The general thrust equation for rockets is: F=mdot*V_{e} + (P_{e} - P_{a})*A_{e} where, F=thrust, mdot=mass flow rate, V_{e}=velocity of the exhaust at the nozzle exit, P_{e}=exhaust pressure at the nozzle exit, P_{a}=ambient pressure, and A_{e}=area of the nozzle exit.When considering the P_{a} term during the low altitude segment of the flight, is that very strictly the general atmospheric pressure at the rocket's altitude? I believe the forward travel of the rocket and backward travel of the high velocity exhaust gasses creates a localized low pressure at the base of the rocket (this causes the plume recirculation seen on some launches, right?). Does this area's lowered pressure need to be taken into account for a higher accuracy calculation of the thrust?

For a rocket engine to be propellant efficient..... using propellants which are, or decompose to, simple molecules with few degrees of freedom to maximise translational velocity

If nozzles point directly backward, then variations in thrust of one engine with respect to another will tend to make the rocket turn a bit. That is eliminated if each engine's thrust vector points at the rocket's center of mass.

This post isn’t really meant to answer the gravity losses question but rather summarize my understanding of what gravity losses are. So if I’ve got something wrong please point it out. Orbit isn’t about altitude per se but rather velocity. You need an angular velocity such that the centrifugal force is equal to the pull of gravity to essentially nullify it and enter orbit. Given that we have an atmosphere on earth, achieving the required velocity is easier/possible if your altitude places you outside or at least in the very very thinnest parts of the atmosphere. In order to achieve that altitude, you have to spend some of your fuel countering gravity to increase altitude rather than devoting all the fuel to increasing tangential velocity, thus you lose something to gravity. If you had a perfectly spherical body with no atmosphere you could orbit just off the surface given you have the necessary angular velocity, and orbit could be achieved with almost no gravity losses. Is that at least a decent conceptualization of gravity losses? If I’m understanding the concept correctly the launch trajectory would affect the amount of gravity losses that occur. A more lofted trajectory would have greater losses versus one that pitches down range earlier in the ascent. But there is a sweet spot between fighting atmospheric drag versus gravity losses that would define the “ideal” trajectory. Different trajectories from the ideal might be chosen for various reasons (such as?). What other factors in rocket design and launch trajectories affect gravity losses. I appreciate the feedback and the tremendous resource NSF is for us amateur armchair rocketeers and space nerds.

Has anyone ever designed a rocket which cross-feeds the fuel from the 2nd stage to the first stage.

Quote from: gin455res on 11/07/2018 09:41 pmHas anyone ever designed a rocket which cross-feeds the fuel from the 2nd stage to the first stage.This entire question makes me feel weird. Just what purpose could this possibly have? Reducing the dry mass of the second stage is absolutely crucial for performance. Why would you ever feed fuel down from there, instead of just making the first stage tanks bigger?

Quote from: Tuna-Fish on 12/03/2018 05:58 pmQuote from: gin455res on 11/07/2018 09:41 pmHas anyone ever designed a rocket which cross-feeds the fuel from the 2nd stage to the first stage.This entire question makes me feel weird. Just what purpose could this possibly have? Reducing the dry mass of the second stage is absolutely crucial for performance. Why would you ever feed fuel down from there, instead of just making the first stage tanks bigger?The point was to remove the fuel tank from the first stage. So only 3 tanks need to be manufactured not 4. And also the tanks would be more similar in size. Perhaps, this would mean all three tanks could be made the same thickness. One long cylindrical oxidiser tank on the first stage, and 2 smaller more spherical-ish tanks of oxidiser and fuel on the upper-stage, perhaps with common bulkheads.

In thermodynamics, the triple point of a substance is the temperature and pressure at which the three phases (gas, liquid, and solid) of that substance coexist in thermodynamic equilibrium.[1] It is that temperature and pressure at which the sublimation curve, fusion curve and the vaporisation curve meet. For example, the triple point of mercury occurs at a temperature of −38.83440 °C and a pressure of 0.2 mPa.