Iran missile power

[Philosopher]

Very slow free falling object.

How did you come to the conclusion it is "very slow"? These clips are always slowed down so we can see the warheads, this is also clearly shown by the explosion.

U seriously think that these tiny flappers can cause 90 degree turn of one ton warhead?

That is not the right warhead. That warhead is from Emad whereas the warhead in that clip is from Khromashahr.

 

[500]

How did you come to the conclusion it is "very slow"? Were you there measuring its speed or are you as usual posting drivel? These clips are always slowed down so we can see the warheads, this is also clearly shown by the explosion.

Based on explosion it is not slowed down. Maximum 50% or so.

That is not the right warhead. That warhead is from Emad whereas the warhead in that clip is from Khromashahr.

Whatever is Khromashahr u wont see there big wings.

 

[Philosopher]

Based on explosion it is not slowed down. Maximum 50% or so.

This is an entry vehicle of a ballistic missiles, it is obviously slowed down. It will obviously not be hypersonic intra-atmosphere, but it is still supersonic. There was a full speed version of that video available and you could not see the warhead, only the impact.

Whatever is Khromashahr u wont see there big wings.

Khormashahr MaRV is around 1500kg. There are multiple ways to create MaRV other than relying on winglets.

 

[500]

Yallah lets make a simple calculation.

Lets suppose that genius engineers thanks to spirit of Soleimani and Chavez managed to make 1500 kg warhead with tiny pinny flappers as maneuverable as 5 gen plane = 10 g or 100 m/s2

V=14 Mach = 5780 m/s

Turn radius = v^2/a = 334 km

 

[Philosopher]

Yallah lets make a simple calculation.

Lets suppose that genius engineers thanks to spirit of Soleimani and Chavez managed to make 1500 kg warhead with tiny pinny flappers as maneuverable as 5 gen plane = 10 g or 100 m/s2

V=14 Mach = 5780 m/s

Turn radius = v^2/a = 334 km

You are resorting to disinformation.
I never stated the speed of the missile inside the atmosphere is mach 14, I quite clearly stated otherwise. More-over, like I said, there are other ways than utilising winglets to meneuver the MaRV. You are using out of context calculations and also using the warheads of a completely different missile system.

 

[500]

You are resorting to disinformation.
I never stated the speed of the missile inside the atmosphere is mach 14, I quite clearly stated otherwise. More-over, like I said, there are other ways than utilising winglets to meneuver the MaRV. You are using out of context calculations and also using the warheads of a completely different missile system.

Ur friend claimed it and u liked his post.

Now u agree with my points but still argue with me? I dont understand what u are trying to say.

 

[Philosopher]

Ur friend claimed it and u liked his post.

I liked it because of his overall comment, especially since he debunked some of your comments.

Now u agree with my points but still argue with me? I dont understand what u are trying to say.

The only part you and I agree on is that the missile does not travel at mach-14 inside the atmosphere. Everything else you claimed was false. You need to understand that kinematic profile of missile changes throughout their journey, K-1/2 may reach mach-14 at some point during its journey, but our focus here is the MaRV manveuer you saw inside the atmosphere.

 

[Mithridates]

Yallah lets make a simple calculation.

Lets suppose that genius engineers thanks to spirit of Soleimani and Chavez managed to make 1500 kg warhead with tiny pinny flappers as maneuverable as 5 gen plane = 10 g or 100 m/s2

V=14 Mach = 5780 m/s

Turn radius = v^2/a = 334 km

interesting calculations, care to explain in detail??

 

[Philosopher]

@Mithridates

@500 is trying to use the centrifugal acceleration formula to calculate the radius the warhead would need in order to make that turn. He is using this formula out of place because the MaRV is not turning at a constant speed at a set turn angle. It performed a relative sharp turn and would have created a great load of G forces. This is acceptable in a MaRV, but not in a plane housing a living entity.

Even being kind enough to allow him to use this formula, he has made two false assumptions in his calculation. 1) He is confusing the mach 14 extra-atmospheric missile speed to the intra-atmospheric speeds, which would be more around then trans-hypersonic range if not low range hypersonic. The exact number depends on various factors. The other false assumption he made is trying to compare the G forces capable of a MaRV with a fighter jet. MaRV are capable of 10s of G forces:

To execute sharp turns at high speeds, the MaRVs control system should be able to provide accelerations of tens of gravities in the desired direction. As noted above, the force of aerodynamic drag reaches magnitudes of the order of 50 gravities.

https://scholar.harvard.edu/files/bunn_tech_of_ballastic_missle_reentry_vehicles.pdf

MaRV can not only use surface control systems to maneuvers, but also the use of propellents.

 

[500]

I liked it because of his overall comment, especially since he debunked some of your comments.



The only part you and I agree on is that the missile does not travel at mach-14 inside the atmosphere. Everything else you claimed was false. You need to understand that kinematic profile of missile changes throughout their journey, K-1/2 may reach mach-14 at some point during its journey, but our focus here is the MaRV manveuer you saw inside the atmosphere.

So 14 mach is nonsense. 90 degree turn is also nonsense. I was right in everything.

