My God, you are an idiot.

[Hank Chinaski]

Originally Posted by Tyrone Slothrop

So Bush showed that you need to have the help of a superpower to overthrow your despotic leader? I'm sure Libyans in Benghazi found that inspirational, and threw flowers at the NATO tanks as they rolled past.

Stick with what Club is saying. It's far-fetched to think that Bush's invasion of Iraq showed Libyans anything new about what was possible, but at least it doesn't undercut your point.

Club and I tend to make the same points, I just make them in a smart assed way, because I have no dream of teaching you anything.

Iraq (as would Iran) took help because the leaders were nastier and the armies stronger. Do you have an eye disease where everything looks two dimensional, and in black and white?

[Hank Chinaski]

Originally Posted by Tyrone Slothrop

OK, but the question I asked (twice!) was, Why would target distance here change the maximum altitude in any material way? (I emphasized key bits you aren't helping with.)

As I said, I get why target distance could affect maximum altitude, but it will still be above the level at which warplanes fly. 200 km above the planet is far, far above warplanes.



Ah, yes, I see. Helpful!



You would have to find them first.

Well i'll answer your question (although it would have been helpful to understand how many variables you're able to integrate too.)

Assume the motion of the projective is being measured from a Free fall frame which happens to be at (x,y)=(0,0) at t=0. The equation of motion of the projectile in this frame ( by the principle of equivalence) would be y = xtan(θ). The co-ordinates of this free-fall frame, with respect to our inertial frame would be y = − gt2 / 2. That is, y = − g(x / vh)2 / 2.

Now translating back to the inertial frame the co-ordinates of the projectile becomes y = xtan(θ) − g(x / vh)2 / 2 That is:

y=-{g\sec^2\theta\over 2v_0^2}x^2+x\tan\theta,

(where v0 is the initial velocity, θ is the angle of elevation, and g is the acceleration due to gravity).
[edit] Range and height
Trajectories of projectiles launched at different elevation angles but the same speed of 10 m/s in a vacuum and uniform downward gravity field of 10 m/s2. Points are at 0.05 s intervals and length of their tails is linearly proportional to their speed. t = time from launch, T = time of flight, R = range and H = highest point of trajectory (indicated with arrows).

The range, R, is the greatest distance the object travels along the x-axis in the I sector. The initial velocity, vi, is the speed at which said object is launched from the point of origin. The initial angle, θi, is the angle at which said object is released. The g is the respective gravitational pull on the object within a null-medium.

R={v_i^2\sin2\theta_i\over g}

The height, h, is the greatest parabolic height said object reaches within its trajectory

h={v_i^2\sin^2\theta_i\over 2g}

[edit] Angle of elevation

In terms of angle of elevation θ and initial speed v:

v_h=v \cos \theta,\quad v_v=v \sin \theta \;

giving the range as

R= 2 v^2 \cos(\theta) \sin(\theta) / g = v^2 \sin(2\theta) / g\,.

This equation can be rearranged to find the angle for a required range

{ \theta } = \frac 1 2 \sin^{-1} \left( { {g R} \over { v^2 } } \right) (Equation II: angle of projectile launch)

Note that the sine function is such that there are two solutions for θ for a given range dh. The angle θ giving the maximum range can be found by considering the derivative or R with respect to θ and setting it to zero.

{\mathrm{d}R\over \mathrm{d}\theta}={2v^2\over g} \cos(2\theta)=0

which has a non trivial solution at 2\theta=\pi/2=90^\circ, or \theta=45^\circ. The maximum range is then R_{max} = v^2/g\,. At this angle sin(π / 2) = 1, so the maximum height obtained is {v^2 \over 4g}.

To find the angle giving the maximum height for a given speed calculate the derivative of the maximum height H = v2sin2(θ) / (2g) with respect to θ, that is {\mathrm{d}H\over \mathrm{d}\theta}=v^2 2\cos(\theta)\sin(\theta) /(2g) which is zero when \theta=\pi/2=90^\circ. So the maximum height H_{max}={v^2\over 2g} is obtained when the projectile is fired straight up.

