I was thinking of how complex the orbital elements are and got to wondering whether you can represent orbits in a more simple way. If you start with an ECI coordinate frame and you know the spherical position of the periapsis. You could then represent the orbit with a latitudinal and longitudinal angular speed at that periapsis.
A roughly circular Polar orbit could have a periapsis at 0 deg,0 deg 6700km (right over the north pole) and a longitudinal angular speed of 0.065 deg/s, and a latitudinal angular speed of 0 deg/s.
An Equatorial circular orbit could be ((0 deg, 90 deg, 6700km), 0 deg/s, 0.065 deg /s)
To make the orbit more eccentric you would increase one or both of the angular speeds. By using a combination of the two angular speeds you control the inclination of the orbit:
A 45 degree inclined orbit could be ((0 deg, 90 deg, 6700km), .0463 deg/s, .0463 deg /s)
The length of the vector (.0463,.0463) is .065 so you would have the same circular orbit as before even though each angular speed is less.
A retrograde (westward) orbit would have a negative latitudinal angular speed (the longitudinal angular speed is always positive.)
The reason this should work is because at the periapsis the radial velocity is always zero. So you can eliminate 1 element. Is there any reason why this would not work?