RCgothic

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Everything posted by RCgothic

  1. Falcon Heavy or New Glenn could probably manage about 15t to TLI. That's still less capable than the Apollo lunar lander because it needs to insert into lunar orbit/decent from trans-earth velocity somehow and on Apollo the service module did that. LMFAOROFL at "that low powered SLS"! Pretty much. SLS block 1 has more nominal payload to TLI than that spacex or blue origin, but it can't dual manifest payloads and Orion is a mandatory component so that extra mass amounts to diddly squat. It will never have the launch cadence to support a rendezvous mission with another SLS payload.
  2. So EM2 is probably going not going to meet that target and the ICPS is rubbish but they can't get EUS prepped in time for either of the planned missions that need it. SLS is ever more clearly a booster to nowhere.
  3. Work is force times distance. If you're going faster you apply the force over a greater distance and so do more work.
  4. No, this isn't what a gravity slingshot is. A gravity slingshot is when you use a gravity well flyby to borrow velocity from the body at the centre of the gravity well. Relative to the planet you haven't gained or lost any speed, but in the context of the global reference frame you've added or subtracted some velocity due to the motion of the celestial body, accompanied by a course change. ------------------------------------------ What we're discussing is minimising gravity losses on take off. If you have a constant TWR of 5/3 then you can burn upwards but you lose 1g to gravity and so get an acceleration of only 2/3g. If instead you burn sideways you hover at 1g upwards with 5/3g on the hypotenuse and a 4/3g horizontal component. You accelerate twice as fast horizontally as you could vertically and because the earth is a ball the surface will eventually fall away from you so that that horizontal velocity becomes vertical anyway. So for the the same fuel, engine and throttle settings, you'll minimise gravity losses by burning horizontally. In KSP you'll often see players turn horizontally immediately on airless bodies. This is why. As long as they sustain just enough vertical thrust to prevent them from crashing back to the surface the most efficient use of thrust is to burn sideways. Often there are other practical considerations that mean you want to fly a more lofted trajectory - terrain avoidance and atmospheric drag.
  5. Suppose your space vehicle has a TWR of 1.5. It can accelerate straight up at 0.5g as 1g goes into just hovering. Now burn at an angle, with 1g up to cancel gravity and the rest on a horizontal vector sideways. By Pythagoras you accelerate at 1.1g sideways. (1.1g horizontal, 1g vertical, 1.5g hypotenuse). This is *much* better than trying to burn directly upwards!
  6. Some further thoughts - some discussion of the energy required, but not much on the mass requirement. By conservation of momentum: The Earth weighs ~6e24kg. If you flung the entire atmosphere (~8.5e20kg) off at the speed of light that's a DV of 42km/s. The entire oceans (~1.4e21kg) gets you 70km/s. The entire crust (~2.5e22kg) gets you 1250km/s. If you can only manage 0.1c then reduce dv by 10. If you can only manage 0.01c reduce by 100. A very good ion thruster might have an exhaust velocity of 300,000m/s, so reduce by 1000. In the case of a very good ion thruster, throwing the entire mass of the crust gets you just 1.25km/s. That's hardly anything and what's left wouldn't be recognisable as earth. *Effects of relativity ignored (as unlikely exhaust will actually approach c) **Rocket equation not used as in all cases the mass of the earth isn't significantly changed by expending the started fuel quantities.
  7. New sum! It's estimated that there is about 100 trillion tonnes of Uranium in earth's crust. Of that, approx 0.72% is fissionable U235. That's approx 720 trillion kg of fissionable U235. U235 releases 83 trillion joules per kg when fissioned. So all the uranium in earth's crust could produce about 60,000 trillion trillion joules. Earth weighs about 6 trillion trillion kg, therefore we've released about 10,000J/kg. 1kg with a kinetic energy of 10,000J would be travelling at 140m/s, so the velocity of earth powered by uranium fission can't exceed this. Therefore NERVAs can't significantly move the earth. Hydrogen bombs and nuclear fusion is required for anything greater.
