General Question CoG in Zero G...HELP!

[paddy2]

I am designing my own vessel and am having a problem understanding the real and orbiter configured Centre of Gravity. I understand the parallelogram of forces acting on an airframe, with its lift, drag, G and the like. I also have a passing understanding of good old Newton

Here is a very simple drawing of my craft. It is for use on and around the Moon, i.e. a vacuum and low to nil gravity. It has a single lift off / hover engine, a single retro braking engine and 3 main engines. The main body is mounted on 4 legs. The actual CoG is nowhere near the middle in any axis and would change with cargo and fuel. If I was a Loadmaster on a C130 I would tell you where to stick your boxes, a Flight Eng on Concorde would pump fuel around. I don't have those options. Not all UCGO cargoes are the same Mass and the weight will change from lunar surface to low orbit and above.

I have the RCS thrust to make changes (well the autopilot would) but where would the main engines force act, both in and out of gravity? Around which point would the roll occur. I get I need to have two jets acting together, like the tracks on a turning tank, but where is the CoG in a zero grav setup. Each time I reason it in my head I end up with a different answer.

Please understand this is from some who does not believe Bees, 747s and Harriers can fly!!!:facepalm:

 

[Urwumpe]

Trimming for varying CoG is done by gimballing the main engine.

The position of the CoG in orbiter is always the same (0,0,0) in local vessel coordinates, what changes is the position of the thrusters relative to the CoG

For calculating where the CoG is, you should take a short excursion into rigid body physics. While it is pretty simple for the CoG (just weighted sum of the vectors to point masses representing your spacecraft structure), it is way more complex for the PMI, the moments of inertia that influence how your vessel will rotate.

The spacecraft will roll always around the CoG. even if you put two thrusters on a large stick and pretend them to rotate around the center between both, in space, the spacecraft will still rotate around the CoG, while also getting some translation by the thrusters. Its a bit hard to explain without drawings.

The important formula for thrusters and rotations is always

Torque(vector) = radius(vector) x Force (vector)

In geometry, this relation means: The radius and the force arrows will span a parallelogram, with the torque vector being perpendicular to the parallelogram and the length of the torque vector being the area of this parallelogram.

(In case you didn't know it yet).

Which of course means: If the force vector and the radius vector are parallel, no torque will appear, and thus no rotation. If the force vector and the radius vector are perpendicular, the torque will be maximal.

So, now it would be good to know - what should be the exact problem to solve? And did you already simply try it in Orbiter and looked what happened?

And of course:

 

[paddy2]

Thanks, digesting this info..

---------- Post added at 02:54 PM ---------- Previous post was at 02:40 PM ----------

So am I right in thinking the CoG is really the CoM(ass) forget gravity

So on a 4 meter cube, if the the COG was 1.0m up, 1.0m in from the back and 1.0 meter from the side, Thrusters on one side would have to work twice as hard as on the other to get it to spin around the mathmaticial centre.

Surely the on board computer, given fuel state and payload data could do the maths to adjust the RCS and Gimball the main....... i.e make the problem go away!!!

Lazy I know but still workable?

 

[Urwumpe]

So am I right in thinking the CoG is really the CoM(ass) forget gravity

Yes. It is badly named, since the CoG is not the CoM unless you assume Galilean gravity.

 

[paddy2]

Thanks for the help in sorting it in my mind.

don't quite get the last bit mind

since the CoG is not the CoM unless you assume Galilean gravity

Understand the idea of 2 iron balls, get the bit about the string/chain inbetween, ball and feather, flat paper crumpled paper but not the Galilean gravity.

 

[BruceJohnJennerLawso]

Yes. It is badly named, since the CoG is not the CoM unless you assume Galilean gravity.

Forgive me for asking, but when do they start to differ?

 

[Urwumpe]

Forgive me for asking, but when do they start to differ?

Understand the idea of 2 iron balls, get the bit about the string/chain inbetween, ball and feather, flat paper crumpled paper but not the Galilean gravity.

The difference between both is, that CoM is plainly the geometric center of the mass, while the CoG is the geometric center of the gravity force, as sum of the gravity forces of all components of your spacecraft.

In Galilean gravity, the gravity force is everywhere the same for every mass, so the center of gravity is the same as the center of mass.

For Newtonian gravity, the gravity force drops by distance, so the parts deeper down the gravity potential experience more gravity force than the parts further "outside". If you have two connected iron balls, the ball closer to Earth gets more gravity force, and thus, the center of gravity is closer to Earth than the center of mass.

It is only important to differ between both in extreme cases though. For orbit calculations, the forces act on the center of gravity. For rotations, the forces act on the center of mass.

 

[paddy2]

I find it difficult to think of a CoG in zero grav. Also if I add a fuel tank to the front of the vessel I can "see" the mass moving no matter how high the craft is!

From what else has been said am I right in thinking that while all the screens I have seen show the RCS are fired in Pairs, i.e. left up / right down a single (bigger) one would still work albeit in a single direction?

 

[Urwumpe]

I find it difficult to think of a CoG in zero grav. Also if I add a fuel tank to the front of the vessel I can "see" the mass moving no matter how high the craft is!

From what else has been said am I right in thinking that while all the screens I have seen show the RCS are fired in Pairs, i.e. left up / right down a single (bigger) one would still work albeit in a single direction?

Actually it is very easy: Think of a spacecraft in the form of a 30 km large tube. One end is 15 km above, the other end 15 km below the CoG.

First question: Now you are floating from the CoG to the upper end, what happens?

Second question: You return to the CoG and move to the lower end. What will happen?

 

[paddy2]

What will happen?

Part of me wants to say the effect of gravity will lessen... but as you move closer to the ends then "that" end will have more effect while the other end lessens i.e. it all balances out. If the 30km was made up of different material (lead bottom gassy top) then I see a delta slope.

yes its sinking in now .. Danke

 

[Urwumpe]

Part of me wants to say the effect of gravity will lessen... but as you move closer to the ends then "that" end will have more effect while the other end lessens i.e. it all balances out. If the 30km was made up of different material (lead bottom gassy top) then I see a delta slope.

yes its sinking in now .. Danke

Actually, you will move to the retrograde wall, when you go up, and to the prograde wall, when you move down. You are yourself in an orbit around Earth.

While you are really weightless in the CoG of the tube, you will experience low microgravity the further you go away.

 

[paddy2]

Just was not thinking along the right lines here. OK in the center, the attraction of the top cancels the attraction of the bottom "weghtless"; as you move to the top the bottom has less "minus" effect so you gain weight.

As for pro and retro grade.. Higher orbit must travel faster... if not you appear to be going slower and so move backwards. So to go "up" within the tube you also have to go faster "around"

Like maintaining your dressing in a body of marching men going round a bend.... ( maybe not everyone will get the reference)


So to come back to the first problem...
given a natural distribution of pipes/ frames /fuel/ cargo the CoG will never be in the "middle" of the craft.
The RCS will act on / around the Centre of Mass

Using a pair of matched thrusters built on a symmetrical frame, one will have more effect as its radius of action is bigger.

By altering the relative thrust values I can still roll around the centre line of the craft or does it simply effect the speed of rotation.

( there is a part of me that's thinking I should just go back to banging sticks together :rofl: