare/scrapped parts bins. The contraption went on tour to some technical training centers such as Lowry AFB in Colorado and Keesler AFB in Mississippi. I handled it. There is no way back in WW II, even with Germany's vaunted technological prowess back then, that the V2 was capable of programming to calculate a gravity turn. Just because the rocket was capable of reaching, barely, what we called 'suborbital' altitude that does not make the rocket comparable to what we have today. All that primitive guidance system did was to shut off the engine at the appropriate time based upon inertial calculations, which was primitive back then. Inertial navigation are vulnerable to drift and that was why the V2 was so militarily ineffective. The engines would shut off in different times at different altitudes and rendered the rockets wildly inaccurate when they descends.

If you insist on getting technical about it in order to salvage your arguments, then fine...ANY parabolic trajectory is a gravity turn, including the shallow arc of a bullet's flight. Try not to be too embarrassed when experienced engineers slaps their knees in laughter when you present them with this argument.

 

[chauism]

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


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.

From your other data, I wouldn't call 10% difference in the change of speed not a factor in determining the thrust profile.

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.

I never argued that gravity loss is not the the dominant factor. However since gravity force is constant and the direction of its action is also the same throughout the flight, where as aerodynamic drag is a changeable variable, aerodynamic drag is the factor that will determine the difference in vehicle's burnout speed and its terminal speed when it reenter the atmosphere.

You want another smack...???

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

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

Now, I wonder if you know why it is small in this figure. Aerodynamic drag is almost non-exist in the early part of launch. The net loss in your figure is the integral of the gravity drag or aerodynamic drag with respect to time, that is why it is in m/s instead of m/s^2 as for acceleration. What it means is that if we draw a graph of the aerodynamic drag and gravity drag, the net loss for each is the area under the graph. Of course the area will be much smaller for aerodynamic drag. A simple example is that, the acceleration for space shuttle is around 3 g after take off, so when it reach a constant velocity where the acceleration is 0. Right after take off, the air drag is 0, so Lift = T -mg, since L here produces 3g acceleration here, 3mg = T -mg, T = 4mg. When it reaches the constant velocity, lift is 0. So T = mg + Drag. The drag is 3mg, so the aerodynamic force is actually 3 time bigger than the gravity force. When it reenter the earth atmosphere, the drag force will give a net 3 g deceleration to the shuttle until it gradually reaches its terminal velocity where drag force equal to gravity force. Of course for a slow moving rocket like Saturn V where it only experience 1.14 g acceleration, the aerodynamic drag is minimal as comparing to others.
g-force - Wikipedia, the free encyclopedia

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.

The reason to assume without air drag is to simplify the math. Fine, let's put the air drag back in the equation D = drag force, a = angle of the projectile
x = v*cos(&#952*t - integral of (D/m)*cos(a)*t with respect to t
y = v*sin(&#952*t - 1/2*g*t^2- intergral of (D/m)*sin(a)*t with respect to t.
However air resistance is a nonlinear function of velocity, so it is time varying. The only way to solve the differential equations of the movement is numerically. You attribute an initial value to the velocity, calculate the drag and from that the accelerations in the horizontal and vertical directions. You give a small time increment and calculate the new value of the velocity. Work iteratively until you find that the height is zero. The horizontal distance at that instant is the range. Luckily I have found a neat program that can calculate projectile motion with drag included. With couple trials you can find that the projectile will travel the same distance when it is a 60 degree and 23 degree, and it will reach its maximum distance still around 45 degree with initial velocity of 50 m/s and 10kg of mass. You can try it yourself.
Projectile Motion

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

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

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.

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

30-06 Springfield

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.

Now let's take the number of bullets into that program. AK-47 has a muzzle velocity of 715 m/s and the bullet weight is 8 gram, let put that number in. The maximum distance it can travel is still reached by aim around 10 degree, not horizontally. The reason for smaller angle is because the mass of the bullet is much smaller, so the overall acceleration or deceleration produced by drag is much greater than others with bigger mass. The bullet will reach the same distance if it is shot at 60 degree angle and 0.5 degree. And it still reaches the maximum attitude when it is shoot at 90 degree. The only reason the the bullet will travel horizontally is because when you shoot a gun, the bullet exit with a initial height. Try to shoot a gun horizontally as close to the ground as possible, if the initial height is zero, see how far the bullet will go. Anyways this is not what I am arguing about.

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.

When I say vertical distance, I meant its displacement in the vertical axis(altitude), not is horizontal axis displacement (range). The distance between an object and the ground is its height or altitude. Tell me where it is wrong in my statement.
Yes, the distance or range will be greater if the gun is shot at a angle other than 90 degrees and at its maximum when it is shot around 10 degree, however its maximum altitude is reached by shot it at 90 degrees. Assuming the initial height is 0, if the bullet will fall to the ground intermediately if it is fired horizontally and may still travel on the ground against the ground friction, but it will not travel far. Likewise, if the bullet is fired when its initial height is much higher like someone shoot it horizontally from a tower. The distance it will cover will be much much greater than someone fire it on the ground.