Eisen Feuer

OK so what I'm getting is a maximum of 100m/s ∆V for starting in a dragless environment, a maximum of 463m/s for launching at the equator instead of the poles, no losses to inclination adjustment by launching in the tropics... How much ∆V is expended in altitude increase rather than contributing to orbital speed? This is fuel being expended when the rocket is at its heaviest, the fuel spent at the beginning is what really makes the rocket heavy when it could be swapping out that fuel for increased payload.

I've seen the radical differences in spacecraft design required to land at sea level or on the high plains of Eve, I was wondering what might be the implications of launching from higher altitudes here on earth, specifically relating to the advantages to be gained by a USA-based space program attempting the build a space station/LEO vehicle assembly station in the equatorial plane of the solar system. So we're talking about establishing a long term goal with many launches, the heavier we can make each launch the better. Less pieces = less assembly = less to go wrong in LEO. Let's say right now we launch from southernmost Texas at 26° latitude, at only 2.5° above the Tropic of Cancer, only 2.5° of adjustment in the orbital inclination would be required to enter the equatorial plane. But, we're at sea level. Let's increase the altitude, which for staying inside the continental USA, requires us to move north. At about ~31.5° and ~10,000ft/2900m we have Miller Peak in far southern Arizona. If a peak is too much, there are flatter areas in the Big Bend of Texas that average 4,000ft/1200m. Then disregarding geopolitics, shipping costs and the challenges of launching from a peak/volcano, there is the ultimate advantage, picking a peak in the tropics (no inclination adjustment) like Pico de Orizaba outside of Mexico City at ~18490ft/5636m, or Mauna Kea at 13795ft/4205m I guess now that I've said all that what I'm really looking for is the math to make a comparison chart of ∆V saved per altitude increase and ∆V required per degree of inclination adjustment.

I managed to figure this out by generating the orbit which was simple to incline around one axis (I had already gotten that far by the time of my post), then transposing the XYZ to spherical coordinates, where it was simple to offset the longitude of the path being generated, and then converting back to XYZ. Your solution is definitely more elegant though and I'd assume less prone to accrue error over time than my back-and-forth so I will definitely incorporate it into version 2. Thanks a lot!

K^2, is there a way to graph the lat/lon traversed over time? I'm having massive problems making an orbit like this in a 3D application/XYZ space using pure trig and I think spherical coordinates may be my way out. As longitudes are traversed more quickly at different latitudes, I have no idea where to start or even where to look for such a specific solution. Anyway, here's the above equation made interactive on Desmos Graphing Calculator if anyone wants to play with it: https://www.desmos.com/calculator/ui8awrisw5 (for those wondering why it looks like a square wave, this is a near polar orbit (87° inclination), which, just like the poles of Earth in an equirectangular projection map, stretch out. A 90° inclination will return an actual square wave, and an equatorial orbit will be a flat line, and a 51° inclination like the ISS will return a nice soft sine-wave looking curve. As for the math to make it mesh with the rotation of the Earth, divide a 24-hour day by the orbital period (let's use 90 minutes) (24x60)/90 = 16 360/16= a loss of 22.5° per orbit, so you'll want to sample the image once every 360-22.5=337.5° and stack the samples on top of each other to see your forthcoming orbits.