[Tyrone Slothrop]

Originally Posted by Hank Chinaski

Club and I tend to make the same points, I just make them in a smart assed way, because I have no dream of teaching you anything.

Iraq (as would Iran) took help because the leaders were nastier and the armies stronger. Do you have an eye disease where everything looks two dimensional, and in black and white?

Interesting. In the six months since things started up in Libya, have you seen any Libyans say that Bush inspired them? Or is it that his influence was so profound that they just can't even recall a Middle East without W's inspirational, pro-democratic stamp on it?

[Tyrone Slothrop]

Originally Posted by Hank Chinaski

Assume the motion of the projective is being measured from a Free fall frame which happens to be at (x,y)=(0,0) at t=0. The equation of motion of the projectile in this frame ( by the principle of equivalence) would be y = xtan(θ). The co-ordinates of this free-fall frame, with respect to our inertial frame would be y = − gt2 / 2. That is, y = − g(x / vh)2 / 2.

Now translating back to the inertial frame the co-ordinates of the projectile becomes y = xtan(θ) − g(x / vh)2 / 2 That is:

y=-{g\sec^2\theta\over 2v_0^2}x^2+x\tan\theta,

(where v0 is the initial velocity, θ is the angle of elevation, and g is the acceleration due to gravity).
[edit] Range and height
Trajectories of projectiles launched at different elevation angles but the same speed of 10 m/s in a vacuum and uniform downward gravity field of 10 m/s2. Points are at 0.05 s intervals and length of their tails is linearly proportional to their speed. t = time from launch, T = time of flight, R = range and H = highest point of trajectory (indicated with arrows).

The range, R, is the greatest distance the object travels along the x-axis in the I sector. The initial velocity, vi, is the speed at which said object is launched from the point of origin. The initial angle, θi, is the angle at which said object is released. The g is the respective gravitational pull on the object within a null-medium.

R={v_i^2\sin2\theta_i\over g}

The height, h, is the greatest parabolic height said object reaches within its trajectory

h={v_i^2\sin^2\theta_i\over 2g}

[edit] Angle of elevation

In terms of angle of elevation θ and initial speed v:

v_h=v \cos \theta,\quad v_v=v \sin \theta \;

giving the range as

R= 2 v^2 \cos(\theta) \sin(\theta) / g = v^2 \sin(2\theta) / g\,.

This equation can be rearranged to find the angle for a required range

{ \theta } = \frac 1 2 \sin^{-1} \left( { {g R} \over { v^2 } } \right) (Equation II: angle of projectile launch)

Note that the sine function is such that there are two solutions for θ for a given range dh. The angle θ giving the maximum range can be found by considering the derivative or R with respect to θ and setting it to zero.

{\mathrm{d}R\over \mathrm{d}\theta}={2v^2\over g} \cos(2\theta)=0

which has a non trivial solution at 2\theta=\pi/2=90^\circ, or \theta=45^\circ. The maximum range is then R_{max} = v^2/g\,. At this angle sin(π / 2) = 1, so the maximum height obtained is {v^2 \over 4g}.

To find the angle giving the maximum height for a given speed calculate the derivative of the maximum height H = v2sin2(θ) / (2g) with respect to θ, that is {\mathrm{d}H\over \mathrm{d}\theta}=v^2 2\cos(\theta)\sin(\theta) /(2g) which is zero when \theta=\pi/2=90^\circ. So the maximum height H_{max}={v^2\over 2g} is obtained when the projectile is fired straight up.

When I cut and paste from Wikipedia, I try to take out the tags that say "[edit]."

[Hank Chinaski]

Originally Posted by Tyrone Slothrop

When I cut and paste from Wikipedia, I try to take out the tags that say "[edit]."

I find it absurd that you would criticize cut and paste posts.

[Cletus Miller]

Originally Posted by Tyrone Slothrop

200 km above the planet is far, far above warplanes.

