Originally Posted by rattman
(Post 11276895)
ATACMS is believed to reach mach 3 during its terminal phase
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Originally Posted by soarbum
(Post 11276893)
Velocity is a vector with components in x,y and z. ORAC clearly referred to zero vertical velocity, ie the z component. If it went up and came back down, then at some point, the vertical component was zero regardless of the shape of the trajectory.
I don't see why not if the object has a small enough cross section, a large enough mass and some fins to keep the pointy end at the front. Also bear in mind that the air density is much lower during the upper part of the trajectory. ps: since you mentioned sky divers, low level divers reach 120mph while spread but Felix Baumgartner reached 843.6 mph (Mach 1.25) on his dive |
That is totally height dependent, the higher you go, the thinner the atmosphere, the faster you drop. |
Originally Posted by NutLoose
(Post 11276738)
Still as a means of avoiding arty, his posture was pure genius, could be new |
I am no marine warfare expert, but that sub (if it is a sub as suggested) seems to be rather close to the shore. I've no idea what are the capabilities of UKR Navy against submarines nowadays, but they did sink Moskva and other vessels.
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Originally Posted by Winemaker
(Post 11276914)
I agree. That velocity is not achieved from simple acceleration due to gravity. A large component of that is its initial velocity, which is what I was attempting to explain.
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For those that want to play there is a calculator at the bottom.
Weight of an ATACMS is 3690lbs and drag coefficient is below 0.7. Have to guess at the cross section, I put in 3ft. Gave a terminal velocity of 3100 f/s - between M2.5 and M3.0. https://www.grc.nasa.gov/www/k-12/rocket/termvr.html For those that doubt that missiles go supersonic on descent I refer you back to the Falcon 9 video and the double crack boom on landing - and that after burning back down range to the launch site and a long deceleration burn during the descent to reduce horizontal speed.to zero and reduce vertical speed prior to the last second landing burn. |
Originally Posted by Winemaker
(Post 11276917)
Again, I agree, except that we don't live in a vacuum. The lower you go the slower you go until you hit the ocean, when you stop. The final velocity of the rocket at impact is not only dependent on its maximum altitude during flight. This is a vector thing.
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military aviation content is a bit hard to come by, but here's a little something to keep the thread on topic.
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6 Minutes in May
Originally Posted by DaveReidUK
(Post 11276840)
About 10 years ago, I had the privilege of seeing the late, great Warren Clarke playing Churchill in "Three Days in May", a dramatisation of events in 1940 when Britain teetered on the brink of giving in to Hitler.
I can't vouch for how historically accurate it was, but by all accounts it was a close-run thing. |
Originally Posted by Winemaker
(Post 11276873)
Since I'm being pedantic, I'll also point out that a falling object in the atmosphere will reach some terminal velocity, determined by its drag characteristics; for a sky diver it's about 120 mph. I seriously doubt any dropped object in the atmosphere will go supersonic.
With the reference to parachuting one should express some sympathy for those whose job entails leaving an aircraft at altitude that flies at an IAS with a parachute that effectively, at opening, respects TAS. Even with an aircraft that can drop safely from (for example) 135kts from the 30-block, with the relative comfort of a static-line delivery, will find themselves at 0.4~0.5M as the canopy inflates. All that energy from the rapid deceleration is taken by the crotch straps. When you have to drop with a delayed canopy, higher, faster and with large amounts of equipment, the careful packing of manhood is as much of a priority as the chutes. |
Originally Posted by Winemaker
(Post 11276914)
That velocity is not achieved from simple acceleration due to gravity. A large component of that is its initial velocity
Firstly, lets assume that at the apex of its flight (that point where its vertical velocity IS zero), the rocket motor has finished burning. The energy in the system is the kinetic energy due to its momentarily horizontally velocity plus the potential energy from having climbed that high. If it travelled 200km over 30min, horizontal velocity averages 400kmph or 111m/s. The kinetic energy due to its velocity at the apex is 0.5mv^2, roughly 6.2e3 x m, where m is the mass of the missile with no fuel left. (the horizontal velocity at the apex won't exactly equal the average velocity but it will be of that order) The energy in the system when it hits the ground at Mach 2.5 (3087kmph) is 367.6e3 x m The initial velocity gets it up there but the velocity at impact is very much caused by "simple acceleration due to gravity" |
Originally Posted by Winemaker
(Post 11276873)
The extreme other example is a projectile (rifle bullet for example) that is fired absolutely horizontally. During its flight the bullet will experience a downward force from gravity of 9.8 m/sec^2. If the bullet is shot from a mountain top, in one second it will fall 9.8 m....
Since I'm being pedantic.... The bullet would experience a force from gravity of 9.8 . (mass in kg). The 9.8 m/sec^s quantity you wrongly claim to be the force acting upon it is in fact the acceleration it would experience. Also, in 1 second, the bullet (or any falling object) would reach a vertical velocity of 9.8m/s, but it would only have fallen only 5m (S = u.t + 1/2.a.t^2), not 9.8m as you claim. |
Gob smacked, Steven Seagal appears to be working for the Russians.
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To be fair, he got this part right:
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So, basically, it was a rant against fake hair?
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Beamr - I think that clip has been miss-reported and what is being seen is the rocket booster phase burning out. The boosters have thrust vectoring, hence the trajectory flown as revealed by the smoke trail.
https://cimg1.ibsrv.net/gimg/pprune....bbc741f8c2.jpg |
Originally Posted by Wokkafans
(Post 11277084)
Beamr - I think that clip has been miss-reported and what is being seen is the rocket booster phase burning out. The boosters have thrust vectoring, hence the trajectory flown as revealed by the smoke trail.
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Is replenishment not to replace a possibly lost item? or just badly worded, or is he talking ammunition for them? In the future, please use "preview post" to detect, and thus avoid, the kind of visual clutter that you created. Thank you for going back and cleaning up the mess. :) T28B |
Originally Posted by NutLoose
(Post 11277114)
h
Is replenishment not to replace a possibly lost item? or just badly worded, or is he talking ammunition for them? To replenish is to refill not replace. Since the question of language has been raised, it would be good if you could sort out the difference between "its" and "it's " e.g. " unaware of what hit it and it’s range.. win win either way." It's very simple and necessary because the more this bloody mistake is repeated, the more difficult it is to eliminate it. |
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