sevenperforce

Other than the treadmill pushing it backward. The treadmill cannot push the plane backward. It cannot push the plane because it is not touching the plane. The treadmill is simply touching the plane's wheels. The underside of those wheels, to be precise. And no amount of pushing against the underside of those wheels will exert any significant force on the chassis of the plane.

The belt is not holding onto the plane because there is no way for the belt to hold onto the plane.

Nah, the dynamic pressure from wind storms isn't nearly high enough to throw it around.

I concur. It looked less burnt when it landed, too.

If a treadmill was placed on a plane, and the treadmill was set to match the speed of the plane, would the plane be able to take off?

Plot twist: You have a treadmill on a plane.

A lovely and energetic discussion (no pun intended) on the use of liquid ozone as a propellant: https://forum.nasaspaceflight.com/index.php?topic=13023.0 Oh, and an exciting bonus: if you mix ozone and fluorine, you can fill your tanks with it and make a turbine engine that will run in any atmosphere in the solar system! (a mixture of ozone and fluorine will combust with equal enthusiasm whether it is mixed with the nitrogen in our air or with the carbon dioxide on Venus/Mars)

/pedant, but while I agree with your conclusion, thrust does not produce lift. Thrust moves the plane forward; the motion of the air over the airfoil produces lift. If the plane is somehow held stationary, then all the thrust in the world will do nothing to lift the plane. Of course, there is no way for the conveyor belt to hold the plane stationary, no matter how fast it goes. So that's that.

The problem doesn't state that the wheels are moving at the same rate as the treadmill. It states that the conveyor belt is designed to match the speed of the wheels. Let us suppose the 747 is stationary on the treadmill, ready for takeoff. The pilot throttles the four engines to 100% and then releases the wheel brakes. Up to this point, the conveyor belt cannot move. If the conveyor belt is designed to match the speed of the wheels, then the conveyor belt cannot move unless the wheels are moving. However, we must acknowledge that the conveyor belt can have NO impact on the plane or its wheels unless it moves. So it cannot inhibit the initial motion of the plane. Therefore, the plane is free to move. If the plane is free to move, then it will begin moving. If it begins moving, it is not stationary. If it is not stationary, then the treadmill will not be able to match the speed of the wheels, even though it will attempt to do so. Thus, the plane will take off. The way it is generally understood is that "the speed of the wheels" targeted by the conveyor belt is defined as the revolution rate multiplied by the outer wheel circumference. Not that people sit down and do the math; that's just what the assumed interpretation typically arrives at. This is also the only interpretation that leads to any problems. If you interpret it as "the linear speed of the hub with respect to the ground" then there is no problem; the conveyor is merely accelerating to match the forward speed of the plane which doesn't produce any issues at all.

I suppose the separation problems. And spontaneous ignition problems.

Oh my god. This subject. Let me see if I can manage to hit the high point properly. There is inferred ambiguity in the way the question is phrased. Without resolving this ambiguity, no conclusion is possible. So, the question: The problem is in one little phrase: "the conveyor belt is designed to exactly match the speed of the wheels". What does this mean? Well, it can mean one of two things: The conveyor belt is designed to measure the tangential speed of the wheels and adjust its speed to match. A series of conditions unspecified in the problem has resulted in the situation where the tangential velocity of the wheels is equal and opposite to the vector of motion of the surface of the conveyor belt. If the correct meaning is (1), then the conveyor belt will attempt to match the speed of the wheels, but will not be able to do so. The plane takes off. This is the most straightforward interpretation. The question says nothing about what will happen; it only says what the conveyor belt is designed to do. Whenever the conveyor belt detects the wheels moving, it will immediately attempt to match their speed, which will increase their speed, which will cause the conveyor belt to accelerate faster and faster in an attempt to meet its designed performance criteria. Meanwhile, the plane merrily takes off without even noticing. If the correct meaning is (2), then circumstances are quite different. Nothing the conveyor belt does can exert a rearward force on the plane greater than the mass of the plane multiplied by the coefficient of kinetic friction in the wheel bearings. Therefore, the only circumstance where the tangential velocity of the wheels is equal and opposite to the vector of motion of the surface of the conveyor belt is one where all velocities are zero. The plane is not moving, but neither are the wheels or the conveyor belt, because everything is turned off. Obviously, I feel (1) is a more accurate interpretation. The question never says anything other than what the conveyor belt is designed to do. It is at this point that many will try to shunt explanation (2) back into (1), saying "But what if the conveyor belt has instantaneous acceleration?" But this is neither specified nor even implied in the original question. The original question only says what the conveyor belt is designed to attempt.

I love this. But yeah, ozone is a monopropellant in its own right, so it has a nasty tendency to do what hydrogen peroxide does, except worse.

I meant that they said they plan on using the BE-3 engine model indefinitely. BE-3 is, I believe, the first-ever fully-operational combustion tap-off engine. Oh, they'll never launch a second Vulcan for 65% of the price of the previous rocket. They've very cleverly pointed out that as far as their raw cost for the booster alone, 65% is inside the engine compartment. So cost per launch reductions are probably closer to 20-30% for reusing the engine pod. Vulcan is still far enough away that test runs are not really feasible.

What, exactly, is the harm in taking the Canal? I mean, that's kind of why Teddy built it.

With the BE-3 being an engine they plan on using indefinitely, rather than limiting to NS, you'd imagine it's already been discussed. Then again, if they have no plans for developing orbital hydrolox transfer, it may not be necessary. No sense developing infinite restarts if you only have so much props per engine. Question -- does anyone know if ACES is planned to come with enough RCS capabilities to perform its own maneuvers for autonomous rendezvous, docking, and proptrans?

I wonder if BE-3 will be redesigned to use augmented spark ignition or if they will just stick with using pyros.

How much are the GEM-60 SRBs currently used for Delta IV? They can't be terribly expensive. Vulcan cannot compete financially with Falcon 9 for LEO, but with SMART reuse recovering 65% of the cost of the booster, Vulcan-ACES can definitely compete for destinations that stretch the capabilities of F9 and FH, like heavy GTO payloads, direct-to-GEO insertions, and anything beyond.

I don't dispute that. But plane changes at perigee are not efficient. Geostationary comsats typically perform a bi-elliptic transfer; the launch vehicle sends them to a supersynchronous transfer orbit (with an apogee higher than their intended destination), then the payload raises its perigee, then lowers its apogee to circularize half an orbit later. This is more efficient than a Hohmann transfer, particularly when you have a plane change to do. So excess propellant capacity on the launch vehicle is always used to raise apogee rather than any sort of perigee plane change maneuver.

Uh, that's vastly different. Unless this was sarcasm.

"Sea level" on Mars is defined as the mean surface radius.

Sounds like it was a bad thing.

Just you. It is always more efficient to burn to a higher apogee and let the payload correct inclination than it would be to try and perform an inclination change at perigee. The payload is performing a bi-elliptic transfer; it will correct inclination, then raise perigee, then circle around and lower apogee to circularize. Importantly: that's one down, four to go before Falcon 9 is man-rated.

MVac TWR was limited to Block 4 thrust levels on this flight. GTO burns are typically quite short. The Falcon 9 upper stage has a really really good mass ratio so it's pushing very little dry mass by the end of the burn.

Elon is hugging everyone.

And we have a good deployment!!