No doubt, but do we have any reason to believe that the interceptors were ramming the scuds? Or were they lauching missiles with much different operational ceilings?

[Tyrone Slothrop]

If you want to give Bush some credit for what happened in Libya, I would look to the idea that we can support local combatants very effectively airpower and spotters on the ground, a la Afghanistan.

[Tyrone Slothrop]

Originally Posted by Hank Chinaski

I find it absurd that you would criticize cut and paste posts.

That wasn't a criticism of your post. It was a deflation of your pretension to superior expertise. I took humanities classes where I learned to tell the difference.

[Hank Chinaski]

Originally Posted by Tyrone Slothrop

Interesting. In the six months since things started up in Libya, have you seen any Libyans say that Bush inspired them? Or is it that his influence was so profound that they just can't even recall a Middle East without W's inspirational, pro-democratic stamp on it?

Not Bush, the Iraq freedoms. And given the cluster fuck of tribes within most of these countries I can certainly see why there is little spoken credit about other countries. We are seeing an amazing, almost universal (within the region) movement of people that had been under a dictator's thumb to a hopeful freedom.

I can see where a poly sci major might argue it coincidence, but I am a scientist, and I cannot deny that there must be some generating cause. Can you name one that makes more sense that Iraq?

[Hank Chinaski]

Originally Posted by Tyrone Slothrop

If you want to give Bush some credit for what happened in Libya, I would look to the idea that we can support local combatants very effectively airpower and spotters on the ground, a la Afghanistan.

again, you are speaking to military tactics, which has little to do with Bush. I am saying the freedom of the Iraq people motivated the movements.

[Tyrone Slothrop]

Originally Posted by Cletus Miller

No doubt, but do we have any reason to believe that the interceptors were ramming the scuds? Or were they lauching missiles with much different operational ceilings?

If this really happened, the most likely scenario (to me) is that NATO warplanes used air-to-air missiles to get the SCUDs in their ascent, rather than at their zenith (too high) or in their descent (too fast).

eta: DoD confirms it happened.

[Hank Chinaski]

Originally Posted by Tyrone Slothrop

That wasn't a criticism of your post. It was a deflation of your pretension to superior expertise. I took humanities classes where I learned to tell the difference!

you deleted the part where I asked how many variable you can integrate up to. I was trying to find an explanation that fit the math backgrounds of the posters here. My textbooks were aimed a bit too high, I fear.

[Tyrone Slothrop]

Originally Posted by Hank Chinaski

Not Bush, the Iraq freedoms. And given the cluster fuck of tribes within most of these countries I can certainly see why there is little spoken credit about other countries. We are seeing an amazing, almost universal (within the region) movement of people that had been under a dictator's thumb to a hopeful freedom.

I can see where a poly sci major might argue it coincidence, but I am a scientist, and I cannot deny that there must be some generating cause. Can you name one that makes more sense that Iraq?

Food prices, the revolution next door in Tunisia, and the protests next door in Egypt, to name three.

As you should know, science makes it impossible to say that Bush's invasion of Iraq had no effect in Libya. I just think you and club are wildly overstating the case.

[Tyrone Slothrop]

Originally Posted by Hank Chinaski

again, you are speaking to military tactics, which has little to do with Bush. I am saying the freedom of the Iraq people motivated the movements.

Just trying to give Bush credit where due. And the military tactics are of political significance, because the Afghans and Libyans justifiably feel that they were the ones doing the fighting.

I decline to engage on the extent to which Iraqis are now free, and to which the costs of that freedom would prompt envy on the part of Libyans.

[Tyrone Slothrop]

Originally Posted by Hank Chinaski

you deleted the part where I asked how many variable you can integrate up to. I was trying to find an explanation that fit the math backgrounds of the posters here. My textbooks were aimed a bit too high, I fear.

Speaking of "trying to find," I also must have missed the part where you said anything helpful in answering my question, or identified corrigenda.