Getting CG forward on a swept wing design

[rakm]

Home designed powered swept wing that flys very well. The only problem is I have to place the battery packs all the way to the rear to get the cg correct. What needs to be done in the design to get the battery packs forward? Do I need to increase the root cord? Reduce the sweep? Thanks for any help.

[Sparky Paul]

Reducing the sweep moves the c.g. and therefore the battery pack forward.
If changing the root chord reduces the sweep...
Adding a tail will move the c.g. forward also.

[vintage1]

ER what?


First of all are the wings swept forward or back?

If you reduce sweepback, all other things being equal it will move the CG *back* with respect to the wing center of lift, at least. So although Sparky is right that the CG moves forward as sweepback is reduced, because some wing mass will move forward, its not going to do what you automatically think 'moving CG forward' would do. Thats because the wing is moving forward faster than the CG is, so to speak!!! :(

If the aircraft is nose heavy, reduce sweepback. Or add a tail!

[rakm]

The wings are swept back. Looks like a small Zagi (28 inch span). Thanks for the advice. It looks as though the answer is to reduce the sweepback. I'm assuming this will also reduce the stability as it's taking away some effective dihedral? Will try a few different angles and get the best trade-off. Right now it's very stable/forgiving and don't really want to loose that.

[Sail 'n Soar]

Better question is, "If it flys well now, why worry about the aft battery location?"

[exempt]

I had almost the same problem with my home made Zagi style wing, but with mine it's tail heavy (i think) when I tried to launch it, it just did a backwards loop?? I dunno, I guess somethings wrong. :rolleyes:

[Robbie d]

you can calculate the cg of any wing easily using nothing but a ruler and a pencil. First measure your root and tip chords.
The red lines on the diagram represent the root chord and the light blue lines are the tip chords. You draw the chord lines top and bottom. Then draw two intersecting lines (the dark blue ones) from the tip of each line and where they cross is the c of g.

[Sparky Paul]

Not quite.
That procedure locates the mean aerodynamic chord.
You must then compute where the c.g. is on that chord, and project that location to the root chord.