Quote from: pagheca on 12/10/2013 02:05 pmreally no one is able to answer the question I asked in my previous message? Sorry to insist, but I need it for my simulations.Thankstry looking for the location of the 'american islander'.
really no one is able to answer the question I asked in my previous message? Sorry to insist, but I need it for my simulations.Thanks
[try looking for the location of the 'american islander'.
That's quite a contrast to at least one of the Falcon 9 v1.0 flights, where the first stage apparently impacted much further away.
Quote from: Proponent on 12/11/2013 01:02 pmThat's quite a contrast to at least one of the Falcon 9 v1.0 flights, where the first stage apparently impacted much further away.Staging velocity for F9 v1.0 was over 3000 m/s, for v1.1 around 2000 m/s; you can expect also more horizontal trajectory to impart more horizontal velocity to second stage and payload.
They also braked for reentry. That would further shorten the flight distance.
BTW, the only two marks publicly available during powered descent show the same velocity, like if the ACS was aiming to a constant descent speed (Thrust = Mg). It could be an interesting "optimal theorem". Do you know if anyone speculated about this in the forum?
Quote from: pagheca on 12/11/2013 05:27 pmBTW, the only two marks publicly available during powered descent show the same velocity, like if the ACS was aiming to a constant descent speed (Thrust = Mg). It could be an interesting "optimal theorem". Do you know if anyone speculated about this in the forum?Here,http://forum.nasaspaceflight.com/index.php?topic=32859.msg1109900#msg1109900follow the discussion and you will find some contribution.
I may be wrong, but here is one reason the "straight up" trajectory may be suboptimal.If the stage falls vertically, atmospheric pressure and the braking force may increase too quickly.By falling more horizontally, you can shed more speed per meter of descent, which would require less aggressive braking burn.
Quote from: hrissan on 12/12/2013 03:21 pmI may be wrong, but here is one reason the "straight up" trajectory may be suboptimal.If the stage falls vertically, atmospheric pressure and the braking force may increase too quickly.By falling more horizontally, you can shed more speed per meter of descent, which would require less aggressive braking burn.Just bear in mind that due to the peculiar landing conditions, the F9R booster may have to be falling at a certain minimum speed in order to be able to land.The peculiarity is that it's landing on an engine whose thrust exceeds the weight of the stage, so that while the engine is firing, it's always accelerating up. Under these circumstances, the only way to land is to time the burn so that upward velocity goes to zero exactly when the stage hits the ground --- or, at any rate, close enough that the legs can do the rest of the braking (or handle the fall after engine cutoff).So, let's say that from velocity v, a burn starting at height h gets you to a safe landing at the lowest throttle setting. If you come in faster, and start the burn at the same height, you can just throttle up to burn off the excess velocity. But if you come in slower, the only thing you can do is start the burn from a lower height, and burn for less time. And short enough burns may get really tricky to time due to engine startup transients.
Quote from: pagheca on 12/12/2013 06:31 pmQuote*before* it slows to terminal velocity in lower atmosphere.Ok but do there is any obvious way to show that such an object would reach terminal velocity faster than the pressure build up during descent? To me it is not so obvious.thanksyes - maybe there is - sorry Heavy and much denser meteors actually reach terminal velocity.pagheca
Quote*before* it slows to terminal velocity in lower atmosphere.Ok but do there is any obvious way to show that such an object would reach terminal velocity faster than the pressure build up during descent? To me it is not so obvious.thanks
*before* it slows to terminal velocity in lower atmosphere.
What you are discussing here is basically the ballistic coefficient of the stage, which will be much lower than than that of, say, a meteor, because a large volume of the stage will be empty tanks, and because legs will be extended at some point.http://en.wikipedia.org/wiki/Ballistic_coefficientThe stage will of course reach terminal velocity, but we don't quite know what that number is, though there has been some speculation here:http://forum.nasaspaceflight.com/index.php?topic=31513.msg1105554#msg1105554In this Russian example posted previously by Lars J, terminal velocity appears to be 130 m/sec.http://forum.nasaspaceflight.com/index.php?action=dlattach;topic=32718.0;attach=544671
What you are discussing here is basically the ballistic coefficient of the stage, which will be much lower than than that of, say, a meteor, because a large volume of the stage will be empty tanks, ...
What hrissan was discussing is the much higher/faster portion of the trajectory when the stage is well above terminal velocity and has yet to shed much of its velocity due to aero drag, and may still have significant horizontal velocity. That is the phase where this discussion of trajectory shape is applicable. And the question there is, how best to shape the trajectory to shed the several hundred m/sec the stage will be carrying at high altitudes *before* it slows to terminal velocity in lower atmosphere.