Rocket Science: CR4 Challenge (11/17/09)

GA from me for a well explained, understandable answer. Nice job dac.

Agreed GA for a quick response. (And horizontally on a frictionless surface to boot).
But we could still treat this as a semi-reasonable* question - define "launch" as "lift off" - i.e. the time (about 145 seconds after ignition) at which the rocket leaves the ground.

*Only semi, as the hot exhaust gases would presumably destroy the launch-apparatus

Another GA to you (but I did only say semi-sensible, and I didn't notice anything in the challenge to suggest that the rocket would be well-designed).

I think this can be launched on a catenary curved ramp that ends vertically, so it starts horizontally and accelerates and the angle changes to suit the remaining mass. As it loses mass, the incline is changed progressively until it has lost enough mass to maintain, say, a 3g acceleration and is heading vertically, but if there are no people on it, it would be more efficient to allow unlimited acceleration until the fuel is gone. I think this would work to get the max velocity or height. Max velocity = infinite horizontal ramp in vacuum.

Mabe the is a tangential launch moon rocket returning to earth?

The answer of 1.97 m/s is actually the maximum (vector) value for the "liftoff"case of post #4.

To achieve it the rocket must have at least 387 kg to burn. Anything less than 362 kg and it will not launch. Anything between 362 and 387 and it will rise - run out of fuel (and thrust) - and follow a projectile path.

This means of course that the rocket may have a negative velocity after 10 seconds. I have calculated that the minimum velocity occurs under the following conditions

Mass of fuel installed on the rocket 384.2 kg mass non-fuel = 615.8

Burn off 362 kg fuel until total weight = thrust

Fuel left = 384.2 kg - 362.0 Kg = 22.2 kg

Fuel burns for 22.2/2.5 s = 8.88s

v(8.88) = 1.516 m/s

Height(8.88) = 4.4 m

Travel as a projectile for 1 12 s

s(1.12) = 1.516*1.12 - ½*9.8*1.12² = -4.4 m

V(1.12) = 1.516 - 9.8*(1.12) = -9.5 m/s

That is the rocket rises 4.4m above the launchpad, runs out of fuel and falls back to the launchpad striking it 10 seconds after launch at 9.5m/s

(If the designers are stupid enough to overfill it so it can't take off, they're stupid enough to underfill it)

A "nice" concept. I'll give you a GA when I return to cookie-land.
Though perhaps you will agree that a more likely explanation is that the rocket-motors are too small. My guess is that NASA's specification for the motors was 5.5-kg/sec (giving a maximum net initial acceleration of about 0.4-g *). Unfortunately, the actual motors delivered were designed for 5.5-lbs/sec.
*Given the effects of atmospheric drag and the mass of the rocket motor itself, is there any point in providing more than this?

Fyz
____________________
The following is self-referential:
"The lowest quote is not always the best"

From http://lsda.jsc.nasa.gov/books/apollo/s2ch5.htm

As an aside I was surprised (probably because I was too stupid to think about it) how much they messed about with the atmosphere (content and pressure) inside the living quarters (almost pure oxygen at about ¹/₃ bar). They changed the launch composition after the fire.

Interesting - 60% oxygen at launch but with no replacement of nitrogen thereafter. And always more total oxygen available than in a normal atmosphere. I can understand the weight advantage of omitting the nitrogen, but the reason for increasing the total available oxygen escapes me - can anyone help?

Fyz

Without re-reading it: what I understood from the paper was that they had decided that the best atmosphere was pure oxygen at the lowest acceptable pressure, but, after they had the fire they decided that pure oxygen was just too big a hazard at launch.

Thanks FYZ. DAC answered so quickly and so well that it was difficult to add anything.

The last two challenges have been "shot down" rather quickly, so I am now working on a presenting a more challenging aerodynamic question that I hope will survive a little longer.

http://cr4.globalspec.com/thread/45769/Why-You-Shouldn-t-List-Your-Email-Address

I need ur contact, if u r interested.

My email is udayghatge@gmail.com

I m a researcher, we wil have some contact.

after 10 seconds the mass of the rocket has decreased by 25kg to 975kg; this is even more than the rocket engine thrust can lift.

6250N can lift a mass of 637,322633111205151606307964493481kg on earth, to decrease the rockets mss to this it needs (1000-637,322633111205151606307964493481)/2.5 seconds = 145,070946755517939357476814202607 seconds; than the rocket will have a lift off

Getting yet dafter:

Obviously the rocket will not take off. So we can assume that it is poorly designed and that its attitude control will not function properly while it is on the ground. Also that the rocket base will sit on an air cushion and soon after the rockets have fired the base restraining tower will disintegrate. Given that the rockets disturb any fragile equilibrium, the first thing that will happen is that the rocket/tower system will start to keel over under gravity.

With a mass of 1-tonne the rocket might be about 6-metres tall. If we assume that the mass of the rocket+tower is evenly distributed along the rocket's length, the radius of inertia about the ground will be 3.5 metres. This is now equivalent to the "pencil-on-its-point" problem. dθ/dt=g.sin(θ).h/I≈g.θ/4 (for small theta). Solving gives θ=constant.exp(g.t/4). Even if we assume that the rocket unbalance only gives 0.001 radians, the rocket will topple and hit the ground within seven seconds.

Thereafter it's velocity depends on the mass of the tower and the coefficient of friction...

It makes no substantive difference, but:
On second thoughts the toppling will probably take appreciably less than seven seconds, because once the base of the rocket is "floating" on a gas cushion the centre of rotation moves from the base of the rocket to the position of the centre of gravity, so effective radius of inertia is only 1.75-metres.

by integrating the resulting acceleration, after 10 seconds the rocket has a velocity of less than -5131.4627324859 m/s (calculated with Windows Calculater) - this results in the same, the rocket sinks into the ground.