  8. A rocket engine requires mass to use as a reaction medium. Simply detonating every nuke in a large combustion chamber would only throw the mass in that chamber. Even accelerated to near light speed that amount would be insignificant compared to the mass of the planet. To change the velocity of the planet by 1km/s, you'd need to throw about 20,000 trillion tonnes of material off at the speed of light. A typical ion thruster has an exhaust velocity in the 20-50km/s range. So let's assume the exhaust leaves Earth's gravity well at 30,000m/s because that's a fraction of the speed of light and makes things easier. The mass required just increased another 10,000 times. 200 quintillion tonnes required. That's about 40 million times the mass of the entire atmosphere by the way. At that scale I'd expect the rocket exhaust to be wide in proportion to the thickness of the atmosphere. Basically the portion of the atmosphere over the rocket exhaust would get blasted off into space and the drag around the exhaust perimeter would be negligible. The energy required would be 1.6e32J. 1 Megaton is roughly 4e15J. 4e16 Megatons of nuke required. We have approximately 6400 Megatons of nuke. Their combined yield of channeled into a perfectly efficient ion thruster could alter the velocity of earth by 0.16nm/s. Or not enough that anyone would notice. I may have misplaced a few factors of a thousand here or there, so it could be worth checking.
  9. Yes, most of Falcon Heavy's theoretical load capacity end up on orbit as Stage 2 propellant. So they have a few options: Direct insertion to GSO. Missions to deep space. Aid recovery of S2.
  10. The best bet is to join TRA and work your way up. I believe they're a little more experimentation friendly than NAR. If you haven't launched an L3 high power hobby rocket chances are you're not going to succeed at anything more ambitious.
  11. Whilst a miniature black hole in itself is probably relatively benign, wouldn't anything falling into its gravity well release a tremendous amount of energy?
  12. Ariane4 is by far the nicest looking of the Ariane family.
  13. Wikipedia's article on augmented rockets claim they weigh as much as 5x as much as a normal rocket engine when you include the intakes, and that the intakes are not a trivial engineering challenge. I would suggest that's the reason we don't see them on first stages right there.
  14. Basically it's a lot of weight and complication for most rockets that spend very little time in the lower atmosphere. More generally useful for missiles.
  15. Also hydrogen can actually diffuse through solid tank walls, so even a perfectly refrigerated tank is going to lose some.
  16. There's nothing in interplanetary space for anything to back scatter off of. Probably the worst place to be would be the Van Allen belts, but they'll only be passing through those for a short time.
  17. Solar storms only come from one direction. It'd be no big deal to manoeuvre to place the bulk of the fuel tank between sun and passenger compartment. Storms don't hit full intensity instantly, and even if they did they'd need exposure longer than it would take to reorient to do serious damage.
  18. Which way the wheels spin is irrelevant. The winch can tow the car sideways if it wants to. The wheels won't enjoy that, but they don't get a say in the matter. With both the winch and the plane the motive power available far outweighs any conceivable resistance of road, runway or treadmill. And the whole point of freely spinning wheels is that they are resistance reducers. Neither the towed car nor the plane need to push against the runway to move. One can pull on a winch and the other can push on the air. Again, the road is irrelevant.
