You are in an A-10 with a maverick on one side. You have an aircraft CG (which the FDM is reporting the position of) and an MRP, which the FDM is also supplying to FlightGear. The MRP is given to FlightGear in lat/lon/alt. The FDM calculates that position because it knows where the CG is absolutely, as well as where the MRP is relatively. Since the 3D model builder and the flight modeler for the aircraft agree on the MRP (we hope!) the aircraft can be placed properly in the scene. Now, say we are stationary on the runway and we drop the missile. The CG shifts instantaneously - however the FDM will not report a different position because no force has caused an acceleration which in turn could not have been integrated to produce ultimately a change in position. This is a flaw that may need to be addressed. In any case, let's assume that the FDM *did* shift the reported lat/lon/alt of the CG as reported to FlightGear by the equal and opposite amount that dropping the Maverick resulted in. The vector to the MRP would be shifted by an equal and opposite amount, and the end result would be that the model would not "jump" and the FDM and the model would agree.
In the case of smooth flight where fuel burns off slowly, it's not so critical.
It is hard to tell from what's said whether you're using the COL as the reference or the CG. COL is the real reference to use in the FDM, the CG is purposefully forward on most craft for stability. The CG swings around the COL pivot point noticably when you change pitch. Also all moment arms are from COL and you actually load the plane to put the CG where you want it, the COL is the real reference point. If you're not already using this, try a very high wing low CG plane. Pivot it around the CG. Then pivot it around the wing's COL. Which looks even close to the real pivot point?
To model a bomb drop correctly you simply use the COL as the suspension point, and calculate all forces off of it on each pass, including the moment arms. With the Maverick gone and no longer having it's moment arm in the equation for CG the next calculation pass the CG will shift automatically, and the new distance between the CG and the COL will take care of any resulting motions on it's own and move the plane till things are back to balance. It would be built into the calculation pass, and move on it's own. Really you can just calculate total CG once or rarely and just move the CG when the bomb drops by subtracting it's moment arm. But the COL should be the motion and suspension point of the craft not the CG. The change in the CG relative to the COL should automatically introduce the forces to change the overall orientation around the COL. The COL can shift too as orientation changes but it's the point to calculate from, no one said an accurate model was easy. Just accurate.
You can use any point as a point of reference as long as you make sure all translations are in all the calculations. You can start from the CG, and then calculate the COL is over here, and use that to move the craft. But the craft should swing around the COL either way, so you have to do extra translations to make the CG reference point move. COL and CG can both move, but COL is the real hang point of the aircraft so using it simplifies calculations for a truly accurate model. That's why weight and balance calculations are relative to it instead of where the starting CG was. The starting CG is used in the calculations, along with it's moment arm from COL. There are less translation adjustments if you reference everything from that point.
Alan
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