Talk:Place-Anywhere 7 Linear RCS Port

Minimal number required

You can have a controllable craft using only 4 of those in vertices of regular tetrahedron. One thruster (back) points directly away from CoM and 3 (front) point away from a point slightly in front of CoM. Back thruster moves forward, 3 front thrusters combined move backward and 1-2 front thrusters plus back thruster provide rotation.

How do you provide roll authority? If you can a graphic/screenshot would be good and a short animation/video showing all rotations be even better. Oh and how do you provide lateral motion? — xZise [talk] 07:54, 25 August 2014 (CDT)

Strictly speaking, one thruster is enough, provided that the vehicle has reaction wheels to point itself (and thus the thruster) in any direction. I suppose that's okay for formation flight, but probably not very convenient for things like docking.

If we want to be able to translate the spacecraft in any direction (again, leaving rotation to the reaction wheels), four thrusters is enough. By analogy, consider the two-dimensional case: a triangle with one outward-facing thruster on each vertex can push itself in any direction over the plane (without rotating) with three thrusters. Getting an arbitrary direction will require pulsing or throttling of two (or possibly three, in the 3D case) thrusters together, but it can be done.

That said, if we want the spacecraft to use thrusters to rotate as well (to get more torque), without any "cross-coupling" (that is, the spacecraft has to be able to translate in any direction without rotating and vice versa) then I do think that six is the minimum. The best arrangement I've come up with is a trio of thrusters on opposite corners of a cube. In the 2-D case, that's pairs of thrusters on opposite corners of a square. I don't think that can be decreased to three because, if we get an off-center thrust, it will cause a torque of some sign (positive, let's say), and we need two thrusters to make the opposite torque without disrupting the thrust. However, if either of those thrusters fires, it will cause a negative torque, so we a need a positive moment, but we need two motors to get a moment without force, so we're back up to four (thus six, in 3D). --Kerbas ad astra (talk) 00:48, 11 February 2015 (CST)