Even at 5 Mach speed it will have 29 km radius turn. Thats assuming that magic Khamenai engineers managed to cause a warhead with tiny flappers to maneuver like 5 gen aircraft with big fully mechanized wings.

 

[Agha Sher]

The warhead is not manoeuvring at this state. As @500 clearly states, firstly, it is impossible to manoeuvre like that at the terminal stage, and secondly, why would you manoeuvre at the terminal stage? The warhead has already penetrated enemy BM defences and its velocity makes it nearly impossible for point-defence systems to intercept it.

The weird shape is likely due to the poor video quality.

 

[Philosopher]

So 14 mach is nonsense. 90 degree turn is also nonsense. I was right in everything.

Even at 5 Mach speed it will have 29 km radius turn. Thats assuming that magic Khamenai engineers managed to cause a warhead with tiny flappers to maneuver like 5 gen aircraft with big fully mechanized wings.

Your claims have been addressed in my previous comment above. Your formula does not work in this case, and even if it did, the tens of g forces capable of by these MaRV would lead to a turn radius which measure in 10 ish KM. Which is more than do able. Given the speed traveling by these systems, that radius is nothing. In reality, the MaRV pulled a very sharp turn as obvious in the clip.

The warhead is not manoeuvring at this state. As @500 clearly states, firstly, it is impossible to manoeuvre like that at the terminal stage, and secondly, why would you manoeuvre at the terminal stage? The warhead has already penetrated enemy BM defences and its velocity makes it nearly impossible for point-defence systems to intercept it.

The weird shape is likely due to the poor video quality.

Because clearly they are demonstrating the ability of the MaRV to maneuver. It was just for demonstration purposes. Moreover, care to explain to me how it is "impossible" to manoeuvre like that (despite the fact you can clearly see it happening)? Use actual science, not your own conjecture please.

 

[500]

Your formula does not work in this case

LOL, why not? You invented a magic way to make turns without acceleration? Sorry but that's too much even for spirit of Chavez and statue of Solemani.

and even if it did, the tens of g forces capable of by these MaRV would lead to a turn radius which measure in 10 ish KM. Which is more than do able.

To make sharp turns u need low wing loading. Thats another physics basic. Which cant be achieved on that warhead.

Given the speed traveling by these systems, that radius is nothing. In reality, the MaRV pulled a very sharp turn as obvious in the clip.

Nothing in this clip. Only silly lines drawn in MS Paint by your friend who knows nothing in physics just like you.

 

[Philosopher]

LOL, why not? You invented a magic way to make turns without acceleration? Sorry but that's too much even for spirit of Chavez and statue of Solemani.

You do not understand that formula. That formula calculates the radius of turn when assuming the body if turning at an usually low angle (usually set) at particular speed. This is not what is happening in this cause. This was a relatively shape turn performed by the MaRV. Your formula would work for example if we are trying to calculate the radius of turn for a SR-71 blackbird traveling whilst tolerating its maximum g which would be done at a set bank angle. Look below for an illustration:

What is the minimum turning radius of an SR-71?

https://aviation.stackexchange.com/questions/8028/what-is-the-minimum-turning-radius-of-an-sr-71#:~:text=That can hardly be called a turn.&text=The Turn Radius of the,80 nautical miles (NM) .

This does not apply to this MaRV case.

To make sharp turns u need low wing loading. Thats another physics basic. Which cant be achieved on that warhead.

Such turns can be performed using on board propellant systems. Nothing to do with "wing loading". You are making assumptions that this turn was made using surface wings without any proof whatsoever.

Nothing in this clip. Only lines drawn in MS Paint by your friend who knows nothing in physics just like you.

The clip shows a MaRV performing very high angle of turn. You tried to "explain" this using a formula that does not apply to this case.

 

[500]

You do not understand that formula. That formula calculates the radius of turn when assuming the body if turning at an usually low angle (usually set) at particular speed. This is not what is happening in this cause. This was a relatively shape turn performed by the MaRV. Your formula would work for example if we are trying to calculate the radius of turn for a SR-71 blackbird traveling whilst tolerating its maximum g which would be done at a set bank angle. Look below for an illustration:

What is the minimum turning radius of an SR-71?

https://aviation.stackexchange.com/questions/8028/what-is-the-minimum-turning-radius-of-an-sr-71#:~:text=That can hardly be called a turn.&text=The Turn Radius of the,80 nautical miles (NM) .

This does not apply to this MaRV case.

In contrast to you I did study a physics, so I understand formula very well. You cant make any turn without acceleration. The sharper turn the more acceleration. There is no any magic exceptions here. Your SR-71 example only proves my point. Seems u dont even understand what u are posting.

Such turns can be performed using on board propellant systems. Nothing to do with "wing loading". You are making assumptions that this turn was made using surface wings without any proof whatsoever.

Sharp turn is acceleration, in order to accelerate 1.5 tonn warhead u need very powerful engines there. That's why no one installs anything like that.

The clip shows a MaRV performing very high angle of turn. You tried to "explain" this using a formula that does not apply to this case. You found a formula on wikipedia and tried to apply it to a case not suited for. Stick to what you know.

There is no any turns in that clip. Just some slow free falling object.