  19. Now watch me confuse the issue. All previous discussion has assumed that the wheels are massless/inertialess and the wheel hub is frictionless. When this is the case any forces on the wheels are infinitesimal. It takes no force from the belt to rotate the wheels freely or to speed them up or slow them down and there is no resistance from the wheel hub. The wheels spin freely to match whatever speed the belt is moving completely decoupled from the plane. The plane doesn't care what the wheels are doing it sees no force through the wheel hub. Now add wheel mass/inertia. The force of the belt on the wheels is not a couple, or pure torque. It exerts a moment so that when the belt is accelerating it spins up the wheels. But the force at the rim must be matched by a force at the hub in order to hold the wheel stationary relative to the plane (Resolving forces). This force does act as a drag on the plane, but it's tiny relative to the forces the plane's engines are capable of exerting because it takes very little force to spin a wheel compared to moving a plane, so inertialess is a good first approximation. And this force only acts whilst the belt is accelerating relative to the plane. At constant speed no force is required to change the wheels' velocity. Now add wheel hub friction. A resistive torque at the wheel hub is opposed by a force from the belt creating a torque that holds the wheel to the speed of the belt. But as we've previously established, force from the belt is not a pure torque. So the force of the belt on the wheel rim is reacted at the wheel hub. It acts as a constant drag on the plane. But wheel hubs are built to have very low coefficients of friction. This rolling resistance is tiny compared to the force required to accelerate a whole plane. Therefore frictionless wheel hubs are a very good first approximation. The belt is capable of exerting tiny drag forces on the plane. The plane's engines are massively more capable of exerting force on the plane. Therefore for any reasonable speed of belt it doesn't matter what the belt is doing. The plane can just apply thrust to overcome it.
  20. No they don't. It depends on the relative motion of the plane and the belt. If the belt matches the speed of the plane, the plane can take off without the wheels spinning at all and the brakes fully applied. If the belt is stationary, the wheels are turned by friction with the belt as the plane accelerates by the force of the engines. If the belt goes backwards the wheels turn but the plane doesn't move. The wheel hubs are effectively frictionless and can't exert any force on the plane to accelerate it.
  21. Unirradiated uranium isn't all that dangerous. It's only once it's been irradiated in an online reactor that it becomes so (due to neutron activation and fision products). As long as any reactor isn't started until it's on orbit there's nothing particularly nasty on the rocket being launched. The biggest issue is that you'd probably use highly enriched fuel for mass reduction so you'd want to be able to recover it if the spacecraft didn't make orbit, but even if it gets incinerated and disperses into the atmosphere it's no big deal. Of course media and public would probably react hysterically, but that's a separate issue.
  22. ISP and acceleration are not directly linked. ISP = Thrust / Rate of Propellant Expenditure Force released per second? Pretty sure that's not a physical quantity. Closest I can think of is change in force per second, which is a measure of throttle response, not ISP. Agree that ISP isn't always king. A stage with a denser propellant and lower ISP may be smaller than an equivalent stage with higher ISP and less dense propellant. Even between stages with the same propellant density, in a choice between a larger rocket with a Be3 or a smaller rocket (because it doesn't require as much fuel) with an RS25 you might choose the cheaper Be3 because extracting maximum efficiency can be expensive.
  23. ISP is thrust per unit of propellant consumed and is the engine's efficiency. It's one of the most important parameters of a rocket engine when comparing different engines. The units can be written as: N/kg/s or Ns/kg. But N can be broken down into kg m/s2. That gives: Ns/kg = m/s m/s is a speed. What speed? Turns out it's exhaust velocity. The higher your exhaust velocity the higher your specific impulse. To complicate further, this is often "normalised" to change its units by dividing by Earth's gravity. Why? Because some fools get confused by working in meters but everyone agrees what a second is. m/s / m/s2 = s There's also a factor of roughly 10 because earth's gravity has a value, but in terms of units that's the why. ISP in seconds is basically how long an engine can burn for whilst producing 1 unit of thrust from 1 unit of fuel. Longer is better because your end velocity will be higher. So that's what ISP is. Now why does it change in an atmosphere? It's because back pressure on the engine bell slows down the exhaust. Also the engine bell can't be as large because back pressure causes instability that causes turbulent flow separation which is destructive to engine nozzles. Smaller engine nozzles aren't as good at accelerating the exhaust, so that's a second factor in why it ends up slower. The higher atmospheric pressure, the more the exhaust is slowed, the lower your efficiency compared to a vacuum.
  24. If I were planning the moon landings today I would go for earth-lunar orbit rendezvous. We're just so much better at rendezvousing than we used to be. The rocket could be much smaller which would do wonders for launch cadence, and as spacex are discovering quantity is a quality all of its own.
  25. A cutting charge around the nozzle extension wouldn't weigh much, true. Not sure how difficult it would be to change the nozzle afterwards once it's back on earth.